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:

<br /> \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 &amp; 1 &amp; 0 &amp; 0 &amp; 0 \\1 &amp; 0 &amp; \sqrt{2}&amp; 0 &amp; 0 \\0 &amp; \sqrt{2}&amp; 0 &amp; \sqrt{3} &amp;0\\ 0 &amp; 0 &amp; \sqrt{3}&amp;0&amp;0 \end{array}<br />

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!