Given the gauge fields $B$ (generator Y) for weak hypercharge and $W_3$ (generator $I_3$) for isospin, we combine the generators as $$Q=I_3+Y/2\tag{1}$$ to represent electric charge. Why does this not imply that we must also combine the fields as $$A=W_3+B/2\tag{2}$$ to produce the electromagnetic field?

Instead, the Weinberg angle enters in the mixing of the fields. How can this be compatible with $(1)$ At the same time, the coupling constants are also related by the Weinberg angle. Is this the trick to compensate for the incompatibility?


The reason is that $W^{3}_{\mu}$ belongs to $SU_{L}(2)$ group, while $B_{\mu}$ belongs to $U_{Y}(1)$. This means that the covariant derivative takes the form $$ D_{\mu} = \partial_{\mu} - ig_{1}\sigma^{j}W_{\mu}^{j} - ig_{2}YB_{\mu}, $$ where $g_{1,2}$ are coupling constants (in general different).

Because of these different coupling constants, in order to extract the operator $eQ$ (and interaction in the form $eQA_{\mu}$), where $Q$ is defined in Your question, You have to introduce the non-trivial mixing linear combination.

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  • $\begingroup$ This would imply that the definition of $Q$ also includes the coupling constants (or equivalently, the Weinberg angle). Indeed in Weinbergs QFT book this is the case. $g_2$ is absorbed into $Y$ for $(1)$ to make sense. $\endgroup$ – old123987 Jan 17 '17 at 16:05
  • $\begingroup$ @nuu : the final definition must be free from couplings (or proportional to $e$). $\endgroup$ – Name YYY Jan 17 '17 at 16:43
  • $\begingroup$ This, exactly, is my question. How can this be achieved? Where is the trick? $\endgroup$ – old123987 Jan 18 '17 at 17:14

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