# $U(1)$ local gauge invariance in QED [duplicate]

While constructing Lagrangian of QED, we don't add the mass term for photon $\dfrac{1}{2} m^{2}A_{\mu}A^{\mu}$ because gauge invariance does not allow. I want to ask, whether "$\bf{Theoretically}$", is this the only reason we don't have mass term. I know why we need this term to vanish by using the experimental facts, but I feel like I am missing something while constructing the lagrangian for QED. Please stay within the domain of Quantum field theory and Lagrangian formulation while answering (if possible).

• – Qmechanic Aug 21 '15 at 12:46

The violation of gauge invariance by this term is the "only" reason why it's never written down – as long as we define the word "only" to include all other reasons that may be shown to be "physically equivalent" to gauge symmetry.

Gauge symmetry is extremely important and its violation would make a similar theory inconsistent, especially at the quantum level. In quantum field theory, the fields $A_\mu$ may be basically interpreted as four Klein-Gordon like fields because $\mu=0,1,2,3$. There are corresponding creation operators $a^\dagger(\vec k)_\mu$ for each momentum $\vec k$. These creation operators carry an extra index $\mu$.

Because the theory is Lorentz-covariant, the inner product of such one-particle states must be $$\langle 0 | a_\mu (\vec k) a^\dagger_\nu(\vec k) | 0 \rangle = -C g_{\mu\nu}$$ The (infinite, delta-function-like etc.) normalization factor was included to $C$ but the dependence on the $\mu,\nu$ indices has to be proportional to the metric tensor because it's the only tensor with these indices that a Lorentz-covariant theory may produce. Well, there's also $k_\mu k_\nu$ but let me ignore that.

What the formula above implies is, among other things, that the time-like and space-like components of the "photon 4-dimensional multiplet" have the opposite squared norm. That would mean that one of these types would predict negative probabilities because the inner products are interpreted as probabilities in quantum mechanics.

This mustn't happen because probabilities cannot be negative. The reason why this threat is not a problem is gauge invariance which guarantees that not just one but actually two components of $A_\mu$ create unphysical modes i.e. particle species that completely "decouple".

This verb means that if those are not present in the initial state, they won't be present in the final state. So we may consistently define a "sub-theory" in which they don't exist at all. The condition that the unphysical modes can't be produced out of the physical particles is a consequence of the gauge invariance. If gauge invariance were broken, the production of the "wrong" polarizations of the photon (the time-like one or the longitudinal one) would become unavoidable, and it would imply that about 1/2 of the predicted probabilities are negative.

Even in the classical theory, the violation of gauge invariance would be a problem. Classical physics doesn't fundamentally talk about probabilities but the wrong-sign components of the 4-vector field would at least imply the energy that is not bounded from below which would make the vacuum unstable much like a pencil vertically standing on its tip.

Gauge invariance is absolutely essential in any Lorentz-covariant theory with fields of spin 1 or higher – because the "wrong" components have to be eliminated. That's why the W-bosons, Z-bosons, and gluons have to be associated with the corresponding, equally strong Yang-Mills symmetries, too. And the metric tensor $g_{\mu\nu}$ in general relativity has to come equipped with the diffeomorphism symmetry, too. A metric tensor theory without the diff symmetry would imply the same existence of polarizations of the graviton that predict negative probabilities.

I cheated a bit and avoided the precise reasons why two of the 4 photon polarizations are made unphysical by the Lorentz symmetry. Well, the creation operator $a^\dagger_\mu (k^\alpha)$ is contracted with the polarization 4-vector $\epsilon^\mu$ in general. The gauge symmetry then implies that the polarization with $\epsilon^\mu \sim k^\mu$ is "pure gauge", has zero norm, and completely decouples (adding this particle to a scattering process yields zero probability because of gauge invariance), while it is possible to demand that $k^\mu \epsilon_\mu = 0$ (it is possible because if it holds in the initial state, it holds in the final state again thanks to the gauge invariance). So only the two transverse spatial polarizations, i.e. $x,y$ for a photon moving in the $z$ directions, survive, have a positive norm, and are physical (and interacting with each other or other particles).

The gauge symmetry is needed for consistency which is also why all quantum field theories have to cancel all of their gauge anomalies, i.e. the quantum effects (usually one-loop effects) that make it impossible to preserve a gauge symmetry that existed in the classical theory. A gauge anomaly means that a quantum theory respecting the gauge symmetry can't exist – which would mean that the theory has to include the wrong-sign fields with the spin 1 or higher.