# Why the photon doesn't acquire mass with the Higgs mechanism?

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.

• Part 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
• 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 4:50
• 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 1 '18 at 21:28
• Possible duplicates: physics.stackexchange.com/q/23161/2451 and links therein. – Qmechanic Dec 3 '18 at 20:34

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.