Electron in dielectric cube (Quantum Mechanics)

[xago]

Homework Statement

[PLAIN]http://img408.imageshack.us/img408/1685/wergp.png

Homework Equations

$$H\psi = E\psi$$

The Attempt at a Solution

For part a) I used $$H\psi = E\psi$$ to get:
$$E = \frac{\widehat{p}}{2m} + \frac{Kx^2}{2} + \frac{e\Phi_o}{a}x$$

and assuming E = Q? and rearranging for K gives:

$$K = \frac{Q - \frac{\widehat{p}}{2m} - \frac{e\Phi_o}{a}x}{x^2/2}$$

Part b) I'm not exactly sure what to do. It tells us to complete the square but i don't see how solving for values of x relates to the allowed energies of the electron.

 

[arkajad]

For 2)

ax^2+bx can be always represented as a(x-x0)^2 + B. You have a and b. Find x0 and B. Substitute x-x0 ->x', notice that d/dx = d/dx', use harmonic oscillator energy levels.