In the standard explanation, the physical Hilbert space of a quantized gauge theory (such as QCD) is given by the cohomology of the BRST charge acting on some larger, unphysical Hilbert space.

More precisely, for each Cauchy surface $\Sigma$, there is an unphysical Hilbert space $\mathcal{H}^\Sigma$, and an operator $Q^\Sigma_{BRST}$ acting on $\mathcal{H}^\Sigma$ obtained by integrating the normal component of the conserved current $j^\mu_{BRST}$ over $\Sigma$. Then the physical Hilbert space is $\mathcal{H}^\Sigma_{phys}:= \ker Q^\Sigma_{BRST}/\text{im }Q^\Sigma_{BRST}.$

Unfortunately, I have no intuition for why this is the correct construction.

Firstly: what is $\mathcal{H}^\Sigma$? Is it the space of wavefunctionals taking classical gauge field configs on $\Sigma$ as their arguments? This is what I'd guess having studied the scalar field.

Secondly: is there an alternate, more intuitive, characterisation of $\mathcal{H}^\Sigma_{phys}$, say in terms of wavefunctionals?

Note: a similar question was asked before, and the top answer says the following:

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... This space is ghost-free.

This promises exactly what I want: an alternate characterisation of the physical states of a gauge theory. However, I don't understand the quoted answer at all, and I can't find a similar explanation anywhere else, so I'm not even sure it's correct.

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    $\begingroup$ The role of wavefunctionals should be the same as what it is in a non-gauge theory. But now physical states are equivalence classes of states that differ by a gauge transformation. $\endgroup$ Commented Dec 12, 2021 at 13:54
  • $\begingroup$ @Connor Behan This sounds very appealing, but I've not seen it stated this way anywhere else. Could you provide a reference? In particular, does this agree with the BRST procedure? $\endgroup$ Commented Dec 12, 2021 at 16:34
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    $\begingroup$ It's more of a general takeaway from various gauge theory chapters than something I saw explicitly in print. But it looks like physics.stackexchange.com/questions/420442/… discusses this. $\endgroup$ Commented Dec 12, 2021 at 17:57
  • $\begingroup$ I find it hard to believe that this really gives us the same Hilbert space as the BRST cohomology way. The method you suggest seems very simple and well motivated, where BRST is totally opaque to me. $\endgroup$ Commented Dec 12, 2021 at 19:05

1 Answer 1


Forget about BRST, gauge theories, etc. and just look at the humble photon: The photon is massless and exists in two polarizations. The free one-particle space can therefore just be built from the usual $\lvert p,\zeta\rangle$, where $p$ is momentum and $\zeta$ a compatible polarization, just like you would build the one-particle space for a massive particle from $\lvert p,\sigma\rangle$ with $\sigma$ the spin. The Fock space on this is just the usual Fock space. There is no trace of gauge symmetry on this space because that's the whole point of constructing the "physical" space - to find the space where all gauge-equivalent states are collapsed into one state and where only gauge-invariant objects exist.

Furthermore, the two polarizations are created from the transversal modes of the gauge field - the gauge symmetry "just" means that the time-like and longitudinal modes don't create anything physically relevant. So this is just like for other free fields, you just have to ignore some modes, and by a general argument you can say that the one-particle wavefunctions obey the same equation as the free gauge field.

This was easy, so what's the deal with BRST? The BRST procedure is far more general than (free Abelian) gauge fields, and in principle allows you to quantize arbitrary constrained Hamiltonian systems. You find it difficult to grasp "intuitively" because it is not concerned with any particular physical situation - it quantizes a generic constrained system, not even necessarily one with "particles". It is the correct quantization procedure because a) you can actually derive it from general considerations, see e.g. "Quantization of Gauge Systems" by Henneaux and Teitelboim, and b) it coincides in specific cases with the "known good" quantizations like Gupta-Bleuler quantization of QED - see this answer of mine for a more explicit demonstration that BRST and Gupta-Bleuler produce the same space.

Note also that my argument here started from the particles. One can actually show that the quantum field associated with all massless vector bosons must be a gauge field, see Weinberg's "Quantum Theory of Fields" Vol. I. And it is only there that the gauge symmetry enters - when constructing the field operator, not when looking at the space of states. But our usual way of thinking about quantization does not start with particles and then constructs the fields (though this is very much Weinberg's approach, and it is worthwhile to look at QFT from this perspective at least once), it starts with the fields and then looks for states and particles. That's what quantization procedures like Gupta-Bleuler for QED and BRST for general gauge theories do - they tell you how to get to the space of states just from the fields and the Hamiltonian.

  • $\begingroup$ Just to clarify: we have two proposals for $\mathcal{H}_{phys}$. Option A is $\mathcal{H}_{phys}=$ im $Q$/ker $Q$, the BRST cohomology. Option B is that $\mathcal{H}_{phys}$ is the space of wavefunctionals taking gauge equivalence classes of gauge fields on $\Sigma$ as their input (as suggested by Connor Behan in the comments). Call these spaces $\mathcal{H}_A$ and $\mathcal{H}_B$. Do you claim that $\mathcal{H}_A=\mathcal{H}_B$? Or that they're isomorphic, perhaps via some obvious isomorphism? $\endgroup$ Commented Dec 13, 2021 at 16:24
  • $\begingroup$ @nodumbquestions I'm not talking about "wavefunctionals" here at all (though you could probably express the $\lvert p,\zeta\rangle$ as wavefunctionals, sure), I'm just saying the "naive" or "experimental" space built from particles $\lvert p,\zeta\rangle$ (which is the Fock space the "top answer" you link in your question) is the same as the BRST space. For a more explicit demonstration that the BRST space is indeed the same as the Gupta-Bleuler space in QED, see this answer of mine $\endgroup$
    – ACuriousMind
    Commented Dec 13, 2021 at 16:48
  • $\begingroup$ Yes, I appreciate that you didn't bring up wavefunctionals in your answer, so strictly speaking this is less a clarification and more a separate follow-up question. If it were the case that $\mathcal{H}_A=\mathcal{H}_B$, this would give me a lot of new intuition and understanding of BRST. But if $\mathcal{H}_A\neq\mathcal{H}_B$, then your answer contradicts Connor Behan's answer in the comments above. My naive guess is that $\mathcal{H}_A$ and $\mathcal{H}_B$ are "getting at the same thing", even if they're not precisely equal. But this is just a hand-wavey guess. $\endgroup$ Commented Dec 13, 2021 at 17:24

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