How to Apply the Hamiltonian to a Wavefunction in Quantum Electromagnetism?

[frogjg2003]

Homework Statement

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

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

Homework Equations

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

The Attempt at a Solution

The Hamiltonian is
<br /> \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}<br />
and the \lambda=0 wavefunction is
<br /> \psi^{(0)} = exp(\mathit{i} \left(\mathbf{k}\cdot\mathbf{r} - \frac{\hbar k^2 t}{2m}\right)<br />

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

Is it
<br /> \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}<br />
or
<br /> \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}<br />

 

[TSny]

The Hamiltonian is
<br /> \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}<br />
and the \lambda=0 wavefunction is
<br /> \psi^{(0)} = exp(\mathit{i} \left(\mathbf{k}\cdot\mathbf{r} - \frac{\hbar k^2 t}{2m}\right)<br />

Note typo: In the first term of the Hamiltonian, there should be a negative sign before the $\mathit{i}\hbar$

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

Is it
<br /> \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}<br />
or
<br /> \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}<br />

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
<br /> \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}<br />

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 $(\mathit{-i}\hbar\mathbf{\nabla} + \frac{e}{c}\mathbf{\nabla}\lambda(\mathbf{r},t))$.

 

[frogjg2003]

Thanks.
I was doing a bit too much work. I didn't need to know the form of \psi^{(0)}, just that it solved the free particle Schrödinger equation. From there, I just applied the Hamiltonian and started simplifying. I got the original free particle Shodinger equation right back.

 

[TSny]

Good deal.