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I've been reading about gauge theory quantization, and understand it mostly. The only thing I don't get is why people talk about "ghost number conservation".

As far as I can tell, the ghost number is a quantity defined to be $+1$ for ghost and $-1$ for antighost fields. It is then claimed that total ghost number is conserved for the action. I don't understand what is means for something to be "conserved for an action". I only know what it means for something to be conserved in classical or quantum evolution. Alternatively I know what it means for an action to be invariant under a transformation.

Is this just bad nomenclature? I'm worried about it for the following reason.

In the quantum theory there is a quantum "ghost number" operator which operates on the Fock space of the theory and tells you the cumulative ghosts $-$ antighosts in your state. I've seen it argued that this must commute with the Hamiltonian because "ghost number is conserved in the action". This seems pretty fundamental, but the logic seems flawed because I don't understand the terminology.

Maybe everyone just means that the ghost number is $0$, which is obvious. But this seems like too simple an explanation (why would people call it conserved, when $0$ is significantly shorter?!)

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"Something is conserved for an action" simply means that the action carries a zero overall value of "something" (for an additive quantity). In this case, the action has $N_{gh}=0$. It follows that the equations of motion derived from the action imply $dN_{gh}/dt=0$. Most typically, they imply $\partial_\mu J^\mu_{gh}=0$ i.e. the local continuity law for a whole current.

The action has $N_{gh}=0$ because each term contains the same number of ghost factors as antighosts. The symmetry generated by $N_{gh}$ is a $U(1)$ symmetry adding a phase $\exp(i N_{gh} \alpha)$.

We use the terminology "the ghost number is conserved" to emphasize that the ghost number generates a symmetry. The proposition "the value of the ghost number is zero" wouldn't make it clear that the quantity is measured in the same way as potentially conserved charges are measured.

The conservation applies to each term in the action (kinetic terms and interaction terms: they produce propagators and vertices in Feynman diagrams, respectively). Saying that something is zero has a different flavor in it. We usually say that something is zero when we talk about properties of a particular state or a configuration, not about properties of the action (properties of the laws of physics). The right words for the property of the laws of physics (or the action/Lagrangian) is "conserved".

Moreover, in those cases, "zero" is just another possible value. That differs from the word "conserved" which is qualitatively different from any other value that would correspond to "not conserved".

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  • $\begingroup$ Ah right - it was the bit about the $U(1)$ symmetry that I didn't really get. But I guess in retrospect it's pretty obvious that any real valued quantity defined on fields which is zero for the action (in part or in full) gives a $U(1)$ symmetry in exactly the way you describe. It's just completely analogous to the concept of number operator for (say) Dirac fermions. So am I correct in thinking that the ghost number operator will just be given by quantizing the conserved charge associated with your $U(1)$ symmetry, just like for Dirac fermions? Cheers! $\endgroup$
    – Edward Hughes
    Commented Oct 13, 2013 at 21:40
  • $\begingroup$ Dear Edward, for the symmetry to be $U(1)$ which is compact, the conserved quantity can't be real-valued. It has to be integer-valued in some units (i.e. an integer multiple of a "quantum")! If it is continuous real-valued, the group is a noncompact version of $U(1)$ which is $R^+$ (multiplication of real positive numbers) or, when converted to an additive group by a logarithm, $R$. Yes. the ghost number is the generator of the $U(1)$, it's the charge, it's always the same operator. $\endgroup$
    – Luboš Motl
    Commented Oct 14, 2013 at 6:35
  • $\begingroup$ Apologies - I meant real-valued in the classical theory. Obviously it's integer-valued in the quantum theory. Just one final question. Do you know if one has to impose $Q_{gh}\psi = 0$ as well as the BRST cohomology to get physical states? In other words does there exist a state which is not BRST exact but nevertheless has nonzero ghost number? I'm not quite sure how I'd construct such an example. Many thanks for your help! $\endgroup$
    – Edward Hughes
    Commented Oct 14, 2013 at 9:04

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