Harmonic oscillator in electric field

[skrat]

Homework Statement

Potential energy of electron in harmonic potential can be described as $V(x)=\frac{m\omega _0^2x^2}{2}-eEx$, where E is electric field that has no gradient.
What are the energies of eigenstates of an electron in potential $V(x)$? Also calculate $<ex>$.

HINT: Use $(x-a)^2=x^2-2ax+a^2$

Homework Equations

The Attempt at a Solution

I am sorry, to say, but I have no idea how start here.

I know that if there were no electric field, the energies would be $E_n=\hbar \omega (n+1/2)$. Since there is Electric field, I assume I have to solve $\hat{H}\psi =E\psi$ for $\hat{H}=\hat{T}+\hat{V}$... but, how on Earth can i do that?

 

[TSny]

The hint is to "complete the square" in the expression $V(x)=\frac{m\omega _0^2x^2}{2}-eEx$.
That is, can you write it as $V(x) = b(x-a)^2 - c$ for certain constants $a$, $b$ and $c$?

 

[skrat]

Ok, I got that but I can't see how this makes my life any easier..

$V(x)=\frac{m\omega _0^2x^2}{2}-eEx=(\sqrt{\frac{m}{2}}\omega _0x-\frac{eE}{\sqrt{2m}\omega _0})^2-\frac{e^2E^2}{2m\omega _0^2}$

 

[TSny]

Ok, I got that but I can't see how this makes my life any easier..

$V(x)=\frac{m\omega _0^2x^2}{2}-eEx=(\sqrt{\frac{m}{2}}\omega _0x-\frac{eE}{\sqrt{2m}\omega _0})^2-\frac{e^2E^2}{2m\omega _0^2}$

Try to factor out something inside the parentheses so that the x has a coefficient of 1 inside the parentheses. Then define a new variable in terms of x and solve the problem in the new variable.

 

[skrat]

Hmm,

$V(x)=(\sqrt{\frac{m}{2}}\omega _0x-\frac{eE}{\sqrt{2m}\omega _0})^2-\frac{e^2E^2}{2m\omega _0^2}$

$V(x)=(\sqrt{\frac{m}{2}}\omega _0x-\frac{eE\sqrt{m}\omega_0}{\sqrt{2}m\omega _0^2})^2-\frac{e^2E^2}{2m\omega _0^2}$ and finally

$V(x)=(\frac{m}{2})^{1/4}\omega_0^{1/2}(x-\frac{eE}{m\omega _0^2})^2-\frac{e^2E^2}{2m\omega _0^2}$

now let's say $u=x-\frac{eE}{m\omega _0^2}$ than $V(u)=(\frac{m}{2}\omega_0^2)^{1/4}u^2-\frac{e^2E^2}{2m\omega _0^2}$

Now $\hat{H}\psi = E_n \psi$

$\hat{V}\psi=((\frac{m}{2}\omega_0^2)^{1/4}\hat{u}^2-\frac{e^2E^2}{2m\omega _0^2})\psi =E_n \psi$

Are the energies of eigenstates than $W_n=(\hbar \omega(n+1/2)+\frac{e^2E^2}{2m\omega _0^2})(\frac{m}{2}\omega_0^2)^{-1/4}$?