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\title{Quantum Mechanics GRE Notes}
\author{Peter~Williams}

\begin{document}

\maketitle

Here is some stuff that you should know about quantum mechanics for
the GRE.

\begin{itemize}

\item {\em Schr\"odinger Equation}. General and time-independent:

  \begin{eqnarray}
    i \hbar \frac{\partial \Psi}{\partial t} &=& \left(-\frac{\hbar^2}{2m}
    \nabla^2 + V\right) \Psi \cr
    \cr
    \cr
    H\psi &=& E \psi \cr
    \Psi &=& \psi e^{-iEt/\hbar}
  \end{eqnarray}
  
\item {\em Formalism and Operator Algebra}.  This is probably actually
  the easiest topic for people who have taken 143a. Some equations of
  use are:

  \begin{eqnarray}
    \braket{\alpha}{\beta} &=& \braket{\beta}{\alpha}^* \cr
    \cr
    \hat{O} &=& \hat{O}^\dagger \mathrm{~if~} \hat{O} 
    \mathrm{~is~Hermitian} \cr 
    \cr
    \bra{\alpha}\hat{O}\ket{\beta} &=& \beta \braket{\alpha}{\beta}
    \mathrm{~if~} \ket{\beta} \mathrm{~is~an~eigenstate~of~} \hat{O}
    \cr
    &=& \int_{-\infty}^{\infty}\ud x\,\alpha^*(x) \hat{O}(x) \beta(x)
    \mathrm{~in~1D} \cr
    &=& A^\dagger \times O \times B \mathrm{~as~matrices} \cr
    \cr 
    [\hat{A}, \hat{B}] &=&
    \hat{A}\hat{B} - \hat{B}\hat{A} \cr \therefore~[\hat{A}, \hat{B}]
    &=& -[\hat{B}, \hat{A}] \cr [\hat{A}\hat{B}, \hat{C}] &=&
    [\hat{A}, \hat{C}]\hat{B} + \hat{A}[\hat{B}, \hat{C}]
  \end{eqnarray}

  A Hermitian matrix has real eigenvalues; observables thus must be
  represented by Hermitian operators.

\item {\em Position and momentum}.

  \begin{eqnarray}
    \hat{p} &=& -i \hbar \nabla \cr
    \cr
    [\hat{x}, \hat{p}] &=& i \hbar \cr
    [\hat{f}(x), \hat{p}] &=& i \hbar \frac{\ud f}{\ud x}
  \end{eqnarray}
  
\item {\em De Broglie equation}. This one always trips me up even
  though it's easy: 

  \begin{eqnarray}
    \lambda &=& h / p \cr
    &=& h / \sqrt{2 m k_B T}
  \end{eqnarray}

  if the particle's energy is thermal. Fun fact: the De Broglie
  wavelength of thermal neutrons is about 1.78 angstroms.

\item {\em Emitted photons}. Yeah. The name of the game:

  \begin{eqnarray}
    E &=& \frac{-\mu e^4}{2 h^2} \frac{Z^2}{n^2} \cr
    &=& \left(-13.61 \mathrm{eV}\right) \frac{Z^2}{n^2} \mathrm{~for~} \mu = m_e \cr
    \therefore\ \lambda^{-1} &\propto& Z^2\left(\frac{1}{n^2} - \frac{1}{n'^2}\right)
  \end{eqnarray}

  This means the key to their beloved positronium problems is that
  $\mu \rightarrow \mu/2$ since the reduced mass is now $(m_e^{-1} +
  m_e^{-1})^{-1}$. So basically you can do the problem as if it were
  hydrogen with 13.6 eV $\rightarrow$ 6.8 eV.

\item {\em Hydrogen Ground State}. A good thing to know:

  \begin{eqnarray}
    \psi(r) &=& \frac{1}{\sqrt{\pi a^3}} e^{-r/a}
  \end{eqnarray}

\item {\em Infinite Square Well}. Straightforward:

  \begin{eqnarray}
    \psi_n &=& \frac{1}{\sqrt{a}} \sin{\frac{n \pi x}{a}} \cr
    E_n &=& \frac{\pi^2 n^2 \hbar^2}{2 m a^2}
  \end{eqnarray}

  Pay attention to how $E_n$ scales with $n$. Also, $n \ge 1$ of
  course.

