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. $H_{prod} = \Sigma_{ia} \Phi_{ia} e_{i} \otimes e_{a}$)
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 $\phi_{i}$
$U(a\phi+b\psi)= aU\phi + bU\psi$ (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
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
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
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
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
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
We want to show that $(\phi_{j’’}^{m’’}O_{j}^{m}\psi_{j’}^{m’}) = C(j’’m’’;j’m’) (\phi || P|| \psi)$
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
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
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
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 $n^{2}$
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 $\delta_{1} \phi_{a} = \Sigma_{b\neq a} \phi_{b} \frac{(\phi_{b}\delta H \phi_{a})}{E_{a}-E_{b}}$
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 $\delta E_{njlm} = \frac{e\hbar g_{jl} B}{2 m_{e} c} m$
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 $\delta_{2} E_{a} = \Sigma_{b\neq a} \frac{|(\phi_{b} \delta H \phi_{a})|^{2}}{E_{a}-E_{b}}$
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
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
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 $c_{n}(t)=0$ 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 $W = E_{1}+\hbar\omega -E_{n}$
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
We can imagine the total probability as the weighted sum of each individual probability (ie. $P(t)=\int P_{if}(t)\rho(E) dE$)
How do you find the equation of state?
Imagine a large box with sides $L_{i}$ that holds the system
Define $p_{i} = \frac{2\pi \hbar n_{i}}{L_{i}}$
Each volume in n space has an associated volume in p space of $d^{3}p = \frac{V}{(2\pi \hbar)^{3}}$
Hence, we can get the total rate by summing $\Gamma(1\rightarrow n)$ over all possible states. This is done by treating $\Gamma$ as some differential and then integrating over the momenta of the free particles in the final state, whose wave functions have the form $\phi = \frac{exp(i p\cdot x)}{(2\pi \hbar)^{\frac{3}{2}}}$
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
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 $f_{nm}$, 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