How Is the BRST Charge Derived in Quantum Field Theory?

[ismaili]

I would like to derive the explicit formula of the BRST charge.
http://en.wikipedia.org/wiki/BRST_formalism
(Bottom of Wiki link, I copy the formula here)
[tex]Q = c^i\left(L_i - \frac{1}{2}f_{ij}{}^kb_jc_k\right)[/tex]
where [tex]c[/tex] is the ghost field, and [tex]b[/tex] is the antighost field, [tex]L_i[/tex] is the gauge group generator.
Actually, in wiki's article, right above the formula of BRST charge, there is a Lagrangian. I tried to use Noether theorem to calculate the charge, but in vain. [tex]J^\mu_a\sim \frac{\partial\mathcal{L}}{\partial\psi_{,\mu}}\delta\psi_a[/tex], replacing the [tex]\psi[/tex] with the ghost [tex]c[/tex], I ends up with
[tex]J^0 = \dot{b}\,\delta c = \dot{b}\left(-\frac{1}{2}f_{ij}{}^kc^ic^j\right)[/tex] whose volume integral looks different as the correct answer.
In there anybody who can help me or give me some hints?
Many thanks!

 

[AndyW]

Thank you for your interest in deriving the explicit formula of the BRST charge. The formula you have provided is indeed the correct one, and it can be derived using the Noether theorem. However, there are a few key points to keep in mind when using the Noether theorem to derive the BRST charge.

Firstly, the BRST charge is a conserved current, not a conserved charge. This means that it is a function of space and time, not just a constant. Therefore, the Noether theorem should be applied to the Lagrangian density, not just the Lagrangian.

Secondly, the BRST charge is associated with a gauge symmetry, and the Noether theorem is usually applied to symmetries that leave the action invariant. However, in the case of gauge symmetries, the action is not invariant, but rather changes by a total derivative. This means that the Noether current is not conserved, but rather has a divergence related to the gauge transformation. In the case of the BRST charge, this divergence is precisely the term involving the ghost and antighost fields in the formula you have provided.

To derive the BRST charge, you can follow these steps:

1. Start with the Lagrangian density given in the Wikipedia article, which includes the ghost and antighost fields.

2. Apply the Noether theorem to the Lagrangian density to obtain the Noether current, which will include the ghost and antighost fields.

3. Use the gauge transformation rules for the fields to simplify the Noether current and eliminate any terms that are total derivatives.

4. The remaining terms in the Noether current will give you the BRST charge, up to a possible normalization factor.

I hope this helps to clarify the process of deriving the BRST charge using the Noether theorem. If you have any further questions, please do not hesitate to ask. Best of luck with your research!

 

[sir-pinski]

The derivation of the BRST charge involves several steps, so I will try to explain it as clearly as possible.

First, we start with the BRST transformation, which is given by:
δϕ = [Q, ϕ]
where Q is the BRST charge and ϕ is any field in the theory. This transformation is meant to be a symmetry of the theory, which means that the action S should be invariant under this transformation.

Next, we introduce the ghost fields c and b, which are fermionic fields, and the antighost field b, which is a bosonic field. These fields are introduced to ensure that the BRST transformation is nilpotent, meaning that applying it twice should give zero:
[Q, [Q, ϕ]] = 0

Now, we can write the BRST transformation explicitly as:
δAμ = Dμc
δc = - 1/2 [c, c]
δb = -b^2
δΨ = -i [c, Ψ]
where Aμ represents the gauge fields, Ψ represents the matter fields, and Dμ is the covariant derivative.

Next, we introduce the gauge group generators L_i, which are the Noether charges associated with the gauge symmetries of the theory. These generators satisfy the commutation relations:
[L_i, L_j] = f_{ij}{}^k L_k
where f_{ij}{}^k are the structure constants of the gauge group.

Now, we can use the Noether theorem to find the BRST charge Q. The Noether current for a continuous symmetry is given by:
J^μ_a∼ ∂L/∂(∂μΨ_a) δΨ_a
where Ψ_a is a field that transforms under the symmetry. In our case, Ψ_a = c, so we have:
J^0 = ∂L/∂(∂_0c) δc
Substituting the BRST transformation for δc, we get:
J^0 = ∂L/∂(∂_0c) (-1/2 [c, c])
Now, we need to calculate the Lagrangian L. This can be done by considering the gauge invariant part of the action, which is given by:
S = ∫d^4x (L_gauge + L_matter)
where L_g