Fall 2003 5 Qm

A particle with charge $q$ and mass $m$ is bound by a three-dimensional harmonic oscillator potential

\begin{align} V(x,y,z) = \frac{1}{2} k ( x^2 + y^2 + z^2 ) \end{align}
  1. What is the degeneracy of the energy eigenstate with $E = \frac{7}{2} \hbar \sqrt{k/m}$?
  2. Now, a (non-constant) electric field in the $\hat{x}$ direction is applied to this system. How much of the above degeneracy is lifted?
  3. Now assume that the electric field is given by $\vec{E} = \hat{x} E_0 x \exp^{-Ax^2}$. Find the change in the energy of the ground state to lowest non-vanishing order in $E_0$.




We can write the Hamiltonian as

\begin{align} H = \frac{p_x^2}{2m} + \frac{1}{2}kx^2 + \frac{p_y^2}{2m} + \frac{1}{2}ky^2 + \frac{p_z^2}{2m} + \frac{1}{2}kz^2 \end{align}

This is just the sum as 3 1D harmonic oscillators, so the solution is a product of 3 1D Harmonic oscillators, with the energy as the sum.

\begin{align} E = \hbar \omega (n_x + n_y + n_z + \frac{3}{2} ) \end{align}

where $n_i \ge 0$

For part 1, in order to get $E = \frac{7}{2} \hbar \omega$, we must have $n_x + n_y + n_z = 2$. There are 6 possibilities (3 combinations each of 1,1,0 and 2,0,0). We therefore have a degeneracy of 6.

For part 2, the change in potential energy is $q \Phi$, where $- \vec{\bigtriangledown} \Phi = \vec{E}$. Since $\vec{E}$ only points in the $\hat{x}$ direction, $\Phi=\Phi(x)$. The change in energy of the state is

\begin{align} \langle n_x n_y n_z | q \Phi | n_x n_y n_z \rangle = \int d^3 \vec{r} |\Psi|^2 q\Phi(x) \end{align}

Since this integral depends only on $x$, the states $n_x n_y n_z$ of 0,2,0 and 0,0,2 will still have the same energy. Also 1,1,0 and 1,0,1 will have the same energy. We therefore split the single energy (6 different states) state into 4 energies with the degeneracies as described om the previous sentence.

For part 3, the ground state of the harmonic oscillator is $N \exp (-\frac{m\omega}{2\hbar}x^2)$. If you don't remember this, you can derive it by knowing that $a = \frac{1}{\sqrt{2}}(\lambda x + \frac{ip}{\hbar \lambda})$, where $\lambda = \sqrt{\frac{m \omega}{\hbar}$, and $a| 0 \rangle = 0$. Solving for the normalization (just need to integrate a gaussian, which we can do), we get the ground state of the 3D oscillator as

\begin{align} \left( \frac{m \omega}{\pi \hbar} \right)^{3/4} \exp \left( - \frac{m \omega}{2 \hbar}(x^2 + y^2 + z^2) \right) \end{align}

If $\vec{E}=\hat{x}E_0 x \exp(-Ax^2)$, then $\Phi = \frac{E_0}{2A}\exp(-Ax^2)$, so the change in energy to the ground state is

\begin{align} \int dxdydz \left( \frac{m \omega}{\pi \hbar} \right)^{3/2} \exp \left( - \frac{m \omega}{2 \hbar}(x^2 + y^2 + z^2) \right) \frac{E_0}{2A} \exp(-Ax^2) \end{align}

The y and z integrals can be done and just get rid of their normalization factor. We then get the change in energy as

\begin{align} \frac{E_0}{2A} \int dx \left( \frac{m \omega}{\pi \hbar} \right)^{1/2} \exp \left( -(\frac{m \omega}{\hbar} + A)x^2 \right) = \frac{E_0}{2A} \left( \frac{m \omega}{\pi \hbar} \right)^{1/2} \left( \frac{\pi}{\frac{m \omega}{\hbar} + A} \right)^{1/2} \end{align}
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