# 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?

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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 $p-qA/c$. – genneth May 25 '12 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. – Luboš Motl May 25 '12 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. – Luboš Motl May 25 '12 at 13:48