# Gauge theories

RedX
Just to review a little bit:

In general, for a gauge field with Yang-Mills Lagrangian

$$\mathcal L=-\frac{1}{4}F^{c}_{\mu \nu}F^{c \mu \nu}$$

for each c it is impossible to find the resulting free Green's function G(k) in momentum space:

$$(g^{\mu \nu}k^2-k^{\mu}k^{\nu})G_{\nu \eta}(k)=\delta^{\mu}_{\eta}$$

since $$(g^{\mu \nu}k^2-k^{\mu}k^{\nu})$$ has an eigenvalue of zero along the direction of k, which means the determinant of the operator is zero and uninvertible.

However, it seems you are allowed to insert a delta function into the path integral and use the FP-method, and for a $$R_\xi$$ gauge this results in trying to invert:

$$[g^{\mu \nu}k^2-(1-\frac{1}{\xi})k^{\mu}k^{\nu}]G_{\nu \eta}(k)=\delta^{\mu}_{\eta}$$

which can readily be checked has an inverse:

$$G_{\nu \eta}(k)=\frac{1}{k^2}\left(g_{\nu \eta}-(1-\xi)\frac{k_\nu k_\eta}{k^2} \right)$$

My question is with regards to gauge symmetry of the action. By inserting a delta function (that contains the information on the choice of gauge) into the path integral, it somehow manages to find its way into the exponential and alters the action itself. Is this new action in general gauge invariant?

Also, what if the choice of gauge is not Lorentz invariant? Will your scattering amplitudes still be Lorentz invariant? That is, if you boost the initial and final states to larger momentum, will the amplitude be the same as the amplitude for the slower initial and final states pre-boost?

Also, what if your choice of gauge is not linear? Can you just choose any equation involving the fields to be a gauge condition? Also, in all the books I've seen, the gauge condition doesn't mix up group indices. Often the gauge condition is one of spacetime, and each gluon field is given this same spacetime restriction. Can your gauge condition mix the different gluons for example?

Homework Helper
I have studied this theoretical model from the BRST perspective and I can't remeber putting any delta into the path integral.

As for the choices of the gauge-fixing terms that will <cure> our path integral, of course they can't be arbitrary, they need to have some properties: in the Lagrangian formulation, since Lorentz covariance is never lost, the gauge fixing term must be Lorentz invariant as well and have a differential order no higher than the original action.

As for the nonlinearity issue, well, it must be linear for the purpose it should serve: eliminating the antifields from the theory through a Legendre transformation wrt the nonminimal spectrum. If the gauge was nonlinear, the Legendre transformation couldn't be reversed and the antifields couldn't be removed.

weejee
A gauge-fixed Lagrangian breaks the gauge symmetry, by construction.
Furthermore, choosing something like the Coulomb gauge breaks the Lorentz invariance, too , at the level of Lagrangian. Still, all the scattering amplitudes must be Lorentz invariant.

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By inserting a delta function (that contains the information on the choice of gauge) into the path integral, it somehow manages to find its way into the exponential and alters the action itself.
It gets into the exponential by writing the detla functional as (and here I'm being schematic) $\delta(G(A)-\omega)$, and then integrating over $\omega$ with a gaussian weight factor.

Is this new action in general gauge invariant?
No! If it was, you would still have the same problems.

Also, what if the choice of gauge is not Lorentz invariant? Will your scattering amplitudes still be Lorentz invariant?
Yes.

Also, what if your choice of gauge is not linear? Can you just choose any equation involving the fields to be a gauge condition?
You are supposed to choose an equation which has one solution for A given the field strength F. Srednicki's text discusses Gervais-Neveu gauge, which is nonlinear.

Also, in all the books I've seen, the gauge condition doesn't mix up group indices. Often the gauge condition is one of spacetime, and each gluon field is given this same spacetime restriction. Can your gauge condition mix the different gluons for example?
Yes. But there doesn't seem to be any advantage to doing so, so no one does it.

RedX
weejee said:
Furthermore, choosing something like the Coulomb gauge breaks the Lorentz invariance, too , at the level of Lagrangian. Still, all the scattering amplitudes must be Lorentz invariant.

How would you prove that? Take an initial state |i>, and a final state <f|. In between is a time evolution operator:

$$<f|i>_H=<f|\int [d\phi] e^{i \int \mathcal L d^4x}|i>$$

Now

$$<f|i>_H=<f|\int d[\phi] e^{i \int \mathcal L d^4x}|i>=<f'|U{\dagger}\int [d\phi] e^{i \int \mathcal L d^4x}U|i'>$$

where U is a Lorentz transformation that takes primed into unprimed. For the amplitude <f|i>H to be equal to <f'|i'>H (where H means Heisenberg picture), doesn't $$U^{\dagger}\mathcal L(x) U=\mathcal L(x')$$, i.e., the Lagrangian is Lorentz invariant?

