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QM Notes

Table of Contents

  1. Postulates and Principles
  2. Operators and Pictures
  3. Quantum Harmonic Oscillator
  4. Angular Momentum and Spin
  5. Identical Particles
  6. 3D Mechanics
  7. Approximation Methods
  8. Scattering
  9. Particles in EM Fields

Postulates and Principles

the underlying assumptions are

  1. Physically realizable states live in projective hilbert space
  2. Observables $\iff$ hermitian operators on this HS
  3. Possible observations given by eigvals of these ops
  4. $P(a_n) = |\braket{a_n}{\psi}|^2$, after measurement $\ket{\psi}\to\hat{P}_{a_n}\ket{\psi} = \ket{a_n}$ where the last equality holds if the eigval is nondegenerate. $\ev{f(A)} = |\braket{a_i}{\psi}|^2f(a_i)$
  5. $i\hbar\partial_t\ket{\psi}=\hat{H}\ket{\psi}$, where eigenfunctions $\ket{\psi_n}$ satisfy $\psi_{n}(x,\,t) = \psi_n(x)\exp(-iE_nt/\hbar)$ and an arbitrary vector can be written as a linear combination of these steady states.
  6. $n$ identical particles are always in states which are eigenfunctions of all exchange operators – eigval of $\pm1$ depending on statistics.

kernel of fourier transform used to move between \(x\) and \(p\) representations:

\begin{gather} \ip{x}{p} = (2\pi\hbar)^{-1/2}\exp(-ixp/\hbar)\\ \psi(p) = \ip{p}{\psi} = \int\dd{x}\ip{p}{x}\ip{x}{\psi} = \int\frac{\dd{x}}{\sqrt{2\pi\hbar}}e^{+ipx/\hbar}\psi(x) \end{gather}

important distinction:

  • expected location: $\ev*{\hat{r}}$
  • most probable location: maximum of $|\psi|^2$
  • most probable radius: $r$ that maximizes $\int\dd{\Omega}r^2|\psi|^2$

probability current:

\begin{equation} j = \frac{\hbar}{2mi}\qty[\psi^\ast(\partial_x\psi) - (\partial_x\psi)^\ast\psi] = \frac{\hbar}{m}\mathrm{Im}(\psi^\ast\partial_x\psi) \end{equation}

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Operators and Pictures

\begin{gather*} \ev*{\hat{A}} = \mel{\psi}{\hat{A}}{\psi} \qquad \dv{t}\ev*{\hat{A}} = \ev{\pdv{\hat{A}}{t}} + \frac{i}{\hbar}\ev*{[\hat{H},\,\hat{A}]}\\ \Delta A = \sqrt{\ev*{(\hat{A}-\ev*{\hat{A}})^2}} = \sqrt{\ev*{\hat{A}^2} -\ev*{\hat{A}}^2} \qquad \Delta A\Delta B \geq \frac{1}{2}|\ev{[A,\,B]}| \end{gather*}

Eherenfest derivable from just performing the derivative (being sure to hit the \(\ket{\psi}\)s) and the uncertainty relation derivable from the schwartz inequality on \(\hat{O}-\ev*{\hat{O}}\)

Heisenberg Dirac Schrodinger
States constant evolve with $V$ evolve with $H$
Operators evolve with $H$ evolve with $H_0$ constant
where operators evolve as $-i\hbar\partial_t\hat{A} = [\bullet,\,\hat{A}]$
and states evolve as $i\hbar\partial_t\ket{\psi} = \bullet\ket{\psi}$

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Quantum Harmonic Oscillator

\begin{equation} \hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2\hat{x}^2 = \hbar\omega\qty(\frac{\hat{P}^2 + \hat{X}^2}{2}) \end{equation}

after adimensionalizing the operators:

\begin{equation} \hat{P} = \frac{\hat{p}}{\sqrt{m\hbar\omega}} \qq{and} \hat{X} = \hat{x}\sqrt{\frac{m\omega}{\hbar}} \qq{for which}[\hat{X},\,\hat{P}] = i \end{equation}

check: \(\hbar\omega\) is an energy, \([mE] = [m]^2[c]^2 = [p]^2\), similarly for the \(\hat{x}\) prefactor

