When comparing the experimental phenomenology with the standard model, one usually takes a combination of the $W^a_\mu$ and $B_\mu$ gauge bosons to obtain the physically observable ones ${W^+}_\mu$, ${W^-}_\mu$, $Z_\mu$ and $A_\mu$. In particular, with the usual choice for the minimum of the Higgs doublet, one relates charged $W$ boson with $W^1_\mu$ and $W^2_\mu$, while the $Z$ boson and the photon arise as a mixture of the $W^3_\mu$ and $B_\mu$ fields.

My question is about what would happen if a different minimum is chosen between all the infinite ones that are possible. Could the mixtures be different? For example, being the $Z$ boson and the photon related to $W^2_\mu$ instead of $W^3_\mu$. What would happen with the masses? My guess is that they should be the same, specially in the sense of hierarchy, where one should get that one of the boson is massless (the redefined photon), and two of the other new bosons should have the same mass (the new mixtures that would correspond to $W^+$ and $W^-$).

  • $\begingroup$ Everything would be same except that the minimum value of the Higgs field would get an extra constant phase factor. $\endgroup$
    – Aman pawar
    Commented May 8 at 17:16
  • 2
    $\begingroup$ Just a note it is Phenomenology not fenomenology...likely just a translation error though. $\endgroup$
    – Triatticus
    Commented May 8 at 17:18

1 Answer 1


All vacua of this potential are equivalent. You couldn't tell the difference.

The generators of the su(2) algebra can be SU(2)-rotated into an su(2) algebra equivalent to your original one. You may use the 2nd Pauli matrix to represent $\tau^3$ under such an isomorphism. The physics will be completely identical.

Formally, you are asking whether the conventional setup $$ \Phi= e^{i\vec \pi\cdot \vec \tau/v} \begin{pmatrix} 0\\ v+h \end{pmatrix} ~~~~\leadsto ~~~~ \langle\Phi\rangle= \begin{pmatrix} 0\\ v \end{pmatrix} $$
is distinguishable from $$ \langle\Phi\rangle= e^{i\vec \theta \cdot \vec \tau } \begin{pmatrix} 0\\ v \end{pmatrix}, \\ \Phi= e^{i\vec \theta \cdot \vec \tau } e^{i\vec \pi\cdot \vec \tau/v}e^{-i\vec \theta \cdot \vec \tau } e^{i\vec \theta \cdot \vec \tau } \begin{pmatrix} 0\\ v+h \end{pmatrix} \equiv e^{i\vec \pi'\cdot \vec \tau/v} e^{i\vec \theta \cdot \vec \tau}\begin{pmatrix} 0\\ v+h \end{pmatrix}. $$

The three goldstons $\vec \pi'$ are a mere isorotation of the three $\vec \pi$ parameterized originally, and absorbed away in the Higgs mechanism, so there is no trace left of them, and hence their difference!

It is a mere change of basis, so changing labels cannot matter.


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