# Obtaining an expression for the Lorentz Force in the Dirac theory [duplicate]

We know that $P = p - \frac{e}{c} A$

How can we obtain a expression for the Lorentz force from the equation above using the Dirac Theory??

Could you please explain this to me step by step?

The only idea I have now is: $i\hbar \frac{dP}{dt} = [P,H]$

## marked as duplicate by Ben Crowell, user10851, ManishearthJun 17 '13 at 3:45

• See Landau and Lifshitz, The classical theory of fields, section 17, "Equations of motion of a charge in a field." – Ben Crowell Jun 14 '13 at 0:28
• Thanks, but Landau doesn't use the Dirac theory for the demostration – Surreal Jun 14 '13 at 1:14
• What exactly do you mean by "the Dirac theory"? The canonical momentum is a classical concept, and the Lorentz force can be derived classically. Do you mean the Dirac equation? But the quantum-mechanics tag likely implies you are thinking nonrelativistically. The fact that you write $c$ explicitly stengthens this. So do you mean canonical quantization? – Michael Brown Jun 14 '13 at 3:23
• @Surreal This question has been already answered multiple times on this website. See e.g. physics.stackexchange.com/q/26845/16689 or physics.stackexchange.com/q/45796/16689 for instance. Tell us if the answer given there are sufficient for you or not. They are not step by step solutions, but perhaps you can understand by yourself, and then answering your question yourself. Don't forget then to publish your solution here of course :-), it may help other users. – FraSchelle Jun 14 '13 at 10:18
• voting to close as a duplicate of physics.stackexchange.com/q/26845/4552 – Ben Crowell Jun 14 '13 at 14:48

$$(i \hbar \gamma^\mu \partial_\mu - mc) \psi = 0$$
$$\partial_{\mu} \rightarrow \partial_{\mu}+ieA_{\mu}$$
• I think it's even simpler: replace $H=\gamma_{0}\left(\gamma^{i}P_{i}-m+qA_{0}\right)$ in the Heisenberg equation, and take care of the commutator $\left[\mathbf{P},\mathbf{A}\right]$ which should give the magnetic field (or a quantity proportional to it, depending on the conventions of the magnetic field sign, and $\hbar=1$ you may choose or not). – FraSchelle Jun 14 '13 at 10:22