Lorentz force in Dirac theory and its classical limit It is well known that in Dirac theory the time derivative of $$P_i=p_i+A_i$$ operator (where $p_i=∂/∂_i$, $A_i$ - EM field vector potential) is an analogue of the Lorentz force:
$$\frac{dP_i}{dt} = e(E_i+[v×B]_i).$$
On the other hand, in classical theory we have the same equation for $p_i$ instead of $P_i$. How comes that the effect of $A_i$ in Dirac theory vanishes in the classical limit?
 A: I found the question completely clear and sort of important for this piece of learning about magnetism in quantum mechanics; the only problem is that it assumes a wrong answer to the question.
Nothing is vanishing; indeed, it would be very bad if a finite term such as $\vec v\times \vec B$ evaporated without a trace while taking the classical limit. However, one must be careful about the quantum mechanical commutators in order to see that everything is valid.
The full quantum mechanical (non-relativistic) Hamiltonian (which may be obtained as the non-relativistic limit of the Dirac equation) is
$$ H = \frac{1}{2m}(\vec p+ \vec A)^2 + V(\vec x)$$
where $V(\vec x)$ primarily contains the electrostatic potential and generates the electric force $e\vec E = -i\nabla \Phi$ in a way you find uncontroversial. Your problem is localized to the terms containing $\vec A$ or its derivatives. I kept the conventions for the normalization and sign of $\vec A$ to be just like yours although it's a bizarre convention: one would usually add the factor of $q$ or $e$ in front of $\vec A$, too.
In these conventions of yours, $\vec p = -i\hbar \vec \nabla$, as you correctly wrote. However, this object isn't the usual $m\vec v$. Instead, the full $\vec P = \vec p + \vec A$ is equal to $m\vec v$, a multiple of the velocity. This is usually denoted as $\vec p$ in non-quantum physics but we obviously need to map it to $\vec P$: in the limit, once again, the usual non-quantum momentum is $\vec P$, not $\vec p$: $\vec P$ is gauge-covariant, $\vec p$ isn't, and the non-quantum momentum of a particle is clearly gauge-covariant so it can't be $\vec p$.
To quantum mechanically calculate the time-derivative of $P_i$ (the momentum as directly calculated from the velocity), we must compute $1/(-i\hbar)$ times the commutator of $P_i$ with the Hamiltonian (the Heisenberg equations of motion). The commutator of $P_i$ with $V(x)$, the electrostatic potential energy, gives us the usual electric force.
However, we must also add the commutator of $P_i$ with the first term of the Hamiltonian which is $P_j P_j / 2m$. It is not zero because the different components of $P_j$ don't commute with one another. Instead,
$$ \frac{1}{2m}  [P_i,P_j P_j] = \frac{1}{m} [P_i,P_j] P_j + {\rm subleading\,\,in\,\,}\hbar   $$
The commutator of $[P_i,P_j]$ is nonzero because $P_i$ depends both on $\vec p$ as well as $\vec x$: those two lowercase objects are the objects with the usual simple commutation relations. We have
$$ [P_i,P_j] = [p_i,A_j] - [p_j,A_i] $$
in your conventions. You can see that the commutators on the right hand side are nothing else than $-i\hbar$ times the components of ${\rm curl} \vec A = \vec B$, more precisely $\epsilon_{ijk}B_k$. This is multiplied by $P_j/m = v_j$ above, so the total term in the commutator clearly gives $\epsilon_{ijk}v_j B_k = (v\times B)_i$, which is – after $-i\hbar$ is cancelled between the commutator and the factor in the Heisenberg equation (I ignored this factor) – exactly the magnetic force. Again, the usual convention would have the charge $q$ in front of it but I followed your conventions for the normalization of $\vec A$.
However, the correctly calculated classical limit obviously does generate and has to generate the full Lorentz force including the magnetic piece. The overall sign of $\vec A$ and the Lorentz force wasn't tracked very carefully above but believe me that it works (and has to work) as well when the calculation is done perfectly.
Because Murod repeated his or her doubts in the relativistic case, let me rerun the derivation above for the full relativistic Dirac equation. Its Hamiltonian is
$$ H = \gamma_0 (P^i \gamma_i - m + A_0) $$
Note that if you multiply it by $\gamma_0$ and move everything to the same side, you get the simple and uniform operator $P^\mu \gamma_\mu -m$ which must annihilate the Dirac spinorial wave function. $A_0$ is the electric potential – normally one would write $q$ explicitly in front of this term as well.
The commutator $[H,P^i]$ which is what determines the change of the momentum of a relativistic classical (non-quantum) particle even though this energy-momentum vector is sometimes called $p_\mu$ in non-quantum relativistic physics of particles. However, this object comes from the limit (and should be identified with) $P_\mu$. Again, the commutator with $A_0$ produces the electric force. The commutator of $P^i$ with $P^j\gamma_j$ produces the Lorentz force except that $\gamma_j$ is here instead of $v_i$. But that's OK, $\gamma^\mu$ acts on the Dirac spinors just like the velocity 4-vector. Note that $\gamma^\mu$ is formally a vector that squares to one – a unit time-like vector – and it must be the velocity because it's the only gauge-invariant spacetime direction picked by the (quickly oscillating) plane wave. So in the relativistic case, the derivation of the non-quantum equation for the particle is almost identical, just with $v^\mu$ expressed as the matrix $\gamma^\mu$ instead of $\vec P / m$, a part of the reason why the Dirac equation manages to be a first-order equation (for the price of having many components and matrices).
