The Basics Hibert Space Review In an infinitely dimensional space, we have some abstract vector $\Psi$. We can choose coordinate axes such that all values can be taken by the position x, so that we can describe $\Psi$ as as set of components $\Psi(x)$ Could have also chosen p instead for our coordinate axes $(v,w) \geq 0$, where (a,b) denotes an inner product This is linear in the right argument, and anti-linear in the first argument (ie. take complex conjugates) (v,v) = 0 implies v = 0 $v = \Sigma_{i} e_{i} (e_{i}, v)$ always holds in Hilbert spaces (ie. can always decompose into an eigenbasis, where $(e_{i}, e_{j}) = \delta_{ij}$) This can be extended to non-orthonormal bases (ie. $(e_{i}, e_{j}) = G_{ij}$), which implies that $(v,w) = \Sigma_{ij} (v_{i}e_{i})G_{ij}^{-1} (e_{j}, w)$ You can add Hilbert spaces of different dimensions together The new Hilbert space has the equivalence relationship of $<v_{1}+v_{2},w_{1}+w_{2}> = <v_{1},w_{1}> + <v_{2},w_{2}>$ The total dimensionality of the new Hilbert space is just the sum of the previous two You can also multiply two Hilbert spaces together The equivalence relation which must hold for the product is that $\lambda <v,w> = <\lambda v,w> = <v,\lambda w>$ You can decompose this product Hilbert space as a sum of tensor products of the eigenbases of each prior Hilbert space (ie. $H_{prod} = \Sigma_{ia} \Phi_{ia} e_{i} \otimes e_{a}$) Observables Can think of any operator as any mapping of the Hilbert space on itself (ie. A transforms $\phi$ such that $A\phi$ is in the same Hilbert space) Real and complex numbers commute with any other operators Real observables are postulated to be linear and Hermetian (ie. self adjoint) Linear means that $A(\phi+\phi’) = A\phi+A\phi'$ the adjoint means that $(\phi’, A^{\dag} \phi) = (A\phi’,\phi)$, which means $A^{\dag} =A$ If we have a complete set of orthonormal basis vectors, we can embed the operator as a matrix: $A_{ij} = (\Phi_{i},A\Phi_{j})$ The eigenvalues for Hermetian observables are real The trace of an operator is defined by $Tr(A) = \Sigma_{i} (\phi_{i},A\phi_{i})$ This is basis independent This is only well defined for finite dimensional Hilbert spaces Projection Operator We can define a special operator $P = |\phi_{i}><\phi_{i}|$ called the projection operator $\Sigma_{i} P_{i} = Id$ You can rewrite any Hermetian operator A with eigenvalues $\alpha_{i}$ and a complete set of orthonormal eigenvectors $\phi_{i}$ as $A = \Sigma_{i} \alpha_{i} (\phi_{i}\phi_{i}^{\dag})$ Derived from seeing that the operator $A - \Sigma_{i} \alpha_{i} (\phi_{i}\phi_{i}^{\dag})$ annihilates any $\phi_{i}$ Symmetries Cannonical Transforms We know that canonical transformations are an important part of classical Hamiltonian mechanics. What is the analog in quantum? We want the probability to be conserved under a canonical transformation! (ie. $P(\Phi) = |(\Phi,\Phi)|^{2}$) We also want to preserve linearity (ie. $(\phi_{1}+\phi_{2})’ = \phi_{1}+\phi_{2}$) The only transformations which satisfy these properties are either Unitary or Anti-Unitary (Wigner’s Theorem) Unitary $U(a\phi+b\psi)= aU\phi + bU\psi$ (can applies either before or after linearity) $UU^{\dag} = I$, where the adjoint of an operator is defined as $(\psi O^{\dag} \phi) = (\phi O \psi)$ Needs to be invertible Operator O transforms like $U^{\dag}O U = O'$ transform inner product, do adjoint on the left, compare Suppose that you have some unitary operator U which is arbitrarily close to 1 (ie. $U = 1+i\epsilon T$) U need to maintain unitarity up to order $\epsilon$, which implies $T = T^{\dag}$, or that T is Hermetian T is called the generator of the symmetry Define $\epsilon = \frac{\theta}{N}$ where $\theta$ is some finite N-independent parameter. Imagine applying the symmetry transformation N times, and letting N go to infinity $(1+\frac{i\theta T}{N})^{N} = exp(i\theta T) = U(\theta)$ Under a symmetry transform $\phi \rightarrow U\phi$, any observable A must transform like $A \rightarrow U^{-1} A U$ For an infinitesimal transformation, A transforms like $A \rightarrow A-i\epsilon[T,A]$ Anti-Unitary $U(a\phi+b\psi)= a^{*}U\phi + b^{*}U\psi$ (ie. must maintain anti-linearity) $(U\phi, U\psi) = (\psi, \phi) = (\phi, \psi)^{*}$ Operators transform like $U^{-1} O U$ Parity Defined as $\Pi^{\dag} \vec{x} \Pi = -\vec{x}$ We also define $\Pi* \Pi = 1$, which implies $\Pi = \Pi^{-1}$ and that the eigenstates are $\pm 1$ Suppose that $\phi$ is an eigenstate of $\Pi$. Then $(\phi x \phi) = 0$ (Apply parity definition, transfer parity operator to states, apply eigen equation $\Pi \psi = \epsilon \psi$), and then note that you get $-q =q $, which implies q=0 As a consequence of this, if you Hamiltonian commutes with parity, then you know that you can write you state as a sum of even and odd states Rotations Defined by $\Sigma_{i} R_{ij}R_{ik} = \delta_{jk}$ Alternatively, any real linear transformation that leaves the scalar product $x\cdot y = \Sigma x_{i}y_{i}$ Alternatively any matrix which satisfies $R^{T}R = Id$ and $det(R)=1$ det(R)=-1 are spatial inversions like parity Imagine you have an operator V representing a vector observable (things like the coordinate vector X or the momentum vector P). A unitary rotation must act on this vector operator like $U^{-1}V_{i}U = \Sigma_{j} R_{ij}V_{j}$ For infinitesimal rotations, we know that unitarity must hold. We can write any infinitesimal rotation as $U(1+\omega) = 1+\frac{i}{2\hbar} \Sigma_{ij}\omega_{ij}J_{ij}$ where $\omega_{ij} = -\omega_{ji}$ and J is some set of Hermetian operators Can use composition of rotations in order to extract commutation relationships for J. In 3 dimensions $[J_{i},J_{j}] = i\hbar \Sigma_{k} \epsilon_{ijk} J_{k}$ More generally: $[J_{i},V_{j}] = i\hbar \Sigma_{k} \epsilon_{ijk} V_{k}$, where $V_{k}$ is any 3 dimensional vector operator Can define $S=J-L$, where S obeys the same commutation rules as J and L, but is independent of position and momentum operators. S also commutes with L, X and P Spin et Cetera Eigenvalues of $J^{2}$ and $J_{3}$ To derive the eigenvalues of $J^{2}$ and $J_{3}$, we have the following prescription: Realize that $J^{2}$ and $J_{3}$ commute with each other, which means they can have simultaneous eigenstates Define raising and lowering operators $J_{\pm} = J_{1}\pm iJ_{2}$ Realize that acting $J_{3}$ on $J_{\pm}\phi_{m}$ shows that $J_{\pm}\phi_{m}$ is an eigenvector of $J_{3}$ with eigenvalue $(m\pm 1)\hbar$ Realize that there is some min/max value of eigenvalues of $J_{3}$ since $J_{3}$ is a part of the vector of $J^{2}$ Define the min and max eigenstates, then act the raising and lowering operators on both of them (NOTE: j and j’ currently have no relationship to each other at this point in time) $(J_{1}+iJ_{2})\phi_{j} = 0$ $(J_{1}-iJ_{2})\phi_{j’} = 0$ Since the raising and lowering operators are atomic, we need to apply an integer number of them to transition from $\phi_{j}$ to $\phi_{j’}$, which implies that $j-j’$ must be a whole number Using the commutator relationships, you can show that $J_{\mp}J_{\pm} = J^{2}-J_{3}^{2}\mp \hbar J_{3}$ Can use the above to find the eigenvalues of $\phi_{j}$ and $\phi_{j’}$. Equating the eigenvalues to each other (since they both represent the same spectrum) yields that either $j’=-j$ or that $j’=j+1$ The second is impossible, since j’ is the minimum eigenvalue, and thus can’t be higher than the maximum Because the distance between j and j’ is an integer, j must be either an integer or a half integer To calculate the normalization of acting the raising and lowering operators on $\phi_{j,m}$, use $J_{\pm} \phi_{j,m} = \alpha_{\pm}(j,m)\phi_{j,m\pm 1}$, take the norm of this expression, use the commutators to show that $J_{\pm} \phi_{j,m} = \hbar \sqrt{j(j+1)-m^{2}\mp m} \phi_{j}^{m}$ The eigenstates for J are just the spherical harmonics Addition of Angular Momenta $\phi_{j,j’,j’’}^{m} = \Sigma_{m’m’’} C_{j’j’’}(jm;m’m’’)\phi_{j’,j’’}^{m’,m’’}$ The only non-vanishing Clebsch–Gordan coefficients occur when $m=m’+m’'$ To construct the Clebsch–Gordan coefficients, start at the highest state, imagine applying the lowering operator for each particle with the appropriate normalization, then recursively do this In practice, look this up in a table Wigner Eckart Theorem Suppose that you have a set of 2j+1 operators $O_{j}^{m}$, with m ranging from -j to j in integer increments Also suppose that this set of operators obeys the angular momemtum commutation relationships: $[J_{3},O_{j}^{m}] = m O_{j}^{m}$ $[J_{\pm},O_{j}^{m}] = \sqrt{j(j+1)-m^{2}\mp m} O_{j}^{m}$ We want to show that $(\phi_{j’’}^{m’’}O_{j}^{m}\psi_{j’}^{m’}) = C(j’’m’’;j’m’) (\phi || O|| \psi)$ This is the Wigner Eckart Theorem. What the hell does it mean? $O_{j}$ is a spherical tensor operator Spherical denotes the fact that we are using a spherical basis of coordinates (think spherical harmonics) Tensor implies that the set of $O_{j}$ transforms like a tensor (ie. applying a rotation on the set just yields a linear combination of the original set) The C are the Clebsch-Gordan coefficients between a given set of spherical harmonics The remaining element is called the reduced matrix element. The important part is that this proportionality factor is independent of m, m’, or which $O_{j}$ you use This allows you to calculate a bunch of matrix elements by only calculating one matrix element, then using the proportionality to “rotate” to another matrix element Indistinguishable Particles For simplicity, assume that we have two particles, whose total Hamiltonian is $H_{0}=H_{1}+H_{2}$. Naively, we would write the total wavefunction as $\phi_{0} = \phi_{1}\phi_{2}$, but if we have a boson or a fermion, this isn’t invariant under particle swap So, we need to create a more symmetric state: $\Phi = \frac{1}{\sqrt{2}}(\phi_{1}\phi_{2} \pm \phi_{2}\phi_{1})$ where the plus is for bosons and minus is for fermions Generalizing to N particles yields the idea of the Slater determinant. For bosons, you just turn all of the minus signs in the determinant to plus signs Calculating the probability density yields that $P = \int d^{3}\vec{x_{1}}d^{3}\vec{x_{2}} |\phi_{1}(x_{1})|^{2}\phi_{2}(x_{2})|^{2}+|\phi_{1}(x_{2})|^{2}\phi_{2}(x_{1})|^{2} \pm 2 Re(\phi_{1}(x_{1})\phi_{2}(x_{2})\phi_{1}^{*}(x_{2})\phi_{2}^{*}(x_{1}))$ The last term arises from the bosonic and fermionic interactions To generalize to N particles: Suppose that each particle has it’s own hamiltonian h and wavefunction $\phi$ which satisfy $h e_{\alpha} = E_{\alpha} e_{\alpha}$ and $\phi = \Sigma_{\alpha} \phi_{\alpha} e_{\alpha}$ We can write the total hamiltonian of N particles as the sum of each particle’s hamiltonian, and the wavefunction as $\psi = \Sigma \phi_{\alpha_{i}…\alpha_{n}} e_{alpha_{1}} \otimes … e_{alpha_{n}}$ If everything was distinguishable, then $\phi_{\alpha_{i}…\alpha_{n}} = \Sigma_{i=1}^{N} \phi_{i}(x_{i})$ What is the size of the Hilbert space of we have indistinguishable particles? We go by induction. With 0 particles, we have 1, represented by $\Omega_{\Sigma_{i=1}^{N} n_{i}}$ How to we add particles? we define a creation operator $a^{\dag}_{i}$ such that a new particle is created in the ith state an anihilation operator such that $a_{i}$ such that a particle is removed from the ith slot. If $n_{i}=0$, then the overall resulting state is 0 We can construct the cannonical bosonic commutator relationships $a_{i}a_{j}-a_{j}a_{i} = 0$ $a_{i}^{\dag}a_{j}^{\dag}-a_{j}^{\dag}a_{i}^{\dag} = 0$ $a_{i}a_{j}^{\dag}-a_{j}^{\dag}a_{i} = \delta_{ij}$ We can motivate these by examining $a_{j}a_{i}^{\dag}\Omega = \delta_{ij}$ and $a_{i}^{\dag}a_{j} \Omega =0$ We can also define the normalization from these commutators: $a_{i}^{\dag} \Omega_{…n_{i}…} = \sqrt{n_{i}+1} \Omega_{…(n_{i}+1)…}$ and $a_{i} \Omega_{…n_{i}…} = \sqrt{n_{i}} \Omega_{…(n_{i}-1)…}$ We can construct an analogous set of anticommutator relationships for the fermions $b_{i}b_{j}+b_{j}b_{i} = b_{i}^{\dag}b_{j}^{\dag}+b_{j}^{\dag}b_{i}^{\dag} = 0$ $b_{i}b_{j}^{\dag}+b_{j}^{\dag}b_{i} = \delta_{ij}$ Hydrogen Spectrum Derivation Classically, there is an additional vector conserved in central potential celestial mechanics called the Runge-Lenz vector $R = \frac{-Ze^{2}x}{r}+\frac{1}{2m}(p\times L - L \times p)$ This commutes with the Coulomb Hamiltonian $H = \frac{p^{2}}{2m}-\frac{Ze^{2}}{r}$ You can use a bunch of commutator relationships between x, p, and L in order to show that $R^{2} = Z^{2}e^{4}+\frac{2H}{m}(\vec{L}^{2}+\hbar^{2})$ Since H is in this expression and since R commutes with H, you can calculate the energies if you know the eigenvalues of $R^{2}$ $[R_{i},R_{j}] = \frac{-2i}{m}\hbar \Sigma_{k}\epsilon_{ijk} H L_{k}$ from a bunch of commutator identities Since R is a vector, we also get that $[L_{i}, R_{j}] = i\hbar \Sigma_{k} \epsilon_{ijk} R_{k}$ We can define raising and lowering operators $A_{\pm} = \frac{1}{2}(L\pm \sqrt{\frac{m}{-2H}} R)$ which obeys the following commutation relationships $[A_{\pm i}, A_{\pm j}] i\hbar \Sigma_{k}\epsilon_{ijk} A_{\pm k}$ $[A_{\pm i}, A_{\pm,j}] = 0$ Since we know that $R\cdot L=0$, we can see that $A_{\pm}^{2} = \frac{1}{4}(L^{2}+\frac{m}{-2H}R^{2})$ From the angular momentum eigenvalue derivation, we know that the eigenvalues of $A_{\pm}^{2}$ take the form of $hbar^{2}a(a+1)$. Defining a principle quantum number as $n = 2a+1 = 1,2,3…$, once can show that $E = \frac{-Z^{2}e^{4}m}{2\hbar^{2}n^{2}}$, just like the standard Bohr atom derivation This also gives the degeneracy at each level to be $n^{2}$ Approximations for Energy Eigenvalues 1st Order Non-degenerate Time Independent Perturbation Theory We have some unperturbed Hamiltonian with some orthonormal eigenvectors $H_{0}\phi_{a} = E_{a}\phi_{a}$ $(\phi_{a},\phi_{b}) = \delta_{ab}$ Suppose that we have some small perturbation to the Hamiltonian called $\delta H$ which is proportional to some $\epsilon$ This produces a change in the state vector $\delta_{1} \phi_{a}$ and some small change in energy $\delta_{1} E_{a}$ Make these perturbations to all of the variable, eliminate terms of order $\epsilon^{2}$ and higher, contract the entire expression with $\phi_{a}$ and eliminate like terms to get $\delta_{i} E_{n} = (\phi_{a}, \delta H \phi_{a})$ This is fine and dandy if you have no degenerate eigenstates. This problem can be made lucid by contracting with $\phi_{b}$ instead of $\phi_{a}$ you end up with $(\phi_{b} \delta H \phi_{a}) = (E_{a}-E_{b}) (\phi_{b} \delta \phi_{a})$ If a and b are degenerate,then you have $(\phi_{b} \delta H \phi_{a})=0$, which is no gaurenteed The work around for this solution is to diagonalize the degenerate subspace in some new orthonormal basis In math terms, this means you diagonalize the subspace such that $(\phi_{b}, \delta H \phi_{a}) = 0$ Given the above $(\phi_{b}, \delta_{1}\phi_{a}) = \frac{(\phi_{b},\delta H \phi_{a})}{E_{a}-E_{b}}$ The above needs to be normalized which gives the condition $0 = (\phi_{a} \delta_{i} \phi_{a})$ Hence, the $\delta_{1} \phi_{a} = \Sigma_{b\neq a} \phi_{b} \frac{(\phi_{b}\delta H \phi_{a})}{E_{a}-E_{b}}$ Zeeman Effect The shift of atomic energies in the presence of an external magnetic field is called the Zeeman effect We can define the peturbation as $\delta H = \frac{e}{2m_{e}c} \vec{B} \cdot (\vec{L} + g_{e} \vec{S})$ Comes from classical contribution to energy $\frac{e}{2m_{e}c} \vec{B} \cdot \vec{L}$ as well as quantum spin term with the g factor which is ~ 2 Can use Wignar-Eichart theorem $(\phi_{nlj}^{m’}, (L+g_{e}S) \phi_{nlj}^{m}) = g_{njl} (\phi_{nlj}^{m’}J \phi_{nlj}^{m})$ where $g_{njl}$ is the Lande g-factor J commutes with $J^{2}$, so you write the eigenvector $J\phi_{njl}^{m}$ as a linear combination of $\phi_{nlj}^{m’’}$ with various m'' In math: $\Sigma_{i} (\phi_{nlj},(L+g_{e}S) \phi_{nlj}^{m}) = g_{njl} \Sigma_{i} (\phi_{nlj}^{m’}J_{i}J_{i} \phi_{nlj}^{m})$ Can use identities that $J = L+S$, and eigenspectrum of J,L, and S to show that $g_{njl} = 1+(g_{e}-1)(\frac{j(j+1)-l(l+1)+\frac{3}{4}}{2j(j+1)})$ Note that $g_{njl}$ is independent of n This means that we need to compute the matrix elements $(\phi_{njl}^{m’}, \delta H \phi_{njl}^{m}) = \frac{eg_{l}}{2m_{e}c}(\phi_{njl}^{m’}, \vec{B}\cdot \vec{J} \phi_{njl}^{m})$ Choose a coordinate system such that the z axis aligns with B. This implies that $BJ_{z} \phi_{njl}^{m} = \hbar m B \phi_{njl}^{m}$ This gives that $\delta E_{njlm} = \frac{e\hbar g_{jl} B}{2 m_{e} c} m$ Second Order Perturbation Theory You need 2nd order perturbation theory if the 1st order vanishes, or if you want more precision Expanding to 2nd order, we find that $H_{0} \delta_{2} \phi_{a} + \delta H \delta_{1} \phi_{a} = E_{a} \delta_{2} \phi_{a} + \delta_{1} E_{a} \delta_{1} \phi_{a} + \delta_{2} E_{a} \phi_{a}$ We know what $\delta_{1}$ and $\delta_{1} \phi_{a}$ Using a similar procedure as 1st order perturbation, we see that $\delta_{2} E_{a} = \Sigma_{b\neq a} \frac{|(\phi_{b} \delta H \phi_{a})|^{2}}{E_{a}-E_{b}}$ You diagonalize degenerate subspaces as needed Variational Method Let $\phi$ be some normalizable wave function $(\phi H \phi) \geq E_{ground}$ This sets an upper bound on the ground state energy Born-Oppenheimer Approximation Consider a molecule with N nuclei and n electrons. The Hamiltonian becomes $H = \Sigma_{\alpha=1}^{N}\frac{p_{\alpha}^{2}}{2M_{\alpha}}+V_{NN}(\vec{R})+\Sigma_{i=1}^{n}\frac{p^{2}}{2m}+V_{eN}(\vec{R},\vec{r}) +V_{ee}(\vec{r})$ $\alpha$ variables are associated with the nuclei and i variables are associated with the electrons We consider the nuclei essentially fixed (ie. $\vec{R}$ is fixed). This gives rise to a Schrodinger equation for the electrons: $(-\frac{\hbar^{2}}{2m}\Sigma_{i=1}^{n} \Delta_{\vec{r_{i}}}^{2} + V_{eN}(\vec{R},\vec{r})+V_{ee}(\vec{r}))\phi_{\vec{R}}^{i}(\vec{r})= E_{e}^{i}(\vec{R})\phi_{\vec{R}}^{i}(\vec{r})$ We can try and write the full ansatz as $\Phi(\vec{R},\vec{r}) = \Sigma_{i} \eta(\vec{R})^{i} \phi_{\vec{R}}^{i}(\vec{r})$ This is too complicated to solve, so we make the simpler ansatz $\Phi(R,r) = \eta(R)\phi(R,r)$ Time Dependent Perturbation Theory (TDPT) Suppose that you have some Hamiltonian $H(t) = H_{0}+H’(t)$ where $H’(t)$ is small compared to $H_{0}$ The eigenvector satisfies $i\hbar \frac{d\phi}{dt} = H(t) \phi(t)$, where $\phi_{n}$ is a orthonormal basis of time-independent unperturbed eigenvectors of $H_{0}$ The time evolution is then $\phi(t) = \Sigma_{n} c_{n}(t) exp(-\frac{-iE_{n}t}{\hbar}) \phi_{n}$ The perturbation acting on $\phi_{n}$ can be expanded as a linear combination of the original Hamiltonian eigenstates $H’(t) \phi_{n} = \Sigma_{m} \phi_{m}(\phi_{m},H’(t) \phi_{n}) = \Sigma_{m}H_{mn}’(t) \phi_{m}$ where $H_{mn}’ = (\phi_{m}, H’(t)\phi_{n})$ Plugging in the above equations for $\phi$ and $H’$ acting on $\phi$ into the Schrodinger equation for the total Hamiltonian, you can Cancel out the terms proportional to $E_{n}$, interchange the labels m and n on the RHS, and equate the coefficients of $\phi_{n}$ to eventually get $i\hbar \frac{dc_{n}(t)}{dt} = \Sigma_{m} H_{nm}’ c_{m}(t) exp(i\frac{(E_{n}-E_{m})t}{\hbar})$ To make a perturbative approximation, we see that the rate of change of $c_{n}$ is proportional to the perturbation (ie. $H_{nm}’$). Hence, we can replace $c_{m}(t) \approx c_{m}(0)$ (ie. $c_{m}$ doesn’t change that fast…). This allows ups to solve the differential equation $c_{n}(t) \approx c_n(0)-\frac{i}{\hbar}\Sigma_{m} c_{m}(0) \int_{0}^{t} dt’H_{nm}’(t’) exp(i\frac{(E_{n}-E_{m})t’}{\hbar})$ TDPT Interaction Picture Same starting point: $H = H_{0}+V(t)$ and $i \frac{d}{dt}H = H \phi$ Assuming $H_{0}$ is time-independent, define the time evolution operator $U(t) = exp(itH_{0})$. This obeys $i\frac{d}{dt}U = H_{0}U$ Act the operator differential equation in $\phi$, then multiply by $U^{-1}$ to get $i \frac{d}{dt}\phi = U^{-1} V U \phi = V_{I}\phi$ Hence, we get a differential operator equation for the interaction potential. We can solve this for U: $U_{I}(T)-1 = -i \int_{0}^{T} dt V_{I} U_{I}$ This is equivalent