Work done on charged particle by magnetic field in quantum mechanics

Classically, we know from $\mathbf{F}=q\mathbf{v}\times \mathbf{B}$ that magnetic field does no work on a charged particle. In quantum mechanics, the Hamiltonian of a charged particle in a magnetic field is given by

$$\hat{H}~=~\frac{1}{2m} \left[\frac{\hbar}{i}\vec\nabla - \frac{q}{c}\vec A\right]^2.$$

How can we deduce from this Hamiltonian whether work done is on the particle?

First, if the Hamiltonian is time-independent, and your Hamiltonian is assuming that $\vec A$ doesn't depend on time, then energy – the Hamiltonian itself – is conserved.
By the Heisenberg equations of motion (or, by an equivalent classical procedure of relating $L$ and $H$ and the canonical velocities with canonical momenta), the speed may be determined from the commutator of $H$ with $x$ because $$i\hbar \frac{d}{dt} \vec x = [\vec x,H]$$ is the Heisenberg equation of motion for $x$. Hats are everywhere. Taking your Hamiltonian, the only building block that refuses to commute with $\vec x$ is $\vec \nabla$. If you apply the Leibniz rule, you will easily see that $$i\hbar \frac{d}{dt} \vec x = [\vec x,H] = \frac{1}{m}\left[\frac{\hbar}i \vec \nabla - \frac{q}{c}\vec A(\vec x)\right]$$ so the speed – the rate of change of the position – is given by the operator that has both the nabla as well as the vector potential, in the very same combination you wrote. It follows that the kinetic energy is $$E = \frac{mv^2}{2} = H$$ i.e. it is exactly equal to your Hamiltonian. The whole Hamiltonian you wrote is the operator of kinetic energy. It commutes with itself so the kinetic energy is conserved (hint: write the Heisenberg equation of motion for the kinetic energy itself) which means that the magnetic field does no work on the charged particle.
• Just to expand on this for the OP: when one deals with minimally coupled gauge theories such as here, one needs to distinguish between canonical and mechanical momentum. As a matter of convention, $p$ is the canonical momentum, meaning that $[x,p]=i\hbar$. This quantity is not the momentum that you know and intuitively understand. That momentum is given in this case by $p-qA/c$. May 25, 2012 at 8:09
• Right, i should have written this thing explicitly, thanks for your addition. The kinetic (or, as you say, "mechanical") momentum is $m\vec v = \vec p - q\vec A/c$ and it's the very thing that is squared to get the OP's Hamiltonian. Well, I have written this thing already; just without the focus on the contrast between the two momenta. Let me add that the canonical momentum $\vec p$ is always $(\hbar/i)\vec\nabla$, which follows from the commutator. May 25, 2012 at 8:18
• Dear @Vijay, as Genneth said, the Hamiltonians are inequivalent. The Hamiltonian in the presence of the magnetic field is still $mv^2/2$ but these are different $v$ operators. In particular, different components of $\vec v$ don't commute with each other in the presence of the magnetic field. So the plane waves are no longer energy eigenstates (as for a free particle) and you have to solve another problem to find $H$ eigenstates that turns out to include a harmonic oscillator and you get Landau levels. You've linked to a very page that explains how. May 25, 2012 at 13:48