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 $c_{n}(0)=0$ 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 $W = E_{1}+\hbar\omega -E_{n}$
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 $\Gamma_{i\rightarrow f} = \int P_{i\rightarrow f}(t_{0}) \rho(E_{f}) dE_{f}$
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 $f_{nm}$ 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
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 $\Gamma(1\rightarrow n) = \frac{2\pi e^{2}}{\hbar^{2}} | \Sigma_{N} <x_{N}>_{nm} |^{2} P(\omega)$
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
$\gamma_{k}(t)$ 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
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-relativistical regime (no spawning additional particles)
Assume potentials only depend on relative distance between particles (so 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 ${d\sigma} = |f(\theta,\phi)|^{2} {d\Omega}$
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 $\tan \delta_{l} = \frac{-B_{l}}{A_{l}}$
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?
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:
Suppose our Hamiltonian now looks like $H_{0}+V(\vec{x})$
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
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 $\phi_{k}(\vec{x}) = (2\pi \hbar)^{-\frac{3}{2}} e^{i\vec{k}\cdot \vec{x}}+ \int d^{3} y G_{k}(x-y) V(y) \phi_{k}(y)$
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
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 $f_{k}(\hat{x}) = -\frac{\mu}{2\pi \hbar^{2}} \int d^{3}y V(y) exp(i(\vec{k}-k\hat{x})\cdot \vec{y})$
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 $q = |\vec{k}-k\hat{x}|$
Suppose that we have some Hamiltonian $H = H_{0}+V$ where V becomes negligible as the particles move away from each other. At $t = \pm \infty$, we can treat the Hamiltonian as free:
$H_{0} \Phi_{\alpha} = E_{\alpha} \Phi_{\alpha}$ where $\alpha$