# Why do Faddeev-Popov ghosts decouple in BRST?

Why do Faddeev-Popov ghosts decouple in BRST? What is the physical reason behind it? Not just the mathematical reason.

If BRST quantization is specifically engineered to make the ghosts decouple, how does this engineering work?

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I am missing any subtlety here. FP ghosts by definition are unphysical states, not part of the physical Hilbert space. BRST quantization is designed specifically to make them decouple. Maybe I am misunderstanding what you are asking. – user566 Feb 10 '11 at 14:11
Sorry to misunderstand things again. This engineering is called BRST quantization, lots of textbooks to read about this. Are you asking about a reference? I don't think any impromptu answer here can compete with some of the excellent sources which exist in the literature. – user566 Feb 10 '11 at 14:48

Dear Student, as Moshe says, the reason why the Faddeev Popov ghosts "decouple" is that they're designed to decouple.

More precisely, they - and all the formulae that depend on them - are designed so that the excitations of these ghosts, as well as unphysical excitations of more ordinary physical fields - such as the time-like and longitudinal components of gauge fields - are cleanly separated from the physical polarizations - such as transverse polarizations of the photons and other gauge bosons.

The addition of the extra BRST ghosts and the BRST charge makes things particularly natural.

The rough reason why the right physical states survive - as BRST cohomologies - while the wrong states don't survive is easily seen in the case of commuting gauge symmetries, generated by generators $L_i$. In that case, the BRST charge is $$Q = \sum_i c^i L_i$$ The ghosts $c^i$ anticommute with each other - and square to zero. You see that the single condition $$Q|\psi \rangle = 0$$ automatically implies that $$L_i |\psi\rangle = 0,$$ among other things, so that the states $|\psi\rangle$ are automatically gauge-invariant. Also, states that are gauge transformations of each other become equivalent if we define $Q$ variations of all states to be unphysical. In particular, if $$|\chi\rangle = \sum_i \lambda^i b_i |\phi\rangle$$ then $$Q|\chi\rangle = \sum_i \lambda^i L_i |\phi\rangle$$ is an arbitrary gauge variation of the $|\phi\rangle$ vector (which was assumed to be annihilated by the $c$'s in this case), which is identified with the vector $0$ in the physical Hilbert space by the BRST cohomology business.

Excitations of $b,c$ themselves are unphysical

The states that contain nothing else than unwanted excitations of $b_i$ fail to be BRST-closed (they are not annihilated by $Q$, much like the components of the gauge field that violate the Gauss's law) while the states that contain just unwanted excitations of $c_i$ are BRST-trivial (much like the pure gauge configurations of the gauge field). So even the excitations of $b,c$ become unphysical, together with two non-transverse excitations of the gauge field.

The whole BRST machinery is only useful for non-Abelian (non-commuting) groups but the machinery above may be easily generalized. A $cbc\cdot f/2$ term ($f$ are the structure constants) has to be added to the BRST charge to preserve $Q^2=0$.

The presentation above is mathematical because the BRST machinery is just a mathematical trick. There is no real physics. Indeed, by the very definition, the whole toolkit added by the BRST formalism is by definition unphysical. It is by definition a mathematical trick. Everyone who is telling you that he or she is discovering new physics by playing with the BRST formalism itself is lying to you; it is not possible. There's no physics in the BRST formalism; it's just a very useful modern tool to do physics.

But let me add one more comment why the unphysical states - the states that are not BRST-closed; and the states that are BRST-exact - decouple. If the initial state at $t=0$ is physical, it remains physics at later moments - because $Q$ commutes with the Hamiltonian $H$.

At $t=0$, it is legitimate to identify physical; states that differ by $Q|\phi\rangle$ for any $|\phi\rangle$ because all these differences of the form $Q|\phi\rangle$ are orthogonal to all physical states $\langle \kappa|$ because $$\langle\kappa|Q|\phi\rangle =0.$$ The latter is true because you may act with $Q$ on the left and the result is zero because $\langle \kappa|$ was BRST-closed by assumption.

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This is an interesting question. And I know that Lubos has already written a nice and complete answer but I think I can add some of my own words here.

I'll concentrate here on the non-abelian YM case, but the results here are quite general. The idea is that the BRST quantum action can be written as:

$$S(qu) = S(cl) + \int\; d^4x\; s\Psi$$

where $\Psi$ is related to a certain surface over which you rotate the antifields of the action, or, as is more commonly stated, to the gauge fixing function. For non-abelian YM, the usual way to write is:

$$\Psi = b_a(F^a+\frac{1}{2}\xi d^a)$$

This terms appears because we insist on describing a massless vector boson as a four component field while there are only two physical degrees of freedom. For external states there is no problem since only the two degrees of freedom appears, and that's ultimately the physical reason why the BRST ghosts are engineered to decouple. For internal lines, ie, propagators, the four components contribute and the extra two non-physical degrees of freedom spoil unitarity. This term above fixes it. However, you don't want your theory to depend on the unphysical choice of $\Psi$.

So, the natural way is to ask that the physical states are in the cohomology of the charge $Q$ associated to the operator $s$, ie, $sF = [F,Q]$ and this implements mathematically the decoupling (the rest follows from what Lubos already said, that one ghost excitation is not BRST-closed and the other's is BRST-exact).

Although everyone calls this action the quantum action, is does not include any quantum term. The argument above will continue to hold quantum mechanically if not only the action but also the measure is BRST invariant. Well, it turns out that the YM measure is not BRST invariant (this is a regulator-dependent statement!!), but one can show that the variation is always $s\chi$, ie, BRST-exact and everything continues to work as in the classical case.

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