to the Schrodinger picture, but now you assign the evolution to the operators, not the state Monochromatic Perturbations Assume that $H’(t) = U exp(-i\omega t) + U^{\dag} exp(i\omega t)$ where U is some unitary operator Think of this as a sinusoidal perturbation with angular frequency $\omega$ Using time dependent perturbation theory (straightforward integrals), you can write $c_{n}(t) = c_{n}(0)+\Sigma_{m} U_{nm} c_{m}(0) (\frac{exp(i\frac{(E_{n}-E_{m}-\hbar\omega)t}{\hbar})-1}{E_{n}-E_{m}-\hbar\omega})+U_{nm}^{*} c_{m}(0) (\frac{exp(i\frac{(E_{n}-E_{m}+\hbar\omega)t}{\hbar})-1}{E_{n}-E_{m}+\hbar\omega})$ Examine the case where $c_{n}(0)=0$ for all n except n=1. Then $c_{n}(t)$ for $n \neq 1$ is $c_{n}(t)U_{n1}\frac{exp(i\frac{(E_{n}-E_{m}-\hbar \omega)t}{\hbar})}{E_{n}-E_{m}-\hbar \omega}+c_{n}(t)U_{n1}^{*}\frac{exp(i\frac{(E_{n}-E_{m}+\hbar \omega)t}{\hbar})}{E_{n}-E_{m}+\hbar \omega}$ At t=0, both terms vanish Both terms increase from time 0 to time $t = \frac{i(E_{n}-E_{1}-\hbar\omega)t}{\hbar}$, after which they just oscillate Suppose that $E_{n} \approx E_{1}+\hbar\omega$ This corresponds to the absorption of one quanta of energy This means that the ramp up time can be very long This causes the 2nd term to fall off much faster than the 1st term After a long enough time: $|c_{n}(t)|^{2} \approx 4 |U_{n1}|^{2}\frac{\sin^{2}(\frac{(E_{n}-E_{1}-\hbar \omega)t}{2\hbar})}{(E_{n}-E_{1}-\hbar \omega)^{2}}$ For very large times, we can approximate $\frac{2\hbar \sin^{2}(\frac{Wt}{2\hbar})}{\pi t W^{2}} \approx \delta(W)$ where $W = E_{1}+\hbar\omega -E_{n}$ Hence: $|c_{n}(t)|^{2} = 4|U_{n1}|^{2} \frac{\pi t}{2\hbar} \delta(E_{1}+\hbar \omega -E_{n})$ We can define the transition rate $\Gamma = \frac{|c_{n}(t)|^{2}}{t}=\frac{2\pi}{\hbar}|U_{1n}|^{2} \delta(E_{n}+\hbar\omega-E_{1})$ This is Fermi’s Golden Rule We can extend this derivation to a continuous final spectrum of the Hamiltonian Imagine putting the system in a big box with sides of length L with periodic boundary conditions (ie. $\phi(x+c) = \phi(x)$). The solution to this system has eigenstates $\phi(\vec{x}) = \frac{1}{\sqrt{L^{3}}} exp(i\vec{k}\cdot \vec{x})$ where $k_{i} = \frac{2\pi n_x}{L}$ The phase space volume is then $d^{3} k = \frac{(2\pi)^{3}}{L^{3}} d^{3}n$ We occasionally want to write $d^{3}n = \rho(E) dE$, where $\rho(E)$ is the density of states of the system (ie. it’s easier to work with hypersurfaces of E than 3 components of momentum) Since $E = \frac{\hbar^{2}k^{2}}{2m}$, we can write the momentum integral in spherical coordinates we can show that $\rho(E) = (\frac{L}{2\pi})^{2} \frac{m}{\hbar^{2}} k d \Omega$ The final transition rate is just summing over this continuum of final states. Each little volume of momentum space has some density associated with it. Hence $\Gamma_{i\rightarrow f} = \int P_{i\rightarrow f}(t_{0}) \rho(E_{f}) dE_{f}$ Ionization by an EM Wave Consider a hydrogen atom in the ground state placed in a light wave. If $\lambda » a$ (ie. the wavelength of light is much larger than the Bohr radius), then the perturbation Hamiltonian only depends on the E field at the location of the atom (neglecting magnetic field on non-relativistic charged particles b/c those are negligible in comparison) We can write this perturbation as $H’(t) = e\Epsilon\cdot X exp(i\omega t) + e\Epsilon^{*} \cdot X exp(i\omega t)$ $\Epsilon$ is a constant, and X is the electron position operator We can make the association $U = e\Epsilon X_{3}$ if we align $\Epsilon$ with the $X_{3}$ direction We want to ionize the electron (ie. rip it from the atom). Hence, our initial state is the ground state of the hydrogen atom $\phi_{1s}(x) = \frac{\exp(\frac{-r}{a})}{\sqrt{\pi a^{3}}}$ and our final state is a free electron with momentum $\hbar k_{e}$ (ie. $\phi_{e}(x) = \frac{exp(ik_{e}\cdot x)}{(2\pi\hbar)^{\frac{-3}{2}}}$) We are assuming that the final electron energy is much larger than the hydrogen binding energy The matrix element takes the form $U_{e,1s} = \frac{e\Epsilon}{(2\pi\hbar)^{\frac{3}{2}}\sqrt{\pi a^{3}}}\int d^{3}x e^{-i\vec{k_{e}}\cdot \vec{x}} x_{3} exp(\frac{-r}{a})$ This can be solved by examining the integral $\int d^{3}x e^{-i\vec{k_{e}}\cdot \vec{x}} f(r)$ Transform to spherical coordinates, integrate out the spherical components to $4\pi$, take the derivative w.r.t. $k_{3}$, then solve the remaining straightforward (if tedious) integrals The final result is $\frac{-4\pi ie\Epsilon k_{e3}}{k_{e}^{3}(2\pi\hbar)^{\frac{3}{2}}}\sqrt{\pi a^{3}} \frac{8k_{e}a^{5}}{(1+k_{e}^{2}a^{2})^{3}}$ You can make the approximation $k_{e}^{2}a^{2} » 1$ since the final electron energy is large In this limit, we have $U_{e,1s} = \frac{-8 \sqrt{2}i e \Epsilon \cos \theta}{\pi \hbar^{\frac{3}{2}}k_{e}^{5}a^{\frac{5}{2}}}$ where $\cos \theta$ represents the angle between $k_{e}$ and the polarization vector of the EM wave The differential ionization rate then becomes: $d\Gamma(1\rightarrow k_{e}) = \frac{2\pi}{\hbar}|U|^{2} \delta(\hbar c k_{\gamma} -E_{e}) \hbar^{3} k_{e}^{2} dk_{e} d\Omega$ You can integrate out $k_{e}$ dependency via the delta function Fluctuation Perturbations Imagine that we have some perturbation which fluctuates randomly in time We define $\overline{H_{nm}’(t_{1})H_{nm}’^{*}(t_{2}) }= f_{nm}(t_{1}-t_{2})$ In words, we assume that the correlation between two points in time only depends on the differences in times Assuming $c_{n}(0) = \delta_{n1}$, we have that $\overline{|c_{n}(t)|^{2}} = \frac{1}{\hbar}\int_{0}^{t} dt_{1} \int_{0}^{t} dt_{2} f_{n1}(t_{1}-t_{2}) exp(i\frac{(E_{n}-E_{1})(t_{1}-t_{2})}{\hbar})$ We can Fourier transform $f_{nm}$ to $F_{nm}$, use Fubini’s theorem to convert the two time integrals into the modulus squared of a single time integral over $t_{1}$, and then use the approximation as in previous sections to get $\Gamma(1\rightarrow n ) = \frac{\overline{|c_{n}(t)|^{2}}}{t} = \frac{2\pi}{\hbar} F_{n1}(\frac{E_{n}-E_{1}}{\hbar})$ where $F_{nm}(\Omega)$ is the Fourier transform Absorption and Stimulated Emission of Radiation $H_{nm}’(t) = e \Sigma_{N} <x_{N}>_{nm} E(t)$ where N denotes the position of each electron in the atom Good approximation since scale of E field variation is much larger than atom radius $<x_{N}>_{nm} = \int \phi _{n}^{*}(x) x _{N} \phi _{m}(x) \Pi _{m} d^{3}x _{M}$ Assume correlation of E field goes like $\overline{E_{i}(t_{1})E_{j}(t_{2})} = \delta_{ij} \int_{-\infty}^{\infty} d\omega P(\omega) exp(-i\omega (t_{1}-t_{2}))$ You can write the average correlation of the perturbation as a fourier transform, with $F_{nm}(\omega) = e^2|\Sigma_{N} <x_{N}>_{nm}|^{2}P(\omega)$ Using $\delta_{ij}$ assumes that there is no preferred direction for the electric field $P(\omega) = P(-\omega)$ via realness of E field This allows us to define the correlation function of the perturbation, which in turn allows us to define $F_{nm}(\omega) = e^{2} | \Sigma_{N} <x_{N}>_{nm} |^{2} P(\omega)$ As per the previous section, we can write the transition rate from 1 to n as $\Gamma(1\rightarrow n) = \frac{2\pi e^{2}}{\hbar^{2}} | \Sigma_{N} <x_{N}>_{nm} |^{2} P(\omega)$ Adiabatic Approximation Suppose that the Hamiltonian depends on some set of parameters s which are slowly varying functions in time (ie. s(t)) We can find the solution of the time-dependent Schrodinger equation by use of the adiabatic approximation The fundamental results of the adiabatic theorem are: The state that you are currently in does not change, but the eigenstates you are in do change You pick up an overall phase from two source: a dynamical phase and a Berry phase For an adiabatic approximation, we want to show that $H(t) |\phi_{n}(t)> = E_{n}(t)| \phi_{n}(t)>$ where the $|\phi_{n}(t)>$ form an orthonormal basis for all t Can think of $\phi_{n}(t)$ as some instantaneous eigenstates which vary with time (this is the state that you are stuck at) Importantly, $\phi_{n}(t)$ does not solve the Schrodinger equation in general. We want to make it approximately solve it though We are also assuming that there is no degeneracy in the problem We can decompose any solution to the Schrodinger equation in this orthonormal basis: $\Phi(t) = \Sigma_{n} c_{n}(t) \phi_{n}(t)$ Plug this into the Schrodinger equation and act with $<\phi_{k}(t)|$ from the left to yield $i\hbar \dot{c_{k}} = (E_{k}-i\hbar <\phi_{k}|\dot{\phi_{k}}> c_{k}) - i\hbar \Sigma_{n\neq k} <\phi_{k}|\dot{\phi_{n}}> c_{n}$ We can show that $<\phi_{k}(t) \dot{\phi_{n}} = \frac{<\phi_{k}|\dot{H}(t)| \phi_{n}>}{E_{n}(t) E_{k}(t)} = \frac{\dot{H_{kn}}}{E_{n}-E_{k}}$ Take the time derivative of $H(t) |\phi_{n}(t)> = E_{n}(t)| \phi_{n}(t)>$, contract with $<\phi_{k}(t)$ from the left, then rearrange as needed Since we are assuming an adiabatic process, then $\dot{H_{kn}}$ is approximately 0. This leaves $i\hbar \dot{c_{k}} = (E_{k}-i\hbar <\phi_{k}|\dot{\phi_{k}} c_{k})$, which can be solved to yield $c_{k}(t) = c_{k}(0) e^{i\theta_{k}(t)}e^{i\gamma_{k}(t)}$ $\theta_{k}(t) = \frac{-1}{\hbar}\int_{0}^{t} E_{k}(t’) dt'$ $\gamma_{k}(t) = i \int_{0}^{t} <\phi_{k}| \dot{\phi_{k}}> dt'$ Berry Phase $\gamma_{k}(t)$ is geometric in nature. It depends on the path through parameter space of the Hamiltonian from s(0) to s(t), but not on the time-dependence of travel along this path So you could take a nanosecond or a year to travel between two points in configuration space. $\gamma_{k}(t)$ is the same in either case! You can see this by using the chain rule to expand $ |\dot{\phi_{k}(\vec{s}(t))}>$ as $\nabla_{s} \phi \cdot \frac{\vec{s}}{dt}$. When you plug this into $\gamma$, the dt cancel, leaving you with an integral over the path in configuration space (call this $\Gamma$) In math: $\gamma_{n}(\Gamma) = \int_{\Gamma} i <\phi_{n}(\vec{s})| \nabla_{s} | \phi_{n}(\vec{s}) \cdot d \vec{s}$ $i <\phi_{n}(\vec{s})| \nabla_{s} | \phi_{n}(\vec{s}) = \vec{A_{n}}(s)$ is called the Berry connection. This is not unique. if we change our eigenstates by a arbitrary phase $e^{i\beta(\vec{s})}|\phi_{n}>$, then the Berry connection goes like $\vec{A_{n}}(s) = \vec{A_{n}}(s)+ \nabla_{s}(\beta)$ So in this configuration space, A transforms like a vector potential This also implies that $\gamma’ = \gamma+ \beta(R_{f})-\beta(R_{i})$ From this, if $\Gamma$ is closed, then this phase becomes observable (since then this choice in wavefunction phase doesn’t show up). This is called Berry’s phase Since $<\phi_{k}| \dot{\phi_{k}}> $ is imaginary, then no Berry’s phase occurs if $\phi_{k}$ is real If we are in 1D, then there is no Berry’s phase Scattering Scattering as Time-Dependent Perturbation We can imagine our base Hamiltonian as $H_{0}+V(x)$ where $H_{0} = \frac{p^{2}}{2m}$ whose eigenvalues are $E_{k} = \frac{\hbar^{2}k^{2}}{2m}$ and whose eigenvectors are plane waves We can view scattering as a perturbation which is “turned on” when the potential is large enough The transition from an initial plane-wave to a final plane wave can be written as: $<n|U_{t}|i> = \delta_{ni}-\frac{i}{\hbar}\Sigma_{m}<n|V|m> \int_{t_{0}}^{t} exp(i\omega_{nm}t’) <m|U_{t}(t’,t_{0})|i> dt'$ Since our initial and final states are asymptotic (ie. at $\pm \infty$), we need to massage the integral a bit to prevent things from exploding. We define the transition matrix T as $<n|U_{t}|i> = \delta_{ni}-\frac{i}{\hbar} T_{ni} \int_{t_{0}}^{t} exp(i\omega_{ni}t’+\epsilon t’)$ where $\epsilon >0$ and $t« \frac{1}{\epsilon}$ Prevents blow up at both extremes From the above, we can define the scattering matrix as $S = lim_{t \rightarrow \infty} \lim_{\epsilon\rightarrow 0 } <n|U(t,-\infty) | i> = \delta_{ni} - \frac{i}{\hbar} T_{ni} \infty_{-\infty}^{\infty} exp(i\omega_{ni} t’) dt’ = \delta_{ni}-2\pi i \delta(E_{n}-E_{i}) T_{ni}$ Transition Rates and Cross Sections The transition rate is defined as $w(i\rightarrow n) = \frac{d}{dt}|<n|U(t,-\infty)|\i>|^{2}$ Plugging in the above definition and ensuring $i\neq n$, you can take the appropriate limits (first $\epsilon$, and then $t$), you eventually get that $w(i\rightarrow n) = \frac{2\pi}{\hbar}|T_{ni}|^{2} \delta(E_{n}E_{i})$ Like with Fermi’s Golden Rule, we can integrate over the density of states: $\rho(E) = \frac{mk}{\hbar^{2}}(\frac{L}{2\pi})^{3} d\Omega$ which results in $w(i\rightarrow n) = \frac{mkL^{3}}{(2\pi)^{2}\hbar^{3}}|T_{ni}|^{2} d\Omega$ The cross section is then just the transition rate divided by the flux. In this case: $j = \frac{\hbar k}{m L^{3}}$ and $\frac{d\omega}{d\Omega} = (\frac{mL^{3}}{2\pi \hbar^{2}})^{2}|T_{ni}|^{2}$ General Scattering Far from the potential, we can write the solution as a superposition of wavepackets: $\phi_{g}(t) = \int d^{3}k g(\vec{k}) exp(\frac{-i \hbar t |k|^{2}}{2\mu}) \phi_{k}^{in}$ where $\phi_{k}^{in}$ is the particular solution of the eigenvalue $H \phi_{k}^{in} = \frac{\hbar^{2}k^{2}}{2\mu} \phi_{k}^{in}$ In the limit as $t\rightarrow \infty$, we have that $\phi_{k}^{in} = \Phi_{k}$, where $\Phi_{k}$ is the eigenvector of the momentum operator $P \Phi_{k} = \hbar k \Phi_{k}$ In this limit, $\Phi_{k}$ is also an eigenvector of the free Hamiltonian $H_{0}$ We want to satisfy the above limit. We can rewrite the Schrodinger equation as an integral equation: $(E|k|-H_{0})\phi_{k}^{in} = V \phi_{k}^{in}$ The solution to this is $\phi_{k}^{in} = \Phi_{k}+(E(|k|)-H_{0}+i\epsilon)^{-1} V \phi_{k}^{in}$ This is known as the Lippmann-Schwinger equation The $\epsilon$ is a math trick to let you do contour integration Solving Lippmann-Schwinger We consider the incident particle as a linear superposition of plane waves: $V \Psi_{k}^{in} = \hbar^{3} \int d^{3}q \Phi_{q}<\Phi_{q}|V\Psi_{k}^{in}>$ where $|\Phi_{q} = exp(i \vec{q}\cdot \vec{x})$ We define the Green’s function of the Lippmann-Schwinger equation as $G(\vec{x}-\vec{y}) = <\Phi_{x}| (E(|k|)-H_{0}+i\epsilon)^{-1} | \Phi_{y}>$. The full solution is then given by $\phi_{k}(\vec{x}) = (2\pi \hbar)^{-\frac{3}{2}} e^{i\vec{k}\cdot \vec{x}}+ \int d^{3} y G_{k}(x-y) V(y) \phi_{k}(y)$ $G(\vec{x}-\vec{y}) = \int \frac{d^{3}q}{(2\pi\hbar)^{\frac{3}{2}}} \frac{exp(i \vec{q}\cdot(\vec{x}-\vec{y}))}{E(k)-E(q)+i\epsilon} = \frac{-2\mu}{\hbar^{2}} \frac{e^{ik|x-y|}}{4\pi |x-y|}$ For potentials which fall off fast enough: $\phi_{k}(\vec{x}) = (2\pi \hbar)^{-\frac{3}{2}} (exp(i\vec{k}\cdot \vec{x})f_{k}(\hat{x})\frac{e^{ikr}}{r})$ $f_{k}(\hat{x})$ is the scattering amplitude, and is given by $f_{k}(\hat{x}) = \frac{-\mu}{2\pi \hbar^{2}}(2\pi\hbar)^{\frac{3}{2}} \int d^{3}y e^{-ik \hat{x}\cdot \vec{y}} V(y) \phi_{k}(y)$ We can get the differential cross section via $\frac{\sigma(\hat{x},\vec{k_{0}})}{d\Omega} = |f_{k}|^{2}$ Basic Setup for Spherical Symmetry Why do we care? If we have a complicated potential, then one way we can tease out information about it is to launch particles at it and see how they scatter off of the target Some restrictions We study processes of the form $a+b \rightarrow a+b$ We will look at elastic scattering (ie. the internal states of the particles does not change) For now, we ignore spin Non-relativistic regime (no spawning additional particles) Assume potentials only depend on relative distance between particles (so V(r)) Our Hamiltonian becomes $H = \frac{p^{2}}{2m}+V(r)$ We send in a free-particle wavefunction, with energy $E= \frac{\hbar^{2}k^{2}}{2m}$ since at very long distances, the effect of the potential vanishes For V=0, we have solutions $\phi = e^{i\vec{k}\cdot \vec{x}}$ We also have a spherically symmetric solution $\phi = f(\theta,\phi) \frac{e^{ikr}}{r}$ for the resulting outward propagation when you are very far away from the center ($\frac{1}{r}$ for conservation of probability, and $f(\theta,\phi)$ for generality) Hence, we expect the solution to this scattering problem to look like a super position between the incident and outgoing waves: $\Phi(r»a) = e^{ikz}+ f(\theta,\phi) \frac{e^{ikr}}{r}$ We align our coordinate system with the z axis $f(\theta,\phi)$ encodes the cross section of the process $\frac{d\sigma}{d\Omega}$ In words, $\sigma(\theta,\phi)$ is the number of particles scattered per unit time into the solid angle $d\Omega$ at $(\theta,\phi)$ divided by the flux of the incident particles For the above problem, we can write the incident particle fux as $\frac{k}{m} Im(\phi^{*} \nabla \phi) = \frac{\hbar k}{m} \hat{z}$ where $\phi = e^{ikz}$ A similar probability current calculation yields the outgoing flux. Can make a physical argument about what it is though Imagine a little pill box at distance r with thickness $dr$ which spans angles $\phi$ and $\theta$ The number of particles inside the box is $|\frac{f(\theta,\phi)e^{ikr}}{r}|^{2} r^{2} d\Omega dr$ (ie. multiply the probability of the outgoing wavefunction by the differential volume of the box) The particles spend $dt = \frac{dr}{v} = \frac{dr}{\frac{\hbar k}{m}}$ inside the box Taking the ratio of the two yields $\frac{dn}{dt} = \frac{\hbar k }{m}|f(\theta,\phi)|^{2} d\Omega$ Hence, we can write the differential cross section as ${d\sigma} = |f(\theta,\phi)|^{2} {d\Omega}$ Solving the Problem We don’t expect any $\phi$ dependency due to spherical symmetry around the z axis For a free particle in spherical coordinates, the radial portion takes the form $\frac{u_{l}}{r} = A_{l}\rho j_{l}(\rho)+ B_{l} \rho n_{l}(\rho)$ $\rho = kr$ j and n and the spherical bessel functions of the 1st and 2nd kind This allows us to write the incoming plane wave in terms of the spherical free particle solution: $e^{ikz} = e^{ikr\cos\theta} = \sqrt{4\pi} \Sigma_{l=0}^{\infty} \sqrt{2l+1} i^{l} Y_{l,0}(\theta) J_{l}(kr)$ We drop $n_{l}$ since those diverge as r goes to 0 For large values of r, we can expand $j_{l}(x) \approx \frac{1}{x} \sin(x-\frac{l\pi}{2})$ for $x=rk$, we have $\frac{1}{k}(\frac{e^{i(kr-\frac{l\pi}{2})}}{r}-\frac{e^{-i(kr-\frac{l\pi}{2})}}{r})$ Based on all of the above, we make the ansatz that the final wavefunction has the form $\Phi(r) = \sqrt{\frac{4\pi}{k}} \Sigma_{l=0}^{\infty} \sqrt{2l+1} i^{l} Y_{l0} \frac{1}{2i} (\frac{exp(i(kr-\frac{l\pi}{2})+2i\delta_{l})}{r}- \frac{exp(-i(kr- \frac{l\pi}{2}))}{r})$, where each outgoing l mode is shifted by some phase $\delta_{l}$ The physical argument for this is that conservation of probability forces the incoming and outgoing waves to have the same amplitudes, but doesn’t constrain the phase at all We can then write $f_{k}(\theta) = \frac{\sqrt{4\pi}}{k} \Sigma_{l=0}^{\infty} \sqrt{2l+1} Y_{l0}(\theta) e^{i\delta_{l}} \sin \delta_{l}$ Plugging the above into the cross section formula $\sigma = \int |f_{k}(\theta)|^{2} d\Omega$, which you can easily solve using the orthogonality of the spherical harmonics: $\sigma = \frac{4\pi}{k^{2}} \Sigma_{l=0}^{\infty} (2l+1) \sin^{2}(\delta_{l})$ AT very large distances, we can calculate these $\delta_{l}$ as $\tan \delta_{l} = \frac{-B_{l}}{A_{l}}$ Example: Hard Sphere We have the potential $V(r\leq a) = \infty$ and 0 else where The general solution outside the sphere is once again $\phi(r,\theta) = \Sigma_{l} (A_{l}j_{l}(kr)+ B_{l} n_{l} (kr)) P_{l}(\cos \theta)$ This wavefunction needs to vanish at $r=a$ due to the hard sphere This implies $(A_{l}j_{l}(ka)+ B_{l} n_{l} (ka)) = 0$ for all l due to the independence of the $P_{l}$ This implies $\tan \delta_{l} = \frac{j_{l}(ka)}{n_{l}(ka)}$ or alternatively $\sin^{2} \delta_{l} = \frac{j_{l}^{2}}{j_{l}^{2}+n_{l}^{2}}$ Optical Theorem This is a general statement, whose name makes more sense in an EM context For simplicity, you have a sphere, and you shine a light on it (re. EM plane waves). This creates a shadow where no light exists. Think in the particle picture: the shadow represents a region where no photons got scattered to. Hence the dearth of photons in the shadow needs to be compensated by an increase in the scattered photons in other regions Going back to quantum, what is the scattering amplitude along this forward direction? $Y_{l0}(\theta) = \sqrt{\frac{2l+1}{4\pi}} P_{l}(\cos(\theta))$. At $\theta=0$, $P_{l}(1) = 1$, hence $f_{k}(\theta=0) = \frac{1}{k}\Sigma_{l} (2l+1) e^{i\delta} \sin \delta$ If you take the imaginary