# A Time Dependent Hamiltonian problem

1. Apr 22, 2014

### Emspak

1. The problem statement, all variables and given/known data

We want to get the time evolution of a wavefunction and the expectation value of the Hamiltonian, and from there we can show that it's the same as the time-independent result. So to be clear: given a wavefunction, get the time evolution of that function and the expectation value of the Hamiltonian.

The (normalized) wavefunction is $\psi = \frac{1}{\sqrt{5}}(1\phi_1 + 2\phi_2)$

2. Relevant equations

The (time dependent) Hamiltonian: $\hat H \psi = i \hbar \frac{\partial \psi(x,t)}{\partial t}$

3. The attempt at a solution

For any wavefunction the time dependent form would be $\psi (x) e^{-i\omega t}$

That means that $\frac{\partial \psi}{\partial t} = -i \omega \phi (x) e^{-i\omega t}$

So for this particular wavefunction $\frac{\partial \psi}{\partial t} = \frac{1}{\sqrt{5}} (-i \omega_1 \phi_1 (x) e^{-i \omega_1 t} -2i \omega_2 \phi_2 (x) e^{-i \omega_2 t})$

(since we are taking a derivative w/r/t time)

Given that if we want the expectation value of the Hamiltonian (from interval 0 to L) it would look like this:

$$\langle \hat H \rangle = \int^L_0 \psi^* \hat H \psi dx = \int^L_0 \psi^* \frac{1}{\sqrt{5}} \hbar ( \omega_1 \phi_1 (x) e^{-i \omega_1 t}+2 \omega_2 \phi_2 (x) e^{-i \omega_2 t}) dx$$
$$= \int^L_0 \frac{1}{\sqrt{5}}(\phi_1^*(x) + \phi_2^*(x)) \frac{1}{\sqrt{5}} \hbar ( \omega_1 \phi_1 (x) e^{-i \omega_1 t}+2 \omega_2 \phi_2 (x) e^{-i \omega_2 t}) dx$$

and multiplying this out we end up with

$$\frac{1}{5} \hbar \int^L_0 (\omega_1 \phi_1^*\phi_1 e^{-i\omega_1 t} + 2 \omega_2\phi_1^* \phi_2 e^{-i\omega_2 t}+\omega_1\phi_1 \phi_2^* e^{-i \omega_1 t} + 2\omega_2 \phi_2^* \phi_2 e^{-i\omega_2 t})dx$$

This is where I am curious if a certain thing is "allowed." As I understand it the terms $2 \omega_2\phi_1^* \phi_2 e^{-i\omega_2 t}$ and $\omega_1\phi_1 \phi_2^* e^{-i \omega_1 t}$ will go to zero, (at least when integrated). That should leave

$$\frac{1}{5} \hbar \int^L_0 (\omega_1 \phi_1^*\phi_1 e^{-i\omega_1 t} + 2\omega_2 \phi_2^* \phi_2 e^{-i\omega_2 t})dx$$

and the term $\phi_1^*\phi_1$ should be equal to $|\phi_1|^2$. If that's the case, then I can use the sinusoidal version of the wavefunction, $\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)$ with the n being 1 and 2, and just do the full on integration w/r/t x from there. That done I find that all the sin terms cancel out and with them the exponentials.

Of course I cold be wrong -- if I am spposed to integrate with respect to t though, it seems I get a similar result, though I would have a bunch of constants multiplied by exponentials with $\frac{1}{i \omega}$ terms.

Anyhow I want to know if I am approaching this correctly or if I have really, really lost the plot somewhere. Thanks in advance.

2. Apr 22, 2014

### vanhees71

You forgot the time-dependence of $\psi^*$ in your integral. Otherwise it looks good. The integrals are trivial given that $\phi_1$ and $\phi_2$ are normalized and orthogonal to each other.

3. Apr 22, 2014

### Emspak

I wasn't sure if the complex conjugte would have that, but yes, you're right -- it should be there -- so that means that $\psi^*$ should be $\psi^*= (\phi_1^*(e^{(-i \omega_1 t)} + \phi_2^*(e^{(-i \omega_2 t)}$), correct? And when I multiply that out I would end up with $(\phi_1^*\phi_1(e^{(-2i \omega_1 t)}) + 2\phi_2^*\phi_2(e^{(-2i \omega_2 t)})$ and integrating that is, as you say, not much of a problem (Interestingly, whether i do it with respect to x or t seems to make no difference; in one case I get sine terms that go to zero. In the other I get exponentials that go to zero as t -> infinity and get smaller in any case. I didn't know what that tells me, but it's a funny thing).