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I was studying the path integral quantization of non-abelian gauge field. After the path integral quantization, the action becomes $$\mathcal{L}=-\frac{1}{4}F^a_{\mu\nu}F^{a\mu\nu}-\frac{1}{2\zeta}(\partial_{\mu}A^{\mu a})^2 +\partial^\mu\bar c^a(D_{\mu}c)^a$$

Feynman rules for antighost-ghost-gauge boson vertex (e.g. $SU(2)$ gauge boson) is $-g \epsilon_{abc}p_\mu$. I can't understand why there is no process with ghost in external line.(For example, two gauge bosons annihilate and become ghost and antighost. I can certainly calculate the amplitude for this diagram according to Feynman rules. )

Certainly I know ghost is unphysical so should not exist in external line. But I want to know whether no external ghost line can result from the theory itself (like the amplitude of this process can be canceled by some other process?) or is just an axiom that we put on this theory?

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Qmechanic is right, but his answer doesn't explain why we can't just consider the ghosts as physical and be done with it.

There are two main reasons why ghosts can't be considered physical.

  1. They violate spin-statistics (ghosts are scalar fermions).
  2. The S-matrix operator as it stands isn't unitary.

The problem can be traced back to the kinematics of gauge-fixing. Remember how in the $U(1)$ case we had a gauge-fixing constraint (e.g. Lorentz gauge condition), which after implementing it as a quantum operator constraint $C$ selected a unique subspace $\text{ker} C$ of physical states? Well, here we deal with a similar situation, plagued by additional technical difficulties because the group is non-abelian.

The Fock space of the gauge+ghost system of the Lagrangian mentioned in your question isn't physical. It contains negative-norm states (just like in the $U(1)$ case). As an example of the negative-norm state, consider a timelike polarized gauge boson $$ a_0^{\alpha\,\dagger} \left| 0 \right>.$$

Just like in the $U(1)$ case this can be solved by implementing a gauge condition constraint as a quantum operator and solving. However, we run into the following complication:

Solutions of the constraint no longer decompose into the physical and spurious (zero-norm) subspaces which can be treated separately, because dynamics of the theory mixes these two subspaces.

This can be traced back to the following fact: the current-conservation law contains a covariant derivative instead of the ordinary one, while the Lorentz gauge condition still operates with an ordinary partial derivative.

This difficulty can be successfully solved with help of BRST quantization technique. Existence of ghosts is essential for the BRST to work.

In conclusion: the S-matrix given by the quantization of the Lagrangian from your question gives the correct quantum dynamics of the quantum gauge field, but only when projected to a subspace of the naive Fock space given by the BRST cohomology. It also has a nice property of not mixing physical and unphysical degrees of freedom, meaning that we can use its full form in practical computations and only project down to the physical subspace afterwards.

That we can't use the extended Fock space is already obvious because of the two reasons given in the beginning of my answer.

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  • $\begingroup$ So do you mean that Faddeev-Popov path-integral quantization itself is not complete? It doesn't require there is no external ghost particle. It's like that path integral quantization of Maxwell field can only give us the Feynman rules but doesn't forbid external longitudinal photon. It's the Gupta-Bleuler method that requires the physical states. In nonabelian case, the BRST is something like Gupta-Bleuler method? $\endgroup$ – user153663 Apr 25 '17 at 18:44
  • $\begingroup$ @fff123123 path integral quantization is never complete on its own. Path integrals are simply a covariant way of writing transition amplitudes between states. They have to be completed with a comprehensive description of the Hilbert space and operators associated to physical observables. Yes, BRST is something like Gupta-Bleuler for the NA gauge theory. $\endgroup$ – Prof. Legolasov Apr 25 '17 at 20:33
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On one hand, the S-matrix does not depend on the gauge-fixing condition. On the other hand, there exist a unitary gauge, where Faddeev-Popov ghosts decouple from the theory.

References:

  1. M.D. Schwartz, QFT and the Standard Model, 2014; Section 28.4.

  2. C. Itzykson & J.B. Zuber, QFT, 1985; Subsection 12-5-5.

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  • $\begingroup$ Thanks for your answer. May you explain more explicitly? Or just tell me some reference where I can find your saying. Really thanks. $\endgroup$ – user153663 Apr 25 '17 at 3:39
  • $\begingroup$ I updated the answer with some references. $\endgroup$ – Qmechanic Apr 25 '17 at 9:41

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