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The covariant form of the Dirac equation is given by

$$(i\gamma^{\mu}\partial_{\mu} - M) \Psi(x) = 0 $$

Einstein's summation is implied here, $x=(x^0,x^1,x^2,x^3)^T$.

I am simply looking for the Dirac theory equivalent to the classic electromagnetic interaction energy

$$ W = \int_V \rho \phi - \frac{1}{c} \mathbf{j\cdot A} \ dV $$

I have shown the Dirac equation's covariance and current conservation, but I can't seem to make the transition to the interaction energy.

Also, I have made myself familiar with the Klein-Gordon equation

$$(\square + M^2) \Psi(x) = 0$$

In that case, the interaction energy was "brought into" the equation by making the transition $ p^{\mu} \rightarrow p^{\mu} - \frac{e}{c} A^{\mu} $, where $p$ denotes the momentum operator, so the equation transforms to the form

$$(i\hbar\partial_{\mu} - \frac{e}{c}A_{\mu})(i\hbar\partial^{\mu} - \frac{e}{c}A^{\mu})\Psi(x) = M^2 \Psi(x)$$

Could it be that simple for the Dirac case, too?

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    $\begingroup$ That equation of motion that you have is for a free Dirac field. That is, there is no photon (electromagnetic) field for the Dirac field to interact with. You would need to add an interaction term to your Lagrangian. Have you done interacting QFT? I believe adding a Maxwell Lagrangian $$ \mathcal{L}_M = -\frac{1}{4} \mathcal{F}_{\mu \nu} \mathcal{F}^{\mu \nu}$$ and promoting your partial derivative to a gauge covariant derivative $$ \partial_{\mu} \rightarrow \mathcal{D}_{\mu} = \partial_{\mu} - i e A_{\mu}(x) $$ should give you what you want. $\endgroup$
    – Flint72
    Commented May 12, 2014 at 23:31
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    $\begingroup$ @Flint72 that should be an answer $\endgroup$
    – David Z
    Commented May 13, 2014 at 0:56
  • $\begingroup$ @DavidZ : Ok, I just wrote it out again as an answer. Sorry, I did not think that it was detailed enough to be an answer before, which is why I left it only as a comment. $\endgroup$
    – Flint72
    Commented May 13, 2014 at 1:21
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    $\begingroup$ Nah, that's quite detailed enough. The "rule" is that anything that answers the question should be posted as an answer, regardless of the level of detail. (Note that this does not include merely linking to another page that answers the question.) $\endgroup$
    – David Z
    Commented May 13, 2014 at 1:25

2 Answers 2

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The probability current for the Dirac equation is $$j_p^\mu = \bar\psi \gamma^\mu \psi$$ and therefore the electromagnetic current is $$j^\mu = -e\bar\psi\gamma^\mu\psi$$ since the charge of the Dirac field is $-e$. So to the Lagrangian we just add the usual term coupling current to vector potential $$\mathcal L = \bar\psi(i\gamma^\mu\partial_\mu - m)\psi - e\bar\psi\gamma^\mu\psi A_\mu.$$ Now the Euler-Lagrange equation for $\bar\psi$ can be read off as $$(i\gamma^\mu \partial_\mu - m -e \gamma^\mu A_\mu)\psi = 0.$$

(Different conventions will change some signs here but that's not really important.)

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  • $\begingroup$ Maybe a trivial question, but why is the electromagnetic current $-e j^{\mu}_p$? $\endgroup$
    – user46208
    Commented May 13, 2014 at 11:15
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    $\begingroup$ $j^\mu_p$ is the particle flux, each particle carries a charge $-e$. $\endgroup$ Commented May 13, 2014 at 14:08
  • $\begingroup$ So if I conclude correctly, the interaction energy then is simply $j^{\mu}A_{\mu} = -e \bar{\Psi}\gamma^{\mu}\Psi A_{\mu}$. Which is similar to the classical case with $\rho \Phi - \frac{1}{c} \mathbf{j}\mathbf{A}$ $\endgroup$
    – user46208
    Commented May 14, 2014 at 18:00
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    $\begingroup$ It's exactly the same as the classical case, it's just that the expression for the current is different. $\endgroup$ Commented May 14, 2014 at 18:18
  • $\begingroup$ I have now also read about bilinear covariants. As you probably know, they behave like a scalar/pseudoscalar/vector/pseudovector/rank 2 tensor under Lorentz transformation. Would it, in theory, be possible to expand the interaction energy in these quantities, similar to a multipole expansion in classical electrodynamics. I know, this should probably be a separate question, but I am new to this platform. Sorry for that. $\endgroup$
    – user46208
    Commented May 18, 2014 at 12:49
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The equation of motion that you have is for a free Dirac field. That is, there is no photon (electromagnetic) field for the Dirac field to interact with. You would need to add an interaction term to your Lagrangian.

Have you done interacting QFT?

I believe adding a Maxwell Lagrangian $$ \mathcal{L}_M = −\frac{1}{4} \mathcal{F}_{\mu \nu} \mathcal{F}^{\mu \nu} $$

and promoting your partial derivative to a gauge covariant derivative

$$ \partial_{\mu} \rightarrow \mathcal{D}_{\mu} = \partial_{\mu} − i e A_{\mu}(x) $$

should give you what you want.

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  • $\begingroup$ No, I have just had relativistic quantum theory, no QFT until now. Why do I need to perform these steps? $\endgroup$
    – user46208
    Commented May 13, 2014 at 11:16
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    $\begingroup$ The mature way to view QED and the standard model is as something called a gauge theory, where interactions are generated by demanding that the Lagrangian should be invariant under $\psi \mapsto e^{i\Lambda(x)}\psi, A_\mu \mapsto A_\mu + i\partial_\mu \Lambda(x).$ If you start with the free Dirac field and add the free Maxwell Lagrangian, you will have to change the derivative operator as in the answer above, and the QED Lagrangian will be generated. This is elegant but it is not really motivated in any way if you don't already have the QED Lagrangian from physical arguments. $\endgroup$ Commented May 14, 2014 at 13:28

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