I did the computation that from $$(0,v)^{T}(\partial_\mu+igA_\mu^a\tau^a+i\frac{g'}{2}B_\mu)(\partial_\mu-igA_\mu^b\tau^b-i\frac{g'}{2}B_\mu)(0,v)$$ with $(0,v)$ being the expectation value of the Higgs field and $\tau, B_\mu$ being the generator of the SU(2) and U(1) group proves the mass term for the $Z_\mu$ and $W^\pm_\mu$ and not for the photon. That said if I rewrite the covariant derivative in terms of the $Z_\mu, W^\pm_\mu$ and $A_\mu$ ($c_1,c_2$ being the constant dependent on $e,\theta_\omega$ for brevity): $$D_\mu=\partial_\mu-c_1[W_\mu^+(\tau^1+i\tau^2)+W_\mu^+(\tau^1-i\tau^2)]-c_2Z_\mu(\tau^3-\sin^2{\theta_\omega}Q)-ieQA_\mu$$ I don't realize how the term in $A_\mu^2$ doesn't give a mass term for the photon. Any hint is well appreciated.
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1Part of the confusion is probably that the $A_\mu$ in your first equation is not the same as the $A_\mu$ in your second. They are the gauge fields for $SU(2)L$ and $U(1){EM}$ respectively. – knzhou Nov 29 '18 at 16:44
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I know that the two $A$ are not the same and I don't think I confused the square with the index (good tip though) – Ringo_00 Nov 30 '18 at 04:50
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The term $e^2Q^2A_\mu^2$ is the problem. I don't understand why it doesn't get mass with the v from the H field – Ringo_00 Dec 01 '18 at 21:28
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Possible duplicates: https://physics.stackexchange.com/q/23161/2451 and links therein. – Qmechanic Dec 03 '18 at 20:34
1 Answers
The short answer is for you to see what the charge operator in your 2×2 matrix notation is when acting on the Higgs doublet, whose upper component is +, and lower component is neutral: of course, the v.e.v. must be chargeless! This is by dint of the hypercharge 1 of the (entire) Higgs doublet.
Thus, sandwiching the charge matrix squared between the Higgs v.e.v.s (o,v) annihilates Q2 and thereby any feared photon mass term.
More explicitly, ignore the plain derivative, since it collapses on the constant v.e.v., and omit the $W^{\pm}$ in the covariant completion, since they amount to terms orthogonal to the photon and Z in the square.
The remnant is the diagonal part of the derivative-completion-squared 2×2 matrix acting on a Higgs doublet, just $$ g^2v^2 ~~(0,1) \operatorname{diag}(^3A_\mu +\tan^2\theta_W ~B_\mu, -^3A_\mu +\tan^2\theta_W ~B_\mu )^2 ~ (0 , 1)^T \\ \equiv \frac{g^2 v^2 }{\cos^2 \theta } (0,1) \operatorname{diag} (A_\mu ^2, Z_\mu^2) ~(0,1)^T = \frac{g^2 v^2 }{\cos^2 \theta } Z_\mu^2, $$ the calculation you said you had no trouble with.
The very same calculation in the physical (propagating) basis involves $$ Q=\operatorname{diag} (1,0), \\ \tau^3 - \sin^2\! \theta_W ~Q= \cos^2\! \theta_W \tau^3 -\sin^2\!\theta_W ~ Y/2\\ = \operatorname{diag}\!(1/2 -\sin^2\! \theta_W ,-1/2 ). $$ Acting on the v.e.v., Q vanishes, decoupling $A_\mu$ from the uncharged vacuum, as already indicated; while the eigenvalue of the neutral current charge is just -1/2, to be squared to multiply your $c_2^2$, namely $4e^2/(\sin 2\theta_W)^2$, to yield the above mass.
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