\item {\em Singlet and triplet states}. These are useful from two
  perspectives. The first point of view is taking them as eigenstates
  of particle exchange in a two-particle system. The second is ways of
  adding two spin-$\frac{1}{2}$ particles to get one spin-1
  particle. From the first perspective, the singlet state has
  eigenvalue -1; from the second, it has $s = m = 0$:

  \begin{eqnarray}
    \ket{0~0} &=& \frac{1}{\sqrt{2}}\left(\ketud -
     \ketdu\right)
  \end{eqnarray}
  
  The triplet states have eigenvalue 1 with regards to parity and have 
  $s = 1$ and $m \in \{-1, 0, 1\}$:

  \begin{eqnarray}
    \ket{1~1} &=& \ketuu \cr
    \ket{1~0} &=& \frac{1}{\sqrt{2}}\left(\ketud +
     \ketdu\right) \cr
    \ket{1~-1} &=& \ketdd
  \end{eqnarray}
  
\item {\em Selection rules}. They seem to like making you remember
  these.

  \begin{eqnarray}
    \Delta l &=& \pm 1 \cr
    \Delta m_l &=& \pm 1 \mathrm{~or~} 0 \cr
    \Delta j &=& \pm 1 \mathrm{~or~} 0
  \end{eqnarray}

  Here $l$ is the orbital angular momentum; $m_l$ is the $z$ component
  of $l$; $j = l + s$ is the total angular momentum. (I'm not exactly
  sure how you can get $\Delta j = 0$ but I read it on the internet so
  it must be true!) Typically $\Delta n = \pm 1$ but there is no
  reason that this must hold. Just remember that the emitted photon is
  spin 1 and so its angular momentum can be described as one of
  $\ket{0~0}$, $\ket{1~1}$, $\ket{1~0}$, or $\ket{1~-1}$.

  The phrase ``electric dipole radiation'' implies that $\Delta l = 0$
  is not physically permissible, even though it is allowed
  mathematically. Electric dipole transitions are ``allowed'', while
  magnetic dipole or electric quadrupole transitions are ``forbidden''
  (but do actually occur).

\item {\em Observed states and expectation values}. The usual format
  of such a problem is

  \begin{eqnarray}
    \ketpsi &=& \sqrt{p_1}\ket{k_1} + \sqrt{p_2}\ket{k_2} +
     \sqrt{p_3}\ket{k_3} + \cdots
  \end{eqnarray}

  The expectation value is $p_1 k_1 + p_2 k_2 + p_3 k_3 + \cdots$ if
  $k_i$ are the eigenvalues associated with eigenkets $\ket{k_i}$, and the
  probability of being in a state $\ket{k_i}$ is just $p_i$. This is all
  assuming the $\ket{k_i}$ are eigenkets of whatever operator you're
  finding the expectation value of.

\item {\em Heisenberg relation}. The general version:

  \begin{eqnarray}
    \Delta A \Delta B &\ge& \frac{1}{2}\left|\left<[A, B]\right>\right|
  \end{eqnarray}

  In particular,

  \begin{eqnarray}
    \Delta x \Delta p &\ge& \hbar / 2 \cr
    \Delta E \Delta t &\ge& \hbar / 2
  \end{eqnarray}

\item {\em Time Independent First-Order Pertubation Theory}. Given a
  perturbation $H'$ to some original Hamiltonian with eigenstates
  $\ket{\psi_n^0}$ and energies $E_n^0$, the perturbed energies and
  states are:

  \begin{eqnarray}
    E_n^1 &=& E_n^0 + \bra{\psi_n^0}H'\ket{\psi_n^0} \cr
    \psi_n^1 &=& \psi^0_n + \sum_{m \ne n}{\frac{\bra{\psi_m^0}H'\ket{\psi_n^0}}{E_n^0-E_m^0}
    \psi_m^0}
  \end{eqnarray}

  Recall that

  \begin{eqnarray}
    \bra{\psi_n}\hat{O}\ket{\psi_m} &=&
    \int_{-\infty}^{\infty}\psi_n^*(x) \hat{O}(x) \psi_m(x) \ud x
  \end{eqnarray}
  
  in one dimension. (You will often be perturbing the infinite square well.)

\item {\em Fundamental Understanding of the Concepts of Quantum
  Mechanics}. Yeah. Whatever.

\end{itemize}
\end{document}