Time dependent perturbation for harmonic oscillator

[Vandmelon]

Homework Statement

I'm looking at the 1d harmonic oscillator
\begin{equation}
V(x)=\frac{1}{2}kx^2
\end{equation}
with eigenstates n and the time dependent perturbation
\begin{equation}
H'(t)=qx^3\frac{(\tau^2}{t^2+\tau^2}
\end{equation}
For t=-∞ the oscillator is in the groundstate n=0
I need to show that for n>3 the states will not get excited and I only need to look at the first order perturbation.

The Attempt at a Solution

What I'm thinking is \begin{equation}P_{a→b}=\vert c_b(t)\vert^2 \end{equation} therefore I need to find when c_b=0
and because
\begin{equation}
c_b'=-\frac{i}{\hbar}∫H_{ba}'(t')e^{i\omega_0t'}dt'
\end{equation}
I need to find
\begin{equation}
H_{ba}'=<\psi_n\vert H' \vert \psi_a >=0
\end{equation}
For n>3
It would make sense if H_ba look a bit like this
\begin{equation}
H_{ba}'=\frac{d^n}{dx^n}H'
\end{equation}
But I get some horrible results if I try to solve H_ba. Is it right what I am doing or is it way off.

 

[fzero]

You want to compute \langle n| H&#039; |0\rangle, which is proportional to \langle n| \hat{x}^3 |0\rangle. If you express \hat{x} in terms of raising and lowering operators \hat{a},\hat{a}^\dagger, you should be able to see how to reach the necessary result.

 

[Vandmelon]

so if I use
\begin{equation}
\hat x=\sqrt{\frac{\hbar}{2m\omega}}(a_+ + a_-)
\end{equation}

then if I solved it right
\begin{equation}
\hat x^3 \psi_0=(\frac{\hbar}{2m\omega})^{\frac{3}{2}}[(a_+)^3+a_+ + a_- + (a_-)^3]\psi_0
\end{equation}
Is it correct if I use a_- on ψ_0 it will just be 0 so
\begin{equation}
\hat x^3 \psi_0=(\frac{\hbar}{2m\omega})^{\frac{3}{2}}[(a_+)^3\psi_0+a_+\psi_0]=(\frac{\hbar}{2m\omega})^{\frac{3}{2}}[\sqrt6 \psi_3+\psi_1]
\end{equation}
So it will end up like this
\begin{equation}
\langle \psi_n \vert\sqrt6 \psi_3+\psi_1 \rangle
\end{equation}
If this is correct I am not sure why the oscillator wouldn't get excited for n>3. Or is it my calculations there are wrong?

*EDIT
Oh is it because
\begin{equation}
\langle \psi_n \vert \psi_m \rangle = 0
\end{equation}
for n ≠ m

 

[fzero]

so if I use
\begin{equation}
\hat x=\sqrt{\frac{\hbar}{2m\omega}}(a_+ + a_-)
\end{equation}

then if I solved it right
\begin{equation}
\hat x^3 \psi_0=(\frac{\hbar}{2m\omega})^{\frac{3}{2}}[(a_+)^3+a_+ + a_- + (a_-)^3]\psi_0
\end{equation}

You'll want to go back through the algebra here. Remember that a_+,a_- don't commute, so you have to keep track of the order of terms.

Is it correct if I use a_- on ψ_0 it will just be 0 so

...

Oh is it because
\begin{equation}
\langle \psi_n \vert \psi_m \rangle = 0
\end{equation}
for n ≠ m

You will have to use both of these relations. You have all of the right ideas, just fix the algebra.