Actually I think I've confused myself because the Lagrangian in the path integral is not an operator but a c-number. O wait, it should be like this:

$$<f|i>_H=\int [d\phi] e^{i \int \mathcal L d^4x}$$

So a Lorentz transformation on the LHS is equivalent to one on the RHS, so focus on the RHS. The field measure doesn't change, since the field is a scalar. The coordinate measure in the exponential doesn't change as a 4-volume is Lorentz invariant. The limits on the coordinates are infinite, so that doesn't change under Lorentz transformation. All that remains is the Lagrangian to be invariant. So if you affect something at the level of the Lagrangian, it can affect the Lorentz invariance of the LHS, or the amplitude.

I have studied this theoretical model from the BRST perspective and I can't remeber putting any delta into the path integral.

Well the book I read has something like this (taking an Abelian field for simplicity):

$$\int [dA^\mu] e^{i\int -\frac{1}{4} F_{\mu \nu}F^{\mu \nu} d^4x}$$

and what they just do is write the path integral instead as:

$$\int [dA^\mu] \delta(\mbox {gauge condition})*(\mbox{FP-determinant})e^{i\int -\frac{1}{4} F_{\mu \nu}F^{\mu \nu} d^4x}$$

The FP-determinant ends up being shifted into the exponential which modifies the Lagrangian (the ghost Lagrangian term). The delta function is gotten rid of similarly by multiplying the path integral by a term that involves the field and integrating, and the net result is that another term is absorbed into the Lagrangian (the gauge-fixing Lagrangian term). So the Lagrangian keeps on getting modified.

Here's the thing. Ultimately whether the new terms are Lorentz invariant should boil down to whether $$\delta(\mbox {gauge condition})*(\mbox{FP-determinant})$$ is itself Lorentz invariant. With a gauge condition that breaks Lorentz invariance, I don't think you can say this term is Lorentz invariant. And if the Lagrangian is not Lorentz invariant, then I think you have problems.

As for the choices of the gauge-fixing terms that will <cure> our path integral, of course they can't be arbitrary, they need to have some properties: in the Lagrangian formulation, since Lorentz covariance is never lost, the gauge fixing term must be Lorentz invariant as well and have a differential order no higher than the original action.

There are gauges that aren't Lorentz invariant such as the Coloumb gauge (3-divergence of vector potential is zero) or an axial gauge (one of the components of the vector field is zero). The one that's most often used is the one where the 4-divergence is zero (I think this is called the Lorentz gauge), which is Lorentz invariant. I heard that Lorentz invariant gauge conditions are the only ones that are renormalizeable. All these gauges are similar in that they are linear in the fields, but other than that, the choice of gauge looks arbitrary. So I'm wondering whether this (linear) gauge condition is allowable:

$$\frac{1}{x}A^3+\frac{\partial A^0}{\partial y}+z\frac{\partial A^0}{\partial y \partial z}=0$$

or in general a condition that is nonlinear in A.

As for the nonlinearity issue, well, it must be linear for the purpose it should serve: eliminating the antifields from the theory through a Legendre transformation wrt the nonminimal spectrum. If the gauge was nonlinear, the Legendre transformation couldn't be reversed and the antifields couldn't be removed.

What's an antifield?

How would you prove that [the scattering amplitudes must be Lorentz invariant]?
By proving the relevant Ward identities. Or, in the BRST approach, it follows from the BRST symmetry of the gauge-fixed action.

RedX
Avodyne said:
You are supposed to choose an equation which has one solution for A given the field strength F. Srednicki's text discusses Gervais-Neveu gauge, which is nonlinear.

For non-Abelian gauge theories, would this be changed to one solution A for the trace of the double contraction of F?

Anyways, since two people mentioned it, I just read about the BRST approach today from Srednicki's book, but I still don't see how Lorentz invariance is preserved.

Srednicki defines the additional term in the Lagrangian to be the BRST derivative of:

$$\bar{c}^a (\frac{1}{2}\xi B^a-G^a)$$

where G is the gauge fixing condition, one for each gluon index a, c overbar is the antifield, and B^a is a scalar defined to be the derivative of the antighost field (is this just arbitrary: that we define the derivative of the antighost field to be a scalar field?)

Taking the BRST derivative for general G:

$$\mathcal L=(\delta_B \bar{c}^a) [(\frac{1}{2}\xi B^a-G^a)]+\bar{c}^a \delta_BG^a =\frac{1}{2}\xi B^aB^a-B^aG^a +\bar{c}^a \frac{\partial G^a}{\partial A^{b\mu}}\delta_BA^{b \mu} =\frac{1}{2}\xi B^aB^a-B^aG^a +\bar{c}^a \frac{\partial G^a}{\partial A^{b\mu}} D^{bc\mu}c^c$$

This expression is still just quadratic in B, so you can substitute in the classical solution for B ($$B^a=\frac{1}{\xi}G^a$$) and you get:

$$\mathcal L=-\frac{1}{2\xi}G^aG^a+\bar{c}^a \frac{\partial G^a}{\partial A^{b\mu}} D^{bc\mu}c^c$$

So this expression is not Lorentz invariant unless $$G^a$$ is.

