In A. Zee's "QFT in a nutshell", there is such a statement about massive spin-1 particles (Chap I.5).

The three polarization vectors $\xi^{(a)}_\mu(k)$ are simply the three unit vectors pointing along the $x$, $y$, and $z$ axes, respectively $(a = 1, 2, 3)$: $\xi^{(1)}_\mu = (0, 1, 0, 0)$, $\xi^{(2)}_\mu = (0, 0, 1, 0)$, $\xi^{(3)}_\mu = (0, 0, 0, 1)$. In the rest frame $k_μ = (m, 0, 0, 0)$.

The amplitude for a particle with momentum $k$ and polarization $a$ to be created at the source is proportional to $\xi^{(a)}_\lambda(k)$, and the amplitude for it to be absorbed at the sink is proportional to $\xi^{(a)}_\nu(k)$.

Why is the amplitude proportional to polarization vectors? Is there an intuitive explanation? Thanks!


1 Answer 1


A short answer is that, to calculate process amplitudes, you have to take in account the interacting Lagrangian part, and use it to establish the rules to calculate Feynman diagrams. So, let's have just a taste of it.

For QED, The interacting Lagrangian density term $L_{int}$ corresponds to expressions:

$j^\mu(x) A_\mu(x) \sim \bar \psi(x) \gamma^\mu \psi(x) ~ A_\mu(x)$

Now consider for simplicity the process described in Zee, Chapter $II.6$ Electron Scattering and Gauge Invariance, where $2$ initial electrons $p_1$ and $p_2$, 2 final electrons $P_1$ and $P_2$, and one "virtual" photon $k$.

We may want to work in momentum space, so an example of the interacting term is in fact :

$\bar u(P_1) \gamma^\mu u(p_1) ~~\epsilon_\mu(k) $, with $P_1=p_1+k$

and this corresponds to one vertex.

Now, you have in fact several possible polarizations $\epsilon_\mu^{(a)}(k)$, and secondly, there is a "propagator term" $\sim \frac{1}{k^2}$, because the photon is on a internal line (it is a "virtual" photon). :

The total amplitude (fig $II.6.1.a$), with $2$ vertex and one "propagator term" may be written :

$A(P_1,P_2) \sim \sum\limits_a \bar u(P_1) \gamma^\mu u(p_1)~\epsilon_\mu^{(a)}(k) ~~\frac{1}{k^2}~~ \bar u(P_2) \gamma^\nu u(p_2)~~\epsilon_\nu^{(a)}(k) \\ \sim u(P_1) \gamma^\mu u(p_1)~~D_{\mu\nu}(k)~~\bar u(P_2) \gamma^\nu u(p_2)$

So the exact form of the photon propagator is simply $D_{\mu\nu}(k) =\sum\limits_a \epsilon_\mu^{(a)}(k) \epsilon_\nu^{(a)}(k) ~~ \frac{1}{k^2}$.

So, this is an example, where the polarizations $\epsilon_\mu^{(a)}(k)$ are participating to the calculus of the amplitude (through the exact expression of the photon propagator). Of course, it is an example where the photon is on an internal line, a "virtual" photon.

If you have a real photon, living on an external line, so a ingoing or outcoming photon, you will have the polarizations $\epsilon_\mu^{(a)}(k)$ used too in the amplitude, but not participating to a photon propagator calculus.

  • $\begingroup$ Trimok, thank you very much for the so detailed explanation and the nice typing. :-) $\endgroup$
    – D-K
    Dec 12, 2013 at 15:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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