- #1
Emspak
- 243
- 1
Homework Statement
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 [itex]\psi = \frac{1}{\sqrt{5}}(1\phi_1 + 2\phi_2)[/itex]
Homework Equations
The (time dependent) Hamiltonian: [itex]\hat H \psi = i \hbar \frac{\partial \psi(x,t)}{\partial t}[/itex]
The Attempt at a Solution
For any wavefunction the time dependent form would be [itex]\psi (x) e^{-i\omega t}[/itex]
That means that [itex]\frac{\partial \psi}{\partial t} = -i \omega \phi (x) e^{-i\omega t}[/itex]
So for this particular wavefunction [itex] \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})[/itex]
(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:
[tex] \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[/tex]
[tex] = \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[/tex]
and multiplying this out we end up with
[tex] \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 [/tex]
This is where I am curious if a certain thing is "allowed." As I understand it the terms [itex]2 \omega_2\phi_1^* \phi_2 e^{-i\omega_2 t}[/itex] and [itex]\omega_1\phi_1 \phi_2^* e^{-i \omega_1 t}[/itex] will go to zero, (at least when integrated). That should leave
[tex] \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 [/tex]
and the term [itex]\phi_1^*\phi_1 [/itex] should be equal to [itex]|\phi_1|^2[/itex]. If that's the case, then I can use the sinusoidal version of the wavefunction, [itex]\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)[/itex] 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 [itex]\frac{1}{i \omega}[/itex] 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.