Functional determinants and gauge invariance

[muppet]

Hi all,

I've been studying the path-integral quantisation of gauge theories in Zee III.4. My understanding is roughly as follows: that one can think of the differential operator in the quadratic tems in the lagrangian as a linear operator between infinite dimensional spaces (morally equivalent to a quadratic form on the cartesian product of the spaces); that the solution to the free theory looks like the reciprocal square root of the corresponding functional determinant; that the functional determinant will be zero if the linear operator annihilates a non-zero field, and consequently our operator will have no inverse (i.e. no propagator); and that the addition of a field thus annihilated to another effectively constitutes a gauge transformation.

Given this (possibly erroneous) understanding, I'm struggling to see why the simple non-interacting scalar klein-gordon isn't a gauge theory, as the vanishing of a field under application of the operator $$(\partial^2+m^2)\phi=0$$ is just the equation of motion, and the "gauge transformation" wouldn't connect physically equivalent states, but would be a restatement of the principle of superposition.

As this seems to be true more or less in general for scalar field theories, I'm thinking that it relates specifically to the fact that Zee is treating a formal matrix of differential operators, and it's actually this level of structure that introduces the non-invertibility. However, Peskin and Schroeder still explicitly identify $$(const.)(\partial^2+m^2)^{-\frac{1}{2}}$$ as the functional determinant, so I'm not sure why the logic doesn't extend to the "1-dimensional" case. If this the reason, is there a similar feature that singles out a gauge symmetry in the scalar case (e.g. conformal invariance of massless phi-fourth theory, or U(1) transformations of complex scalar fields)?

Thanks in advance.

 

[eljose79]

Hi there,

First of all, it's great to see that you're studying path-integral quantisation of gauge theories. It's a fascinating and challenging topic. I can see that you have a good understanding of the concepts involved, but let me clarify a few things for you.

You are correct in thinking that the differential operator in the quadratic terms of the Lagrangian can be thought of as a linear operator between infinite dimensional spaces. This is because the fields in a gauge theory are infinite-dimensional objects, and the operator acts on these fields to give us the equations of motion. However, the solution to the free theory is not always the reciprocal square root of the functional determinant. This is only true in certain cases, such as the non-interacting scalar Klein-Gordon theory that you mentioned. In general, the solution to the free theory is given by the Feynman propagator, which is the inverse of the differential operator.

Now, let's address your question about why the simple non-interacting scalar Klein-Gordon theory isn't considered a gauge theory. The reason is that in a gauge theory, the gauge transformation must connect physically equivalent states. In the case of the Klein-Gordon equation, the gauge transformation does not do this. It simply transforms the field from one solution of the equation to another. This is not the case with true gauge theories, where the gauge transformation connects physically equivalent states.

You are correct in thinking that the non-invertibility of the differential operator is related to the gauge symmetry in the theory. However, this is not the only factor. Another important factor is the presence of constraints in the theory, which arise from the gauge symmetry. These constraints are what make the operator non-invertible, and they are what distinguish a true gauge theory from a non-gauge theory.

In the case of scalar field theories, there are indeed features that single out a gauge symmetry. For example, the conformal invariance of massless phi-fourth theory and the U(1) transformations of complex scalar fields are both examples of gauge symmetries. These symmetries lead to constraints in the theory, which make the operator non-invertible and give rise to the gauge symmetry.

I hope this helps clarify some of your questions. Keep up the good work with your studies!