# Modes of a QFT and irreducible representation of the gauge group

This is in reference to the calculation in section 3.3 starting page 20 of this paper.

• I came across an argument which seems to say that the "constraint of Gauss's law" enforces gauge theory on compact spaces to be such that physical states over which the partition function sums over be gauge invariant.

I would like to hear of explanations of the above argument.

• Also the above seems to lead (quite non-obviously to me) to the conclusion that these physical states correspond to traces of products of operators acting on the Fock space vacuum. It is not clear to me as to how this trace is defined such that even after tracing it remains an operator.

{Very often one seems to want these operators to be in the "adjoint of" the gauge group. The meaning and motivation of this demand is not clear to me. (I am familiar with the notion of adjoint representation of Lie groups)}

• Related to the above is another claim I see which seems say that massless modes will be absent for any gauge group Yang-Mill's theory and any matter content if the theory is on a compact space. Is the above correct? Why (whether yes or no)?

• In such scenarios is the terminology of "basic excitations" of a theory the same thing as single particle states? How are these single particle states in general related to the physical states constructed above?

Which of these is what is referred to when people talk of "modes" of a QFT?

• If there is a gauge symmetry by definition it will commute with the Hamiltonian and hence the states of the theory at every energy level will form a representation of the gauge group. Can something be said about its reducibility or not?

The claim seems to be that if the there are say $n_E$ quanta at the energy level $E$ (transforming under say representation $R_E$ of the gauge group) then when counting its contribution to the partition function the boltzman factor has to further weighted by the number of $1$ dimensional representations ("singlets" ?) in the $n$-fold symmetric (for bosons) or the anti-symmetric (for fermions) tensor power of $R_E$.

I would be glad to know of explanations of the above.

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Maybe I'm missing something obvious but don't physical states in a gauge theory always have to be gauge invariant? –  Tim van Beek Jan 31 '11 at 19:39

It's a lot of questions but they have pretty easy answers, so here they are:

• Gauss's law is just $\mbox{div }\vec D=\rho$ in electrodynamics. Note that it contains no time derivatives so it's not really an equation describing evolution: it's an equation restricting the allowed initial conditions. More generally, it's the equation of motion that you get by varying the Lagrangian with respect to $A_0$, the time component of the gauge field, so the corresponding equation of motion counts the divergence of the electric field minus the electric sources (charge density) that must be equal to it. This difference is nothing than the generator of the overall $U(1)$ group, or any other group, if you consider more general theories, so the classical equation above is promoted to the quantum equation in which $(\mbox{div }\vec D-\rho)|\psi\rangle=0$ which just means that the state $|\psi\rangle$ is gauge-invariant.

• The traces you are encountering here - in non-Abelian theories - are just traces over fundamental (or, less frequently, adjoint) indices of the Yang-Mills group. They're different than traces over the Hilbert space. You must distinguish different kinds of indices. Tracing over some color indices doesn't change the fact that you still have operators.

• "Adjoint of a gauge group" is clearly the same thing as "Adjoint representation of the Lie group that is used for the gauge group."

• It's not true that all compact spaces eliminate all massless fields - for example, the Wilson line of a gauge field remains a perfectly massless scalar field on toroidal compactifications in supersymmetric theories - but in most other, generic cases, it's true that compactification destroys the masslessness of all fields. All the Fourier (or non-zero normal) components of the fields that nontrivially depend on the extra dimensions - the Kaluza-Klein modes - become massive because of the extra momentum in the extra dimensions. But even the "zero modes" become massive in general theories because of the Casimir-like potentials resulting from the compactification.

• Basic excitations are not "the same thing" as one-particle states. In fact, we want to use the word "excitation" exactly in the context when the goal is to describe arbitrary multi-particle states. But the basic excitations are just the creation operators (and the corresponding annihilation operators) constructed by Fourier-transforming the fields that appear in the Lagrangian, or that are elementary in any similar way.

• You haven't constructed any "specific states above" so I can't tell you how some other states you haven't described are related. In this respect, your question remained vague. All of them are some states with particles on the Hilbert space - but pretty much all states may be classified in this way.

• A mode of a quantum field is the term in some kind of Fourier decomposition, or - for more general compactifications and backgrounds - another term (such as the spherical harmonic) that is an eigenstate of energy i.e. that evolves as $\exp(E_n t/i\hbar)$ with time. So for example, a field $X(\sigma)$ for a periodic $\sigma$ may be written as the sum below. The individual terms for a fixed $n$ - or the factor $X_n$ or the function that multiplies it (this terminology depends on the context a bit) - are called the modes. $$\sum_{n\in Z} X_n \,\exp(in\sigma - i|n|\tau)$$

• Gauge symmetry has to commute with the Hamiltonian because we want to ban the gauge-non-invariant states, and by banning them in the initial state, they have to be absent in the final state, too. So it has to be a symmetry. On the other hand, the representation theory is trivial because, as I said at the very beginning, we require physical states to be gauge-invariant; that was the comment about Gauss's law: states have to be annihilated by all operators of the type $\mbox{div }\vec D-\rho$ which are just generators of the gauge group at various points. In other words, all of the physical states have to be singlets under the gauge group. That's why the term "symmetry" is somewhat misleading: some people prefer to call it "gauge redundancy". So the Hilbert space is surely completely reducible - to an arbitrary number of singlets. You may reduce it to the smallest pieces that exist in linear algebra - one-dimensional spaces.

• I just explained you again why all the physical states are singlets under the gauge group. The fact that $n$-particle states with identical particles are completely symmetric or completely antisymmetric reduces to the basic insight that in quantum field theory, particles are identical and their wave function has to be symmetric or antisymmetric (for bosons and fermions) because the corresponding creation operators commute (or anticommute) with each other.

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@Lubos Thanks a lot for this beautiful explanation. You seem to argue that gauge invariance forces each of the "physical states" (over which the partition function sums over) to lie in 1 dimensional sub-representations of the gauge group on appropriately symmetrized tensor power of one-particle Hilbert spaces. right? Can you kindly confirm whether my understanding is right? –  user6818 Feb 1 '11 at 11:47
@Lubos So a collection of n fermions at a certain energy level can lie in any of the 1-dimensional sub-representations of the gauge group on the n-fold anti-symmetric tensor power of the Hilbert space of states of any one of the fermions. So these different singlets of n fermions are indistinguishable in every way? –  user6818 Feb 1 '11 at 11:50
@Lubos Can you explain how this notion of gauge invariant "physical states" (which can contain arbitrary number of particles) different from the terminology of "excitations" which also seem to contains multi-particle states. –  user6818 Feb 1 '11 at 11:53
@Lubos Also if you could give an example of the proper way of writing the indeces of the QFT operators taking into account their gauge indices and Hilbert space indices so that it is clear that tracing over the first kind still leaves us with operators. –  user6818 Feb 1 '11 at 11:57
Dear @Anirbit, I hope that your interpretation is right although you may hide something I can't see. ;-) You just decompose any would-be representation of the gauge group into irreducible reps and physical states may only be the singlets. I don't quite understand what the gauge invariance has to do with the identical nature of fermions, these are two independent rules. Gauge invariance constraints physical states "locally" but when you create several fermions at different places, the total wave functional is still antisymmetric. –  Luboš Motl Feb 1 '11 at 18:31