part of the above, you pick up a $\sin^{2}\delta_{l}$ in the sum, which let’s you relate the cross section to the forward scattering amplitude: $\sigma = \frac{4\pi}{k} Im(f_{k}(\theta=0))$ Born-Approximation This approximation holds for “weak” potentials. More precisely, if relevant matrix elements of the potential V are much less than typical matrix elements of the kinetic energy $H_{0}$, then we can approximate the incoming particle as a free particle wave function $\phi_{in} = \frac{1}{(2\pi\hbar)^{-\frac{3}{2}}}exp(i\vec{k}\cdot \vec{x})$ This implies that $f_{k}(\hat{x}) = -\frac{\mu}{2\pi \hbar^{2}} \int d^{3}y V(y) exp(i(\vec{k}-k\hat{x})\cdot \vec{y})$ Yukawa Potential Example Assume that $V(r) = \frac{Z_{1}Z_{2}e^{2}}{r}exp(-\kappa r)$ is our potential of interest Like a shielded Coulomb potential In the Born approximation, we need to solve: $f_{k}(\theta, \phi) = -\frac{2\mu Z_{1}Z_{2}e^{2}}{q\hbar^{2}}\int_{0}^{\infty} dr exp(-\kappa r)\sin (qr)$ where $q = |\vec{k}-k\hat{x}|$ Eikonal Approximation Fundamental assumption is that V(x) varies very little over some length scale $\lambda$. V doesn’t need to be small like in Born This allows use to represent the wave function as $|\Phi> = Exp(\frac{i S(x)}{\hbar})$ where S satisfies the Hamilton Jacobi equation $\frac{(\nabla S)^{2}}{2m} + V = E = \frac{\hbar^{2}k^{2}}{2m}$ We can calculate S by assuminig that the “trajectory” of the particle is a straight line, which holds for small deflections at high energies Aligning our coordinate system with the z-direction, we have that $\frac{S}{\hbar} = \int_{\infty}^{z} (k^{2}-\frac{2m}{\hbar^{2}}V(\sqrt{b^{2}+z’^{2}})) dz’ +C$ b is the scattering parameter (ie. minimum distance from the origin to the trajectory) C is chosen such that $\frac{S}{\hbar} = kz$ as V approaches 0 so as to recover the plane wave form in the zero potential limit If $E = \frac{\hbar^{2}k^{2}}{2m}$, you can use the high energy limit to expand the square root We can then write the scattering amplitude by calculating $<k’|V|\Phi>$ ie. $f(k’,k) = -\frac{1}{4\pi} \frac{2m}{\hbar^{2}}\int d^{3}x’ exp(-i k’ x) V(\sqrt{b^{2}+z^{2}}) exp(ikx’) exp(-\frac{im}{\hbar^{2}k}\int_{-\infty}^{z’} V(\sqrt{b^{2}+z’’^{2}}) dz’’)$ This integral can be done by going to cylindrical coordinates $d^{3} x’ = b db d\phi_{b} dz'$ making the approximation $(k-k’) \cdot x’ \approx k’ \cdot b$ (small deflections) constraining the scattering to the xz plane Using the identities $\int_{0}^{2\pi} d\phi_{b} exp(-ikb\theta cos\phi_{b}) = 2\pi J_{0}(kb\theta)$ $\int_{-\infty}^{\infty} dz V exp(\frac{-im}{\hbar^{2}k}\int_{-\infty}^{z} V dz’) = \frac{-\hbar^{2}k}{m}exp(\frac{-im}{\hbar^{2}k}\int_{-\infty}^{z} V dz’)|_{-\infty}^{\infty}$ End result is: $f(k,k’) = -ik \int_{0}^{\infty} db b J_{0}(kb\theta) (exp(2i\Delta(b))-1)$ where $\Delta(b) = \frac{-m}{2k\hbar^{2}}\int_{-\infty}^{\infty} V(\sqrt{b^{2}+z^{2}}) dz$ Low-Energy Scattering and Bound States In the low energy regime, only the l=0 spherical wave matters (centrifugal potential severely dampens all of the others) Imagine very low energy ($k \approx 0$). If you are sufficiently far away, then for l=0, we have that $\frac{d^{2}u}{dr^{2}}=0$ A straight line solves this (ie $u = c(r-a)$). This can be imagined as the infinitely long wavelength limit of the standard radial function: $lim_{k\rightarrow 0} \sin (kr+\delta_{0})$ a is the scattering length. This can differ by orders of magnitude from R (ie. the range of the potential) For an attractive potential, it possible for the scattering length to be far greater than the potential’s range In this limit, the wavefunction kind of acts like a decaying exponential (ie. the wavefunction is very flat). This is the wavefunction of a bound state! We equate the logarithmic derivative (z’/z) of the standard bound state with that of this low-energy scattering state $\frac{\kappa exp(-\kappa r)}{exp(-\kappa r)} = \frac{1}{r-a}$. In the limit where R« a, we have that $\kappa = \frac{1}{a}$ The binding energy of this bound state then becomes $-E_{bound} = \frac{\hbar^{2}\kappa^{2}}{2m} \approx \frac{\hbar^{2}}{2ma^{2}}$ Resonances In some systems, certain spherical waves experience cause a peak in the cross section. These are called resonances because of the contrifugal term, potentials can look like attractive potentials followed by a repulsive barrier at larger distances. Hence, a particle can get trapped in the attractive portion (ie. a quasi-bound state) with some chance of tunnelling through the barrier Look at the scattering phase shift of this situation. $\delta_{l}$ rises through the value $\frac{\pi}{2}$ as the incident energy rises through that of the quasi-bound state ($\frac{\pi}{2}$ comes from the $\sin \delta_{l}$ in the cross section formula) Recall that $f_{l}(k) = \frac{1}{k\cot \delta_{l}-ik}$. At resonance, $\cot(\delta_{l})=0$. If we Taylor expand around $E_{R}$ (ie. the resonance energy), w have that $\cot \delta_{l} = -c(E-E_{R})$ we define $\frac{d \cot \delta_{l}}{dE} = -c = -\frac{2}{\Gamma}$ where $\Gamma$ can be though of as the FWHM of the cross section around the resonance Assuming that one spherical wave dominates the cross section, we have that $\sigma_{l} = \frac{4\pi}{k^{2}}\frac{(2l+1)(\frac{\Gamma}{2})^{2}}{(E-E_{r})^{2}+\frac{\Gamma^{2}}{4}}$ This is called the Breit-Wigner formula

Linear Regression You have a set of data $(x_{i},y_{i},\sigma_[i])$, where x is the independent variable, y is the dependent variable, and $\sigma$ is the uncertainty in the dependent variable You want to fit a line to the data ($\hat{y} = mx+b$) You also want the uncertainties of the parameters Why would you want to do this? To know the parameters of the model To extrapolate/interpolate values of the dataset that you don’t have How you do this depends on your viewpoint a frequentist approach would try to maximize the likelihood estimate (information theory) Suppose that you have a probability density for the data $y_{i}$ ranging from 1 to N Treat this density as a function of the parameters. What parameters maximizes this density? A Bayesian approach would be posterior estimates (Cox theorems) Likelihood Function The model we are trying to fit for Linear Regression is $y=mx+b+\pi(0,\sigma)$ We assume that x has no uncertainty that the noise function $\pi$ is Gaussian with mean 0 (why? Central Limit Theorem, and it’s easy to work with) Assume that your variances are correct (big assumption) Each data point is independent from each other (iid) $N = \frac{1}{\sqrt{2\pi \sigma^{2}}} exp(-\frac{(y-\mu)^{2}}{2\sigma_{2}})$ (1D Gaussian) With this, the likelihood function can be written as $\mathbb{L} = \frac{-1}{2}(\vec{Y}-\vec{X}\cdot \vec{a})^{T}C^{-1}(\vec{Y}-\vec{X}\cdot \vec{a})$ Y is the data vector X is the design matrix (first column is the independent variables, 2nd column is all ones) a is the parameter vector $C^{-1}$ is a diagonal matrix, whose diagonals are $\frac{1}{\sigma_{i}^{2}}$ the most likely values of m and b are those which maximize this likelihood (ie. when the derivative w.r.t. a is 0) Dig through this pamphlet to do that calculation Freqentist Week Some Definitions aleatoric uncertainty (noise) and epistemic uncertainty (systematic error or uncertainty in the form of model assumptions) The lower bound on your total uncertainty is the inverse of the Fisher matrix (ie. covariance matrix) This lower bounds is reached only if all of the assumptions are correct (which is normally pretty naive) Freqentist Rules You can never have a measure in the parameter space, only the data space “parameters are set by God. The data are created by us” - D.W.H. This means that statements like “It is more probable that $H_{0}>69$ than $H_{0}<69$” are OK, but “the data are more consistent with $H_{0}>69$ than $H_{0}<69$” are not OK p-value of p<0.05: If the null model were true, we’d get data this “deviant” is less than 0.05 percent of the time The null model is the model where the thing that you are trying to prove is false. So if the likelihood of the null hypothesis is low, then the chance of your hypothesis is high What is $\chi^2?$ You have some data (a set of unordered vectors) and a model that tries to describe this data We can defined $\chi^{2} = \sigma_{i=1}^{N} (\frac{y_{i}-f(x_{i},\theta)}{\sigma_{i}})^{2}$ You can also relate the chi-squared to the log-likelihood via $ln(p(y|\theta)) = \frac{-1}{2}\chi^{2}+const$ This assumes a zero-mean gaussian with known variances, It also assumes that the noise distributions of each point are iid What value do you expect for $\chi^{2}$? If your model describes your data well, then you expect that each data point is offset from the model by the variance of that point. so $\frac{y_{i}-f(x_{i},\theta)}{\sigma_{i}}=1$, so naively, you expect that $\chi^{2} \approx N$ In reality, $\chi^{2} \approx N -dof = N_{dof}$, where dof are the number of degrees of freedom. The handwaving explanation for this is that you aren’t using the true parameters, you are using the best-fit parameters. This causes you to overfit a bit, which juices the $\chi^{2}$ numbers, hence the $-dof$ In the limit of large $N_{dof}$, the $\chi^{2}$ approaches a Gaussian with mean $N_{dof}$ and variance of $2N_{dof}$ A consequence of this is that for sufficiently large N, all models get rejected (“all models are wrong; some are useful” ) If you have multiple models, then you can talk about $\Delta \chi^{2}$ Nuisance Parameters Some parameters in a model you care about. Others you need to include, but don’t want to For example, instrumentation might introduce these extra parameters Suppose that our likelihood function distribution goes like $L = \ln p([y]|\theta,\alpha)$, where $\alpha$ is/are your nuisance parameter(s) You can define a function $max_{\alpha} L(\theta, \alpha) = L_{p}(\theta)$ (ie. you find what $\alpha$ maximizes the likelihood) You didn’t have to specify a prior on $\alpha$ $L_{p}$ is called the profile log likelihood This profiling removes the nuisance parameters from the likelihood, so you can just maximize $L_{p}$ instead of L Uncertainty Estimation Robustness A “robust” method is insensitive to your noise model The previous uncertainty estimates where not robust (ie. they heavily depended on the noise being Gaussian) Fair Sampling Assume that $x_{i}$ is drawn from some distribution $p(x)$. Then your sampling of x is fair if $\sigma_{i=1}^{k} f(x_{i}) \approx \int f(x) p(x) dx$ $p(x)$ must satisfy $\int p(x) dx = 1$ and $p(x) \geq 0$ Bootstrapping Assume that your data $y_{i}$ are fair samples drawn from $p(z|\theta)$ Assumes that your data is “similar”: ie. has similar errors Create an estimator $\hat{\theta}(y_{i})$ ranging from i=1 to i=N Draw K sets of N values from $y_{i}$ with replacement Estimate $\hat{\theta}$ for each of the K samples The variance of $\theta$ then becomes $\frac{1}{K}\Sigma_{j=1}^{K} (\hat{\theta_{j}}-\hat{\theta})^{2}$, where $\hat{\theta}$ if the estimator when using all the data The covariance matrix is $C_{\theta} = \frac{1}{K}\Sigma_{i=1}^{k} (\hat{\theta_{j}}-\hat{\theta})(\hat{\theta_{j}}-\hat{\theta})^{T}$ Jackknifing Assume that your data $y_{i}$ are fair samples drawn from $p(z|\theta)$ Assumes that your data is “similar”: ie. has similar errors Create an estimator $\hat{\theta}(y_{i})$ ranging from i=1 to i=N, and that the estimator holds on a smaller subset of y Take all of the data from y, but drop one datapoint and call this subsample $Y_{j}$ (so you have a subsample of size N-1). Do this for all possible choices of j Define $\hat{\theta}_{j}$ for all of the subsamples $\sigma_{\theta}^{2} = \frac{N-1}{N} \Sigma_{j=1}^{N} (\theta_{j}-<\theta>_{j})^{2}$ $<\theta>_{j}$ is the mean across all of the jackknife estimators Bayesian Statistics Bayes is a theory of prediction, frequentism is a theory of measurement Based on Cox theorems In Bayes, our likelihood function is a probability distribution $p(y|\theta, \alpha, I)$, where $\theta$ are your parameters of interest, $\alpha$ are your nuisancee parameters, and I are your assumptions All of Bayes depends on having a proper set of priors which represents your belief about $\theta$, $\alpha$ and I before your see y Bayes Rule: $p(\theta,\alpha| y, I) = \frac{p(\theta \alpha|I)p(y|\theta, \alpha, I)}{Z}$ where Z is the normalization (or marginalization) equal to $p(y|I)$ Marginalization Suppose that we have some prior $p(\theta,\alpha | I)$ which is seperable ($p(\theta|I) p(\alpha|I) = p(\theta,\alpha | I)$) This can arise if your belief about different set of parameters are distinct from each other (ex. the assumptions about the Higgs mass are seperate from the assumptions about the calorimeter sensitivity) You can integrate out the nuisance parameters $\alpha$. This is called marginalization The marginal likelihood is given by $p(y|\theta, I) = \int p(y|\theta, \alpha, I) p(\alpha | I) d\alpha$ The marginal posterior is $p(\theta|y,I) = \int p(\theta,\alpha|y,I) d\alpha$ MCMC Markov Chain Monte Carlo Universal way of calculating posterior distributions Kind of a crappy way of do so, since no nice convergence theorems Suppose that we have some posterior distribution $p(\Omega| y,I)$ A good sampling of this posterior should be able to approximate the posterior Metropolis Hastings We define some probability distribution $q(\Omega’| \Omega)$ called the proposal distribution (ie. a distribution which generates a new sample given an old sample) We assume that $q(\Omega’| \Omega) = q(\Omega| \Omega’)$ (called detailed balance) Suppose that after drawing from q, you calculate the change in the log-likelihood between the old and the new sample Roll a uniform random number r between 0 and 1. Accept the new sample if $ln(r) < ln(\delta)$ Otherwise, you keep the previous sample In the limit of infinite samples, you will get a fair sampling of the posterior Nearby samples are not independent This is modulated by the “correlation time”, which roughly speaking, is the number of samples you have to take before you get a new independent sample This correlation time contributes to the number of truly independent samples: $N_{eff} = \frac{K}{T}$, where K is the total samples drawn and T is the correlation time How do you choose q? You need to choose a q which equally trades off between exploration and exploitation You could try to measure the autocorrelation time, but that’s hard to compute efficiently You could also try to aim for a target acceptance rate (~0.4) Model Selection Suppose that you have some set of models that you want to test You can select which model you want via AIC and BIC (ie. likelihood ratios) in a frequentist approach With a Bayesian approach, you can find your full marginalized likelihood (integrate over your model parameters) This approach is heavily dependent on your prior (since $p(Y|I) = \int p(y|\theta, I) p(\theta|I) d\theta $) K-fold Cross Validation You leave k datapoints out of your training data. You have your model predict the held out data points. The model which yields the best prediction is empirically the best Appropriate for choosing “nuisance models” Model independent In pseudocode: You hold out k data points Run your model on this subsample of the data You predict the outputs of the withheld data points For frequentists, you predict the probability of getting the withheld data given the subsampled model For Bayesian, you calculate the fully marginalized log likelihood Interpolation AKA overparameterization (p>n) Want to fit to points in between your data There are a couple of types of interpolation linear bilinear cubic For polynomial interpolation, you have roughly the same number of parameters as number of points When interpolation, “sharp edges” in your dataset has a tendency to get smooth out by the interpolation Interpolations are also “too faithful” ie. they stick to the data too much Lot’s of data is “band-limited”, which means that there is a fundamental distance scale below which you can’t get good resolution (think a camera) Noise can in principle have very high frequency noise 3 primary methods for interpolation methods exact pass through the data (think all the flavors of splines) Some wizardry allows these algorithms to run in polynomial time fit a parameterized model to some data with noise cubic time kernel methods (ie. you define some metric, and take a weighted average of the closest neighbors based on the kernel) Over Parameterized Models Dealing with p parameters and n data points where p >n To prevent the values of these parameters from getting too large, we add some regularization Idea is that under some norm, we want to find $argmin ||a||$ such that $y = Xa$ Common regularization schemes are The L2 norm: $\Sigma_{i=1}^{p} a_{i}^{2}$ Called ridge regression The L1 norm: $\Sigma_{i=1}^{p} |a_{i}|$ called lasso regression Has a tendency to zero out irrelevant parameters We have the following model for overparameterization: $Q = ||y-Xa||^{2} + \lambda ||a||$ We take the limit $\lambda\rightarrow 0 $, which yields the argmin on the norm of a Can think of the problem in a Bayesian sense $Q = \chi^{2}+\lambda||a||_{2}^{2}$ $(y-Xa)^{T}C^{-1}(y-Xa)+a^{T}\Lambda a$ This can be solved by minimizing with respect to a The predicted values are then $\hat{y’} = X’\Lambda^{-1}X^{T}(X\Lambda^{-1}X^{T} +C)^{-1} \vec{y}$ where ’ denotes the extrapolated x,y points Gaussian Processes (GPs) Kernels We can view the overparameterized solution as $K’(K+C)^{-1}\vec{y}$, where K is called the kernel kernel matrices have the form $K_{ij} = k(t_{i}, t_{j})$. Typically, most problem have the form $K_{ij} = k(|t_{i}-t_{j}|)$ At $\delta t=0$, K describes the variance of the GP There is also some $\tau$ which describes the correlation time Some common forms for the kernel $K_{ij} = V exp(\frac{-|\delta t|^{2}}{\tau^{2}})$ (squared exponential) $K_{ij} = V exp(\frac{-|\delta t|}{\tau})$ Kernel functions must be semi-positive definite Hard to do in general. Easy way to generate a kernel is to have some strictly positive function in frequency space, then Fourier transform it to real space How do you choose a kernel function? Compute the likelihood function of the data This is done via calculating $\ln p(y_{i}|\mu(t),K)$. Written out: $L = \frac{-1}{2}(\ln \det{2\pi(K+C)})-\frac{1}{2}(\vec{y}-\vec{\mu})^{T}(K+C)^{-1}(\vec{y}-\vec{\mu})$ This allows you to compare different kernel functions, and use the Neyman-Pearson lemma to select the MAP This assumes that the distribution is Gaussian, which is not always a valid assumption Use cross validation Leave one out to get: $\hat{y_{i}} = K_{i}(K_{/i}+C_{/i})^{-1}y_{/i}$ and $\hat{\sigma_{i}^{2}} = k(0)-K_{i}(K_{/i}+C_{/i})^{-1}K_{i}^{T}$ You can then compare this against the actual data and choose the kernel which maximizes that likelihood PCA Imagine that we have our design matrix. We want to decompose $X= USV^{T}$ where S is your singular value matrix and U and $V^{T}$ are your left and right eigenvectors This decomposition creates a spectrum where each eigenvector is weighted by their associated singular value. You can thus get a good approximation of your data by only keeping the top q eigenvectors When is PCA Good? What properties are needed for a good PCA decomposition? When the data is “well aligned” The variation in the data space eis the variation you care about Silly example: trying to differentiate between smiley faces images and sad face images at different positions You want to place all of your faces at the center of your image, so that you don’t waste a PCA eigenvector on locating the average locations of all of the faces When the data has high signal to noise When there are few outliers Assuming that the above holds, then PCA will help de-noise your data (assuming that you choose a sufficient q eigenvectors) You can use leave one out cross validation to help determine your best q Interpolate over missing/bad data Detect anomalies/outliers Misc Writing a Data Analysis Paper Trivial Things Given the paper, people within your field should