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A Time Dependent Hamiltonian problem

  1. Apr 22, 2014 #1
    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 [itex]\psi = \frac{1}{\sqrt{5}}(1\phi_1 + 2\phi_2)[/itex]

    2. Relevant equations

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

    3. 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.
     
  2. jcsd
  3. Apr 22, 2014 #2

    vanhees71

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    You forgot the time-dependence of [itex]\psi^*[/itex] in your integral. Otherwise it looks good. The integrals are trivial given that [itex]\phi_1[/itex] and [itex]\phi_2[/itex] are normalized and orthogonal to each other.
     
  4. Apr 22, 2014 #3
    I wasn't sure if the complex conjugte would have that, but yes, you're right -- it should be there -- so that means that [itex]\psi^*[/itex] should be [itex]\psi^*= (\phi_1^*(e^{(-i \omega_1 t)} + \phi_2^*(e^{(-i \omega_2 t)}[/itex]), correct? And when I multiply that out I would end up with [itex](\phi_1^*\phi_1(e^{(-2i \omega_1 t)}) + 2\phi_2^*\phi_2(e^{(-2i \omega_2 t)})[/itex] 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).
     
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