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When it is said that the photon is ["a mixture of W and B"][1] ($B$ being a gauge field associated with the $U(1)$ hypercharge)

I have one question on this:

  • Why there isn't a boson directly generated by this $B$ gauge?
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  • $\begingroup$ link to part I: physics.stackexchange.com/q/477283/955 $\endgroup$ – lurscher May 2 at 17:58
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    $\begingroup$ because $SU(2)_L\times U(1)_Y$ is broken down to $U(1)_{em}$. When the former symmetry is restored (early universe) there will be massless B boson, and also 3 massless W bosons. $\endgroup$ – Kosm May 2 at 19:55
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As Kosm suggests in his comment, there technically is a massless boson directly generated by the $B_\mu$ field. The reason we do not see it in (say) particle accelerators is because our universe is "too cold," and hence spontaneous symmetry breaking in the Higgs field generates the $A_\mu$ (photon), $Z_\mu$, and two $W^\pm_\mu$ fields.

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