be able to reproduce your work The writing should be unambiguous The notation should not reuse the same symbol Be consistent with naming things Try to mix language and symbols for instance “wavevector $\vec{k}$. Not “wavevector”. Not $\vec{k}$ Non-trivial Things You should have a list of explicit assumptions and choices Visualize your results in the space of your data Frame your discussion section by the assumptions Causal Separation You have some data, and you want to know what portions of the data are explained by what mechanism Example: We have n cells, observed at t time slices, and you use m microscopes Suppose that over time, the size of the cells appear to grow. Is this due to the cells actually growing, or is it due to the different calibrations for all of the microscopes We have a set of observations $y_{l}\pm \sigma_{l}$ in cell i @ time j with microscope k Some microscopes have a smaller size compared to others. How do you decoupled the effect of the microscope from the actual cell size increase You could calibrate all of your images with an object of known size (external calibration) You could also interleave which microscope you observes the cells with at different times (ie. prevent any covariance buildup between the observables) This is hard to do. A sufficient (but worse) substitution is randomly assignment This problem can then be encoded as a design matrix Baby Machine Learning Two main groups of methods: unsupervised and supervised Unsupervised: Find patterns in your data PCA Kernel density estimation K-means Supervised: Presupposes that your data has some labels associated with it linear regression Gaussian Processes K-nearest neighbors SVM (support vector machines) All of the above can be kernelized (ie. do a transformation to your data to linearize it in some basis) You split your data into 3 partitions: the training set: what you learn your model parameters on the validation set: what you learn your optimal hyperparameters on the test set: what you evaluate your final model against K-Nearest Neighbors Choose the k “closest” feature vectors in the training set to test your feature vector, and then assign the “average” of those k labels With regards to “closest”, you need to define a metric distance in feature space that obeys The triangle inequality and must be non-negative With regards to “average”, you can do a lot of different polling strategies (depends on your problem) You gain biases at the “edges” of the label boundaries K-Means We have a bunch of features and no labels We define k points in our feature space (our centroids) We ask for each feature vector: which of the centroids is the feature vector closest to? After doing the assignment for all of the data, we move all of the centroids to the average of the nearby vectors We iterate on this process until convergence Data Management data and housekeeping data should be stored in the same data format data: your actual measurements housekeeping data (metadata): data surrounding your measurements Don’t store these as a filename or directory tree (seems silly, but it happens) The code that reads in the data should be considered as housekeeping data To store in binary or store in text binary data is more compact, is lossless, and is inscrutable to humans text data takes up more space, can lose precision on conversion, and is human-readable Security: Don’t use .pkl files. Just don’t

Logistics Thermodynamics 0th Law 1st Law Response Functions 2nd Law Entropy 3rd Law Thermodynamic Potentials Enthalpy Helmhotz Free Energy Gibbs Free Energy Grand Potential Probability Definitions Important Probability Distributions Kinetic Theory of Gases Liouville’s Theorem Bogoliubov-Born-Green-Kirkwood-Yvon Hierarchy (BBGKY) Boltzmann Equation H-Theorem Classical Statistical Mechanics Microcanonical Ensemble Two State System Example 0th Law 1st Law 2nd Law Ideal Gas Mixing Entropy The Canonical Ensemble Grand Canonical Ensemble Conserved Ensemble Notes Interacting Particles Cumulant Expansion Mean Field Theory of Condensation Critical Points Quantum Statistical Mechanics Dilute Polyatomic Gases Quantized Vibrational Modes Quantized Rotational Modes Blackbody Radiation Formalism Identical Particles Grand Canonical Formulation Degenerate Fermi Gas $f_{m}^{\eta}(z)$ Denegerate Bose Gas Logistics Book(s) Landau and Lifshitz Kardar Pathria Thermodynamics 0th Law Thermodynamic equilibrium means that each object in thermal contact with each other have the same temperature Ideal gases, as you cool them down, all converge to a single point, which defines absolute zero. For Kelvin,, the triple point of water defines the size of each step on the scale (273.16 K) 1st Law $dE = \delta W + \delta Q$ The change is energy can either come from work, or heat $dE = E_{f}-E_{i}$ Energy is a function of state (path independent) $\delta W$ and $\delta Q$ are path dependent Can expand out $\delta W$ into more parts: $dE = -p dV + \delta Q$, where the negative sign comes from the fact that the system does work on the outside environment You don’t need to be restricted to pressure and volume. In general, you can have some “displacements” $\chi_{i}$ (ie. extensive quantities) and their conjugate forces $J_{i}$ (ie. intensive quantities) to get that $dW = \Sigma_{i} J_{i} d\chi_{i}$ Response Functions Response function characterize the macroscopic behavior of a system Heat Capacity: $C_{V} = (\frac{dQ}{dT})_{V}$ $C_{P} = (\frac{dQ}{dT})_{P}$ This is always large than $C_{V}$ since some of the heat is used up in the work done in changes in volume Force constants the isothermal compressibility of gas : $\kappa_{T} = -(\frac{\partial V}{\partial P})_{T}/V $ the susceptibility of a magnet $\chi_{T} = (\frac{\partial M}{\partial B})_{T}/V$ Thermal Responses: Things like the expansitivity of a gas, given by $\alpha_{p} = \frac{\partial V}{\partial T}|_{P}/V$ 2nd Law Efficiency of an ideal heat engine is defined by $\eta = \frac{W}{Q_{H}} = \frac{Q_{H}-Q_{C}}{Q_{H}} \leq 1$ The figure of merit for an ideal refrigerator is $\omega = \frac{Q_{C}}{W} = \frac{Q_{C}}{Q_{H}-Q_{C}}$ Kelvin formulation of 2nd Law: No process is possible whose sole result is the complete conversion of heat into work Clausius’s Statement: No process is possible whose sole result is the transfer of heat from a colder to a hotter body The above two are equivalent: Can show that violation of one implies a violation of another via hooking up outputs of an ideal engine to an ideal refrigerator Carnot’s Theorem: You can’t beat the Carnot engine in terms of efficiency A Carnot cycle is defined by two isotherms at $T_{H}$ and $T_{C}$, and two adiabatic curves linking the isotherms The efficiency of a Carnot engine is defined by $\eta = \frac{T_{H}-T_{C}}{T_{H}}$. This can be derived hooking up two different Carnot engines at 3 temperatures. The heat from one gets dumped to the next. The overall efficiency is then the product of the two, which implies that the efficiency is some ratio of temperatures $1-\eta = \frac{Q_{2}}{Q_{1}} = \frac{T_{2}}{T_{1}}$ NOTE: When dealing with engines, try to keep all quantities positive (ie. a positive amount of heat leaves the hot reservoir, a positive amount of work leaves the engine, and a positive amount of heat enters the cold reservoir) Entropy Clausius’s Theorem: For any cyclic transformation (reversible or not) $\oint \frac{dQ}{T} \leq 0$, where $dQ$ is the heat increment supplied to the system at temperature T Imagine dividing the cycle into a series of infinitesimal portions, where the system receives energy from dQ and dW. Imagine that dQ gets funneled into some Carnot engine which is attached to some reservoir at a fixed temperature $T_{0}$ Since the sign of dQ is unspecified, the Carnot engine must be able to run in both directions. In order to do this, the engine must extract some heat $dQ_{R}$ from the fixed reservoir This implies that $dQ_{R} = \frac{T_{0}}{T} dQ$ The net effect of this process is that some heat $Q_{R} = \oint dQ_{R}$ is extracted from the reservoir and converted to external work W. By the Kelvin formulation of the 2nd law, $Q_{R} = W \leq 0$ which implies $\oint \frac{dQ}{T} \leq 0$ For a reversible cycle, we know $\oint \frac{dQ_{rev}}{T} = 0$ This implies that for a reversible cycle, this integral is independent of the path. We can use this to define the entropy S: $S(B)-S(A) = \int_{A}^{B} \frac{dQ_{rev}}{T}$ Imagine that you make an irreversible change from A to B, but make a reversible regression from B to A: $\int_{A}^{B} \frac{dQ}{T}+ \int \frac{dQ_{rev}}{T} \leq 0$ which implies $\int_{A}^{B} \frac{dQ}{T} \leq S(B)-S(A)$ This is the statement that entropy always increases or stays the same For a reversible (and/or quasi-static) process, we can write $dQ = TdS$, where T and S and conjugate variables Equilibrium exists when entropy is maximized! For a cyclic process, entropy does not change! 3rd Law Entropy of a closed system in thermodynamic equillibrium approaches a constant value as temperature approaches absolute 0 (typically 0) This also implies that heat capacities and thermal expansivities must go to 0 as T approaches 0 Thermodynamic Potentials The energy gets extremized in equillibrium for an adiabatically isolated system If you are in an out of equillibrium system that is not adiabatically isolated and subject to external work, then you can define other thermodynamic potentials which are extremized in equillibrium Enthalpy If there is no heat exchange, and the system comes to mechanical equilibrium subject to a constant external force, then enthalpy is the appropriate potential Can be though of as minimizing the energy, plut the work from the external agent $H = E- \vec{J}\cdot \vec{x}$ $dH = dE -d(J\cdot x)$ (Just the legendre transform w.r.t. J and x) Helmhotz Free Energy For isothermal transformations in the absence of mechanical work (dW=0) F = E-TS (Legendre transform w.r.t. T and S) Gibbs Free Energy For isothermal transformations involving mechanical work at constant external force $G = E-TS-J\cdot x$ Legendre transform in both T,S and J,x Grand Potential What if the number of particles in the system changes? we can define the chemical work as $dW = \vec{\mu} \cdot d\vec{N}$, where each species has it’s own chemical potential We can introduce this work by simply adding this work to a potential Probability Definitions Random variable x has a set of outcome S (can be continuous or discrete) An event is a subset of outcomes from S, and is assigned a probability Probabilities must obey the following rules They must be greater than or equal to 0 They must be additive if the events subsets are disjoint They must be normalizable (ie. probability of S should be 1) a cumulative probability function (CPF) denoted P(x) is the probability of an outcome with any value less than x P(x) must be a monotonically increasing function of x $P(-\infty) = 0$ and $P(\infty) = 1$ a probability density function (PDF) is defined by $p(x) = \frac{dP(x)}{dx}$ This must satisfy $\int p(x) dx = 1$ The expectation value of any function F(x) of the random variable is $<F(x)> = \int_{-\infty}^{\infty} dx p(x) F(x)$ Easily extendable to multiple dimensions The moments of a PDF are expectation values for power of the random variable The nth moment is $m_{n} = <x^{n}> = \int dx p(x) x^{n}$ The characteristic function is the generator of moments of the distribution. It’s just the Fourier transform of the PDF $\tilde{p}(k) = \int dx p(x) exp(-ikx)$ The PDF can be recovered from the characteristic function via the inverse Fourier transform $p(x) = \frac{1}{2\pi} \int dk \tilde{p}(k) exp(ikx)$ The moments can be calculated by expanding $\tilde{p}(k)$ in powers of k: $\tilde{p}(k) = \Sigma_{n=0}^{\infty} \frac{(-ik)^{n}}{n!}<x^{n}>$ Moments of the PDF around any $x_{0}$ can also be generated by substituting x for $x-x_{0}$ in the moment generation The cumulant generation function is the logarithm of the characteristic function You can generate cumulants of the distribution by expansions in powers of x: $ln \tilde{p}(k) = \Sigma_{n=1}^{\infty} \frac{(-ik)^{n}}{n!}<x^{n}>_{c}$ Moments can be related to cumulants by using $\ln(1+\epsilon) = \Sigma_{n=1}^{\infty} (-1)^{\epsilon+1} \frac{\epsilon^{n}}{n}$ to expand out $\ln\tilde{p}(k)$ A Joint PDF is the probability density of many variables. If all variables are independent, this is just the product of the individual pdfs Stirling’s approximation for N! holds for very large N. This states that $ln(N!) \approx N ln N - N + \frac{1}{2}ln(2\pi N)$ Important Probability Distributions Gaussian: $p(x) = \frac{1}{\sqrt{2\pi \sigma^{2}}} exp(- \frac{(x-\lambda)^{2}}{2\sigma^{2}})$ The characteristic function is $\tilde{p}(k) = exp(-ik\lambda-\frac{k^{2}\sigma^{2}}{2})$ Only the first two cumulants are non-zero Multidimensional version is $p(x) = \frac{1}{\sqrt{(2\pi)^{N}det(C)}}exp(\frac{-1}{2}\Sigma_{mn}C_{mn}^{-1}(x_m-\lambda_{m})(x_{n}-\lambda_{n}))$ C is the a positive definite symmetric matrix Binomial: $p_{N}(N_{A}) = \binom{N}{N_{A}} p_{A}^{N}p_{B}^{N-N_{A}}$ Characteristic function is $\tilde{p}(k) = (p_{A}e^{-ik}+p_{B})^{N}$ Poisson Distribution This is the limit of the binomial distribution. We want to find the probability of observing N decays in time interval T. Subdivide the interval into $N = \frac{T}{dt} » 1$ In each interval, the chance of an event occuring os $p_{A} = \alpha dt$ and the chance of no event in the interval is $p_{B} = 1-\alpha dt$ This implies the characteristic function is $\tilde{p}(k) = (p_{A}exp(-ik)+p_{B})^{N} = lim_{dt\rightarrow 0}(1+\alpha dt (e^{-ik}-1))^{\frac{T}{dt}} = exp(\alpha(e^{-ik}-1)T)$ Taking the inverse Fourier transform yields the Poisson PDF: $p(x) = \int_{-\infty}^{\infty} \frac{dk}{2\pi} e^{ikx} \Sigma_{M=0}^{\infty} \frac{(\alpha T)^{M}}{M!}e^{-ikM}$ (using power series for the exponential) Using the identity $\int_{-\infty}^{\infty} \frac{dk}{2\pi} exp(ik(x-M)) = \delta(x-M)$, which implies $p_{\alpha,T}(M) = e^{-\alpha T}\frac{(\alpha T)^{M}}{M!}$ Kinetic Theory of Gases We want to be able to connect macroscopic properties of systems to the microscopic picture of particles Imagine a 6N dimensional phase space, where time evolution of each particle is governed by Hamilton’s equations $\dot{q_{i}} = \frac{\partial H}{\partial p_{i}}$ $-\dot{p_{i}} = \frac{\partial H}{\partial q_{i}}$ We can consider some probability density $\rho(p,q,t)$ which describes the density of particles in that region of phase space We can compute macroscopic values for functions $O(p,q)$ via $<O> = \int \Gamma \rho(p,q,t) O(p,q)$ Liouville’s Theorem Concisely, Liouville’s Theorem states that $\rho$ behaves like an incompressible fluid The proof is as follows: Imagine that you Taylor expand out all the q and p’s as a function of time The volume of phase space before is simply $\Pi_{i} dq_{i}dp_{i}$ The volume of phase space after is then the dot product of the Taylor expanded momenta and position. You can use Hamilton’s equations to eliminate the time dependent portion of the expansion, which implies that the phase space volume is unchanged A consequence of this is that $\frac{d\rho}{dt} = \frac{\partial \rho}{\partial t}+\Sigma_{\alpha=1}^{3N} (\frac{\partial \rho}{\partial p_{\alpha}}\frac{dp_{\alpha}}{dt}+\frac{\partial \rho}{\partial q_{\alpha}}\frac{dq_{\alpha}}{dt}) = 0$ Alternatively: $\frac{\partial \rho }{\partial t} = -\{\rho, H\}$, where $\{\}$ is the Poisson bracket Another consequence of this is that $\frac{d<O>}{dt} = <\{O,H\}>$ straightforward to prove (integration by parts, and using Hamilton equations gives it) If a density has reached equillibrium, this implies that it’s independent of time. This implies that $\{\rho_{eq},H\}= 0$ If $\rho_{eq}$ is a function of the Hamiltonian, then this equation holds This implies that $\rho$ is constant of surfaces of constant energy in phase space This is the basic assumption of statistical mechanics! You have equal probabilities at constant energies Bogoliubov-Born-Green-Kirkwood-Yvon Hierarchy (BBGKY) For multiple particle, you can generate the s-particle density via $f_{s}(p_{1}…q_{s},t) = \frac{N!}{(N-s)!}\int \Pi_{i=s+1}^{N} dV_{i} \rho(p,q,t) = \frac{N!}{(N-s)!}\rho_{s}(p_{1},…,q_{s},t)$ For the one particle case, this can be interpreted as the probability that any one of the N particles has the specified p and q values (easily generalized to s particles) The normalization is $\frac{N!}{(N-s)!}$ (comes from permutations of s particles amongst N indistinguishable particles) and $p_{s}$ is the unconditional PDF for the coordinates of s particles Can imagine a Hamiltonian which is $H = \Sigma_{i=1}^{N} (\frac{p_{i}^{2}}{2m}+U(q_{i}))+\frac{1}{2}\Sigma_{i,j=1}^{N} V(q_{i}-q_{j})$ where U is some external potential and V is a two-body interaction Imagine splitting this hamiltonian into 3 parts $H_{s} = \Sigma_{n=1}^{s}(\frac{p_{n}^{2}}{2m}+U(q_{n}))+\frac{1}{2}\Sigma_{n,m=1}^{N} V(q_{n}-q_{m})$ $H_{N-s} = \Sigma_{i=s+1}^{N}(\frac{p_{i}^{2}}{2m}+U(q_{i}))+\frac{1}{2}\Sigma_{i,j=s+1}^{N} V(q_{i}-q_{j})$ $H^{’} = \Sigma_{n=1}^{s}\Sigma_{i=s+1}^{N} V(q_{n}-q_{i})$ You can write the time evolution of $\rho_{s}$ as $\frac{\partial \rho_{s}}{\partial t} = -\int \Pi_{i=s+1}^{N} dV_{i} \{\rho, H_{s}+H_{N-s}+H^{’}\}$ If you take the Poisson bracket of each term, you find that $\frac{\partial f_{s}}{\partial t} = -\{H_{s}, f_{s}\} = \Sigma_{n=1}^{s} \int dV_{s+1} \frac{\partial V(q_{n}-q_{s+1})}{\partial q_{n}}\frac{\partial f_{s+1}}{\partial p_{n}}$ Note the hierarchy here: $\frac{\partial f_{s}}{\partial t}$ depends on $\frac{\partial f_{s+1}}{\partial t}$ To terminate the hierarchy, we need to be able to neglect some terms. Since all terms have units of inverse time, we can seperate terms according to time scale. $\frac{1}{\tau_{U}}\approx \frac{\partial U}{\partial q}\frac{\partial}{\partial p}$ is some extrinsic time scale, which can be made arbitrarily small by increasing the system size. On the order of 1E-5 sec for length scales of 1E-3 m ($\tau \approx \frac{d}{v}$) $\frac{1}{\tau_{c}}\approx \frac{\partial V}{\partial q}\frac{\partial }{\partial p}$ has some collision time, on the order of 1E-12 sec Longe range interactions are harder to get a collision time There are also collision terms which depend on $f_{s+1}$ $\frac{1}{\tau_{x}} \approx \int dV \frac{\partial V}{\partial q}\frac{\partial}{\partial p}\frac{f_{s+1}}{f_{s}}$ where $\tau_{x}$ is the mean free time $\tau_{x} \approx \frac{\tau_{c}}{nd^{3}}$ Boltzmann Equation The Boltzmann equation assumes that the density is much smaller than 1. This allows us to drop the mean free time terms from the equation. The first two equations in the BBGKY hierarchy are $(\frac{\partial}{\partial t} - \frac{\partial U}{\partial q} \frac{\partial}{\partial p }+\frac{p_{1}}{m}\frac{\partial}{\partial q}) f_{1} = \int dV_{2} \frac{\partial V}{\partial q_{1}}\frac{\partial f_{2}}{\partial p_{1}}$ $(\frac{\partial}{\partial t}-\frac{\partial U}{\partial q_{1}}\frac{\partial }{\partial p_{1}}-\frac{\partial U}{\partial q_{2}}\frac{\partial}{\partial p_{2}}+\frac{p_{1}}{m}\frac{\partial}{\partial q_{1}}+\frac{p_{2}}{m}\frac{\partial}{\partial q_{2}}-\frac{\partial V(q_{1}-q_{2})}{\partial q_{1}}(\frac{\partial}{\partial p_{1}}-\frac{\partial}{\partial p_{2}}))f_{2} = \int dV_{3} (\frac{\partial V(q_{1}-q_{3})}{\partial q_{1}}\frac{\partial}{\partial p_{1}}+\frac{\partial V(q_{2}-q_{3})}{\partial q_{2}}\frac{\partial}{\partial p_{2}}) f_{3}$ With the dilute gas approximation, you can truncate the BBGKY hierarchy by setting the RHS of the 2nd equation to 0 It’s also reasonable to expect that at very large distances, the particles become independent (ie. $f_{2}(p_{1},q_{1},p_{2},q_{2},t) \rightarrow f_{1}(p_{1},q_{1},t)f_{1}(p_{2},q_{2},t)$) The final closed form of Boltzmann’s equation is $\frac{\partial f_{1}}{\partial t}-\{H,f_{1}\} = \int d \Gamma_{2} d\Omega \frac{|p_{2}-p_{1}|}{m} \frac{d\sigma}{d\Omega}[f_{1}(\Gamma_{1},t)f_{1}(\Gamma_{2},t)-f_{1}(\Gamma_{1},t)f_{1}(\Gamma_{2},t)]$ H-Theorem The Theorem: If $f_{1}(\vec{p},\vec{q},t)$ satisfies the Boltzmann equation, then $\frac{dH}{dt} \leq 0$, where $H(t) = \int d^{3}\vec{p}d^{3}\vec{q} f_{1} ln f_{1}$ H is very closely related to the information content of a one particle PDF The proof is straight forward Take the time derivative of H, and use approximation $\ln (f_{1})+1 \approx \ln f_{1}$ Use Boltzmann equation Use some integration by parts to eliminate the streaming terms (ie. $\ln f_{1}(\frac{\partial U}{\partial \vec{q_{1}}}\cdot\frac{\partial f_{1}}{\partial p_{1}}-\frac{\vec{p_{1}}}{m}\cdot \frac{\partial f_{1}}{\partial \vec{q_{1}}})$) Observe that the equation must hold if you swap particle 1 for particle 2. Do the swap, then average the two expressions Make the change of integration variables from the initiators of the collision $(p_{1},p_{2},b)$ to the products of the collision $(p_{1}’,p_{2}’,b’)$ We know that the Jacobian is unitary from time symmetry We realize that we can swap the initiators and the products. Do this swap, then average the two equations again The final results is something like $\frac{dH}{dt} = -\frac{1}{4}\int d\Gamma x$ where x is some strictly positive integrand. Hence $\frac{dH}{dt}\leq 0$ This is a big deal, since it gives a microscopic explanation to how entropy arises. Make the definition $S = -k_{b} H(T)$, or equivalently, on a microscopic scale: $\sigma(\Gamma,t) = -k_{b} \ln f_{1}(\Gamma,t)$ Classical Statistical Mechanics Microcanonical Ensemble The microstates are defined by points in phase space, and their time evolution is governed by the Hamiltonian. Since the Hamiltonian is energy conserving, all micro-states are confined to a constant energy surface in phase space This implies that all points on the surface are mutually accessible, which gives the central postulate of stat mech $p(E,x) (\mu) = \frac{1}{\Omega(E,x)}* \delta(H(\mu)-E)$ where $\mu$ enumerate the microstates ie. on the surface, all microstates are equally likely. If you are off the surface, you can’t access it The normalization $\Omega$ is the area of the surface of the constant energy E in phase space. Typically, we define $E-\Delta \leq H(\mu) \leq E+\Delta$, where $\Delta$ is uncertainty in the energy. This allows us to have the normalization be a spherical shell The entropy of this uniform probability distribution is $S = k_{b} \ln \Omega$ The overall allowed phase space is the product of individual ones (assuming independent systems) Another consequence of this is that in the microcanonical ensemble, $dE=0$ since we are prescribing a fixed energy Two State System Example Consider N impurity atoms trapped in a solid matrix. Each impurity can either have energy 0 or $\epsilon$ $H = \epsilon\Sigma_{i=1}^{N} n_{i} = \epsilon N_{1}$ where $N_{1}$ is the total number of excited impurities $p(n_{i}) = \frac{1}{\Omega}\delta_{\epsilon,\Sigma_{i}n_{e}E}$ $\Omega$ is the number of ways of choosing $N_{1}$ excited levels among the available N atoms. This is just ${N}\choose{N_{1}}$ Entropy is just $S = k_{b} \ln \frac{N!}{(N-N_{1})!N_{1}!}$ You can use Stirling’s formula to approximate $\ln N! \approx N \ln N -N$ You can then get the temperature from $\frac{\partial S}{\partial E} = \frac{1}{T}$, and invert this to get the erngy as a function of temperature You can figure out the occupancy of a particular level by imagining a a small subsystem that is decreased by the energy of that state, with particle number that is one less (ie. $p_{n_{1}} = \frac{\Omega(E-n_{1}\epsilon, N-1)}{\Omega(E,N)}$) 0th Law $\frac{\partial S}{\partial E} = \frac{1}{T}$ Think of this as a statement of thermal equilibrium 1st Law $dE = T dS + \vec{J}\cdot d\vec{x}$ Alternatively, you can write $\frac{\partial S}{\partial x_{i}} = - \frac{J_{i}}{T}$ 2nd Law $\delta S = (\frac{1}{T_{1}}-\frac{1}{T_{2}})\delta E_{1} \geq 0$ This is only stable if $\frac{\partial^{2}S_{1}}{\partial E_{1}^{2}}+\frac{\partial^{2}S_{2}}{\partial E_{2}^{2}} \leq 0$ Ideal Gas The Hamiltonian of this system is $H = \Sigma_{i=1}^{N} \frac{p_{i}^{2}}{2m}+U(q_{i})$ where U is some external potential. For now, assume that U=0 Assuming some fixed E, $\Omega$ is just the spatial components ($V^{N}$, since each particle can be anywhere in the volume) times momentum components (this is just the surface of a hypersphere of dimension 3N and radius $\Sigma_{i=1}^{N} p_{i}^{2} = \sqrt{2mE}$) The area of a hypersphere is given by $A_{d} = S_{d}R^{d-1}$ Can calculate $S_{d}$ via $I_{d} =( \int_{\infty}^{\infty} dx e^{-x^{2}})^{d}$ This equals $ \pi^{\frac{d}{2}}$ This integral is the product of d one-dimensional gaussians. This is spherically symmetric. Making that change in variable ($dV_{d} = S_{d}R^{d-1}dR$), then make another change of variable $y=R^{2}$, then use the integral form of the Gamma function $\int_{0}^{\infty} dy y^{\frac{d}{2}-1}e^{-y} = (\frac{d}{2}-1)!$ This implies that $S_{d} = \frac{2\pi^{\frac{d}{2}}}{(\frac{d}{2}-1)!}$ Hence $\Omega = V^{N} \frac{2\pi^{\frac{d}{2}}}{(\frac{d}{2}-1)!} (2mE)^{\frac{3N-1}{2}}\Delta R$ where $\delta R$ is the thickness of the hypersphere The entropy (after Sterling approximation) is then $S = Nk_{b} \ln (V (\frac{4\pi e m E}{3N})^{\frac{3}{2}})$ $\frac{1}{T} = \frac{3}{2}\frac{Nk_{b}}{E}$ Can recover ideal gas law from $\frac{P}{V} = \frac{\partial S}{\partial V}$ Can calculate the probability of finding a particle of momentum $\vec{p_{1}}$ via $p(\vec{p_{1}}) = \frac{V \Omega(E-\frac{p_{1}^{2}}{2m},V,N-1)}{\Omega(E,V,C)}$ After some algebra and Sterling’s approximation, you get the normalized Maxwell-Boltzmann distribution $p(p_{1}) = (\frac{3N}{4\pi m E})^{\frac{3}{2}}exp(\frac{-3N}{2}\frac{p_{1}^{2}}{2mE})$, where $E = \frac{3Nk_{b}T}{2}$ Mixing Entropy The problem with the above ideal gas entropy is that it’s not extensive! (ie. $S(\lambda E, \lambda V, \lambda N) = \lambda S $) What happens if you mix two gases together? Initially, $S_{i} = N_{1}k_{b} \ln(V_{1})+N_{2}k_{b} \ln(V_{2})$ After mixing $S_{f} = N_{1}k_{b} \ln(V_{1}+V_{2})+N_{2}k_{b} \ln(V_{1}+V_{2})$ Which means the change in entropy is $\Delta S = -N k_{b} (\frac{N_{1}}{N}\ln\frac{V_{1}}{V}+\frac{N_{2}}{N}\ln\frac{V_{2}}{V})$ One problem with this is that if you are “mixing” two gases. Imagine that you have a partition separating the gas. Removing the partition shouldn’t change the entropy, but the above argument shows that you have some entropy change The resolution to this is that with indistinguishable particles, you are over counting the number of states by a factor of N!. The above entropy works for for distinguishable particles The Canonical Ensemble Instead of fixing the energy E and varying the temperature T, you fix the temperature T and vary the energy E Consider two systems: one which is the system you care about (S) and another which is sufficiently large such that its temperature is not changed by interactions with S (call this system R) From a microcanonical ensemble POV, the joint probability of micro-states ($\mu_{S}\otimes \mu_{R}$) satisfied $p(\mu_{S}\otimes \mu_{R}) = \frac{1}{\Omega} \delta_{E,E_{tot}}$ where $E=H_{s}(\mu_{s})+H_{R}(\mu_{R})$ The unconditional probability for S just marginalizes over all of the energies of R If $\mu_{s}$ is specified, we can view S as some microcanonical ensemble of the reservoir with energy $E_{tot}-H_{S}$ All of the above implies $p(\mu_{s}) = \frac{\Omega_{R}(E_{tot}-H_{S}(\mu_{s}))}{\Omega_{S\otimes R}(E_{tot})}$ Since we assume that the energy of S is much smaller than the energy of R: $S_{R}(E_{tot} -H_{s}) \approx S_{R}(E_{tot}) -H_{s}\frac{\partial S_{R}}{\partial E_{R}} = S_{R}(E_{tot})-\frac{H_{S}}{T}$ The above yields that $P(\mu,T) = \frac{e^{-\beta H(\mu)}}{Z}$ where Z is the normalization $Z = \Sigma_{\mu} e^{-\beta H(\mu)}$ This Z is called the partition function We can make the association $F = -k_{b}T \ln Z(T,x)$ Suppose that Z is a function of $\beta$ (ie. T) and position (x) $d(\ln Z) = \frac{\partial \ln Z}{\partial \beta}d \beta +\frac{\partial \ln Z}{\partial x} dx = -U d\beta + \beta X dx$ where $X = \frac{1}{\beta} \frac{\partial \ln Z}{\partial x}$ We can solve for dU, which we can rearrange to be dF You can write the n’th cumulant of H as $<H^{n}> = (-1)^{n} \frac{\partial^{n} ln Z}{\partial \beta^{n}}$ This includes things like the average and the variance… $C_{V} = -T \frac{\partial^{2} F}{\partial T^{2}}$ Grand Canonical Ensemble We now allow chemical work on the system. In an analogous manner to deriving the canonical ensemble, we can give the probabilities of being in a particular energy $p(E) = \frac{exp(\beta \mu N - \beta H(E))}{Q}$ where $Q=\Sigma_{E}exp(\beta \mu N(E)-\beta H(E))$ You can get the average number of particles via $\frac{\partial}{\partial (\beta \mu)}\ln Q$ and the variance ($\frac{\partial^{2}}{\partial (\beta\mu)^{2}}\ln Q$) You can define a grand canonical potential $\mathbb{G }(T,\mu, x) = E-TS-\mu N = -k_{b}T \ln Q$ in an analogous way to the canonical ensemble Conserved Ensemble Notes For each thermodynamical potential, there is an associated ensemble where that potential is constant For microcanonical, dE = 0 For canonical, dF = 0 There are similar ensembles for H and G. In any case, you can define $X = -\beta \ln Z$ where X is the conserved potential. Simply add on $\mu N$ to the exponential to recover the associate grand potentials Interacting Particles Cumulant Expansion Imagine a general Hamiltonian $H = \Sigma_{i=1}^{N} \frac{p_{i}^{2}}{2m}+U(q_1,… q_N)$ We can write the total partition function as $Z(T,V,N) = \frac{1}{N!} \int \Pi_{i=1}^{N} (\frac{d^{3}p_{i}d^{3}q_{i}}{h^{3}}) exp(-\beta \Sigma_{i} \frac{p_i^{2}}{2m}) exp(-\beta U(q_{1}, q_{n})) = Z_{0}(T,V,N)<exp(-\beta U)>^{0}$ where $Z_{0} = (\frac{V}{\lambda^{3}})^{N}\frac{1}{N!}$ and $\lambda = \sqrt{\frac{h^{2}}{2\pi m k_{b} T}}$ and $<O>^{0}$ is the expectation value of O computed with with probability distribution of the non-interacting system We can write the above in terms of the cumulants $\ln Z = \ln Z_{0} + \Sigma_{i=1}^{\infty} <U^{i}>_{c}^{0}$ Considering that the $q_{i}$ are uniformly and independently distributed within the box V, we have the moments $<U^i>^{0} = \int \Pi_{j}^{N} \frac{d^{3}q_{j}}{V} U(q_{1},…q_{N})^{i}$ Mean Field Theory of Condensation Phase transitions are characterized by discontinuities in various state functions, and correspond to singularities in the partition functions Start with the same integral for the partition functions of an dilute gas $Z = \int \frac{\Pi_{i=1}^{N} d^{3}p_{i}d^{3}q_{i}}{N! h^{3N}} exp(-\beta\Sigma_{i}^{N} \frac{p_{i}^{2}}{2m} -\beta \Sigma_{i<j} V(q_{i}-q_{j}))$ For a non-ideal gas particles take up some space There is some potential between each particle We can approximate the potential via an average attraction energy, where we assume some uniform density $n = \frac{N}{V}$ $\bar{U} = \frac{1}{2}\Sigma_{i,j}V_{attr}(\vec{q_{i}}-\vec{q_{j}}) = \frac{1}{2}\int d^{3}\vec{r_{1}}d^{3}\vec{r_{2}} n(\vec{r_{1}}) n(\vec{r_{2}}) V_{attr}(\vec{r_{1}}-\vec{r_{2}}) \approx \frac{n^{2}}{2}V \int d^{3} r V_{attr}(\vec{r}) = -\frac{N^{2}}{2V} u$ We can approximate the volume integral via $\Pi_{i=0}^{N-1} (V-i\Omega) \approx (V-\frac{N\Omega}{2})^{N}$ Can take the log to turn the product into a sum, then use the (very handwavy) argument that on average $i=\frac{N}{2}$, and there are N terms, so you have $\Pi_{i=0}^{N-1} \ln(Vi-\Omega)\approx N \ln(V-\frac{N\Omega}{2})$. Invert the log to get the final result Hence, the total partition function is $Z \approx \frac{(V-\frac{N\Omega}{2})^{N}}{N! \lambda^{3N}} exp(\frac{\beta u N^{2}}{2V})$ You can rederive the Van-derWals equation of state by finding the free energy, then using $P = -\frac{\partial F}{\partial V}$ Critical Points In a P,T plot, there is a line of coexistence along with two phases of matter can coexist. This line terminates at the so called “critical point” The critical point can be found by taking the equation of state ($P = P(N,V,T)$), setting it’s first derivative and second derivatives to 0 The first derivative constraint comes from the fact that the critical point is the limit of flat coexistence portion of the isotherms The second derivative constraint comes from stability reasons Solve this system of equations for $P_{c},T_{c},V_{c}$ One can use these critical values to rescale the equation of state into a material independent equation of state (ie $P_{r} = \frac{P}{P_{c}}$ and similar variables) Quantum Statistical Mechanics Dilute Polyatomic Gases The Hamiltonian for each molecule of n atoms is $H = \Sigma_{i=1}^{n} \frac{p_{i}^{2}}{2m}+V(q)$ If the atoms have different masses, you can rescale the coordinates by scaling $\vec{q_{i}}$ by $\sqrt{\frac{m_{i}}{m}}$ and scaling $\vec{p_{i}}$ by $\sqrt{\frac{m}{m_{i}}}$ Ignoring the interactions between molecules, you can define the partition function as $Z(N) = \frac{1}{N!} (\int \Pi_{i=1}^{N} \frac{d^{3}p d^{3}q}{h^{3}} exp(-\beta \Sigma_{i=1}^{n} \frac{p_{i}^{2}}{2m}-\beta V))^{N}$ If the temperatures are smaller than than the dissociation energies ($\approx 1E4 K$), then there are only small deformations. The procedure to find the contributions of the deformations to the single particle partition function goes like: Find the equilibrium positions by minimizing V Do a small perturbation of $\vec{q_{i}} = \vec{q_{i*}} +\vec{u}$ around these equilibrium positions. This expand V around the minimum via $V+ V_{*}+\frac{1}{2}\Sigma_{i,j=1}^{n}\Sigma_{\alpha,\beta = 1}^{3} \frac{\partial ^{2} V}{\partial q_{i,\alpha} \partial q_{j,\beta}} u_{i,\alpha} u_{j,\beta}$ The second derivative matrix is a 3nx3n positive definite, which means that you can diagonalize it and do a change of basis from $u_{i}$ to $u_{s}$. This allows you to write the deformation Hamiltonian as $H_{1} = V_{*} + \Sigma_{s=1}^{3n} (\frac{p_{s}^{2}}{2m} + \frac{K_{s}}{2} u_{s}^{2})$ What is the average energy of each molecule? This is the expectation value of the deformation hamiltonian. Fro equipartition, there are 3n quadratic momentum degrees of freedom and $m\leq 3n$ number of modes with non-zero $K_{s}$ Some of these eigenmodes are forced to be 0 from symmetries Translational symmetry: the original potential is invariant under translation by c. This means that no energy is stored in the center of mass coordinate $\vec{Q} = \Sigma_{\alpha} \frac{\vec{q_{\alpha}}}{n}$, which means the 3 $K_{trans}$ associated with this coordinate are 0 Rotational symmetry: There is no potential energy associated with rotation. There can be at most 3 degrees of freedom in rotations. Some molecule shapes will have less (a rod will have 2, since rotation along the axis is not unique) This implies that $m = 3n-3-r$ (assuming no unique molecule shape). These are the vibrational modes If you calculate $\gamma = \frac{C_{p}}{C_{v}}$ for various gases and compare to the experimental values, you get an overestimation. This gets solved if you quantize the vibrational modes Quantized Vibrational Modes The vibrational partition function becomes $Z_{vib} = \Sigma_{n=0}^{\infty} exp(-\beta \hbar \omega (n+\frac{1}{2})) = \frac{e^{\frac{-\beta \hbar \omega}{2}}}{1-e^{-\beta \hbar \omega}}$ The expectation value of the energy is then $E_{vib} = -\frac{\partial \ln Z}{\partial \beta} = \frac{\hbar \omega}{2} + \hbar \omega \frac{e^{-\beta \hbar \omega}}{1-e^{-\beta \hbar \omega}}$ Quantized Rotational Modes The classical Hamiltonian is $H_{rot} = \frac{\vec{L^{2}}}{2I}$. The quantized version of angular momentum is $L^{2} = \hbar^{2} l(l+1)$ Hence, the partition function is $Z = \Sigma_{l=0}^{\infty} exp(-\frac{\beta \hbar^{2} l(l+1)}{2I}) (2l+1)$, where the 2l+1 comes from the degeneracy of the m quantum number Defining some characteristic function $\theta_{rot} = \frac{\hbar^{2}}{2I k_{b}}$, the partition function can be rewritten as $Z = \Sigma_{l=0}^{\infty} exp(-\frac{\theta_{rot} l(l+1)}{T}) (2l+1)$ Doing this sum in hard, but in high and low temperatures limits, you can approximate the sum as an integral and only keep the first few terms in the sum respectively Blackbody Radiation Imagine EM waves with wave-number $\vec{k}$ and two polarizations $\alpha$. The Hamiltonian for the EM field can then be written as a sum of harmonic oscillators $H = \frac{1}{2}\Sigma_{k,\alpha} (\tilde{p_{k,\alpha}})^{2} + \omega_{\alpha}(\vec{k})^{2} |\tilde{u_{\alpha}}|^{2}$ There is no limit on the size of the Brillouin zone for k, so you can have an arbitrarily large number in the sum. This causes the ultraviolet catastrophe In more details: you can have an arbitrary number of high frequency modes and each of those modes has an energy of $k_{b}T$ associated with it. This implies there is an infinite amount of energy This is solved by quantizing the allowed value of the EM energy $H = \Sigma_{k,\alpha} = \hbar c k (n_{\alpha}(k) + \frac{1}{2})$ where $n_{\alpha}(k) = 0,1,2,…$ The internal energy can be calculated as per usual $E = <H> = \Sigma_{k,\alpha} \hbar c k (\frac{1}{2}+\frac{e^{-\beta \hbar c k}}{1-e^{-\beta \hbar c k}}) = VE_{0} + \frac{2V}{(2\pi)^3}\int d^{3}\vec{k}\frac{\hbar c k}{e^{\beta \hbar c k}-1}$ $VE_{0}$ is the infinite zero point energy, but only energy differences can be measured, so this cancels out The integral can be calculated by making the change of variable $x = \beta \hbar c k$ and using the identity $\int_{0}^{\infty} \frac{dx x}{e^{x}-1} = \frac{\pi^{2}}{6}$ Hence: $\frac{E}{V} = \frac{\pi^{2}}{15}(\frac{k_{b} T}{\hbar c})^{3} k_{b}T$ Imagine that you poke a hole in the side of the box. The energy flux per unit area per unit time is then $\phi = v_{\perp} \frac{E}{V}$ $v_{\perp}$ for the photons is $v_{\perp} = c * \frac{1}{4\pi} \int_{0}^{\frac{\pi}{2}} 2\pi \sin \theta d\theta \cos \theta = \frac{c}{4}$ Hence $\phi = \sigma T^{4}$ where $\sigma = \frac{\pi^{2}}{60}\frac{k_{b}^{4}T^{4}}{\hbar^{3}c^{2}}$ is the Stefan-Boltzmann constant Formalism The micro-state of a quantum system is completely specified by a unit vector $|\Phi>$ This can be decomposed into its components along any suitable orthonormal basis: $|\Phi> = \Sigma_{n} <n|\Phi> |n>$ Observables become operators $O(\vec{p_{i}},\vec{q_{i}})$, where we have the commutation relation: $[p_{j},q_{k}] = \frac{\hbar}{i}\delta_{j,k}$ These operators must be Hermitian The time evolution operator is defined as $\Psi(t)> = U(t,t_{0}) | \Psi(t_{0})>$ U can be related to the Hamiltonian: $U(t,t_{0}) = exp(-\frac{i}{\hbar}H(t-t_{0}))$ Macrostates are now ensemble values: $\bar{O(p_{i},q_{j},t)} = \Sigma_{\alpha} p_{\alpha}O = \int \Pi_{i=1}^{N} d^{3}p_{i} d^{3}q_{i} O \rho$ $\rho = \Sigma_{\alpha} \Pi_{i}^{N} \delta^{3}(\vec{q_{i}}-\vec{q_{i}(t)}) \delta^{3}(\vec{p_{i}}-\vec{p_{i}(t)}_{\alpha})$ is the ensemble density We can then write the expectation value of O as $\bar{O} = \Sigma_{m,n}<n|\rho|m> <m|O|n> = tr(\rho O)$ We can define the density matrix as $<n|\rho(t)|m> = \Sigma_{\alpha} p_{\alpha} <n|\Phi_{\alpha}(t)><\Phi_{\alpha}(t)|m>$ or alternatively: $\rho(t) = \Sigma_{\alpha} p_{\alpha} | \Phi_{\alpha}(t)>< \Phi_{\alpha}(t)|$ The density matrix has some nice properties $tr(\rho) = 1$ due to normalization of states $\rho$ is Hermitian $\rho$ is positive definite Being in equillibrium demands that $\frac{\partial \rho}{\partial t} = 0$, which implies via Heisenberg equation of motion that $i\hbar \frac{\partial \rho}{\partial t} = [H,\rho]$ Given the above, we can define some ensembles: Micro: $\rho(E) = \frac{\delta(H-E)}{\Omega(E)}$. This allows the density matrix to obey the constraint of a fixed energy E Canonical: A fixed temperature $T = \frac{1}{k_{b}\beta}$ can be assigned by being in contact with a reservoir $\rho(\beta) = \frac{exp(-\beta H)}{Z(\beta)}$, and from the normalization condition on $\rho$, we have that $Z = tr(e^{-\beta H})$ (remember, Z is a scalar, not an operator!) Grand Canonical: Similarly, if we vary the particle number, we have that the grand canocial ensemble is $Q = tr(exp(-\beta H \beta \mu_{i}N_{i}))$ Identical Particles If we have N