Hilbert Space of (quantum) Gauge theory

Question

Since quantum Gauge theory is a quantum mechanical theory, whether someone could explain how to construct and write down the Hilbert Space of quantum Gauge theory with spin-S. (Are there something more rich/subtle than just saying the Hilbert Space are composed by the state space of infinite many sets and infinite many modes of harmonic oscillators? - i.e. more rich/subtle than usual spin-0 scalar fields' Hilbert space?) Is the quantum Gauge theory's Hilbert Space written in a tensor product form or not (eg. thinking about put this gauge theory on the lattice)?

Whether there are differences for this procedure construction of Hilbert space for these three cases:

(1) spin-1 quantum Gauge theory with Abelian $U(1)$ symmetry

(2) spin-1 quantum Gauge theory with non-Abelian (such as $SU(N)$) symmetry

(3) spin-2 quantum Gauge theory (Gravity? or anything else)

Also, whether gauge-redundancy plays any roles? Is there similar thing like Faddeev-Popov ghosts happened in the path integral formalism, when one dealing with gauge-redundancy?

Answer 1

In the noninteracting case, the Hilbert space appropriate for a gauge field theory of any spin is a Fock space over the 1-particle space of solutions of the classical free gauge field equations for the same spin. (For spin 1, the associated particles would be noninteracting gluons if these would exist.) This space is ghost-free. The differences for different spins just lie in the different structure of the classical field equations.

This Hilbert space can be described in many different but equivalent forms.

As the other answer mentions, it is usually represented by means of BRST cohomology, since this gives the most tracable renormalized perturbation theory. Here ghosts appear since the BRST Hilbert space is embedded in a bigger indefinite-inner-product space. (The ghost-free physical Hilbert space is recovered as the quotient of the kernel of the BRST charge $Q$ by the image of $Q$. Since $Q$ satisfies $Q^2=0$, it is analogous to the exterior derivative $d$, which satisfies $d^2=0$, and gives rise to BRST cohomology in the same way as $d$ gives rise to traditional de Rham cohomology.)

For an abelian gauge field theory, there are more elementary descriptions of the noninteracting Hilbert space. For example, a free quantum electromagnetic field resides in the Hilbert space of square integrable functions $A(p)$ of light cone momentum $p$ ($p^2=0,p_0>0$) in a degenerate but positive semidefinite inner product where all functions with $A(p)$ parallel to $p$ have norm zero. This gives the standard description of photons in quantum optics. (See, e.g., the entry ''What is a photon?'' in Chapter B2: Photons and Electrons of my theoretical physics FAQ at http://arnold-neumaier.at/physfaq/physics-faq.html .

The Hilbert space appropriate for an interacting gauge field theory is unknown. This is not surprising as it is unknown for any interacting field theory in 4D. The little that is known about the situation in this case can be found in a recent book by F. Strocchi, An introduction to non-perturbative foundations of quantum field theory, Oxford Univ. Press, 2013.

Answer 2

The Hilbert space of a gauge theory is defined by BRST symmetry or to be precise, BRST cohomology.

In the path integral formalism, it is necessary to introduce ghosts in order to fix the gauge for a non-abelian theory. This theory now contains states of negative norm, hence it is a pseudo-Hilbert space. The Lagrangian of this theory has an additional symmetry, i.e. a symmetry in which the ghost fields act as infinitesimal parameters. Associated with this symmetry are both a Noether current and a Noether charge, the latter is referred to as the BRST charge. The BRST charge is a nilpotent operator, i.e. $Q^2=0$. This behaviour allows one to define a cohomology, which can be understood as follows:

Since the BRST charge is a quantum operator, we can ask ourselves what happens when we let it act on some state $|\Psi\rangle$. Since the operator $Q$ is nilpotent, $Q|\Psi\rangle=0$ if $|\Psi\rangle$ can be written as $|\Psi\rangle=Q|\Phi\rangle$, i.e.

$$Q|\Psi\rangle=Q^2|\Phi\rangle=0.$$

But there is also the possibility of states vanishing under the action of the BRST charge without them being defined by $Q|\Phi\rangle$. Such states are said to be in the cohomology of the charge operator. They are identified as the physical states of the theory and do not contain ghosts or antighosts. Furthermore, one can argue that the cohomology does not change under unitary time evolution due to the fact that the Hamiltonian commutes with $Q$.

The BRST formalism also works for string theory, which contains spin 2 particles, i.e. gravitons.