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 cannonical transformations are an important part of classical Hamiltonian mechanics. What is the analog in quantum?
    • We want the probability to be conserved under a cannonical 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 $(\phi_{j’’}^{m’’}O_{j}^{m}\psi_{j’}^{m’}) = C(j’’m’’;j’m’) (\phi || P|| \psi)$
    • 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

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

  • End result is that
    • $(\Sigma_{N} (\frac{-\hbar^{2}}{2M_{N}})\nabla_{n}^{2} + \epsilon_{a}(X)) f_{a}(X) = E f_{a}(X)$
    • $\epsilon_{a}(X)$ is the energy of the electronic state with fixed nuclear coordinates X acting as a potential for the nuclei

Time Dependent Perturbation Theory

  • 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
    • $c_{n}(t) \approx c_n{0}-\frac{i}{\hbar}\Sigma_{m} c_{m}(0) \int_{0}^{t} dt’H_{nm}’(t’) exp(i\frac{(E_{n}-E_{m})t’}{\hbar})$
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 $c_{n}(t)=0$ 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 $W = E_{1}+\hbar\omega -E_{n}$
      • 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
    • 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 $\phi = \frac{exp(i p\cdot x)}{(2\pi \hbar)^{\frac{3}{2}}}$
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 $f_{nm}$, 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 follows that of previous section
  • You can write the average correlation of the perturbation as a fourier transform, with $F_{nm}(\omega) = e^2|\Sigma_{N} <x_{N}>_{nm}|^{2}P(\omega)$