Gauge theory symmetry breaking in L&B

[joneall]

TL;DR Summary

Problem understanding Lagrangian

I’m reading Lancaster & Blundell, Quantum field theory for the gifted amateur (even tho I”m only an amateur...) and have a problem with their explanation of symmetry breaking from page 242. They start with this Lagrangian:

$\mathcal{L} = (\partial_{\mu} \psi^{\dagger} - iq A^{\mu}\psi^{\dagger}) (\partial_{\mu} \psi + iq A^{\mu}\psi) + \mu^{2} (\psi^{\dagger} \psi) - \lambda^{2} (\psi^{\dagger} \psi)^2 - \tfrac{1}{4\pi} F^{\mu \nu} F_{\mu \nu}$

which along with
$A_{\mu} \rightarrow A_{\mu} -\tfrac{1}{q} \partial_{\mu} \alpha(x)$

is locally invariant under the transformation

$\psi \rightarrow \psi e^{i \alpha(x)}$

So far, so good. But then they state that this Lagrangian describes “… a world made up of two sorts of oppositely charged massive scalar particles with energies $E_p = (p^2 + \mu^2 )^{\tfrac{1}{2}}$ and two sorts of transversely polarized photons with energies $E_p = |p|$, which are, of course, massless.”

I think they may be using something defined several pages earlier
$\psi = \psi_1 + i \psi_2$ and $\psi^{\dagger} = \psi_1 - i \psi_2$

which could explain the two scalars. But I do not at all see two photons in there. Can anyone point them out to me?
Thanks in advance.

Btw, how do you make a dagger (for a Hermitian adjoint, e.g.) in your version of latex?

 

[Gaussian97]

The term $F^{\mu\nu}F_{\mu\nu}$ describes in the free theory a massless vector particle, these are the photons, also a vector particle usually has 3 degrees of freedom, but being massless is restricted to only 2.

Also to put a dagger you can simply use the \dagger ($\dagger$).

 

[joneall]

Thanks for the tip about the dagger. I know about the $F^{\mu \nu}F_{\mu \nu}$ term being a photon, but can it represent two of them? How do we know?

 

[Gaussian97]

As I said because the $A$ field is massless you can see that it has only 2 degrees of freedom, this is the polarization, you can interpret this as there exists two photons, a right one and a left one.

 

[joneall]

Ok, that sounds good. Thanks a lot.

 

[vanhees71]

It's a bit strange to express it in this way. I guess what the author wants to say is that $\psi$ as a charged massive scalar fields describes two kinds of scalar particles, i.e., a particle of charge $q$ and an antiparticle of charge $-q$, where particle as usual in QFTs means an aysmptotic free 1-quantum Fock state. Both have of course the same mass $\mu$, if you'd put the correct sign for a mass term, i.e., $-\mu^2 \psi^{\dagger} \psi$.

The electromagnetic field is described by a massless vector field and as such has two polarization degrees of freedom (not three as a massive vector boson would have).

Now from the thread title I suppose that the "wrong" sign on the mass term is intentional, because what seems to be the goal is the description of the Higgs mechanism (applied to Abelian gauge theory to have a less complicated case to start with).

To get the physical degrees of freedom and their interpretation as particles with their corresponding masses you have to do however a bit of more work. The first notion should be that due to the "wrong sign" of the mass term you cannot do perturbation theory around the vacuum expectation value $0$ for the scalar field. Rather you have to find the minima of the field potential. If there where no em. field you'd get spontaneous symmetry breaking, i.e., the ground state is not symmetric under the phase transformation $\psi \rightarrow \exp(\mathrm{i} \alpha)$ (with $\alpha=\text{const}$ in this case!), but the dynamics of course is, and this implies that you get a massive scalar field-degree of freedom and a massless one (the latter is the Nambu-Goldstone mode and must occur whenever a continuous global (!) symmetry is spontaneously broken). The occurance of a a massless Goldstone boson is due to the degeneracy of the ground state.

Now introduce the em. field again. Then the above global symmetry under phase changes of the scalar field becomes local, and that changes the physical consequences drastically, and that's very important to understand what's behind the physics of the (electrweak) standard model: Because now the apparently spontaneously broken symmetry is a local symmetry, it turns out that it is not (and in fact mathematically can not, which is known as Elitzur's theorem) spontaneously broken at all and that the ground state is not degenerate any more. This implies that the massless scalar Nambu-Goldstone mode is NOT part of the physical spectrum anymore. You can rather choose a particular gauge such that it is completely incorporated to the gauge-field degrees of freedom, providing a third polarization state and making the gauge boson massive. So the counting of field degrees of freedom is fulfilled, the "would-be Nambu Goldstone mode" is "eaten up" by the gauge field, making it massive and providing the necessary 3rd polarization state of a massive (rather than massless) vector field. The good thing is that now you have a massive gauge field without breaking the local gauge symmetry in any way, i.e., the theory stays physically sane, and you have no massless scalar particles in the game (which was a relief for the physicists trying to find a gauge theory of the weak interactions, where they needed massive gauge bosons without breaking gauge symmetry and had no use of any massless Goldstone bosons, because no such creature is observed). What remains in the physical spectrum from the scalar field is one massive scalar bosons, which is the famous Higgs boson of the Higgsed local symmetry.

 

[joneall]

Thanks for you explanation, vanhees71. In fact, the sign of the mass term is intentionally chosen "wrong" so as to go thru what you explain. You give somewhat more detail, tho, so thanks.

 

[PeterDonis]

because the $A$ field is massless you can see that it has only 2 degrees of freedom, this is the polarization, you can interpret this as there exists two photons, a right one and a left one.

This is not correct. Photon number is different from polarization. You can have a single photon in a state which is a superposition of the two polarizations, and you can have a state with multiple photons all with the one polarization and no photons with the other polarization.

 

[PeterDonis]

I know about the $F^{\mu \nu}F_{\mu \nu}$ term being a photon, but can it represent two of them?

They didn't say "two photons". They said "two sorts of photons". In other words, there are two photon basis states, which represent two orthogonal polarizations that photons can have. That's not the same kind of thing as the two oppositely charged scalar particles: the scalar particles are antiparticles of each other, which is why one has a "dagger" on it, and they have no spin (since they're scalar); whereas the photon is its own antiparticle (which is why there are no "daggers" on the photons), but it has two different orthogonal spin (polarization) states. The polarizations don't show up differently in the Lagrangian; you just have to know that the vector particle $A_\mu$, since it's massless, has two (it would be three if it had nonzero mass) orthogonal polarization states.

 

[vanhees71]

The photon polarizations show up in the decomposition of the free-field operator in terms of annihilation and creation operators wrt. an arbitrary set of orthonormal wave modes. The usual choice are plane-wave solutions (corresponding to single-particle momentum eigenstates).

For the em. field you need to fix the gauge, when working in the most simple "canonical operator formalism". The most simple, though not manifestly covariant, gauge is the Coulomb gauge with the additional condition, valid for free em. fields, $A^0=0$ (the socalled radiation gauge). Then you get, for each wave-number $\vec{k}$ (corresponding momentum $\vec{p}=\hbar \vec{k}$ of a single photon and energy $E=c|\vec{p}|$, which reflects the masslessness of the photon) two linearly independent polarization states. From the "particle" point of view the most natural and covariant choice for the corresponding basis are the helicity eigenstates, where helicity is the projection of the total (sic!) angular momentum to the diretion of $\vec{k}$. Since photons are massless spin-1 particles the two helicity eigenvalues are $\pm 1$, corresponding to left- and right-circular polarized em. plane waves. Any other polarization state can be written as a superposition of such states.