particles, we have N! permutations P which form a group $S_{N}$. We can label the identity permutation by the ascending positive integers (ie. if N=3, the identity is (1,2,3)) We can define the parity of a permutations as $(-1)^{P} = \pm 1$ with the plus being associated with an even number of exchanges from ascending order, and minus being associated with an odd number of exchanges Aside: Draw lines connecting the initial and final locations of each integer. The parity is (-1) raised to the number of intersections of these lines Two types of particles: Bosons, which are even under permutation: $P|\Psi(1,…N)> = |\Psi(1,…N)>$ Fermions, which are odd under permutation: $P|\Psi(1,…N)> = (-1)^{P}|\Psi(1,…N)>$ Suppose that we have a Hamiltonian of N non-interacting particles in a box of volume V. This gives $H = \Sigma_{\alpha=1}^{N} (-\frac{\hbar^{2}}{2m}\nabla^{2}_{\alpha})$ Assuming distinguishable particles: $<x_{1}…x_{n}|k_{1}…k_{n}> = \frac{1}{V^{N/2}} exp(i\Sigma_{\alpha=1}^{N} \vec{k_{\alpha}}\cdot \vec{x_{\alpha}})$ and $H|k_{1}…k_{n}> = (\Sigma_{\alpha=1}^{N} \frac{\hbar^{2}}{2m} k_{\alpha}^{2}) |k_{1}…k_{n}>$ If we are to make the particle indistinguishable, then depending on their nature, only a subspace of the distinguishable Fock space We can bundle both cases like $|\vec{k} >= \frac{1}{\sqrt{N_{\eta}}} \Sigma_{P} \eta^{P} P | \vec{k}>$ where $\eta$ is 1 for bosons and -1 for fermions Each state is uniquely specified by a set of occupation numbers ${n_{\vec{k}}}$ such that $\Sigma_{\vec{k}} n_{\vec{k}} = N$ For fermions, $|\vec{k}> = 0$ unless $n_{k}$ is 0 or 1 For bosons, any k may be repeated $n_{k}$ times The normalization for both fermions and bosons is N! Grand Canonical Formulation $Q_{n}(T,\mu) = tr(exp(-\beta H)) = \Sigma_{n}^{N} \Pi_{\vec{k}} exp(-\beta(\Epsilon(\vec{k}) -\mu) n_{\vec{k}})$ For fermions, $n_{\vec{k}}$ can be 0 or 1, while bosons can range from 0 to $\infty$. In either case, we have $ln Q_{\eta} = -\eta \Sigma_{k} \ln (1-\eta exp(\beta\mu-\beta\Epsilon(k)))$ with $\eta=-1$ for fermions and $\eta=+1$ for bosons Just do the sum of two terms for fermions, and for bosons, do the infinite geometric series The average occupation number of a state of energy $\Epsilon(k)$ is given by $<n_{\vec{k}}> = -\frac{\partial \ln Q_{\eta}}{\partial (\beta \Epsilon(\vec{k}))} = \frac{1}{z^{-1}exp(\beta \Epsilon(\vec{k}))-\eta}$ The average particle number and internal energy are then just $N_{\eta} = \Sigma_{\vec{k}} <n_{\vec{k}}>$ and $E_{\eta} = \Sigma_{\vec{k}} \Epsilon(\vec{k}) <n_{\vec{k}}>_{\eta}$ Both of these quantities can be extended to a continuum spectrum via the substitution $\Sigma_{\vec{k}} \rightarrow \int d^{3}k \frac{1}{(2\pi)^{3}}$ Degenerate Fermi Gas Fermi occupation is $<n_{\vec{k}}> = \frac{1}{exp(\beta(\Epsilon(k)-\mu))+1}$ At T=0, $\mu=\epsilon_{F}$ or the Fermi energy At T=0, all one-particle states of energy less than $\epsilon_{F}$ are occupied. For an ideal gas, we can $\Epsilon(k) = \frac{\hbar^{2}k_{F}^{2}}{2m}$ where $k_{F}$ is the Fermi wavenumber $k_{F}$ Using the above definition for average N, you can define the number density as $n = \frac{N}{V}$. For an ideal gas, we have that $\Epsilon(k) = \frac{\hbar^{2}}{2m}(\frac{6\pi^{2}n}{g})^{\frac{2}{3}}$ wher g is the spin degeneracy $f_{m}^{\eta}(z)$ The function $f_{m}^{\eta}(z) = \frac{1}{(m-1)!}\int_{0}^{\infty} \frac{dx x^{m-1}}{z^{-1}e^{x}-\eta}$ shows up a lot when dealing with degenerate gases. Namely, the number density is proportional to this function. $\eta = \pm 1 $ for bosons and fermions respectively and $z = exp(\beta \mu)$ $f_{m}^{\eta}(z) = \Sigma_{\alpha=1}^{\infty} \eta^{\alpha+1} \frac{z^{\alpha}}{\alpha^{m}}$ in the high temperature (small z), low density limit There is a recursion relationship for this function: $\frac{d}{dz}f_{m}^{\eta} = \frac{1}{z} f_{m-1}^{\eta}(z)$ Denegerate Bose Gas In the low T limit, we have that z approaches unity. Hence, we need to find $f_{+}^{1}(1)$ in order to calculate the maximum number density $n_{max}$ For m<1, the integral diverges, while for m>1, the integral is finite Hence, this sets an upper bound on the density of excited states. For instance, in a bosonic, non-relativistic gas, we have $n \leq \frac{g}{\lambda^{3}}f_{\frac{3}{2}}^{+}(1)$ where $\lambda = \frac{h}{\sqrt{2\pi m k_{b}T}}$ The critical temperature $T_{c}$ is when the above is at equality. For $T<T_{C}$, z gets pinned to 1. The remaining density $n_{0} = n-n_{max}$ occupied the lowest energy state This is Bose-Einstein condensation

IEEE Floating Point Standard Arrays Numerics Random Numbers LCM (Linear Congruent Method) Testing RNG Nonuniform Distributions Non-Analytic Nonuniform Distributions Interpolation Linear Interpolation Fitting Basis Coefficients Polynomial fits Fourier Basis Integration Trapezoid Rule Simpson’s Rule Rescaling Integrals Differentiation Scaling Linear Algebra LU SVD Eigensystems Uses Principal Component Analysis (PCA) Dimensional Reduction Root-Finding Minimization Quadratic Fit Golden Section Search Multi-Dimensional FFTs ODEs PDEs Initial Condition Problems Boundary Condition Problems MCMC (Markov Chain Monte Carlo) Markov Chain Metropolis Hastings Burn-in and Proposal Distribution IEEE Floating Point Standard $\pm 1+ 2^{e-127}\Sigma_{i=0}^{22} f_{i} 2^{-(i-23)}$ e is the exponent number $f_{i}$ represents the floating bits Some edge cases: e = 0, $f \neq 0$ are called subnormal numbers, where “1” above is replaced with 0 and $e-127$ is replaced by $e-126$ e=0 and f=0 is a signed zero e=255, f=0 is a signed $\inf$ e=255, $f\neq = 0$ is NaN Arrays TL;DR if you want to go fast, then use numpy arrays, since they are more contiguous in memory (and hence cache friendly) compared to python lists You also don’t have to worry about type checking since Python needs to check if operations There are also some built-in functions that numpy provides which are faster than Python built-ins Numerics TL;DR Keep numbers from being too small and too large and you’re good $E[\delta (f(a))^{2}] \approx E[[\frac{\partial f}{\partial a}]^{2} \delta a^{2}]$ Assuming similar errors for each operations, you find that the round-off error scales as $\sqrt{N}$ Making finer time steps yields a better approximation, but the round-off error degrades this. There is some optimal step size which yields the most accurate result Random Numbers Probability distributions have the defining property that $\int_{-\infty}^{\infty} dx P(x) = 1$ LCM (Linear Congruent Method) $x_{i+1} = (ax_{i}+c) mod M$ a,c and M are parameters. After M random numbers, you repeat yourself a and M are choosen such that a*M is less than the highest integer available Numpy uses Mersenne Twister, which has a longer period (ie. number of cycles before generator repeats cycles back to the start) Testing RNG The expectation value of the number of values to fall in each bin is $\frac{N}{n_{bins}} = \bar{N_{bin}}$ With enough bins and enough data, each bin should follow a Poisson distribution. Hence, we expect the variance of each bin to be $\bar{N_{bin}}$ Nonuniform Distributions You can generate these from uniform ones. A generic procedure is as follows You have the uniform distribution $f(r) dr = dr$ for $0<r<1$ and you want to find $f(x) dx$ For each little differential, $|f_{r} dr| = |f_{x} dx|$ Rearranging yields: $|\frac{dx}{dr}| = \frac{1}{f(x)}$ Solving the aoove for $x(r)$ allows one to transform the uniform distribution to the desired nonuniform one The above doesn’t always work. A notable example is gaussians. You get a non-invertible function. You do some strange transformations (Box-Muller method) to allow you to circumvent this Non-Analytic Nonuniform Distributions You try and write down the cumulative distribution of your desired non-analytic distribution: $F(x) = \int x dx’ f(x’)$ which maps to the range $[0,1]$ You can then generate points from f(x) by drawing from a uniform distribution and then using the formula $x = F^{-1}(r)$ where the inverse is calculated numerically via interpolating a tabulation of F(x) Interpolation Interpolation means given some data, how do you generate estimates for data points that are not in the set In a linear framework, we want to find $\hat{f}(x) = \Sigma_{i} W(x, x_i) f(x_{i})$ Linear Interpolation Dead simple $\hat{f}(x) = f(x_{i}) + [f(x_{i+1}) \frac{x-x_{i}}{x_{i+1}-x_{i}}]$ This corresponds to a weight of $W(x,x_{i}) = 1-|x-x_{i}|$ Outside the range $-1 < (x-x_{i}) < 1$, w is 0 Triangular function Fitting Basis Coefficients Can think of $\hat{f}(x) = \Sigma_{i} a_{i} B_{i}(x_{j})$ where B_{i} is some set of basis functions with coefficient $a_{i}$ Can think of this as a matrix equation: $\vec{f} = B\cdot \vec{a}$ Solving yields: $\vec{a} = B^{-1} \cdot \vec{f}$ You can invert B if the basis vectors are independent and have support where your samples are Hence $\hat{f}(x) = \Sigma_{i} B_{i}(x) \Sigma_{j} B_{ij}^{-1} f(x_{j})$ $W(x,x_{i}) = B_{i}(x) \Sigma_{j} B_{ij}^{-1}$ For linear interpolation, B is just the identity matrix Polynomial fits Your basis functions could be other things. Take polynomials for instance: $B_{0} = 1$ $B_{1} = x$ $B_{2} = x^{2}$ etc To help with numerical stability, you can rescale x to x’ by a constant so that powers of x’ are close to unity You can also rescale to the range -1 to 1 Might as well take the polynomials to be the Legendre polynomials, which are defined on that interval For very high order, the matrix behaves poorly numerically Fourier Basis The weights for the Fourier basis are $W(x,x_{i}) = \frac{1}{2\pi} \int_{-pi}^{\pi} d\omega exp[i\omega(x-x_{i})] = \frac{\sin[\pi(x-x_{i})]}{\pi(x-x_{i})}$ Integration The goal is to approximate $\int_{a}^{b} dx f(x) = \Sigma_{i=1}^{N} f(x_{i}) w_{i}$ Trapezoid Rule Each term in the sum takes the form $\frac{1}{2} \delta x (f_{i} + f_{i+1})$ Simpson’s Rule The main idea: we try to approximate the function locally around the points i-1, i, i+1 as a quadratic: $f(x) = \alpha’+\beta’ x + \gamma’ x^{2}$ Make the transformation to $f(x) = \alpha + \beta (\frac{x-x_{i}}{dx})+ \gamma (\frac{x-x_{i}}{dx})^{2}$ evaluate f at each point yields $f_{i-1} = \alpha -\beta +\gamma$ $f_{i} = \alpha$ $f_{i+1} = \alpha + \beta + \gamma$ Inverting the above yields $\alpha = f_{i}$ $\gamma = \frac{f_{i+1} + f_{i-1}}{2} - f_{i}$ $\beta = \frac{f_{i+1} - f_{i-1}}{2}$ You apply this approximation to two segments at a time, which require that N is odd Rescaling Integrals You can rescale an integrals bounds: $I = \int_{a}^{b} dx f(x) \frac{b-a}{b’-a’} \int_{a’}^{b’} dx’ f(x(x’))$ $x’ = (x-a) \frac{b’-a’}{b-a}+a'$ Setting a’ to -1 and b’ to 1 is very useful. This yields $I = \frac{b-a}{2} \int_{-1}^{1} dx’ f(x(x’))$ If you make the substitution: $x = q \frac{1+x’}{1-x’}$, then you can rescale an infinite range to a finite one: $I = \int_{0}^{\infty} dx f(x) = \int_{-1}^{1} dx’ \frac{2q}{(1-x’)^{2}} f(x(x’))$ q is a choice to be made, and x=q when x’ =0 You want to choose q somewhere near the halfway point in the integral Differentiation $\frac{df}{dx} = \frac{f(x+\frac{dx}{2})-f(x-\frac{dx}{2})}{dx}$ The central difference approximation of the derivative of f at x That’s it. There are other schemes that you can use (like forward difference and backwards difference), but those are more niche To extend to multivariable, you just fix the rest of the variables and only vary one at a time There is ostensibly an optimal choice for dx, which relates to the machine precision and the third derivative, but for all practical purposes, this is irrelavent since if you know the third derivative, why would you be approximating anything? There is something called jax, which lets you programatically calculate the derivative of a function using the chain rule The idea is that you calculate the analytical derivative of the function in parallel with the function itself. This process can be recursive, hence allowing the iteration Scaling You can rescale all of the linearly independent units of a problem in order to accomplish two things Maximize the dynamic range by keeping numbers close to unity Allowing one to apply the same equations to different scales Take Poisson’s equation for gravity: $\nabla^{2} \Phi = 4\pi \rho G$ We can redefine $r’ = \frac{r}{R}$ $m’ = \frac{m}{M}$ $t’ = \frac{t}{t_{0}} = \frac{t}{\sqrt{\frac{GM}{R^{3}}}}$ The above let’s use rewrite Poisson’s equation in a manner which is scale invariant Linear Algebra LU Central problem: Solve Ax=b where A is a matrix and b is a vector Can use LU decomposition: A=LU where L is a lower triangular matrix and U is an upper triangular matrix L is a lower triangular matrix whose diagonal elements are 1 (this follows since there are $N^{2}+N$ values to set in L and U, but only $N^{2}$ contraints in A) This factorization is done by using Gaussian elimination with pivoting Once you have the LU factorization, you can easily find the determinant of A, since det(L) = 1 and det(U) = $\Pi_{i} U_{ii}$, and det(A) = det(L)det(U) Numerically, L and U are easier to work with than A (more 0s and 1s) SVD If A is very close to being singular (causing inverses to be unstable), or Ax=b is not unique or doesn’t exist, you can use the SVD to get an approximate solution You can write any matrix as $A = U W V^{T}$ U is an MxN matrix whose columns are orthonormal W is a NxN diagonal matrix $V^{T}$ is the transpose of an orthonormal matrix You can think of the SVD as telling you the range of A (ie. the subspace that A can map into) and the null space of A (ie. the space for which Ax = 0) The columns of U for the orthonormal basis of the range of A The columns of V correspond to the basis which spans the null space W tells you the rank of the matrix (the number of non-zero $w_{j}$) From the SVD, you can get the condition number of a matrix (ie. $C = \frac{max(W)}{min(W)}$) A singular matrix would have C be infinity. if 1/C approaches machine precision, then the matrix is singular enough that doing numerics on it directly is unstable You can easily get the inverse of A from the SVD: $A^{-1} = V W^{-1} U^{T}$ $W^{-1}$ is trivial to find (take the inverse of all the diagonal elements) Eigensystems Solve the equation $Ax = \lambda x$ Technically, these are right eigenvectors. The left eigenvectors are defined as $xA = \lambda x$. These left ones have the same eigenvalues as the right Analytically, you just solve $det(A-\lambda I)=0$, but this is an awful way of doing this numerically. You actually use QR factorization. We don’t talk about this in class though The idea is that you find the QR decomposition, where Q is an orthogonal matrix (ie. $Q^{T} = Q^{-1}$) and R is an upper triangular matrix analytically, you use Gram-Schmidt, but this is numerically unstable. More numerically stable methods, like Givens rotations and Householder reflections, find ways to stuff zeros in the intermediate matrices to prevent round-off errors You then define the matrix B = RQ. Why? Observe that $B = RQ = Q^{-1}QRQ = QAQ = Q^{T}AQ$ This means that B is a similar matrix to A, and hence has the same eigenvalues You apply this process iteratively, and since all the intermediate matrices are similar, they all have the same eigenvalues Note that this process doesn’t preserve the eigenvectors of A. You can find those by solving the characteristic equation ($Ax= \lambdax$) for the particular eigenvalue (potentially solving for multiple lambda at once with the block power method) Uses Principal Component Analysis (PCA) For a given real symmetric, positive definite matrix (this kind of thing shows up a lot in physics, like in the moment of inertia tensor), you can find the three principal axes of the system via eigendecomposition Dimensional Reduction PCA is also useful for reducing the dimensionality of a dataset, which allows a lower dimensional approximation of a high dimensional dataset (for compression or interprability reasons) Suppose that you have some dataset $\vec{q_{i}}$ with mean $\vec{q_{0}}$ You can define the residuals as $\vec{r-{i}} = \vec{q_{i}}- \vec{q_{0}}$ and the Covariance matrix as $C = \vec{r_{j}}\vec{r_{i}}^{T}$ The eigenvalues and vectors tells you the coordinate system which diagonalizes this matrix Since eigenvectors with larger eigenvalues have large variance, they are more important to describing the shape of the data. Hence, you can throw away the smaller eigenvalues and associated eigenvectors, and still keep the majority of the information Root-Finding Solve the equation $f(x) = 0$ For many methods, you need to bracket the root (ie. define a region on the domain of x which has the root in it). Typically, you just start with an initial guess for a bracket, then grow the bracket symmetrically by some factor until you have the root. This doesn’t always work though The brain-dead way of finding a root is bisection. This is just binary search, with your stopping condition being if the new bracket’s width is on the order of machine precision There is a less-sure fire, but faster method called Brent’s method, which fits a parabola to the left, right and midpoint of your bracket, then steps towards the nearest root of the parabola If the root falls outside the bracket, or the update step is very tiny, then Brent defaults to bisection The exact details do some massaging of terms to make things numerically stable when solving the quadratic You you have access to the derivative of the function, you can use Newton-Raphson’s method You iteratively find $x_{i+1} = x_{i}-\frac{f(x)}{f’(x)}$ If you are nowhere near a root, then Newton’s method can not converge, so it’s usually best to use for really precise determinations of a root, due to it’s quadratic convergence Minimization Quadratic Fit Assume that you can evaluate your function f(x), and you have defined your bracket from [a,c], and you choose some point b in the bracket Basic idea: Try fitting a quadratic of the form $\tilde{f}(x) = \alpha x^2 \beta x +\gamma$ to the points a,b,c (ie. solve for the parameters via solving a matrix equation). Define your new bracket according to the minimum of this quadratic. This isn’t perfectly stable, since there is a change that the minimum falls outside of your current bracket Golden Section Search Another idea: We want to shrink the bracket by a consistent amount each step Define the fractional position of x=b as $w = \frac{b-a}{c-a}$ Suppose that the the new trial x is greater than b. Define the fractional position of the trail x as $z = \frac{x-b}{c-a}$ We want the new bracket length ($z+w$) to be the same as the original bracket length $1-w$ If we apply this requirement iteratively, we want the fractional position of the new trail x to be the same as b was in the old bracket In math, this means $\frac{z}{1-w} = w$ Solving for w, we see that $w = \frac{3-\sqrt{5}}{2}$ A more standard 1D minimization routine combines Quadratic fitting and Golden Section Search If the parabolic step falls outside the bracket, or the step is larger than the previous step, you fall back on GSS this is called Brent’s method, confusingly enough… Multi-Dimensional Idea: We use the Hessian matrix ($A_{ij} = \frac{\partial^{2} f}{\partial x_{i}\partial x_{j}}$) to approximate the function locally as a multi-dimensional quadratic, then step to the minimum Problem: You normally don’t know the Hessian of a function Idea: If we know the eigenvectors of the Hessian, then you can minimize along the direction of each eigenvector (see here) If you don’t know the Hessian, then you can use Quasi-Newtonian methods like Broyden-Fletcher-Goldfarb-Shanno FFTs Definition of Fourier Transform: $H(f) = \int_{-\infty}^{\infty} dt h(t) exp(2\pi i ft)$ $h(t) = \int_{-\infty}^{\infty} dt H(f) exp(-2\pi i ft)$ Discretization: Suppose that your set of samples at equally spaced locations $t_{k} = k\Delta$, where k runs from 0 to N-1 Let $f_{n} = \frac{n}{N\Delta}$, where n ranges from $-\frac{N}{2}$ to $\frac{N}{2}$ $H(f_{n}) \approx \Delta \Sigma_{k=0}^{N-1} h(t_{k}) exp(\frac{2\pi i kn}{N})$ $h_{k} = \frac{1}{N} \Sigma_{n=0}^{N-1} H_{n} exp(\frac{-2\pi