I am puzzled by the $W^{\pm}$ vertex factor in weak interactions. In Griffiths' textbook "Introduction to Elementary Particles", the $W^{\pm}$ vertex factor is given by (10.92) on page 324: \begin{equation} \frac{-ig_{w}}{2\sqrt{2}}\gamma^{\mu}(1 - \gamma^{5}) \tag{10.92} \end{equation} However, In Srednicki's textbook "Quantum Field Theory", Problem 88.6 (on page 538) asks us to compute rates for the decay processes $W^{+}\rightarrow e^{+} \nu_{e}$, ... etc. The answer is given in the solutions manual. On page 146 of the solutions manual, it is stated

Consider a massive vector field $Z^{\mu}$ and a Dirac fermion field $\Psi$ with $\mathcal{L}_{int} = Z^{\mu}\overline{\Psi}(g_{v} - g_{A}\gamma^{5})\Psi$; then the amplitude for $Z\rightarrow e^{+}e^{-}$ is $\mathcal{T} = \varepsilon^{*\mu}\overline{v}_{2}\gamma_{\mu}(g_{v} - g_{A}\gamma^{5})u_{1}$. ... ... The amplitude is the same if $\overline{\Psi}$ is a different Dirac field that is unrelated to $\Psi$, so it also holds for a process like $W^{+}\rightarrow e^{+} \overline{\nu}$.

My question is: Why is there no $\varepsilon^{*\mu}$ in (10.92), whereas there is an $\varepsilon^{*\mu}$ (which seems to account for $W^{+}$) in the amplitude $\mathcal{T} = \varepsilon^{*\mu}\overline{v}_{2}\gamma_{\mu}(g_{v} - g_{A}\gamma^{5})u_{1}$ for the decay process $W^{+}\rightarrow e^{+} \overline{\nu}$?


1 Answer 1


Those two things are related, but different.

Your equation $(10.92)$ indicates the value of a vertex, while $\mathcal T$ in Srednicki's book represents an amplitude.

Basically, the vertex is one of the two building blocks of the Feynmann diagrams. A diagram is a multiplication of vertices and propagators, and becomes a complex amplitude for the process when you multiply that amplitude with the external particle factors, such as $\epsilon^\mu$.

An example: Feynmann's QED vertex is given by $-ie\gamma^\mu$ (the sign depends on conventions, I'll follow Michele Maggiore's textbook). Now, let's take the typical first order contribution to the process $e^-\gamma\to e^-$: the relevant graph is

enter image description here

(here future and past are messed up, that graph here is just for reference). Now, the graph is composed by a vertex and three external legs: the vertex has value $-ie\gamma^\mu$, and the amplitude can be written as $$ \mathcal T=\epsilon_\mu(k)\bar u(p_1)(-i e\gamma^\mu)u(p_2), $$ where $k$ is the momentum of the photon, $p_1$ the momentum of the incoming electron and $p_2$ the momentum of the outgoing electron. The modulus of the amplitude squared, $|\mathcal T|^2$, is proportional to decay lengths and cross sections (more in general, in the $S$-matrix), and is used to understand "how much" a process happens.

P.s.: as a nice addendum, note that, if you consider the process $\gamma\to e^+e^-$, you can use the same rotated graph, so you have to change some external legs factor, obtaining $$ \mathcal T=\epsilon_\mu(k)\bar {u}(p_1)(-ie\gamma^\mu)v(p_2). $$ If you calculate the square of this amplitude, you obtain a value that is different from zero. But, from elementary considerations about 4-impulse conservation, you know that this process can't happen, as there is no way to sum the timelike momenta of the matter particles to obtain a lightlike momentum. So the amplitude can be different from zero even if a process is not observed: in this case, the $\delta$ that expresses momentum conservation in the $S$ matrix takes care of that, and the process $\gamma\to e^+e^-$ cannot happen, even if the amplitude is non zero.

  • $\begingroup$ Isn't this taken care of by the fact that the decay amplitude has kinematic factors, or is this the same thing? What you write is the matrix element, not the decay amplitude. The decay amplitude would be multiplied by a factor which is automatically zero for decay of a photon, so the amplitude vanishes even if the matrix element doesn't (as expected for a massless particle). $\endgroup$
    – Tom
    Jul 20, 2023 at 13:40

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