A: The question(v1) seems to be caused by a misunderstanding. Let $\vec{p}^{kin}=\gamma m_0 \vec{v}$ denote the kinetic momentum (also known as the mechanical momentum), and let 
$$\vec{p}^{can}~=~\vec{p}^{kin}+q\vec{A}$$ 
denote the canonical momentum. Quantum mechanically, 
$$\vec{p}^{can}~=~\frac{\hbar}{i}\vec{\partial}.$$
The Lorentz force law
$$ \frac{d\vec{p}^{kin}}{dt}~=~ q(\vec{E} + \vec{v} \times \vec{B}) $$
applies also for the classical case, see e.g., Landau and Lifshitz, Vol.2, The Classical Theory of Fields, Chapter 3.
A: As Qmechanic already pointed out: In order to obtain the kinetic momentum you have to take the derivatives (which give you the canonical momentum) and then subtract the interaction with $A^\mu$. 
So everything is ok and the Dirac equation exactly reproduces the classical result. You can gain a deeper understanding of this if you write the Lorentz force in a more advanced way by using the electromagnetic field tensor.  
$\frac{\partial j^\mu}{\partial \tau} ~~=~~ \frac{q}{mc}\,F^{\mu}_{~\nu}\,j^\nu~$
Which couples the E field with the boost generators K and the B field with the rotation generators J
$     F^{\mu}_{~\nu} ~~=~~  \Big(\,\mathsf{E}^i\,\hat{K}^i + \mathsf{B}^i\,\hat{J}^i\,\Big) \ =\ \left(
\begin{array}{rrrr}
 ~\         0\ \        &  ~~\mathsf{E}_x & ~~\mathsf{E}_y & ~~\mathsf{E}_z \ \\
 ~ \mathsf{E}_x & \          0\ \ & ~~\mathsf{B}_z  & - \mathsf{B}_y \ \\
 ~ \mathsf{E}_y & - \mathsf{B}_z  & \          0\ \ & ~~\mathsf{B}_x \ \\
 ~ \mathsf{E}_z & ~~\mathsf{B}_y  & - \mathsf{B}_x  & \          0\ \ \
\end{array}
    \right)$
For spinors the equivalent interaction generator of time evolution is:
${\cal F}^\mu_{~\nu}\,\varphi ~=~ \left(\,\vec{E}\cdot\hat{\mathbb{K}} + \vec{B}\cdot\hat{\mathbb{J}}\,\right)\varphi$
$\mathbb{K}^i ~=~ -\tfrac12\,\gamma^i\gamma^o, ~~~~~~~~~~ \mathbb{J}^i ~=~ \tfrac{i}{2}\,\gamma^5\gamma^i\gamma^o$
Again the electric field boosts while the magnetic field rotates.
The classical time evolution due to the classical electromagnetic field tensor $F$ operating on the current is exactly the same as when the Spinor field tensor ${\cal F}$ operates on the spinor.
$\exp(F^{\mu}_{~\nu}\,t)\,\bar{\varphi}\,\gamma^\nu\varphi ~~=~~ \overline{\Big(\exp({\cal F}^\mu_{~\nu}\,t)\varphi\Big)}\,\gamma^\mu \, \Big(\exp({\cal F}^\mu_{~\nu}\,t)\varphi\Big)$
If you work out the series expansion of the exponential functions you can find all kind of beauties like.
$\begin{aligned}
&\dot{\bar{\varphi}}\gamma^\mu\dot{\varphi}         &=~~~ &\tfrac12\,T^\mu_{~\nu}~\bar{\varphi}\,\gamma^\nu\varphi         \\
&\dot{\bar{\varphi}}\gamma^5\gamma^\mu\dot{\varphi} &=~~~ &\tfrac12\,T^\mu_{~\nu}~\bar{\varphi}\,\gamma^5\gamma^\nu\varphi \\
&\dot{\bar{\varphi}}~\mathbb{K}^\mu\,\dot{\varphi}  &=~~~ &\tfrac12\,T^\mu_{~\nu}~\bar{\varphi}\,\mathbb{K}^\nu\,\varphi   \\
&\dot{\bar{\varphi}}~\mathbb{J}^\mu\,\dot{\varphi}  &=~~~ &\tfrac12\,T^\mu_{~\nu}~\bar{\varphi}\,\mathbb{J}^\nu\,\varphi   \\
\end{aligned}$
Where T is the symmetric stress energy tensor of the electromagnetic field.
The term ${\cal F}^\mu_{~\nu}\,\varphi$ is just the extra term which occures if you square the Dirac equation with interaction. (although its role there is generally poorly interpreted) The squared Dirac equation contains the second order derivative in time so it should include a term which accounts for the spinor boosts and spinor rotates due to the electromagnetic field. The Klein Gordon equation does not need such a term because it describes a scalar field and scalars are per definition Lorentz invariant.
Regards, Hans
