How Does a Perturbation Affect the Ground State Energy of a Harmonic Oscillator?

[malawi_glenn]

Homework Statement

Sakural modern quantum.. ch 5 problem 1

A simple one dimensional harmonic oscillator is subject to a perturbation:

V = bx, where b is a real constant.

Calculate the energy shift in ground state to lowest non vanishing order.

Homework Equations

You may use:

$$\langle k \vert x \vert n \rangle = \sqrt{\dfrac{\hbar}{2m\omega}}\left( \sqrt{n+1}\delta_{k,n+1} + \sqrt{n}\delta_{k,n-1} \right)$$

where |n> is eigentkets to unperturbed harm. osc

Energy shift:

$$\Delta _{n} \equiv E_n - E^{(0)}_n = \lambda V_{nn} + \lambda^{2} \sum _{k\neq n} \dfrac{\vert V_{nk}\vert^{2}}{E^{(0)}_n - E^{(0)}_k} + . . .$$

Lamda is order, V_nn is matrix elements.

Energy levels for harm osc

$$E_N^{(0)} = \hbar \omega (1/2 + N)$$

The Attempt at a Solution

I first do the matrix representation of V = bx

$$V_{nk} \doteq b\sqrt{\hbar / (2m \omega)}\left( \begin{array}{ccccc} 0 & 1 & 0 & 0 & 0 \\1 & 0 & \sqrt{2}& 0 & 0 \\0 & \sqrt{2}& 0 & \sqrt{3} &0\\ 0 & 0 & \sqrt{3}&0&0 \end{array}$$

Then I choose n = 0, since ground state.

$$\Delta _{0} \equiv E_0 - E^{(0)}_0 = \lambda V_{00} + \lambda^{2} \sum _{k\neq 0} \dfrac{\vert V_{0k}\vert^{2}}{E^{(0)}_0 - E^{(0)}_k} + . . .$$

I notice that $$V_{00} = 0$$ and $$V_{0k}$$is zero for all k except 1; so that:

$$V_{01} = b\sqrt{\hbar / (2m \omega)}$$

And

$$E^{(0)}_0 - E^{(0)}_1} = \hbar \omega$$

So that
$$\Delta _{0} = -b^2 / (2m \omega ^2)$$

I have no answer to this problem, does it look right to you?
Thanx!