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In our lecture, we arrived at the following expression for the Lagrange density of the $\gamma$ and the $Z^{0}$ boson:

$$\mathcal L_{\gamma, Z^{0}} = -\sum_{e\nu}\left\{ \bar{\psi}\gamma^{\mu}\left( g\sin\theta_{W}I_3 + g'\cos\theta_{W}\frac{Y}{2}\right)A_{\mu}\psi + \bar{\psi}\gamma^{\mu} \left( g\cos\theta_{W}I_{3} - g'\sin\theta_{W}\frac{Y}{2} \right) Z_{\mu}\psi\right\}. \tag{1.1}$$

This comes from the electroweak unification.

And then we said:

enter image description here

The first Eq., i. e. $eQ = g\sin\theta_{W}I_{3} + g'\cos\theta_{W}\frac{Y}{2}$ is perfectly clear to me. However, where does the second Eq., i. e. $$eQ = eI_{3} + e\frac{Y}{2}\tag{1.2}$$ come from? Is this through comparison with the QED Lagrangian $$\mathcal L_{\text{QED}} = \bar{\psi}\left( i\gamma^{\mu}\partial_{\mu} - m\right)\psi - \bar{\psi}\gamma^{\mu}eQA_{\mu}\psi - \frac{1}{4}F_{\mu\nu}F^{\mu\nu}. \tag{1.3}$$

If so, then I don't see how we arrive at (1.2) by comparing $(1.1)$ and $(1.3)$.

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  • $\begingroup$ It's considered good form to actually provide the reference for the screenshot you are displaying. If anything it specifies what you have learned and what you have not. $\endgroup$ Commented Feb 13, 2021 at 14:19
  • $\begingroup$ @CosmasZachos Unfortunately, the source is not publicly available.. $\endgroup$
    – user248824
    Commented Feb 13, 2021 at 19:32
  • $\begingroup$ @CosmasZachos I think I get my confusion. I think one cannot arrive at (1.2) by comparing (1.1) and (1.3) - even though it says so in our notes - but rather (1.2) is a definition we already introduced. That's how I understand it, at least. $\endgroup$
    – user248824
    Commented Feb 13, 2021 at 19:32
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    $\begingroup$ Okay, thank you. $\endgroup$
    – user248824
    Commented Feb 14, 2021 at 7:00

1 Answer 1

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The point is that we define the new coupling parameter e as $$e:=g\sin\theta_W=g'\cos\theta_W$$ in order to avoid carrying around these longer expressions

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  • $\begingroup$ I am sorry, but this does not answer my question.. :) $\endgroup$
    – user248824
    Commented Feb 13, 2021 at 13:50
  • $\begingroup$ @MathisFun. I think it does. In what way does it fail you? It is a convenient definition to produce a parameter-less Q which checks with everything you know about EM couplings. Have you paid due diligence? $\endgroup$ Commented Feb 13, 2021 at 14:27

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