Quantum EM field hamiltonian

1. Oct 29, 2012

frogjg2003

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

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

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

2. Relevant equations

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

3. The attempt at a solution

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

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
$$\hat{H}\psi = -\frac{\hbar^2}{2m}\nabla^2\psi + \frac{\mathit{i}e\hbar}{2mc} \left(\nabla\cdot(\psi\nabla\lambda) + (\nabla\lambda)\cdot(\nabla\psi)\right) + \frac{e}{2mc}\psi\nabla^2\lambda + \frac{e}{c} \psi\frac{\partial\lambda}{\partial t}$$
or
$$\hat{H}\psi = -\frac{\hbar^2}{2m}\nabla^2\psi + \frac{\mathit{i}e\hbar}{2mc} \left(\nabla^2(\psi\lambda) + (\nabla\lambda\cdot\nabla)\psi\right) + \frac{e}{2mc}\nabla^2(\lambda\psi) + \frac{e}{c} \frac{\partial(\lambda\psi)}{\partial t}$$

Last edited: Oct 29, 2012
2. Oct 29, 2012

TSny

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
$$\hat{H} = \frac{(\mathit{-i}\hbar\mathbf{\nabla} + \frac{e}{c}\mathbf{\nabla}\lambda(\mathbf{r},t))({-i}\hbar\mathbf{\nabla} + \frac{e}{c}\mathbf{\nabla}\lambda(\mathbf{r},t))}{2m} + \frac{e}{c} \frac{\partial\lambda(\mathbf{r},t)}{\partial t}$$

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

3. Oct 29, 2012

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 Schrodinger equation. From there, I just applied the Hamiltonian and started simplifying. I got the original free particle Shodinger equation right back.

4. Oct 29, 2012

Good deal.