I consider the following Hamiltonian $$H=\frac{p^2}{2m}+\frac{m\omega^2}{2}x^2+\Theta(t)Fx,$$ where $F$ is an external constant force. So the Hamiltonian describes an unperturbed harmonic oscillator if $t<0$ and a constant perturbed harmonic oscillator if $t>0$.

My aim is to calculate the exact transition probability from the initial unperturbed groundstate to a final new excited state.

What I already know: The ecxited eigenvalues ar $\bar{E}_n=E_n-\frac{F^2}{2m\omega^2}$

I need the transition amplitude $\langle \bar{n}|0\rangle$ to calculate the probability.

The time evolution for the new excited states are $|\bar{n}\rangle=e^{-\frac{i}{\hbar}\bar{E}_nt}|n\rangle$. The state $|n\rangle$ I can express with the ladder operators.

Now I have some problems. I calculate the amplitude, $e^{\frac{i}{\hbar}\bar{E}_nt}\langle n|0\rangle=0$, because $|n\rangle$ and $|0\rangle$ are orthogonal. Of course this schould be nonzero.

So where is the mistake?

  • 2
    $\begingroup$ i'm not sure $\bar n$ and $n$ differ just by a phase factor. It seems to me that the corresponding Hermite polynomials have different centres of mass. $\endgroup$
    – Phoenix87
    Jan 22, 2015 at 9:42
  • $\begingroup$ Right, they have shifted arguments. $\endgroup$ Jan 22, 2015 at 22:51

1 Answer 1


The time evolution for the new excited states are $|\bar{n}\rangle=e^{-\frac{i}{\hbar}\bar{E}_nt}|n\rangle$.

No, it is $|\bar{n}\rangle=e^{-\frac{i}{\hbar}\bar{E}_{\bar{n}}t}|\bar{n}\rangle$. Although numerically $\bar{n}$ can be equal to $n$, the states $|\bar{n}\rangle$ and $|n\rangle$ are different. $|n\rangle$ is a superposition of many $|\bar{n}\rangle$; in particular, $|0\rangle$ is a superposition of many $|\bar{n}\rangle$. $|0\rangle$ and $|\bar{n}\rangle$ are not orthogonal. For example, $\psi_n(x)$ and $\bar{\psi}_n(x)=\psi_n(x+a)$ are different.


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