Leading non-vanishing term of the groundstate for ##H_0##

[guyvsdcsniper]

Homework Statement

Find the non leading vanishing term to the ground state

Relevant Equations

Perturbation theory

The eigenvalue for this $H_0$ is given by $\hbar \omega(n+1) ; (n_x+n_y = n)$

At the ground state, $nx = ny = 0$ so the eigenvalue is simply $\hbar\omega$

Next we turn the perturbation potential on and I know that the first order shift in the energy is the expectation value of the perturbing Hamiltonian in the unperturbed state corresponding to that energy.

$E_n^1 = \braket{nx,ny|\hat H_1|nx,ny} = \frac {\lambda \hbar}{2m\omega}[\braket{0,0|a_xa_y+a_xa_y^\dagger + a_x^\dagger a_y + a_x^\dagger a_y^\dagger|0,0} = 0$

From here, I am to calculate the second order energy shift of the ground state.

I am having trouble applying the formula,

$\sum_{k\neq n} \frac{|\braket{k|\hat H_1|n}|^2}{E_n^0-E_k^0}$

For this problem nx and ny = 0 in the ground state
$\sum_{k\neq 0,0} \frac{|\braket{k|\hat H_1|0,0}|^2}{E_{0,0}^0-E_k^0}$
I can express $\hat H_1$

$\sum_{k\neq 0,0}(\frac {\lambda \hbar}{2m\omega})^2 \frac{|\braket{k|\hat x \hat y|0,0}|^2}{E_{0,0}^0-E_k^0}$

I have trouble understand what to do next in this problem. Im not really sure what K would be, I know it just be 0,0 which is what n is.

 

[vanhees71]

It would tremendously help us to help you, if you could give a complete statement of the problem under consideration!

 

[guyvsdcsniper]

It would tremendously help us to help you, if you could give a complete statement of the problem under consideration!

My apologies, I thought I attached a screenshot of the problem but I might have accidentally deleted it. I figured out the problem though. Thank you!