- #1

Ben Niehoff

Science Advisor

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I have some confusion as to how to get from the QED Lagrangian to ordinary electrodynamics. Beginning with the QED Lagrangian (density):

[tex]\mathcal L = \bar{\psi} (i\gamma^{\mu} D_{\mu} - m) \psi - \frac14 F_{\mu\nu} F^{\mu\nu}[/tex]

I expand the covariant derivative [itex]D = \partial - iqA[/tex] to obtain

[tex]\mathcal L = \bar{\psi} i\gamma^{\mu} \partial_{\mu} \psi + q \bar{\psi} \gamma^{\mu} \psi A_{\mu} - m \bar{\psi} \psi - \frac14 F_{\mu\nu} F^{\mu\nu}[/tex]

From here, one can either add a total divergence, or take the sum

[tex]\frac12 (\mathcal L + \mathcal L^{\dagger}) = \frac12 \bar{\psi} i \gamma^{\mu} \overleftrightarrow{\partial_{\mu}} \psi + q \bar{\psi} \gamma^{\mu} \psi A_{\mu} - m \bar{\psi} \psi - \frac14 F_{\mu\nu} F^{\mu\nu}[/tex]

which I kinda like because it is more symmetrical. Anyway, using gauge invariance and Noether's theorem, we can get the conserved current

[tex]j^{\mu} = q \bar{\psi} \gamma^{\mu} \psi[/tex]

This couples to the EM field in the usual way,

[tex]\mathcal L = j^{\mu}A_{\mu} - \frac14 F_{\mu\nu} F^{\mu\nu}[/tex]

But now the question is, how do I include the kinetic part of the Lagrangian? I would like to end up with something like

[tex]\mathcal L = m \sqrt{\dot x^{\mu} \dot x_{\mu}} + q \dot x^{\mu} A_{\mu} - \frac14 F_{\mu\nu} F^{\mu\nu}[/tex]

but I can't seem to do anything with my current [itex]j^{\mu}[/itex] that resembles the kinetic part of the QED Lagrangian. The 4-divergence identically vanishes (by Noether's theorem), so the only scalar I can form is

[tex]j^{\mu} j_{\mu} = q^2 (\bar{\psi} \gamma^{\mu} \psi) (\bar{\psi} \gamma_{\mu} \psi)[/tex]

which I can't see how to simplify. Maybe I am missing something. Maybe what I am trying to do can't actually be done? Basically, I'd like to rewrite the QED Lagrangian in terms of [itex]j^{\mu}[/itex], and (hopefully) recover the classical Lagrangian. Any suggestions?

Edit:

Ah, I see one thing I can try. I might be able to make something of it by exploiting

[tex]i\partial_{\mu} = (i \partial_t, i \vec \nabla) = (E, - \vec p) = p_{\mu}[/tex]

I'll look into this later.

[tex]\mathcal L = \bar{\psi} (i\gamma^{\mu} D_{\mu} - m) \psi - \frac14 F_{\mu\nu} F^{\mu\nu}[/tex]

I expand the covariant derivative [itex]D = \partial - iqA[/tex] to obtain

[tex]\mathcal L = \bar{\psi} i\gamma^{\mu} \partial_{\mu} \psi + q \bar{\psi} \gamma^{\mu} \psi A_{\mu} - m \bar{\psi} \psi - \frac14 F_{\mu\nu} F^{\mu\nu}[/tex]

From here, one can either add a total divergence, or take the sum

[tex]\frac12 (\mathcal L + \mathcal L^{\dagger}) = \frac12 \bar{\psi} i \gamma^{\mu} \overleftrightarrow{\partial_{\mu}} \psi + q \bar{\psi} \gamma^{\mu} \psi A_{\mu} - m \bar{\psi} \psi - \frac14 F_{\mu\nu} F^{\mu\nu}[/tex]

which I kinda like because it is more symmetrical. Anyway, using gauge invariance and Noether's theorem, we can get the conserved current

[tex]j^{\mu} = q \bar{\psi} \gamma^{\mu} \psi[/tex]

This couples to the EM field in the usual way,

[tex]\mathcal L = j^{\mu}A_{\mu} - \frac14 F_{\mu\nu} F^{\mu\nu}[/tex]

But now the question is, how do I include the kinetic part of the Lagrangian? I would like to end up with something like

[tex]\mathcal L = m \sqrt{\dot x^{\mu} \dot x_{\mu}} + q \dot x^{\mu} A_{\mu} - \frac14 F_{\mu\nu} F^{\mu\nu}[/tex]

but I can't seem to do anything with my current [itex]j^{\mu}[/itex] that resembles the kinetic part of the QED Lagrangian. The 4-divergence identically vanishes (by Noether's theorem), so the only scalar I can form is

[tex]j^{\mu} j_{\mu} = q^2 (\bar{\psi} \gamma^{\mu} \psi) (\bar{\psi} \gamma_{\mu} \psi)[/tex]

which I can't see how to simplify. Maybe I am missing something. Maybe what I am trying to do can't actually be done? Basically, I'd like to rewrite the QED Lagrangian in terms of [itex]j^{\mu}[/itex], and (hopefully) recover the classical Lagrangian. Any suggestions?

Edit:

Ah, I see one thing I can try. I might be able to make something of it by exploiting

[tex]i\partial_{\mu} = (i \partial_t, i \vec \nabla) = (E, - \vec p) = p_{\mu}[/tex]

I'll look into this later.

Last edited: