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\begin{document}
\begin{center}
  {\bf University of Kentucky, Physics 520 \\
    Homework \#11, Rev. B, due Friday, 2015-01-02 }
\end{center}

\vspace{.3 in}

{\bf 0.} Griffiths [2ed] Ch.~4
\#2, %p135  cube well
\#3. %p139  construct Ylm
%\#5. %p140 construct Ylm

{\bf 1.} We will see that {\bf central potentials} have effectively the same
Schr\"odinger equation in all dimensions.  We will use this fact to solve the
2-d and 3-d harmonic oscillators using the solution of the 1-d harmonic
oscillator from H07.
    
{\bf \quad a)} Calculate the 2-d and 3-d Laplacian in cylindrical and
spherical coordinates using
\[
\nabla^2 = \frac{1}{h_1h_2h_3}\sum_i\frac{\partial}{\partial q^i}\frac{h_j
  h_k}{h_i} \frac{\partial}{\partial q^i}
  \qquad \mbox{[$i,j,k$ cyclic]}.
\]

{\bf \quad b)} Substitute eigenvalues $m^2$ and $\ell(\ell+1)$ for the angular
parts and derive the formulas
\[
  \nabla^2
  ~=~ \frac{1}{\rho}\frac{\partial}{\partial\rho}\rho\frac{\partial}{\partial\rho}
   - \frac{m^2}{\rho^2} 
  ~=~ \frac{\partial^2}{\partial\rho^2}+\frac{1}{\rho}\frac{\partial}{\partial\rho} 
   - \frac{m^2}{\rho^2}
  ~=~ \frac{1}{\sqrt\rho}\frac{\partial^2}{\partial\rho^2}\sqrt\rho 
   - \frac{(m-\frac12)(m+\frac12)}{\rho^2}
\]
in cylindrical coordinates and
\[
  \nabla^2 
  ~=~ \frac{1}{r^2}\frac{\partial}{\partial r}r^2\frac{\partial}{\partial r}
   - \frac{\ell(\ell+1)}{r^2}
  ~=~ \frac{\partial^2}{\partial r^2}+ \frac{2}{r}\frac{\partial}{\partial r}
   - \frac{\ell(\ell+1)}{r^2}
  ~=~ \frac{1}{ r}\frac{\partial^2}{\partial r^2} r 
   - \frac{\ell(\ell+1)}{r^2}
\]
in spherical coordinates.

{\bf \quad c)} Substitute
$\psi(\rho,\phi)=\frac{u(\rho)}{\sqrt{\rho}}e^{im\phi}$ and
$\psi(r,\theta,\phi)=\frac{u(r)}{r}P_\ell^{|m|}(\cos\theta)e^{im\phi}$ into
the 2-d and 3-d Schr\"odinger equations $-\frac{\hbar^2}{2m}\nabla^2\psi +
V\psi = E\psi$ for central potentials $V(\rho)$ and $V(r)$ to obtain the 1-d
Schr\"odinger equation for $u(\rho)$ or $u(r)$ with the same potential plus a
centrifugal term involving $m$ or $\ell$.  The factors of $\sqrt{\rho}$ and
$r$ account for the wavefunction spreading out in space.

{\bf \quad d)} Comparing the equations for the 2-d and 3-d harmonic
oscillator, $V(\rho)=\frac12 m\omega^2 \rho^2$ and $V(r)=\frac12 m\omega^2
r^2$, with the 1-d Hamiltonian and its parity solutions in H07,
\[
\mathcal{H}_x = -\frac{\hbar^2}{2m}\left[\frac{d^2}{dx^2} -
  \frac{(p-1)(p)}{x^2} \right] + \frac12 m\omega^2 x^2,\qquad
\psi(\xi)=\xi^pe^{-\xi^2/2}L_k^{(p-\frac12)}(\xi^2), \quad n=2k+p,
\]
where $x,\rho,r =\sqrt{\hbar/m\omega}\,\xi$, derive the radial functions
$u(\rho)$ and $u(r)$, respectively, and calculate the energy levels of the 2-d
and 3-d harmonic oscillators.  The centrifugal term $(p-1)(p)/x^2$ is zero
since $p=0$ (even) or $1$ (odd), but it is included to help determine the
relation \mbox{between $p$, $m$, and $\ell$.}

{\bf \quad e)} The 2-d harmonic oscillator Hamiltonian separates into the sum
of two 1-d Hamiltonians $\mathcal{H}=\mathcal{H}_x+\mathcal{H}_y$, with the
solution $\psi_{n_x n_y}(x,y) = \psi_{n_x} \psi_{n_y}$ for independent quantum
numbers $n_x$ and $n_y$.  Compare the wave functions for $n_x+n_y \leq 2$ with
the wave functions from part d) for $n,|m|\leq 2$ and show the two sets are
linear combinations of each other.

{\bf \quad f)} The same symmetry between dimensions holds for all central
potentials, including the free particle.  Use part c) to show that the 1-d,
2-d, and 3-d free particle solutions can all be written in terms of $J_\nu(x)$,
where $\nu$ is a function of $p$, $m$, or $l$.

\end{document}