Let's look at the form of the term whose BRST derivative we take:

$$\bar{c}^a (\frac{1}{2}\xi B^a-G^a)$$

If you have a c overbar on the left side, then in the parenthesis B can only be to one power or the expression won't be quadratic in B when you take the BRST derivative, and you need to integrate out B since you don't want to even mess with what B is as a field. So I think this expression for the object on which a BRST derivative will be applied, is general. The $$\xi$$ stands for linear as in $$R_\xi$$ gauge.

So my conclusion is even with the BRST approach instead of FP approach, only Lorentz invariant gauge conditions preserve the Lorentz invariance of amplitudes.

I'm interested in linear gauges, of the form $$n^\mu A_\mu=0$$. But not necessarily Lorentz invariant, so $$n^\mu$$ might not be a 4-vector like the $$(\partial_0,\partial_i)$$ but for example the axial gauge (0,0,0,1) which is not Lorentz invariant. However, I just don't see how amplitudes are preserved if your Lagrangian breaks Lorentz invariance.

Anyways, to me the FP approach is way more intuitive, as it has the interpretation of not integrating over degeneracies due to gauge freedom. The BRST approach, which I just read today, seems to come out of nowhere, and the idea is to add the ghost fields from the outset and augment the Lagrangian in a way that preserves BRST symmetry.

So my conclusion is even with the BRST approach instead of FP approach, only Lorentz invariant gauge conditions preserve the Lorentz invariance of amplitudes.

The amplitudes are gauge invariant, so if Lorentz-invariant gauge conditions exist (and they do), then the amplitudes are Lorentz invariant.

Also, what if the choice of gauge is not Lorentz invariant? Will your scattering amplitudes still be Lorentz invariant? That is, if you boost the initial and final states to larger momentum, will the amplitude be the same as the amplitude for the slower initial and final states pre-boost?
First of all you fix the gauge symmetry, you don't break it, that's a difference. One can show in the Hamiltonian approach that this gauge fixing is like fixing the c.o.m. frame in a two-particle system with V(r1-r2). Of course the symmetry of translating the system as whole get's lost, but it is not broken but fixed. The translational symmetry acts in the unhysical sector of the Hilbert space, but the spectrum remains invariant under this fixing. Of course it's not obvious, that's why you need BRST and the Ward identities in the PI approach.

Yes, as said the Lorentz invariance is preserved, but again that's not directly visible. Doing the same in the Hamiltonian approach leads to rather complicated expressions for the generators of the Poincare symmetry. One has to check explicitly that the gauge fixed generators (using a non-Lorentz invariant gauge condition like A°=0 and e.g. Coulomb gauge) still fulfill the Poincare algebra. It takes some sheets of paper to show that but they do.

The biggest problem is that (afaik) there is no gauge fixing condition which is defined globally. This is the famous Gribov problem. Gauge fixing means that for each physical gauge field configuration all gauge transformations are reduced to unity. But this is not possible due to the complicated geometry of the fibre bundle in non-abelian gauge symmetries. That means there is a residual symmetry, i.e. you get discrete copies of a physical field configuration. You don't see that in perturbation theory, but for non-perturbative calculations it must be taken into account.

RedX
I'm trying to look at things in the Hamiltonian approach, but I'm a little confused about the basic concepts.

If you have an initial Heisenberg state |i>H, and a final one |f>H, then Lorentz transformation takes |i>H into U|i>H and |f>H into U|f>H, so the amplitude is preserved:

<f'|i'>H=$$<f|_HU^{\dagger}U|i>_H$$=<f|i>H

But what if you work in the Schrodinger picture?

$$<f|i>_H=<f|e^{-iHt}|i>$$

So if you Lorentz transform the initial and final states:

$$<f'|i'>_H=<f|U^{\dagger}e^{-iHt}U|i>\neq <f|e^{-iHt}|i>$$

since in general, the Hamiltonian transforms as the time component of a 4-vector under U.

So the Heisenberg and Schrodinger pictures don't give the same answer.

Actually, I think:

$$<\phi_2,x_2,t_2|\phi_1,x_1,t_1>_H$$

is most analogous to

$$<q_2,t_2|q_1,t_1>_H$$

So the Heisenberg state includes position and time, and to get the Schrodinger state you need the full momentum 4-vector and not just the Hamiltonian:

$$|\phi_1,x_1,t_1>_H=e^{iHt_1-iPx_1}|\phi_1>_S$$

So then it's obvious:

$$<\phi_2',x_2',t_2'|\phi_1',x_1',t_1'>_H=<\phi_2'|_Se^{-iH't'+iP'x'}|\phi_1'>_S= <\phi_2|_SU^{\dagger}e^{-iH't'+iP'x'}U|\phi_1>_S=<\phi_2|_Se^{-iHt+iPx}|\phi_1>_S= <\phi_2,x_2,t_2|\phi_1,x_1,t_1>_H$$

since now the exponential is a scalar and is Lorentz invariant, instead of just isolating time.

I think it's quite clear I need to review the Hamiltonian/operator approach, as it's been all path integrals for awhile and I have a poor memory.

Thanks everyone!

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