I'm currently studying the GWS-model and am a bit confused with the matrix (def. via the Weinberg mixing angle) "mixing" photon and $Z$ boson mass eigenstates. It is defined as

$$ \begin{pmatrix} Z^0 \\ A \end{pmatrix} = \begin{pmatrix}\cos\theta_W & -\sin\theta_W\\ \sin\theta_W & \cos\theta_W\end{pmatrix} \begin{pmatrix} A^3 \\ B \end{pmatrix} $$

(Peskin and Schröder p.702)

So in the GWS-model the photon is massless, as it should be, but why is it then useful to mix the mass eigenstates of the massive $Z^0$ boson and the massless photon? Do they not also transform in different groups?

  • $\begingroup$ If I remember correctly, it's in order to diagonalize the mass matrix and make the lagrangian density look nicer. $\endgroup$ Commented Sep 3, 2022 at 15:16

1 Answer 1


As your text (any text!) reminds you, you may change your basis from the theoretical, "clean" basis of gauge bosons $(A^a,B)$ living in different groups, SU(2) and hypercharge U(1), respectively, to the "physical", ugly/messy basis $W^\pm, Z^0, A)$ of mass eigenstates.

As equivalent descriptions of the same couplings, they are completely equivalent. The clean basis displays the couplings simply, so you may check their normalizations and write simple group expressions, whereas the "physical" basis displays the mass eigenstates you observe in the lab, and, crucially, their highly counterintuitive couplings to fermions: the photon, A, couplings are clean and conventional/memorable, but countless students get faked out by the peculiar/quasi-magical neutral current couplings of the Z boson, involving both chiralities! Cf. (10.6a,b) in (10.2) of the SM model PDG summary, worth memorizing and understanding.

A summary of how the respective couplings are related is this right triangle, enter image description here

with the electric charge representing its height when laid on its hypotenuse.

  • 1
    $\begingroup$ Thank you for your fast and detailed answer! :) $\endgroup$
    – physicsCat
    Commented Sep 3, 2022 at 16:03

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