Given is a two-dimensional harmonic oscillator with a pertubation:

$$H = \frac{1}{2m}(p_1^2 + p_2^2) + \frac{m \omega^2}{2}(x_1^2 + x_2^2) + \gamma \frac{m^2 \omega^3}{\hbar} x_1^2 x_2^2$$

My question:

How to find the first order energy correction of the first excited state and the zeroth order of this excited state?

My idea:

The first excited state can be noted as $|n = 1\rangle$ with $n = n_1 + n_2$. So it is twofold degenerated, because both $|n_1 = 1; n_2 = 0\rangle$ and $|n_1 = 0; n_2 = 1\rangle$ correspond to this state.

Thus, we have to apply degenerate pertubation theory.

We have to evaluate the matrix $V_{ij} = \langle\psi_i|\gamma \frac{m^2 \omega^3}{\hbar} x_1^2 x_2^2|\psi_j\rangle$ with $|\psi_1\rangle = |n_1 = 1; n_2 = 0\rangle$ and $|\psi_2\rangle = |n_1 = 0; n_2 = 1\rangle$.

Calculation gives

$V = \begin{pmatrix} 3\gamma \hbar \omega & 0 \\ 0 & 3\gamma \hbar \omega \end{pmatrix} $

The eigenvalue of $V$ is $\lambda = 3\gamma \hbar \omega$.

So, the energy of the first excited state is $E_1' = 2\hbar \omega + 3\gamma \hbar \omega$.

Can one solve this problem like this?

And how to find the first excited state in zeroth order?


This is the easiest way I'm familiar with -

  1. Find the eigenvalues and eigenstates of the unpertubated hamiltonian ( $H=\hbar\omega(\hat{a_1}^\dagger \hat{a_1}+\hat{a_2}^\dagger \hat{a_2}+1)$ ). The eigenstates are just the tensor product of eigenstates of regular 1D harmonic oscillator.
  2. Convert the pertubation term ($H_I=\gamma\frac{m^2 \omega^3}{\hbar} x_1^2 x_2^2$) to representation with creation and anhilation operators, in your case $x_1 \propto \hat{a_1}+\hat{a_1}^\dagger$ and so on.
  3. Find the correction to each level, for example - for ground state by $\langle\phi_1(n_1=0)|\langle\phi_2(n_2=0)|H_I(\hat{a_1},\hat{a_1}^\dagger,\hat{a_2},\hat{a_2}^\dagger)|\phi_1(n_1=0)\rangle |\phi_2(n_2=0)\rangle$

Just to be clear, the treatment above calculates corrections to at least second order. The first order corrections are just the diagonal terms in the $H_I$ matrix, which is easily constructed with creation and anhilation operators.

  • $\begingroup$ Thank you for your answer, it helps me a lot! Do you also know how to find the first excited state in zeroth order? $\endgroup$ – Peter123 Oct 17 '16 at 10:27
  • $\begingroup$ Does your algorithm also hold for states which are excited even if they're degenerate, for example $|n=1>$, which can be represented by $|n_1=1;n_2=0>$ and $|n_1=0;n_2=1>$? $\endgroup$ – Peter123 Oct 17 '16 at 10:58
  • $\begingroup$ The original eigenstates hold up to 1st order. This method is the formal perturbation theory (I just explained how to easily construct the $H_I$ matrix) so it holds to degenerate states as well. $\endgroup$ – Alexander Oct 18 '16 at 8:50

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