i kn}{N})$ This is slow to calculate (scales as $N^{2}$). FFT can give $N log N$ scaling instead Basic Idea: Assumes that N is a power of 2 ( can pad out dataset to make it a power of 2). Split the series into even and odd terms. Recursively apply this splitting of even and odd terms. That’s it (lot’s of bookkeeping, but that’s it) ODEs an Nth order differential equation can be reduced to N 1st order differential equations by letting $\frac{dx}{dt} = y$ (and similar constructs can be used for higher derivatives) To calculate first order derivatives, you should use Runga-Kutta methods This is good enough for most applications If you care about energy conservation, you can use things like Leapfrog If you have multiple time scales in a system, then you need to make surer that your step-size is small enough to capture the relevant phenomena PDEs Initial Condition Problems Consider PDEs of the form $\frac{\partial \vec{q}}{\partial t} = -\frac{\partial \vec{F}(q)}{\partial x}$, where F is the conserved flux. For simplicity, let’s look at the example $\frac{\partial \vec{q}}{\partial t} = -v\frac{\partial q}{\partial x}$ The naive way of solving this numerically is discretizing space and time into equal spaces: $x_{j} = x_{0}+j\delta x$ $t_{j} = t_{0}+n\delta t$ This gives an update scheme like: $\frac{q_{j}^{n+1}-q_{j}^{n}}{\delta t} = -v \frac{q_{j+1}^{n}-q_{j-1}^{n}}{2\delta x}$ If we are given q defined for all j at n=0, then we can propagate outwards to solve for when n=1, n=2, etc. This is not stable. We can see this by Taylor expanding around the difference equation solution (which is $q_{j}^{n} = \epsilon^{n}exp(ikj\delta x)$) Solving for $\epsilon$, we see that the magnitude of $\epsilon$ is greater than 1 for all k, which implies instablity The Lax method simply replaces the estimate of $q_{j}$ at step n with the average of the neighboring points This is stable, as long as $\frac{\delta x}{\delta t} < v$, which is calculated by the same sort of stability analysis as before (called von Neumann Analysis) This is called the Courant condition. Physically, this means that the time step of the simulation must be smaller than the propagation speed of the signal across a single cell This has a problem, in that higher order wavenumbers get dampened out You can see this by rewriting the lax equation to tease out a term which looks like $\frac{\delta x^{2}}{2\delta t}\frac{\partial^{2}q}{\partial x^2}$ (like a viscosity term) Implicit methods also exist, where we utilize future mesh points in the difference equations in such a way that they can’t be uncoupled For instance, the diffusion equation ($\frac{\partial q}{\partial t} = \frac{\partial }{\partial x}(D \frac{\partial q}{\partial x})$) can be discretized like $q_{j}^{n+1} = q_{j}^{n}+\alpha(q_{j+1}^{n}-2q_{j}^{n}+q_{j-1}^{n})$ These are just a linear system of equations with N unknowns (the $q_{j}^{n+1}$) Crank-Nicolson methods are a combination of the two (ie. you take a half step explicitly, and a half step implicitly). The overall method is still implicit, but you get better stability properties and conservation of small-scale structure Boundary Condition Problems If you have constant coefficients, you can get away with Fourier methods Look at Poisson’s equation: $\nabla ^{2} q = \rho$ The Fourier transform of $\nabla^{2} q = -k^{2} q$. Hence, we can write q as $q = -FT(\frac{\rho}{k^{2}})$ After discretizing, we can apply a discrete fourier Transform to all the involved variables $q_{j} = \frac{1}{J}\Sigma_{n=0}^{J-1} \tilde{q}_{n}exp(\frac{-2\pi i j n}{J})$ and similar expansion for $\rho$ Plugging in and rearranging, we can match coefficients to find that $\tilde{q_{n}} = \frac{\delta^{2}\tilde{\rho_{n}}}{2\cos(\frac{2\pi n}{J})-1}$. The original variables can be calculated via inverse Fourier Transforms Another category of methods are relaxation methods, where, if you are dealing with elliptic PDEs (ie. $Lq = p$ where L is an elliptic operator), you can instead solve the diffusion equation $\frac{\partial q}{\partial t} = Lq-p$. Static solutions to the above satisfy the original eqaution, so if you start with a random q, then iteratively apply the finite difference method of the diffusion equation until q doesn’t change anymore (you are “relaxing” to the stable solution) MCMC (Markov Chain Monte Carlo) Suppose that you have some data $\vec{d}$ and a set of model parameters $\vec{theta}$. MCMC allows you to generate a distribution of $\theta$ (called the posterior) which is peaked at the “best” choice of parameters, whose spread is associated with uncertainties about the correct parameters This makes no assumption on wheather $\theta$ is properly modelling the data From Baye’s Theorem: $P(\vec{\theta}| \vec{d}) = \frac{P(\vec{d}|\vec{\theta})P(\vec{\theta})}{P(\vec{d})}$ In words: The probability distribution of the real $\theta$ given the data (posterior) equals the probability distribution of the data given the model parameters (likelihood) multiplied by the distribution of the model parameters (prior) divided by the distribution of the data (marginalization) in fewer words: posterior = likelihood*prior/marginalization Intuitively, you can think of your current best guess for distribution of the model parameters (say, based on past data you have collected) The likelihood represents our current understanding of how the model parameters describe the data (ie. our model) The marginalization is the distribution of data sets over all possible choices of $\theta$ You don’t normally need to calculate this, but if you had to, it’s just an integral over all $\theta$ $P(\vec{d}) = \int d^{N} \vec{\theta} P(\vec{d}|\vec{\theta}) P(\theta)$ Markov Chain A Markov Chain is a sequence of points $\vec{\theta_{n}}$ such that the distribution of point n+1 only depends on point n We want this sequence to have the property that it is distributed according to the posterior For any desired distribution $\pi(\vec{\theta})$, we can get a Markov chain that follows $\pi$ if $\pi(\vec{\theta_{1}})p(\theta_{1}|\theta_{2}) = \pi(\vec{\theta_{2}})p(\theta_{2}|\theta_{1})$ Proof: Ask the question: Given a set of points $\theta_{1}$, what is the distribution of the next set of points $\theta_{2}?$ $p(\theta_{2}) = \int d^{N} \vec{\theta_{1}} p(\theta_{2}|\theta_{1}) \pi(\theta_{1})$ Can use Markov chain property to show that $p(\theta_{2}) = \pi(\theta_{2})$ So if the Markov property holds, then $\pi$ is an equilibrium distribution Metropolis Hastings Algorithm used to generate a Markov Chain For each step, we generate some $\theta_{i}$, at which we evaluate the posterior $\pi(\theta_{i})$ We generate a candidate $\theta_{i+1}$ from some proposal distribution $q(\theta_{i+1}|\theta_{i})$ (Gaussian is a common choice) We calculate the acceptance probability $\alpha(\theta_{i+1}|\theta_{i}) = min(1, \frac{\pi(\theta_{i+1})q(\theta_{i}|\theta_{i+1})}{\pi(\theta_{i})q(\theta_{i+1}|\theta_{i})})$. With randomly accept the candidate based on this probability Burn-in and Proposal Distribution Your starting point in MCMC biases what your final distribution looks like You also need a finite number of steps before the equilibrium is reached So you need to figure out how many steps you need, and how may of these steps are independent. Typical tools for this are the autocorrelation function and the autocorrelation time You choice of proposal distribution also affects the burn-in time/convergence of the system a proposal distro which generates too large of steps might not converge a distro which generates small steps might take forever to converge

Taken from Experimental Techniques in Modern High-Energy Physics: by Hanagaki, Tanaka, Tomoto, Yamazaki Why? Two basic categories of HEP experiments Find any new particle by converting collision energy to particles Done through finding resonances in the final state Investigate the nature of a particular interaction Done through precise measurements of other physical observables In theory, you work with a small number of states to keep problems tractable. Experimentally, you can get thousands of final states which you have to parse through The number of particles goes as $ln \sqrt{s}$ where s is the square of the center of mass energy of the collision This the angular density to go up, which create “jets” (ie. collimated bunches of particles) You need to be able to distinguish particles within a jet from each other Hence, you need to cover the full solid angle ($4\pi$) in order to get all the interactions (this is called a hermetic system) Because the cross-section goes as $\frac{1}{E^2}$, you need to compensate with a higher collision rate This causes the background rate to increase because of “soft” collisions (ie. when particles don’t collide head-on). This is called pileup You also have a tougher time taking in data You should also be able to reliably detect all sorts of stable particles (neutrinos, electrons, pions etc.) with sufficient resolution Measurement We can’t observe elementary particle interacting directly, so we do indirect observation by tabulating various distributions Cross Section and Luminosity Particles are not localized. They have some sort of spread to them (ie. they fuzzy) To deal with this fuzziness, we pretend that the particle have some fixed cross sectional area in order to define the probability of hitting the target (ie. we pretend that particles are hard instead of fuzzy). This is called the cross section $\sigma$ We define the instantaneous luminosity L as $L = \sigma f_{coll} \frac{n_{1}n_{2}}{a}$ $n_{1}$ and $n_{2}$ are the number of particles in beam 1 and beam 2, a is the area where the particles get injected in, and $f_{coll}$ is the frequency of collisions The integrated luminosity is then $L_{int} = \int L dt$ $e^{+}e^{-}$ Example In order to identify an event, we first need to identify the type of final state particles (this is called particle identification or pid (Not the UNIX thing…)) For now, assume that we can identify particles with a confidence level Even if you identify particles, you need to think about what other debris there is in the detector For instance, electrons emit soft photons (ie. low energy photons) due to their small mass. You can’t avoid those photons, and you can’t really differentiate them since their energy is so low Hence, you compromise: you fold in the energy of these soft photons into the electron when you cannot distinguish them There is also the fact that most interactions are very forward. This means that most final states end up (near) parallel with the beams Hence, since most detectors are not hermetic (since the beam has to enter from somewhere!) you can miss potential final states Therefore, suitable criterion to identify $e^{+}e^{-}$ events are a pair of electron like particles that are sufficiently far enough from the injection site No other significant energy deposits observed Each electron has a large proportion of the beam energy You could also defined more relaxed conditions (called inclusive event selection) as a pair of electron like particles that are sufficiently far enough from the injection site Each electron has a large proportion of the beam energy The above criteria include other events, which can (hopefully) be removed in post Different selection criteria yield different acceptance A (ie. $A = \frac{N_{sel}}{N}$) and also different efficiencies $ie. \epsilon = \frac{N_{det\bigcap sel}}{N_{sel}}$ $N_{sel}$ is the number of events that passed the selection criteria, imposed on true four-momentum of particles $N_{det\bigcap sel}$ denotes the number of events passing the selection criteron imposed in the measured momenta $N_{sig} = L_{int} \sigma A \epsilon = N_{obs}-N_{bkgd}$ is the total number of events that you take as your signal Coordinate System for Hadron-Hadron Collisions The z-axis is along the beam line, with the positive direction pointing from beam A The x-axis points towards the center of the accelerator ring the y axis is choosen to form a right-handed coordinate system The transverse plane is the xy plane perpendicular to the beam direction The transverse momentum is $p_{T} = \sqrt{p_{x}^{2}+p_{y}^{2}} = p\sin \theta$ where $\theta$ is the polar angle in the xy plane Typically, transverse momenta is characterized by $(p_{T}, \theta)$ This is used because $p_{T}$ is conserved in hadron-hadron collisions, but longitudinal momentum is not(b/c with partons, the $p_{parton} = x*p_{beam}$, and x is not the same for bot beam, which causes the center of mass frame of the two beams to be boosted with respect to the lab frame, hence the longitudial momentum gets squished, but the transverse remains unchanged) We defined the rapidity as $y = ln \frac{E+p_{z}}{m_{T}}$ where $m_{T} = \sqrt{m^{2}+p_{T}^{2}}$ Why? $E\pm p_{z}$ is resistant to particle losses escaping the beam Hence, we define a new coordinate that utilizes these more easily measurable quantities rapidity is that differences in rapidity is preserved under a Lorentz boost In the massless limit (which is a good enough approximation when your beam energy is 14 TeV!), you can define the pseudo-rapidity as $\eta = -\ln \tan (\frac{\theta}{2})$ The phase space for a particle is defined as $d^{4}p\delta(p^{2}-m^{2}) = \frac{d^{3}p}{E} = \pi dydp_{T}^{2}$ Apparatus To make particles, you need to convert energy to mass. Hence, why high energy particles get smashed together In order to generate these high energies, we use at the LHC a sequence of increasingly little accelerators Once the particles get up to speed, they maintain their energy in the main synchrotron via RF (radio frequency) cavities which offsets the radiation losses. At the LHC, the bunch crossing occur every 25 ns bunch crossing refers to when two clusters of particles cross a small region This is not the same as the collision: this refers to the particles that happen to interact with each other when the bunch crossing occurs It’s possible that no collisions happen during a bunch crossing Detectors We call the various expected ways to measure a particles existence “channels” For instance, top quark annihilation event can yield Two leptons a lepton and a jet all hadrons Particle Signatures Electrons with high energy create electromagnetic showers upon hitting dense materials. They also undergo curved trajectories when in the magnetic spectrometer Photons also create electromagnetic showers, but, since they are neutral, don’t leave a curved path in the charged particle tracking system Muons, at the relavent energies, don’t leave EM showers. Hence, if you see long tracks, then you probably have a muon Hadrons: jets Neutrinos: No. Or less faciciously, you can’t detect them since the cross section is absurdly small. You can infer them from the missing transverse momentum ATLAS Arbitrarily choosen since high level design is roughly the same for multi-purpose detectors You have a barrel with two end caps and a beam channel There are solenoid and toroid magnets to provide a magnetic field to measure particle momentum Surrounding the interaction point is the charged particle tracker which consists of pixel and strip silicon detectors the TRT (transition radiation tracker, not the drug) is a gas which produces ions when hit by electrons. These get drifted by the field in the cathode ray tubes up to a mesh of wires You can then reverse engineer the path that a particle took Most particles penetrate the tracking detectors and then hit the electromagnetic calorimeter composed of a sandwich of lead and liquid argon (LAr) Electrons and photons produce showers in the lead (ie. the absorber) The electrons created by the showers deposit ini the LAr (ie. the detector) There is also a hadronic calorimeter outside the EM calorimeter. This is mostly just a plastic scintillator and iron sandwich (although this might change depending on the region in the calorimeter) The outermost layer is the muon spectrometer, which is a gas detector like TRT with a high timing resolution Triggers The ATLAS detector generates 1 petabyt of data every second 40 MHz bunch crossing at 2E8 channels (ie. output signals) This can be reduced by x100 with some cleverness, but it’s still too much You need to calculate on the fly if a piece of data is worth keeping or not. This process is called a “trigger” ATLAS has two levels of triggers: L1 and HLT (higher level trigggers) Statistics Basic Definitions $\int P(x) dx = 1$ $E[x] = \int x P(x) dx$ $\sigma^{2} = E[x-E[x]]^2$ Uncertainty Every physical observable has two uncertainties associated with it statistical uncertainties (ie. stochastic fluctuations in random processes) Systematic uncertainties: fundamental deviations from ideals which creates errors Distributions Binomial $P(n) = \frac{N!}{n! (N-n)!} p^{n} (1-p)^{N-n}$ p is the “hit” probability $\mu = Np$ $\sigma = Np(1-p)$ For large N and small p, can approximate by a Poisson distribution with $\mu = Np$ For large N and moderate p, can approximate by normal distribution Poisson $P(n) = \frac{\mu^{n}e^{-\mu}}{n!