A more discussion of current-velocity relationship in quantum mechanics When considering the simple Hamiltonian $$\hat{H}=\left(\hat{p}-e\vec{A}\right)^{2}+e\phi,$$ if we try to calculate the current, then the two treatments, both seen in textbooks,
$\hat{\jmath}=e\hat{v}=e\frac{\partial \hat{H}}{\partial \hat{p}}$ and $\hat{\jmath}=-\frac{\partial \hat{H}}{\partial \vec{A}}$, would give the same results.
However I am unclear whether there exist more complicated Hamiltonians where these two treatments may not be consistent.
I think the key problem lies here: Whether we should view $\hat{H}$ as $\hat{H}\left(\hat{r},\hat{p}\right)$, or $\hat{H}\left(\hat{r},\hat{p},\vec{A}\right)$. The former would make $\hat{\jmath}=e\hat{v}$ as a more basic definition, and the latter would make $\hat{\jmath}=-\frac{\partial \hat{H}}{\partial \vec{A}}$ more basic.
Personally I prefer $\hat{H}\left(\hat{r},\hat{p}\right)$ to $\hat{H}\left(\hat{r},\hat{p},\hat{A}\right)$, because if $\hat{A}$ is seen as a variable of the Hamiltion, so should be  an operator $\hat{C}$ satisfying $[\hat{A},\hat{C}]=i$ be included. However such an operator is missing.
Yet I am not sure arguments above is correct. Could anyone help to explain?
 A: Whenever the electromagnetic coupling is minimal, then the two definitions are equivalent.  Minimal coupling means that the canonical momentum $\vec{p}$ and the vector potential $\vec{A}$ only enter in the combination $\vec{p}-q\vec{A}$.  (I am adopting natural units in which $\hbar=c=1$.) For a structureless charged particle, the electromagnetic coupling always takes this form.
This applies both nonrelativistically and relativistically. For a pure Dirac particle (and electron is a pure Dirac particle to a very good approximation, although there are small corrections—most notably the anomalous magnetic moment—that arise from the virtual interactions of the electron with its own electromagnetic field), the Dirac equation
$$\left[\gamma^{\mu}\left(i\partial_{\mu}-qA_{\mu}\right)-m\right]\psi=0,$$
where the $\gamma^{\mu}$ are the Dirac matrices satisfying $\{\gamma^{\mu},\gamma^{\nu}\}=2\,I_{4\times 4}\,\eta^{\mu\nu}$, can be written in the form of a Schrödinger equation $i\partial_{0}\psi=H\psi$ with a Hamiltonian
$$H=\gamma^{0}\left[\gamma^{k}\left(i\partial_{k}+qA_{j}\right)+m\right]+eA_{0},$$
with the index $k$ running over the spatial components, $k=1,2,3$. The velocity operator is always given by the classical Hamilton equation $\vec{v}=\partial\vec{x}/\partial t=\partial H/\partial\vec{p}$ or the quantum-mechanical Heisenberg equation of motion, $\partial\vec{x}/\partial t=i[H,\vec{x}]$. (That the $i[H,\vec{x}]=\partial H/\partial\vec{p}$ equivalence follows from the fact that $[x_{k},p_{l}]=i\delta_{kl}$ is a standard quantum mechanics homework problem.) Since the Hamiltonian only contains $\vec{p}$ and $\vec{A}$ in the combination $\vec{p}-q\vec{A}$, it automatically follows that $\partial H/\partial\vec{A}=-q\,\partial H/\partial\vec{p}$, which means that $\vec{v}=q\,\partial H/\partial\vec{p}$ is what the electromagnetic field couples to whenever there is minimal coupling.
This does, however, gloss over a significant complication that occurs in the Dirac theory. For the nonrelativistic Schrödinger equation, with
$$H=\frac{1}{2m}\left(\vec{p}-q\vec{A}\right)^{2}+qA_{0},$$
the velocity takes the form $\vec{v}=\vec{p}/m$, which is the same as in the classical theory. However, the velocity derived from the Dirac Hamiltonian is
$$\dot{\vec{x}}=i[H,\vec{x}]=-\gamma^{0}\vec{\gamma},$$
which is a momentum-independent matrix operator in the spinor space. While its time-averaged expectation value can be related to the momentum in the usual way, the
instantaneous expression for $\dot{\vec{x}}$ has some very peculiar properties. Because of its matrix structure, the only eigenvalues of each $\dot{x}_{k}$ are $\pm 1$, so if it were possible to measure the instantaneous value of velocity of a Dirac particle in a particular direction, it would always be found to be moving at the speed of light.  However, the components of the velocity do not commute with one another, nor do they commute with the Hamiltonian. The components of the velocity thus cannot be fixed simultaneously, nor are they constant in time; instead, on top of the mean (classical), there is an additional high-frequency Zitterbewegung (quivering motion). In terms of wave packets, this corresponds to the fact that a tightly localized Dirac particle wave function must include both positive- and negative-energy plane wave components, and the Zitterbewegung arises from interference between the positive and negative energies.  (An excellent treatment of these issues can be found in Sakurai's Advanced Quantum Mechanics.)  With all these issues, interpreting $\partial H/\partial\vec{p}$ as the particle velocity in the Dirac theory is somewhat fraught.  However, for the purposes of this question, these complications are actually advantageous, since taking $\vec{\jmath}=q\,\partial H/\partial\vec{p}=-\partial H/\partial\vec{A}$ allows $\vec{\jmath}$ to capture both the current associated with the motion of the charged particle and its spin current.
On the other hand, it is easy to see that in the presence of a more general coupling—the most important example being an anomalous magnetic moment—it cannot be the case that the current is simply $\vec{\jmath}=e\vec{v}$.  At low energies, the electromagnetic coupling of a neutron is predominantly through its magnetic moment. The charge of the neutron is $q_{n}=0$, but this clearly does not preclude the neutron from having an electromagnetic interaction, nor is the interaction vanishing for a stationary neutron. The Pauli Hamiltonian for the neutron (using $\vec{B}=\vec{\nabla}\times\vec{A}$),
$$H=\frac{\vec{p}^{2}}{2m_{n}}-\mu_{n}\vec{\sigma}\cdot\left(\vec{\nabla}\times\vec{A}\right),$$
is not minimally coupled, and so the two notions of current are not equivalent.  The convective current carried by a neutron in motion is identically zero, but there is a spin current, because there is a coupling to $\vec{A}$.  The situation does not
change appreciably if we instead consider the single-particle Dirac Hamiltonian for the neutron,
$$H=\gamma^{0}\left(i\gamma^{k}\partial_{k}+m_{n}+\frac{1}{2}\mu_{n}\sigma^{\mu\nu}F_{\mu\nu}\right),$$
where $\sigma^{\mu\nu}=\frac{i}{2}[\gamma^{\mu},\gamma^{\nu}]$ is the spin tensor and $F^{\mu\nu}=\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}$ is the electromagnetic field strength tensor. The coupling to the four-vector potential $A^{\mu}$ is still anomalous in the same way; the electromagnetic field does not couple to $q\,\partial H/\partial\vec{p}$.
