I know the electron-photon interaction Lagrangian from QED: ($-ie\bar{\Psi}\gamma^{\mu}A_{\mu}\Psi$). However, this interaction seems to describe the interaction of electron and positron with a photon (thus there is a $\bar{\Psi}$ for positron). This makes sense when an electron radiates a virtual photon (and becomes a positron) and then reabsorbs the photon again (and becomes an electron again). What about the case when an electron emits a real photon (and never becomes a positron) like the case of synchrotron radiation? What is the Lagrangian in that case since we wouldn't have any electron-positron vertex?

  • $\begingroup$ I'm not well versed in QFT, but in your example that "makes sense", are you implying that the virtual photon is charged? Otherwise how is charge conserved when the electron becomes a positron? $\endgroup$ – probably_someone Jul 28 '17 at 0:30
  • $\begingroup$ The term conserves charge. It describes an electron emitting a photon, or a positron emitting a photon, or an electron absorbing a photon, ... etc.., or a photon producing an electron-positron pair, or an electron and positron annihilating to an electron, ... Have you studied the Dirac equation? $\endgroup$ – Cosmas Zachos Jul 28 '17 at 0:48
  • $\begingroup$ So, let me phrase it another way: if $\Psi$ represents electron, wouldn't $\bar{\Psi}$ represent positron? What if I just want to show an electron-electron-photon vertex (like an electron emitting a photon)? $\endgroup$ – physics_2015 Jul 28 '17 at 5:14
  • 1
    $\begingroup$ $\bar\psi$ refers to the technical way of forming a scalar in the Lagrangian that is invariant to Lorentz transformations together with $\psi$; its definition involves a complex-conjugation but like the answer below states, you should not really think of $\bar\psi\psi$ as a specification for particles-in/antiparticles-out or so, but rather as a technical way of comparing two fieldvalues at a point in an invariant way. The particle/antiparticleness is contained within $\psi$'s composition itself. $\endgroup$ – BjornW Jul 29 '17 at 9:49

First, the examples you state that "make sense" are completely phenomenologically wrong. An electron that radiates a photon can't ever become a positron because the photon isn't charged and the interaction conserves charge (and similarly for the positron).

Now to get to the point of your question, the Lagrangian doesn't change at all. The important thing to note is that $\Psi$ (and thus $\bar{\Psi}$) contains two distinct pieces. One corresponds to the electron, and the other corresponds to the positron. This is what lies behind the solutions to the Dirac equation: that $\Psi$ contains four components, two of which are for the electron and two for the positron. This means the vertex $\bar{\Psi}\gamma^\mu A_\mu\Psi$, isn't for an electron-photon-positron vertex, it's for all of the possible combinations of electron or positron and photon.

Any reference on the Dirac equation should help out if you have more confusion.


To be more precise and clear about how the Dirac field contains both matter and antimatter parts to it, consider the Dirac equation:

$$(i\gamma^\mu\partial_\mu - m)\Psi = 0.$$

This can be solved through the Fourier expansion of $\Psi$:

$$\Psi(x) = \int \frac{d^3p}{(2\pi)^3} \ a(p)u(p)e^{ipx} + b^\dagger(p)v(p)e^{-ipx}.$$

Where $a$ and $b^\dagger$ are annihilation and creation operators after canonical quantization, respectively, and $u$ and $v$ are four component spinors. Now the important thing to note here is that in general this spinor field is a complex field, so the two terms I've summed here need not be related, and so the sets of creation/annihilation operators correspond to different kinds of excitations.

This is what I mean by saying the Dirac field contains both the positron and the electron: since the sets of creation/annihilation operators correspond to different particles coming from the same field. Now due to the sign difference in the exponentials, one of the particles has positive energy and one has negative energy. In this case$-$metric signature $(-+++)$$-$the creation operator $b^\dagger$ is associated with the negative energy solution, and so in fact, the state $\Psi(x)|0\rangle$ is the state with one positron.

The upshot is that because you have one particle and one antiparticle term each in $\bar{\Psi}$ and $\Psi$, when you expand the interaction it looks more like

$$\gamma^\mu A_\mu(a^\dagger a + a^\dagger b^\dagger + b a + b b^\dagger),$$

which corresponds to interactions involving all combinations of electrons and positrons.

| cite | improve this answer | |
  • $\begingroup$ OK. Thanks. Unfortunately, I'm still not clear. Let's consider the Dirac spinor in the case of massless electron in the Weyl basis. The Dirac spinor $\Psi$ will have two components: $\kappa$ and $\eta$ where $\kappa$ represents the left handed electron and $\eta$ the right handed electron. The Lagrangian would then yield: $\kappa^{\dagger}\sigma^{\mu}a_{\mu}\kappa$ + $\eta^{\dagger}\sigma^{\mu}a_{\mu}\eta$ So, are you saying that $\kappa^{\dagger}$ is both the left handed electron and right handed positron at the same time? $\endgroup$ – physics_2015 Jul 28 '17 at 18:33
  • $\begingroup$ I have edited my answer, which hopefully will clear up your confusions. $\endgroup$ – Jared Dziurgot Jul 29 '17 at 6:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.