Ladder operators

Nice in terms of adim ops:

\begin{gather} \hat{a} = \frac{1}{\sqrt{2}}(\hat{X}+i\hat{P}) \qquad \hat{a}^\dagger = \frac{1}{\sqrt{2}}(\hat{X}-i\hat{P})\\ [\hat{H},\,\hat{a}] = -\hbar\omega\hat{a}\qquad [\hat{H},\,\hat{a}^\dagger] = +\hbar\omega\hat{a}^\dagger \qquad [\hat{a},\,\hat{a}^\dagger] = 1\\ H = \hbar\omega(\hat{a}^{\dagger}\hat{a} + \frac{1}{2}) \qq{and} \hat{a}^\dagger\ket{n} = \sqrt{n+1}\ket{n+1},\,\hat{a}\ket{n} = \sqrt{n}\ket{n-1} \end{gather}

working in the dimensionless spatial coordinates \(X\) and \(-i\partial_X\), it’s not too bad to work out wavefunctions starting from

\begin{equation} \psi_0(X)=\ip{X}{0} = N_0 e^{-X^2/2}\qquad \psi_n(X)=H_n(X)\psi_0(X) \end{equation}

Coherent states

indexed by \(\alpha\in\mathbb{C}\) – this is their eigenvalue under \(\hat{a}\); set initial position and velocity with the two dof:

\begin{equation} \ev*{\hat{X}}=\sqrt{2}\mathrm{Re}[\alpha] \qquad \ev*{\hat{P}} = \sqrt{2}\mathrm{Im}[\alpha] \end{equation}

as time goes on, they stay coherent but \(\alpha\) evolves (see eherenfest)

\begin{align*} \ket{\alpha} &= e^{-|\alpha|^2/2}e^{\alpha\hat{a}^{\dagger}}\ket{0}\\ &=\exp(\alpha\hat{a}^\dagger - \alpha^\ast \hat{a})\ket{0} \\ &=\exp(i\sqrt{2}(\hat{X}\,\mathrm{Im}[\alpha] - \hat{P}\,\mathrm{Re}[\alpha]))\ket{0} \end{align*}

can use BCH to split the exp into two; recall

\begin{equation} e^{\hat{A}}e^{\hat{B}} = e^{\hat{A}+\hat{B}+\frac{1}{2}[\hat{A},\,\hat{B}]+\dots} \end{equation}

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Angular Momentum and Spin

Take some operators \(\hat{J}_i\) that satisfy \(SU(2)\) commutation \([\hat{J}_i,\,\hat{J}_k]=i\hbar\epsilon_{ijk}\hat{J}_k\)

we pick a basis \(\ket{jm}\) that simultaneously diagonalizes \(\hat{J}^2\) and \(\hat{J}_z\):

\begin{equation} \hat{J}^2 \ket{jm} = \hbar^2j(j+1)\ket{jm}\qquad \hat{J}_z\ket{jm} = \hbar m \ket{jm} \end{equation}

these imply \(j\) is (nonnegative) integer or half-integer and \(m\in[-j,j]\)

ladder between the \(m\)s with

\begin{equation} \hat{J}_\pm = \hat{J}_x\pm \hat{J}_y \qquad [\hat{J}_z,\,\hat{J}_\pm] = \pm\hbar \hat{J}_\pm \qquad [\hat{J}^2,\,\hat{J}_\pm] = 0 \end{equation}

for normalized \(\ket{jm}\) states,

\begin{equation} \hat{J}_\pm \ket{jm} = \hbar\sqrt{j(j+1) - m(m\pm 1)}\ket{j,\,m\pm1} \end{equation}

remember signs: we kill the extreme \(m\) states

spherical harmonics, eigs of the orbital angmom \(\hat{L}^2\), only get integer \(j\)s (\(\ell\)s)

for spin (internal dof) half integer \(j\) is okay.