}$ $\mu = mu$ $\sigma^{2} = \mu$ For large $\mu$, the Poisson approaches a normal distribution Normal $P(x) = \frac{1}{\sqrt{2\pi}\sigma}exp(-\frac{(x-\mu)^{2}}{2\sigma^{2}})$ $erf(x) = \frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{t^{2}} dx$ FWHM: Full width half maximum = $2\sqrt{2 \ln 2 \sigma}$ Uniform $P(x) = \frac{1}{b-a}$ $\mu = \frac{1}{2} (a+b)$ $\sigma^{2} = \frac{1}{12}(b-a)^2$ Useful in position measurements Breit-Wigner $BW(x;M,\Gamma) = \frac{1}{\pi} \frac{\frac{\Gamma}{2}}{(M-x)^{2}+(\frac{\Gamma}{2})^{2}}$ Exponential Useful with unstable particles of lifetime of $\tau$ $P(x) = \frac{1}{\tau}e ^{-\frac{x}{\tau}}$ $\mu = \tau$ $\sigma^{2} = \tau^{2}$ Chi Square Assuming N observables $x_{i}$ which are drawn from a series of normal distribution. We defined the chi-squared value as: $\chi^{2} = \Sigma_{i=1}^{n} \frac{(x_{i-\mu_{i}})^{2}}{\sigma_{i}^{2}}$ If the hypothesis is correct, then the variance should equal $(x-\mu)^{2}$. Hence, $\frac{\chi^{2}}{n}$ is expected to be 1 Figure of Merit figure of merit refers to how sensitive the measurement can be expected to be A common one is $\frac{N_{signal}}{\sqrt{N_{obs}}} = \frac{N_{signal}}{\sqrt{N_{signal}+N_{back}}}$ The square root arises since $\sigma = \sqrt{N_{obs}}$ shows the statistical uncertainty of total number of events If the background is large enough, you can drop the signal term in the bottom Error Propagation Suppose that observable u is a function of several variables: $u = f(x_0,x_1…x_{n})$ The errors in measuring each variable can add up and affect the error of the overall function. It follows that $\sigma_{u}^{2} = \Sigma_{i=0}^{n} (\frac{\partial f}{\partial x_{i}})^{2} \sigma_{i}^{2}$ Maximum Likelihood We define the likelihood function as the probability to have a set of measurement $x_0,x_1…x_{n}$ $L(x;\theta) = \Pi_{i=1}^{n} f(x_{i};\theta)$ Where $\theta$ represents the unknown physical values One would expect that the real values of $\theta$ would maximize the likelihood function Hence, once can solve the equation $\frac{\partial L}{\partial \theta_{i}} = 0$ to find the most probable values of $\theta$ A little trick is to instead solve $\frac{\ln \partial L}{\partial \theta_{i}} = 0$, since the extrema don’t change This is where $\hat{\mu} = \frac{1}{n}\Sigma_{i=1}^{n} x_{i}$ $s^{2} = \frac{n}{n-1}\hat{\sigma}^{2} = \frac{1}{n-1}\Sigma_{i=1}^{n} (x_{i}-\mu)^{2}$ This is where the stupid n-1 comes from! Hypothesis Test A null hypothesis $H_{0}$ is the hypothesis under consideration The test hypothesis $H_{1}$ is the hypothesis that the null is conpared against In practice, $H_{1}$ is the hypothesis we would like to see, while $H_{0}$ is the hypothesis we would like to reject We define the ratio $\frac{L(H_{0})}{L(H_{1})}$ as the test statistic t. By Neyman-Pearson lemma, this is an optimal discriminator (Why? IDK. I’m sure I could understand if I read the proof, but that’s not relevant for what I’m doing) TBD Detector Calibration Detector calibration is the process of converting ADC values to physical observables Remember, ADC values are just numbers unless they can be correlated to observables Detector Alignment A tracking device is aligned if all 6 degrees of freedom can be pinned down to a reasonable tolerance After the coarse placing of a device, one method to fine tune alignment is with charged particles Suppose that you have a tracking algorithm that is agnostic to which layer is being aligned. You can compare the expected layer hit to the actual particle track, and then try and realign the layers so that the residual (ie. the difference between the two) goes to 0 In practice, this quicks gets tedious. Also, since there are millions of channels, using a layer agnostic tracking algorithm becomes nigh impossible Instead, the normal track reconstruction algorithm is used with looser quality requirements in order to minimize the bias arising from the usage of hit information The residual is also redefined as the sum of the residuals in each individual layer The alignment process occurred in a hierarchical manner (ie. align the big things together, then did down into the constituents) Momentum Resolution Suppose a charge particle travels in a magnetic field B with the radius R. You can place 3 sensors along it’s path (D1, D2, D3) to detect its presence We define the sagitta as $s = R(1-\cos\frac{\alpha}{2})$ The uncertainty in the momentum is given by $\frac{\sigma_{PT}}{p_{T}} = \frac{\sigma_{x} p_{T}}{0.3 BL^{2}}\sqrt{\frac{720}{N+4}}$ N is the number of sensors along the arc In light of the above, better resolution is achieved with more detectors in a wider space witha greater field strength Another calibration technique is to use the “world average value” of the invariant mass of well known particles like the Z boson in order to fine-tune scaling for each energy bin There are some experimental trade-offs you have to make, like what particle to pick? Can you generate a large enough dataset? etc. Energy Calibration Two main methods Calibrate each cell/channel Calibrate the energy of the particle incident to the calorimeter (ie. the energy of the shower after clustering) Cell-By-Cell Calibration The goal of cell-by-cell is to find the relationship between the energy deposit and the ADC count for each channel muons are very common for this purpose since they behave as minimum ionising particles (MIPs) that deposit a constant energy per path length Other particles either cause EM or hadronic showers, or don’t produce constant energy deposits Muons deposit energy just by ionization loss, which causes the constant energy loss Muons decaying from Z bosons are one of the cleanest samples you can get They have no other particles near them they have very high momentum (which means smaller scattering angle) $J/\Phi \rightarrow \mu^{+}\mu^{-}$ is used as a lower momentum calibration source Muons also come from cosmic rays, which are free and available everywhere in the world (except deep underground) Alternatively, one can send in test pulses (say light, or square waves) into each channel and then measure the response in order to make a correlation between input and output signals Energy Clustering Calibration of EM Showers It’s hard to extrapolate muon data to higher energies There could be dead material in the detector To fix these issues, can utilize other auxiliary detectors to get a measurement of momentum of an electron, and the use a $\frac{E}{p}$ distribution to map onto energy You can measure decay of particles whose masses are precisely known It’s kind of hard to find a reaction that can generate enough statistics though… Another difficulty is that it’s hard to produced good photon calibration reactions To side step this, you typically calibrate on electrons, and then say that photons showers are roughly the same as electrons The primary difference between the two is the starting shower depth (photons are deeper) If you want greater precision, you can use MC simulations to dial in the truth depth of the photon shower After setting up an experiment, rarer processes can be used as validation (say $Z \rightarrow \mu^{+}\mu^{-}\gamma$) Energy Clustering Calibration of Hadronic Showers Hadronic energy measurements are harder, since a lot of the energy gets “lost” due to nuclear interaction length Hadronic showers also have a larger variance due to ocassionally producing heavier particles like pions and stuff (EM is mostly just electrons and light) This discussion will be deferred to jet reconstruction, since it better fits there… PID (Particle IDentification) Tracking refers to the reconstruction of a trajectory of a charged particle, or a track. It consists of 3 parts measurement of the space hit points of the particles in the detector the pattern recognition to the hit points to make a candidate track (ie. track finding) the fitting for the candidate track to get a smooth track (best guess fo the true particle trajectory) In the real world, there are detector inefficiencies (which causes phantom hits, particles can penetrate multiple segments at once, multiple particles could deposit in the same segment) To circumvent this, clustering is performs on the raw hits in order to group hits with similar characteristics together Particle position can then be inferred from these clusters Hit Clustering Ideally, one would set a (low) threshold for each channel to trigger a single hit We don’t live in an ideal world, as much as physicists love their spherical cows There can be dead channels, which don’t register anything There can also be noisy channels, which register garbage or are constantly on For dead channels, we can pretend that it registered a hit if the surrounding channels are registered a hit For noisy channels, we could just mask them out (ie. ignore them) One could also demand that channels adjacent to the noisy one registers a hit Hit Point Determination Given a series of hits, can take a weighted average w.r.t. the measured charge to find the most likely position Older detectors utilized wires to glean this information. Needed to offset wires and a stereo angle in order to extend to higher dimensions Stereo angle of 90 degrees very uncommon since geometric constraints typically place readout electronics at one end of detector This reduces positional resolution along the wire direction Tracking Finding In the good ol’ days, we used our eyes to track particle through bubble chambers and we liked it! Modern detectors necessitate computer usage to keep up with data acquisition rate In either case, once a reconstructed track is produced, a chi-squared analysis is used to determine the goodness of the reconstruction Local Method You greedily match tracks until you can’t any more, then repeat You could start inside out, or outside in inside out has a higher number of potential tracks, so you use a lot more CPU time and exhasutively search the parameter space outside in is faster due to the smaller number of potential tracks It also allows reconstruction of tracks from a secondary vertex more easily ATLAS does both of these things in parallel Global Methods Histogramming: You can make a histogram of your hits with respect to some coordinate system such that you are sampling along a constant cooridinate. You can then run a peak finding algorithm to determinte the tracks Hough Transform: the classical version is mapping a writing the equation of a line in polar coordinates. Multiple points a long a line will produce marginally different sine waves, but those waves will overlap at some $r_{0}, \theta_{0}$ value which defines the line of best fit This idea can be extended to other, more relevant paths, such as helices, which are pertinent to identifying charged particle motion Vertex Finding The collision point if often called the primary vertex Decays can produce secondary vertices Intersecting tracks yield vertices Doing a best-fit analysis allows one to deduce the most likely vertices Jet Identification If it looks collimated and is away from other stuff, it may be a jet related to an underlying parton Jet algorithms need to have the following properties The algorithm has to be independent of the type/count of particles The Algorithm should be able to reconstruct high momentum partons The algorithm should be insensitive to soft emmissions Cone Based Moves around a window of jet area defined as a circle in $\eta-\theta$ space with dimensions $\delta r, \delta \eta, \delta \theta$ We iteratively find an energetic cluster such that $\delta r = \sqrt{\delta \eta ^{2} \delta \phi^{2}} < R$ for some cone radius R We also demand that the transverse momentum is maximized This allows one to identify hard jets Unidentified clusters get thrown away Cluster Based All particles, no matter how ghostly, gets assign to a jet If a particle is not assigned a jet, it gets added to the closest one according to some distance metric

Notes adapted from MIT Digital Electronics Lab Also from here Boolean Algebra Theorems Truth Tables Specific Functions 1-Bit Functions 2-bit Functions XOR Overview Verilog Basics Variables Parameters Operators and Loops Enums Syntax Oddities Combinational Logic Continuous Assignment Procedural Assignment Glitches! Gotchas Nesting Modules Combinational Versus Sequential Logic Latches Characteristic Equations Registers Verilog for Sequential Blocks Sequential Always Blocking and Non-Blocking Assignment External Signals (Metastability) FSMs FSM State Encoding Converting to Verilog Clocks PLLS The Treachery of Clocks Skew Jitter Proper usage Timing Timing Parameters Slack Video Black and White TV Color TV VGA HDMI Memory History of Memory Key parameters of memory FPGA SRAM Fabric Case Studies Off FPGA Memory EPROM Family DRAM Buffering DRAM FSM Modularization Pipelining Math Complexity Division CORDIC AXI AXI Signals Signed Number DSP48 CRCs Polynomial Arithmetic Boolean Algebra Theorems Let all variables be boolean in nature (ie. 0 or 1). $+$ Denotes an OR gate, and $\cdot$ denotes and AND gate ...

Notes taken while reading “The Art of Electronics”. I realized that my electronics knowledge is a bit spotty and more theoretically inclined, hence why I’m doing this. The Basics, Because You Have to Start Somewhere Resistors Resistor Rule’s of Thumb Voltage Dividers Thevenin Equivalence Misc Use Cases Capacitors Common Applications RC Time Constants Frequency Filters Differentiators Integrators Inductors Applications Voltage Converters Transformers Diodes Applications Rectifiers Dual Rail Power Supply Voltage Multipliers Biasing Logic Gates Clamps Limiters Logarithm Inductive Kickback Transistors (BJTs) BJTs (Bipolar Junction Transistors) Transistor Switches and Saturation Switching From High Pulse Generator Schmitt Trigger Impedences of Sources and Loads Emitter Follower Emitter Follower Biasing Case Study Canceling Offset Transistor Current Sources Common-emitter Amplifier Unity Gain Phase Splitter Phase Shifter Transconductance Ebers-Moll equation Rules of thumb Emitter Follower Redux Bias Stability Current Mirrors The Basics, Because You Have to Start Somewhere There are two quantities that we care about with circuits: Voltage and Current ...

Logistics Principle of “Least” Action Symmetries Gauge Invariance Example Reduce to Quadratures Hamiltonian Mechanics Phase Portraits Poisson Brackets Properties Jacobi Proof Sketch Poisson’s theorem Canonical Transformations Why We Care Liouville’s Theorem Canonical Invariants Entropy Tangent Infinitesimal CTs Hamiltonian Noether’s Theorem Hamilton Jacobi Equation Action Variables Multiperiodic motion Planetary motion Liouville’s Integrability Theorem Cannonical Peturbation Theory Adiabatic Motion Quantum Mach Zender Hilbert Space Linear Algebra Review Time Evolution Heisenberg Formulation Compatible Observables Symmetries Uncertainty Relationship Phase Space Interpretation Entanglement And Mixed States Mixed States Von Neumann Entropy Partial Trace (“Tracing Out”) Infinite Spaces Translation Operator Scattering Probability Current Delta Function Path Integral Free Particle Saddle Point Approximation Path Integral Proof WKB Logistics Matthew Klebon [email protected] Office Hours: Grading: TBD TA: [email protected]. Room 943 Principle of “Least” Action N particles in 3D There are 6N coordinates We want to find $\vec{q}(t)$ of the system where q represents the coordinates in position for each particle $q = {x_{0},x_{1},… x_{n}}$ As long as the equations of motions are 2nd order, there exists a unique solution for the system Let’s examine a simple system: the double pendulum Naively, one would expect 6 equations of motions for the 6 coordinates. However, there are 4 constrains of the system $z_{1} = z_{2} = 0$ $x_{1}^{2}+y_{1}^{2} = l_{1}^{2}$ $x_{2}^{2}+y_{2}^{2} = l_{2}^{2}$ With these constraints, you can reduce the system to a function of two coordinates: $\theta_{1}$ and $\theta_{2}$ holonomic: when the constraints of a system are solely of a position $f_{k}(q_{1}…q_{n}) = 0$ Not all constraints are holonomic: Rolling without slipping, inequalities etc. To incorporate these constraints, you can either use Lagrange multipliers, or use the constraints to eliminate degrees of freedom Assuming no dissipative forces, you can describe the system by a Lagrangian $\mathcal{L(q, \dot{q}, t)}$ This is typically $\mathcal{L} = T-V$ In polar coordinates, we have that $\dot{r}^{2}+(r\dot{\theta})^{2}$ In spherical coordinates, we have that $\dot{r}^{2}+(r\dot{\theta}\sin \phi)^{2} + (r\dot{\phi})^{2}$ From this Lagrangian, you can define something called the action of the path: $S(q_{i}(t)) = \int_{t_{1}}^{t_{2}} \mathcal{L}(q(t), \dot{q(t)}, t) dt$ This is a functional (ie. it’s arguments are other functions) Units of Plank’s constant and angular momentum You want to find the stationary points of the action function as a function of time. Pretend that the endpoints of the path are fixed at two points in time: $q(t_{1}) == q(t_{2})$ This implies that $\delta q_{i}(t_{1}) = \delta q_{i}(t_{2}) = 0$ Taking the total derivative of the action, use integration by parts, utilize the boundary conditions and rearrange to get $\delta S = \int_{t_{1}}^{t_{2}} dt (\frac{\partial L}{\partial q_{1}}-\frac{\partial}{\partial t}(\frac{\partial L}{\partial \dot{q_{i}}})\delta q_{i}) = 0$ This must hold for all possible $q_{i}$, so if we imagine placing delta functions everywhere, this implies that the bracketed term must be 0 these are the Euler-Lagrange equations: $\frac{\partial L}{\partial q_{1}}-\frac{\partial}{\partial t}(\frac{\partial L}{\partial \dot{q_{i}}}) = 0$ Sometimes, it’s easier to plug in $q+\delta q$ directly into the action and reduce up to first order terms You can add a total derivative $\frac{d}{dt}f(q,t)$ (note lack of dependence on $\dot{q}$) to a Lagrangian and leave the equations of motion unchanged Simply do variational principle with additional term The converse is not necessarily true The principle of least action is invariant under coordinate changes Symmetries When there is a symmetry, there is a conserved quantity (Noether’s Theorem) For instance, if the Lagrangian is time independent (ie. $L(q,\dot{q},t) = L(q,\dot{q})$), then via the Euler Lagrange equations, energy conservation follows Taking the total derivative of the Lagrangian w.r.t. time, applying the time-independent constraint, and then rearranging yields that that the following is conserved. This is the energy (usually)! $\frac{d}{dt}(\frac{\partial L}{\partial \dot{q}}\dot{q}-L) = 0$ A trivial symmetry is when $\frac{\partial L}{\partial q} = 0$. This is called a cyclic coordinate Another example is $L(q_{i}+\delta q_{i}) = L(q_{i})$. If q is a position, then the associated momentum is also conserved Yet another is conservation of angular momentum: $L = \frac{1}{2} m \dot{r}^{2}- V(|\vec{r}|)$ $\frac{\partial \vec{r}}{\partial \theta_{i}} = \hat{n_{i} \times \vec{r}} = \epsilon_{jkl} n_{k} r_{l}$ Can reduce this to $P_{\theta_{i}} = \epsilon_{ilj} r_{l} (m\dot{r}_{j})$, which is just $\vec{r}\times \vec{p}$ This can be done by utilizing the fact that $\hat{n_{ik}} = \delta_{ik}$ that the Levi-Civita symbol is invariant under cyclic permuations of it’s indicies the definition that $p_{\theta} = \frac{\partial L}{\partial \dot{r_{j}}}\frac{\partial \dot{r_{j}}}{\partial \dot{\theta_{i}}}$ More formally, a transformation $q_{i} => f(\epsilon, \bar{q}); f_{i}(0) = q_{i}$ (where $\epsilon$ is the implicit variable and $\bar{q}$ represents all of the coordinates) is a symmetry of the system if the Lagrangian is unchanged (ie. $\delta L = \frac{d}{dt}(\frac{\partial L}{\partial q_{i}} \delta q_{i}) = 0$) If the symmetry holds, then the conserved quantity obeys the following equation $\frac{\partial L}{\partial q_{i}}\delta q_{i} = \epsilon \frac{\partial}{\partial q_{i}}\frac{\partial f_{i}}{\partial \epsilon}|_{\epsilon = 0}$ Gauge Invariance We know that $L’ = L + \frac{d\Phi}{dt}$ holds in general for some scalar field (gauge invariance) This modifies Noether’s therom: $\frac{d}{dt}(\frac{\partial L}{\partial q_{i}}\dot{q_{i}}-\Phi) = 0$ Hence $\frac{\partial L}{\partial q_{i}}\dot{q_{i}}-\Phi$ is conserved Example Let’s couple a charge particle with some fields: $L = \frac{1}{2} m\vec{\dot{x}^{2}}-qV+\frac{q}{c}\vec{A}\cdot \vec{x}$ There is a gauge invariance from the transformation $\vec{A} => \vec{A}+c\nabla \Lambda$ and $V => V - \frac{\partial}{\partial t} \Lambda$ Plugging in this transformation yields that $L’ = L + \frac{\partial}{\partial t}(q\Lambda)+\frac{\partial}{\partial x_{i}}(q\Lambda) \vec{\dot{x_{i}}} = L + \frac{d}{dt} \Phi$ Reduce to Quadratures Instead of solving a PDE to find equations of motion, you can solve a different PDE in terms of conserved quantities For instance: $L = \frac{1}{2}m\dot{x}^{2} -V(x)$ has $h = \frac{1}{2}m\dot{x}^{2} +V(x)$ as a conserved quantity You could solve $m\ddot{x} = 9\frac{\partial V}{\partial x}$ You could also solve $(\frac{dx}{dt})^{2} = \frac{2}{m} (h-V(x))$ Hamiltonian Mechanics The Legendre transform of the Lagrangian is the Hamiltonian Graphically, you can take the slope of the initial function, and project that onto the y-axis. In symbols $v \frac{\partial L}{\partial v}-L = H(p)$ where v is the x axis coordinate and p is the y axis coordinate This assumes that $\frac{dL}{dv} = p(v)$ We assume that p is invertible (ie. p(v) implies v(p)) $L’’(v) \neq 0$ This one allows H to be a well defined function With the above in mind, we define the Hamiltonian as $H(q_{i},p_{i},t) = \Sigma_{i=1}^{3} p_{i}\dot{q_{i}}-L(q_{i},\dot{q_{i}},t)$ $p_{i} = \frac{\partial L}{\partial \dot{q_{i}}}$ Taking the total derivative of the Hamiltonian, applying the equation $p_{i} =\frac{\partial L}{\partial \dot{q_{i}}}$ and applying the EL equations yields $H = \dot{q}dp_{i}-\dot{p_{i}}dq_{i}-\frac{\partial L}{\partial t} dt$ $\frac{\partial H}{\partial q_{i}} = -\dot{p_{i}}$ $\frac{\partial H}{\partial p_{i}} = \dot{q_{i}}$ $\frac{\partial H}{\partial t} = \frac{\partial L}{\partial t}$ You solve double the number of equations, but they are 1st order now Phase Portraits Can plot momentum and position of a particle. Closed loops are called librations Unbounded paths are called rotations The boundary between the two is called the separatrix Poisson Brackets Suppose that you have some observable $f(q_{i}, p_{i},t)$ Utilizing the total time derivative and Hamilton’s equations, you show $\frac{df}{dt} = [f,H]+ \frac{\partial f}{\partial t}$ $[A,B] = \frac{\partial A}{\partial q_{i}}\frac{\partial B}{\partial p_{i}}- \frac{\partial A}{\partial p_{i}}\frac{\partial B}{\partial q_{i}}$ This is the Poisson bracket Properties $[\alpha f_{1} + \beta f_{2}, g] = [\alpha f_{1}, g] + [ \beta f_{2}, g]$ $[f,g] = -[g,f]$ $[f,[g,h]]+[h,[f,g]]+[g,[h,f]] = 0$ (Jacobi identity) $[f,gh]= [f,g]h + g[f,h]$ (Leibniz’s rule) Any object which satisfies the above properties is called a Lie Algebra. ...

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