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Quantum EM field hamiltonian

  1. Oct 29, 2012 #1
    1. The problem statement, all variables and given/known data

    Consider a charged particle of charge e traveling in the electromagnetic
    potentials
    [tex]
    \mathbf{A}(\mathbf{r},t) = -\mathbf{\nabla}\lambda(\mathbf{r},t)\\
    \phi(\mathbf{r},t) = \frac{1}{c} \frac{\partial \lambda(\mathbf{r},t)}{\partial t}
    [/tex]
    where [itex]\lambda(\mathbf{r},t)[/itex] is an arbitrary scalar function.

    Show that the wavefunction is
    [tex]
    \psi(\mathbf{r},t) = exp\left(-\frac{\mathit{i}e}{\hbar c} \lambda(\mathbf{r},t)\right) \psi^{(0)}(\mathbf{r},t)
    [/tex]
    where [itex]\phi^{(0)}(\mathbf{r},t)[/itex] is the solution to the Schrodinger equation for the case [itex]\lambda(\mathbf{r},t)=0[/itex].
    of ¸(r, t) = 0.

    2. Relevant equations

    [tex]\hat{H} = \frac{(\mathbf{p}-\frac{q}{c}\mathbf{A}(\mathbf{r},t))^2}{2m} + q\phi(\mathbf{r},t)[/tex]

    3. The attempt at a solution

    The Hamiltonian is
    [tex]
    \hat{H} = \frac{(\mathit{i}\hbar\mathbf{\nabla} + \frac{e}{c}\mathbf{\nabla}\lambda(\mathbf{r},t))^2}{2m} + \frac{e}{c} \frac{\partial\lambda(\mathbf{r},t)}{\partial t}
    [/tex]
    and the [itex]\lambda=0[/itex] wavefunction is
    [tex]
    \psi^{(0)} = exp(\mathit{i} \left(\mathbf{k}\cdot\mathbf{r} - \frac{\hbar k^2 t}{2m}\right)
    [/tex]

    I'm stuck at applying the Hamiltonian to the function. For the [itex]\nabla^2\lambda\psi[/itex] term, do we apply the Laplacian to just [itex]\lambda[/itex] or to [itex]\lambda\psi[/itex]? Similarly for the cross term and the time derivative.

    Is it
    [tex]
    \hat{H}\psi = -\frac{\hbar^2}{2m}\nabla^2\psi + \frac{\mathit{i}e\hbar}{2mc} \left(\nabla\cdot(\psi\nabla\lambda) + (\nabla\lambda)\cdot(\nabla\psi)\right) + \frac{e}{2mc}\psi\nabla^2\lambda + \frac{e}{c} \psi\frac{\partial\lambda}{\partial t}
    [/tex]
    or
    [tex]
    \hat{H}\psi = -\frac{\hbar^2}{2m}\nabla^2\psi + \frac{\mathit{i}e\hbar}{2mc} \left(\nabla^2(\psi\lambda) + (\nabla\lambda\cdot\nabla)\psi\right) + \frac{e}{2mc}\nabla^2(\lambda\psi) + \frac{e}{c} \frac{\partial(\lambda\psi)}{\partial t}
    [/tex]
     
    Last edited: Oct 29, 2012
  2. jcsd
  3. Oct 29, 2012 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Note typo: In the first term of the Hamiltonian, there should be a negative sign before the ##\mathit{i}\hbar##
    It's the first form that's correct (with a negative sign for the second term). However, it's much easier to apply the hamiltonian in the form
    [tex]
    \hat{H} = \frac{(\mathit{-i}\hbar\mathbf{\nabla} + \frac{e}{c}\mathbf{\nabla}\lambda(\mathbf{r},t))({-i}\hbar\mathbf{\nabla} + \frac{e}{c}\mathbf{\nabla}\lambda(\mathbf{r},t))}{2m} + \frac{e}{c} \frac{\partial\lambda(\mathbf{r},t)}{\partial t}
    [/tex]

    where you first apply the right hand factor ##(\mathit{-i}\hbar\mathbf{\nabla} + \frac{e}{c}\mathbf{\nabla}\lambda(\mathbf{r},t))## to the wave function and simplify before applying the second factor of ##(\mathit{-i}\hbar\mathbf{\nabla} + \frac{e}{c}\mathbf{\nabla}\lambda(\mathbf{r},t))##.
     
  4. Oct 29, 2012 #3
    Thanks.
    I was doing a bit too much work. I didn't need to know the form of [itex]\psi^{(0)}[/itex], just that it solved the free particle Schrodinger equation. From there, I just applied the Hamiltonian and started simplifying. I got the original free particle Shodinger equation right back.
     
  5. Oct 29, 2012 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Good deal.
     
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