for a fixed \(j\) the \(m\)s give us \(2j+1\) states to work with

Addition: Clebsh-Gordon coefficients

say we have two particles with \(\ket{j_1m_1,\,j_2m_2}\) in which \(J_1^2,\,J_{1z},\,J_2^2,\,J_{2z}\) are diagonalized. we can also consider \(\ket{jmj_1j_2}\) in which \(J^2,\,J_z,\,J_1^2,\,J_2^2\) diagonalized. the obstruction comes about because \(J^2\) and \(J_{iz}\) arent simultaneously diagonalizable.

usually we fix $j_1$ and $j_2$ (e.g. fixing total spins of particles, which are determined by species) and then try to find $\ket{jm}$ in terms of $\ket{m_1m_2}$. in this case, $j$ can range from $|j_1-j_2|$ to $j_1+j_2$. We start with the max $j$ and max $m$ and then ladder down, since the ladders for the total is the sum of the individual ladders. we now know $\ket{j,\,m=j-1}$, so we find a state orthogonal to it to use as $\ket{j-1,\,m=j-1}$

Example: \(j_1 = \ell = 1\) and \(j_2 = s= 1/2\).

\begin{align} \textstyle\ket{j\frac{3}{2},\,m\frac{3}{2}} &\textstyle= \ket{ {m_\ell}1,\,{m_s}\frac{1}{2}}\\ \textstyle\ket{j\frac{3}{2},\,m\frac{1}{2}} &\textstyle= N(L_-+S_-)\ket{ {m_\ell}1,\,{m_s}\frac{1}{2}} = \sqrt{\frac{2}{3}}\ket{m_\ell 0,\,m_s\frac{1}{2}} + \frac{1}{\sqrt{3}}\ket{m_\ell 1,\,m_s-\frac{1}{2}}\\ \textstyle\ket{j\frac{1}{2},\,m\frac{1}{2}} &\textstyle= \frac{1}{\sqrt{3}}\ket{m_\ell 0,\,m_s\frac{1}{2}} - \sqrt{\frac{2}{3}}\ket{m_\ell 1,\,m_s-\frac{1}{2}} \end{align}

these are written in CG tables, which give \(j,\,m\) on each column and \(m_1,\,m_2\) on each row. the coefficients are given squared, so you need to take the square root before using them.

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Identical Particles

this is pretty straightforward: if the particles are indistinguishable, we have statistics effects (fermion or boson). keep in mind that if the particles have a spin and a spatial wavefunction then one part of it might already take care of the (anti)symmetry that you need!

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3D Mechanics

\begin{gather} \hat{H} = -\frac{\hbar^2}{2m}\frac{1}{r}\pdv{r}(r^2\pdv{r}) + \frac{\hat{L}^2}{2mr^2} + V\\ \hat{L} = -\hbar^2\qty[\frac{1}{\sin\theta}\pdv{\theta}(\sin\theta\pdv{\theta}) + \frac{1}{\sin^2\!\theta}\pdv[2]{\phi}] \end{gather}

gives us solutions of the form

\begin{equation} \psi_{n\ell m}= R_{n\ell}(r)Y_{\ell m}(\theta,\,\phi) \end{equation}

Angular part: the spherical harmonics are eigens of both \(\hat{L}^2\) and \(\hat{L}_z\) with

\begin{equation} \hat{L}^2 Y_{\ell m} = \hbar^2\ell(\ell + 1)Y_{\ell m} \qq{and} \hat{L}_z Y_{\ell m} = \hbar m Y_{\ell m} \end{equation}

up to normalization, they look like

\begin{equation} Y_{\ell m} = e^{im\phi}P_\ell^m(\cos\theta) \end{equation}

with the \(\ell = 0\) and \(\ell = 1\) ones given explicitly as:

\begin{equation} Y_{00} = \frac{1}{\sqrt{4\pi}} \qquad Y_{11} = -\sqrt{\frac{3}{8\pi}}\sin\theta e^{i\phi} \qquad Y_{10} = \sqrt{\frac{3}{4\pi}}\cos\theta \qquad Y_{1-1} = \sqrt{\frac{3}{8\pi}}\sin\theta e^{-i\phi} \end{equation}

it might be handy to know

\begin{equation} \sin\theta e^{i\phi} = \frac{x+iy}{r} \qquad \cos\theta = \frac{z}{r} \qquad \sin\theta e^{-i\phi} = \frac{x-iy}{r} \end{equation}

which helps explain the extra \(\sqrt{2}\) on the \(1\pm1\)s.

Radial part: with the \(\ell\) fixed on the sphharm side, our equation reads

\begin{equation} -\frac{\hbar^2}{2m}\frac{1}{r}\partial_r(r^2\partial_rR_{n\ell}) + \qty[V(r) + \frac{\hbar^2\ell(\ell+1)}{2mr^2}]R_{n\ell} = E_{n\ell}R_{n\ell} \end{equation}

where the trick is to swap to \(u = rR\) where the equation reads

\begin{equation} \qty[-\frac{\hbar^2}{2m} \partial_r^2 + V + \frac{\hbar^2\ell(\ell+1)}{2mr^2}]u_{n\ell} = E_{n\ell} u_{n\ell} \end{equation}

which is like TISE with hard well at \(x=0\) (\(u\) must be zero when \(r= 0\))


Take \(V\) to be Coulomb:

\begin{equation} V(r) = -\frac{Ze^2}{4\pi\epsilon_0 r} \end{equation}

then the bound states are ordered by an \(n=1,\,2,\dots\) and \(\ell = 0, \dots, n-1\).

Energy independent of \(\ell\) (and goes like \(\alpha^2\)):

\begin{equation} E_{n\ell} = E_n = \frac{-13.6 \text{eV}}{n^2} = -\frac{1}{n^2} \frac{\mu Z^2e^4}{2(4\pi\epsilon_0)^2\hbar^2} \end{equation}

A few wavefuntions:

\begin{gather} \psi_{100} = \frac{e^{-r/a_0}}{\sqrt{\pi a_0^3}} \qquad \psi_{200} = \frac{e^{-r/2a_0}}{4\sqrt{2\pi a_0^3}}\qty[2-\frac{r}{a_0}] \\ \psi_{210} = \frac{e^{-r/2a_0}}{4\sqrt{2\pi a_0^3}}\frac{r}{a_0}\cos\theta\qquad \psi_{21\pm1} =\frac{e^{-r/2a_0}}{8\sqrt{\pi a_0^3}}\frac{r}{a_0}\sin\theta e^{\pm i\phi} \end{gather}

Relativistic corrections a fine structure correction (energy goes like \(\alpha^4\)): take next term in relativistic kinetic energy expansion and perturb hydrogen’s degenerate states (\(L^2,\,L_z\) commute with \(p^2\) and thus with the \(p^4\) terms we get), we have a nice basis to perturb with. end up needing some \(\ev{r^{-n}}\)s; the \(n=1\) you can get with virial theorem, and the \(n=2\) with feynman-hellmann theorem. end up getting a perturbation that depends on \(n\) and \(\ell\).

Sporbit coupling fine structure: if \(\ell \neq 0\) then the orbital angular momentum couples to the spin of the electron and affects the energy depending on aligned/antialigned. kramer’s relation gives further \(n\)s from above.

Lamb shift (energy goes like \(\alpha^5\)) QED vacuum perturbs the atom

Spin-spin coupling hyperfine structure (energy goes like \(\alpha^8m_e/m_p\): coupling between the spin of the electron and the proton

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Approximation Methods


Use: we have a hammy that looks like \(H = H_0 + \lambda V\) for some small parameter \(\lambda\) where we know how to solve \(H_0\)

No degeneracy:

\begin{align*} E_n^{(1)} &= \mel*{\psi_n^{(0)}}{V}{\psi_n^{(0)}} \equiv V_{nn}\\ \psi_n^{(1)} &= -\sum_{m\neq n} \frac{\psi_m^{(0)}V_{mn}}{E_m^{(0)}-E_n^{(0)}}\\ E_n^{(2)} &= -\sum_{m\neq n} \frac{|V_{mn}|^2}{E_m^{(0)}-E_n^{(0)}} \end{align*}

Degeneracy: pick a basis for the degenerate subspace of \(H_0\) such that \(V\) is diagonalized (e.g. if we know eigenstates of \(V\) already that \(H_0\) doesnt care about); these diagonal elements are the first-order corrections to the energy. in this basis, the second-order corrections are given by the same sum, just with a restriction on the sum so that it doesnt blow up.


Idea: invert the schrodinger equation and write

\begin{align} \psi(t) &= \mathcal{T}\exp(-\frac{i}{\hbar}\int_0^tH(t')\dd{t'})\psi(0)\\ &= \psi(0) - \frac{i}{\hbar}\int_0^tH(t_1)\dd{t_1}\psi(0) - \frac{1}{\hbar^2}\int_0^t\dd{t_2}\int_0^{t_2}\dd{t_1}H(t_2)H(t_1)\psi(0) + \dots \end{align}

(note: the bounds on the inner integrals eat the \(1/n!\) from the expansion and preserve the time-ordering)

we typically do this in the interaction picture, so \(\psi\) evolves with just the perturbation \(V\), and we’re also typically interested in transitions between eigenstates of \(H_0\). these assumptions give us the matrix element

\begin{equation} \mathcal{M}_{i\to f} = -\frac{i}{\hbar}\int_0^t\dd{t}\exp(i(E_f-E_i)t/\hbar)\mel{f}{V(t)}{i} \end{equation}

and probability of transition \(|\mathcal{M}|^2\)

Fermi’s golden rule

\begin{gather*} P_{i\to f} \approx \frac{2\pi t}{\hbar}|\mel{f}{F}{i}|^2\delta(E_f-E_i-\hbar\omega) + \frac{2\pi t}{\hbar}|\mel{f}{F^\dagger}{i}|^2\delta(E_f-E_i+\hbar\omega)\\ \Gamma = \dv{P}{t} = \frac{2\pi}{\hbar}|\ev{F}|^2\delta(\Delta E - \hbar\omega) + \frac{2\pi}{\hbar}|\ev{F^\dagger}|^2\delta(\Delta E + \hbar\omega) \end{gather*}


  • often take a $\bra{f}$ continuum with density of states $\rho(E_f)$, i.e. $\rho(E_f)\dd{E}$ is the number of accessible states with energy $\in [E_f,\,E_f+\dd{E_f}]$. We basically then just swap delta for this rho:
    \begin{equation} \Gamma_\text{tot} = \frac{2\pi}{\hbar}\rho(E_i+\hbar\omega)\qty|\mel{f_{(E_i+\hbar\omega)}}{F}{i}|^2 + \frac{2\pi}{\hbar}\rho(E_i-\hbar\omega)\qty|\mel{f_{(E_i-\hbar\omega)}}{F^{\dagger}}{i}|^2 \end{equation}
  • if $\omega = 0$, we must take
    \begin{equation} \Gamma = \frac{2\pi}{\hbar}\qty|\mel{f}{V}{i}|^2\delta(E_i-E_f) \end{equation}


  • $V(t) = Fe^{-i\omega t} + F^{\dagger}e^{i\omega t}$
  • "$t$ large" in that $\rho(E_f)|\ev{F}|^2$ shouldnt change very much for the energy range $\hbar/t$; we need the time to be large relative to [$\hbar$ over the characteristic energy scale for $\rho(E_f)$ to vary]
  • "$t$ not too large" in that there must be many states available within the energy scale; we need the time to be small relative to [$\hbar$ over the characteristic energy spacing] for the delta function to make sense
\begin{equation} \frac{\hbar}{E_{\text{typical}}} \ll t \ll \frac{\hbar}{\varepsilon_\text{spacing}} \end{equation}

Variational principle

Use: finding ground state energy

It is a fact that

\begin{equation} E_0 \leq \frac{\mel{\psi}{\hat{H}}{\psi}}{\ip{\psi}} \end{equation}

for any wavefunction \(\psi\). thus, we create a family of \(\psi_\sigma(x)\) parameterized by some \(\sigma\) (e.g. width of a gaussian or something) and minimize over \(\sigma\) to get a best estimate for \(E_0\)

WKB approximation

Use: this is a semiclassical \(\hbar\qty|\dv{p}{x}|^2 \ll |p(x)|^2\) approximation.

for a slowly-varying \(V(x)\), we extract \(p(x)\) as if it were classical:

\begin{equation} p(x) = \sqrt{2m(E-V(x))} \implies \psi(x) \approx \frac{A}{\sqrt{p(x)}}\exp\Big(\pm \frac{i}{\hbar}\int_0^x\dd{y}p(y)\Big) \end{equation}

this works best in the cases where:

  • $p(x)$ real $\iff E > V$ — "classical region" of oscillatory $\phi$
  • $p(x)$ imag $\iff E < V$ — "forbidden region" exp growth/decay

to patch together in the intermediate regions, try to match typical boundary conditions and the continuity of the phase.

