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

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 3K views
Mayan Fung
Messages
131
Reaction score
14
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.
 
Physics news on Phys.org
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.
 
  • Like
Likes   Reactions: Mayan Fung
vanhees71 said:
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.
 
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)##.
 
  • Like
Likes   Reactions: dRic2 and Mayan Fung
vanhees71 said:
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