Is the Heisenberg Picture Better for a Time-Dependent Hamiltonian?

[Mayan Fung]

Homework Statement

Solve the analytically the time-dependent Schrodinger equation with Hamiltonian:
$$\hat{H} = \frac{\hat{p}^2}{2} + \frac{\hat{q}^2}{2} + \alpha(t) \hat{q}$$
where $\alpha(t) = 1-e^{-t}$ for $t \leq 0$ and 0 for $t <0$. With initial wave function in coordinate space:
$$\Psi(q, t= 0) = Nexp(-\frac{q^2}{4})$$

Relevant Equations

\begin{align}
E_n &= \hbar(n+\frac{1}{2})\\
|\Psi(t)> &= exp(-i\frac{\hat{H}}{\hbar}t)|\Psi(0)>
\end{align}

What I have tried is a completing square in the Hamiltonian so that

$$\hat{H} = \frac{\hat{p}^2}{2} + \frac{(\hat{q}+\alpha(t))^2}{2} - \frac{(\alpha(t))^2}{2}$$

I treat $t$ is just a parameter and then I can construct the eigenfunctions and the energy eigenvalues by just referring to a standard harmonic oscillator solution by shifting some constants. And then I used

\begin{align}
|\Psi(t)> &= exp(-i\frac{\hat{H}}{\hbar}t)|\Psi(0)> \\
|\Psi(t)> &= \sum_{n=0}^\infty exp(-i\frac{\hat{H}}{\hbar}t) |n><n|\Psi(0)> \\
|\Psi(t)> &= \sum_{n=0}^\infty exp(-i\frac{E_n}{\hbar}t) <n|\Psi(0)> |n>\\
\end{align}
where $<n|\Psi(0)> = \int \Psi_n^*(t)\Psi(t=0) \, dx$

Because of the $\alpha(t) = 1-e^{-t}$, I can't compute $\int \Psi_n^*(t)\Psi(t=0) \, dx$. I think the problem can be more easily solved because the initial wave function $\Psi(q, t= 0) = Nexp(-\frac{q^2}{4})$ looks like that it is deliberately chosen. I am also not sure if I can extend the harmonic oscillator solution for a time varying constant.

 

[vanhees71]

Hint: You cannot do this so easily in this way, because the Hamiltonian is explicitly time-dependent. Also what you write as $E_n$ are the energy eigenvalues of the harmonic oscillator Hamiltonian without the extra time-dependent term.

 

[Mayan Fung]

Hint: You cannot do this so easily in this way, because the Hamiltonian is explicitly time-dependent. Also what you write as $E_n$ are the energy eigenvalues of the harmonic oscillator Hamiltonian without the extra time-dependent term.

I tried to follow the steps suggested in this notes:

https://ocw.mit.edu/courses/chemist...pring-2009/lecture-notes/MIT5_74s09_lec02.pdf

By writing
$$|\Psi(t)>= \sum_{n=0} c_n(t) |n>$$, where $|n>$ is the harmonic oscillator wave function. And I still found that the final expression for $c_n$ doen't look analytically solvable.

 

[vanhees71]

If I remember right, the trick is to use the Heisenberg picture and evaluate the time evolution for the annihilation-creation (or ladder) operators and then calculate the overlap of the time-dependent (!) eigenvectors of $\hat{H}_0$ with the state, which by definition is time-independent in the Heisenberg picture and in your case given as a pure state represented by the wave function $\exp(-q^2/4)$.

 

[Mayan Fung]

If I remember right, the trick is to use the Heisenberg picture and evaluate the time evolution for the annihilation-creation (or ladder) operators and then calculate the overlap of the time-dependent (!) eigenvectors of $\hat{H}_0$ with the state, which by definition is time-independent in the Heisenberg picture and in your case given as a pure state represented by the wave function $\exp(-q^2/4)$.

This is a clever approach. Thanks! I would have a try