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Note: we also use TDPT in 3D scattering but i put it in the approximation methods section. FGR holds for weak scattering potentials

1D Scattering

Let \(V_I = \min[V(-\infty),\,V(+\infty)]\) and \(V_{II}\) be the max. Bound states are discrete and require \(\inf V(x) <E< V_I\). We have one scattering state for each \(E\in [V_{I},\,V_{II}]\) and two scattering states for each \(E > V_{II}\)

Take \(\psi\) as wave from \(-\infty\) being reflected and transmitted:

\begin{align} \psi(x\to-\infty) &= \underbrace{A e^{ik_Lx}}_\text{incoming}+\underbrace{Be^{-ik_Lx}}_\text{outgoing}\\ \psi(x\to+\infty) &= C e^{ik_Rx} \end{align}

we have coefficients

\begin{equation} R = \frac{|B|^2}{|A|^2} \qq{and} T = 1-R = \frac{k_R|C|^2}{k_L|A|^2} \end{equation}

we typically use boundary conditions

  • continuity of $\psi$
  • continuity of $\psi'$

however, if we have an infinite potential, e.g. \(V = \lambda\delta(x)\), we dont have the latter (which is the condition for differentiability). Instead, we integrate both sides of the SE to get

\begin{equation} -\frac{\hbar^2}{2m}[\psi_R' - \psi'_L] + \lambda \psi(0) = 0 \end{equation}

Born Approximation

Easiest derivation: start with FGR, scattering in a periodic box

\begin{equation} \psi_i(\vb{r}) = L^{-3/2}e^{i\vb{k}_i\cdot \vb{r}} \qq{to} \psi_i(\vb{r}) = L^{-3/2}e^{i\vb{k}_f\cdot \vb{r}} \end{equation}

where \(\vb{k}\) is quantized as \(2\pi\vb{n}/\hbar\). Replacing the delta in FGR with the dos

\begin{equation} \rho(E)\dd{E} = n^2\dd{n}\dd{\Omega} \implies \rho(E) = \qty(\frac{L}{2\pi\hbar})^3m\sqrt{2mE}\dd{\Omega} \end{equation}

allows us to calculate

\begin{equation} \dv{\Gamma}{\omega} = \frac{\dd{P}}{\dd{t}\cdot A}\frac{A}{\dd{\Omega}}= j_\text{inc}\dv{\sigma}{\Omega} \end{equation}

and thus

\begin{equation} \boxed{\dv{\sigma}{\Omega} = \qty|-\frac{m}{2\pi\hbar^2}\int e^{i(\vb{k}_i-\vb{k}_f)\cdot\vb{r}}V(r)\dd[3]{\vb{r}}|^2} \end{equation}

which is formula obtained via the born approximation.

the actual approximation itself is a bit more complex. we turn the TISE into the helmholtz equation and find a greens function solution that matches our BCs of outgoing waves. we get that the final wavefunction depends on an integral over the final wavefunction, so we plug back in to itself and get an interated integral that we truncate to first order in \(V\) and then compare to the asymptotic free-plus-scattered expected form

\begin{equation} \psi = \psi_\text{in} + \psi_\text{scattered} \approx e^{i\vb{k}_i\vb{r}} + \frac{e^{i\vb{k}_f\vb{r}}}{r}f(\theta,\,\phi) \qq{where} \dv{\sigma}{\Omega} = |f(\theta,\,\phi)|^2 \end{equation}

if \(V\) is spherically symmetric, we can go ahead and do these integrals:

let \(\vb*{\kappa} = \vb{k}_i-\vb{k}_f\) so that \(\kappa = 2k\sin\theta/2\) and

\begin{equation} f(\theta,\,\phi) = -\frac{2m}{\hbar^2\kappa}\int_0^\infty\dd{\bar{r}}\bar{r}V(\bar{r})\sin(\kappa\bar{r}) \end{equation}

