There is a point which confuses me about BRST procedure. One shows that, if we define physical states as the ones that are annihilated by BRST charge $Q$, the scattering amplitudes don't depend on gauge fixing function. But doesn't charge $Q$ depend on the auxiliary field $B$, and therefore (after integrating it out) on gauge fixing function?
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The short answers are:
Yes, the non-minimal BRST charge $Q$ does depend on the Lautrup-Nakanishi (LN) auxiliary field $$.
No, the BRST charge $Q$ does not depend on the gauge-fixing condition (which instead is part of the gauge-fixing fermion $\psi$).
One of the benefits of the BRST formalism, is that it shows formally that the path integral $Z$ doesn't depend on the gauge-fixing fermion $\psi$, cf. e.g. this post.
Integrating out the $B$-field implies that $Q^2$ only vanishes on-shell. However, integrating out fields cannot spoil the gauge-fixing independence of pt. 3.
Qmechanic
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1-) Can you explain how integrating out Lautrup-Nakanishi field is realized in the operator formalism ? What does it really mean in the operator formalism ? 2-) After integrating out the auxiliary field, why don't we have an expression for BRST charge which depends on gauge fixing ? More precisely, in the path integral, if I change gauge-fixing fermion, and then integrate out the auxiliary field, why don't the expression for BRST charge (in terms of gauge fields and (anti)ghost fields) doesn't change ? – Bronsteinx Feb 18 '19 at 23:08
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@Bronsteinx in the operator formalism the equations of motion are automatically satisfied (in Heisenberg picture), so the $B$ field will be identical to some expression of the other fields. This identity comes from the equations of motion of the other fields in the Lagrangian. In the case of string theory this $B$ field will be identically the energy momentum tensor, and this identity comes from the equation of motion for the world-sheet metric. – Nogueira Feb 18 '19 at 23:20