Partial wave expansion

recall the 3D spherical SE

\begin{equation} \qty[-\frac{\hbar^2}{2m} \partial_r^2 + V + \frac{\hbar^2\ell(\ell+1)}{2mr^2}]u_{n\ell} = E_{n\ell} u_{n\ell} \end{equation}

In the far field, we’ve been ignoring angular momentum since it perturbs the effective potential as \(r^{-2}\). What if we stop ignoring it? then \(\psi \sim j_\ell(kr)Y_{\ell m}\) or \(\sim n_\ell(kr)Y_{\ell m}\) (bessels and von neumans), which go like sincs and cosincs respectively; nice for bound states but not for scatterings. We instead look at hankels \(h_\ell^{(1)}=j_\ell + in_\ell\) which asymptotically look like out- and in-going spherical waves

\begin{equation} h_\ell^{(1)}\sim \frac{(-i)^{\ell+1}}{x}e^{ix}\qquad h_\ell^{(2)}\sim \frac{(i)^{\ell+1}}{x}e^{-ix} \end{equation}

the strategy is to write the incoming plane wave in terms of hankels, which corresponds to different angular momenta. we match these to the large-\(r\) boundary conditions to solve for \(u_\ell\). Taking \(r\) to infinity then puts it back into a form where we can recognize \(f(\theta)\)

this is nice because in the low-energy (\(kr_\ast \ll 1\)) limit, we can just take the first \(\ell\), called \(S\)-wave scattering:

  1. for $r>r_\ast$, we take the far field and write $\psi = Ae^{ikz} + B \frac{e^{ikr}}{r}$
  2. for $r\gtrsim r_\ast$, we use the approximation: $\psi \sim A + \frac{B}{r}$
  3. for $r < r_\ast$, we can no longer ignore $V$: solve $(-\frac{\hbar^2}{2m}\nabla^2 + V)\psi \approx 0$ with the boundary condition at $r_\ast$

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Particles in EM Fields

\begin{equation} H = \frac{(p-qA)^2}{2m} + q\phi \end{equation}

common situations:

  • uniform, time-independent $B$: usually want to choose a rotationally symmetric $A=\frac{1}{2}B\times r$, eg $A = B/2(-y,\,x,\,0)$ or $A=B(0,\,x,\,0)$
  • time independent $E$: just say $E=-\nabla \phi$ as usual and work with the hammy

keep in mind \(\phi\) is not gauge invariant:

\begin{equation} A\to A+\nabla\chi,\,\phi\to\phi-\partial_t\chi \implies \phi\to e^{iq\chi/\hbar}\phi \end{equation}

the current density also changes:

\begin{equation} j= \frac{1}{m}\mathrm{Re}[\psi^\ast(p-qA)\psi] = \frac{\hbar}{m}\mathrm{Im}[\psi^\ast(\nabla- \frac{iq}{\hbar}A)\psi] \end{equation}

Zeeman effect

put the Hydrogen atom in a \(B\) field; take \(A = \frac{1}{2}B\times r\) and the hammy becomes

\begin{equation} H = \frac{p^2}{2m} - \frac{e^2}{4\pi\epsilon_0} + \underbrace{\frac{e}{2m}B\cdot L}_\text{cross term} + \underbrace{\frac{e^2}{8m}|B\times r|^2}_\text{$$A^2$$ term} \end{equation}

we usually take the \(B\) to be small so we can ignore \(A^2\) and just perturb the angular momentum coupling

Landau levels

take a \(B\) field in the \(z\) direction and let \(\vb{A} = Bx\vu{y}\). we can coax this into having plane waves in \(y\) and harmonic oscillator in \(x\)

Other situations

Aharanov Bohm effect: phase shift due to \(A\)

particle on ring with B field through it

spin in \(B\) field; \(H\sim B\cdot m\)

stark effect, which is like zeeman with \(E\) and antisemitism

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