BRST Transformation: Notation & Nilpotency of s Operator

[topsquark]

First some basic notation. I am referencing from Weinberg's "The Quantum Theory of Fields" volume II, pg 29. I'm hoping that the notation is relatively standard.

The question refers to the nilpotency of the s operator defined as follows:

The BRST symmetry transformation is parametrized by an infinitesimal constant $$\theta$$ that anticommutes with all ghost ( $$\omega$$ ) and fermionic ( $$\psi$$ ) matter fields.

A fermionic matter field transforms as $$\delta _{ \theta } \psi = i t_ {\alpha} \theta \omega _{ \alpha } \psi$$ and we define the operator s by $$\delta _{ \theta } F \equiv \theta s F$$ where F is a smooth functional of fermionic matter, gauge, ghost, and Nakanishi-Lautrup fields. (In this post I am only going to address the case of F as a single fermionic matter field $$\psi$$. ) Also, the $$t_{ \alpha }$$ form a representation of the Lie algebra of the associated gauge group.

The problem is to show that $$ss \psi = 0$$ . Weinberg starts with the following equation:
$$\delta _{ \theta } s \psi = i t_{ \alpha } \delta _{ \theta } ( \omega _{ \alpha } \psi$$ )

If this equation is taken as a given I can go on to complete the proof. However I have two issues with this equation:
1) I am presuming a typo here. The operator $$\delta _{ \theta } s$$ makes no sense to me. I can only assume that Weinberg meant to write $$\delta _{ \theta } (s \psi )$$. But Weinberg's books have very few typos, so I am not certain about this being an error.

2) By its definition s is clearly a kind of derivative operator. So s acting on a matter field implies that $$s \psi$$ is also a matter field and thus transforms as one under a BRST transformation. Thus I say:
$$\delta _{ \theta } ( s \psi ) = i t_{ \alpha } \theta \omega _{ \alpha } ( s \psi )$$

$$= i t_{ \alpha } \theta \omega _{ \alpha } ( \theta ^{-1} \delta _{ \theta } \psi )$$

$$= i t_{ \alpha } \omega _{ \alpha } \delta _{ \theta } \psi$$

Now
$$\omega _{ \alpha } \delta _{ \theta } \psi = \delta _{ \theta} ( \omega _{ \alpha } \psi ) - ( \delta _{ \theta } \omega _{ \alpha } ) \psi$$

From the BRST transformation of the ghost field $$\omega _{ \alpha }$$ we obtain:
$$\omega _{ \alpha } \delta _{ \theta } \psi = \delta _{ \theta } ( \omega _{ \alpha } \psi ) - \left ( - \frac{1}{2} \theta C_{ \alpha \beta \gamma } \omega _{ \beta } \omega _{ \gamma } \right ) \psi$$
where the $$C_{\alpha \beta \gamma }$$ are the structure constants for the Lie algebra.

Putting it all together gives, finally:
$$\delta _{ \theta } ( s \psi ) = i t_{ \alpha } \delta _{ \theta } ( \omega _{ \alpha } \psi ) + \frac{i}{2} t_{ \alpha } \theta C_{ \alpha \beta \gamma } \omega _{ \beta } \omega _{ \gamma } \psi$$

The first term on the right is what Weinberg states. The second one is a problem. Am I right in saying that the last term is zero due to a combination of the anti-commutivity of the ghost fields and the antisymmetry of the $$C_{ \alpha \beta \gamma }$$ in the $$\beta$$ and $$\gamma$$?

Thanks in advance for the help!

-Dan Boyce

 

[Avodyne]

1) Operators are always assumed to be associative, so $\delta_\theta s\psi = \delta_\theta (s\psi)$.

2) $\delta_\theta$ is a differential operator that obeys the product rule: $\delta_\theta(AB)=(\delta_\theta A)B+A(\delta_\theta B)$. You are missing one of these terms, and when you include it, it will cancel the unwanted term in your result (which does not vanish by symmetry).

 

[topsquark]

1) Operators are always assumed to be associative, so $\delta_\theta s\psi = \delta_\theta (s\psi)$.

Thank you.

2) $\delta_\theta$ is a differential operator that obeys the product rule: $\delta_\theta(AB)=(\delta_\theta A)B+A(\delta_\theta B)$. You are missing one of these terms, and when you include it, it will cancel the unwanted term in your result (which does not vanish by symmetry).

$$\omega _{ \alpha } \delta _{ \theta } \psi = \delta _{ \theta} ( \omega _{ \alpha } \psi ) - ( \delta _{ \theta } \omega _{ \alpha } ) \psi$$

This is the relation I used above. It has all the correct terms. Thank you for letting me know that the asymmetry argument won't work. Back to the drawing board I guess.

-Dan Boyce

 

[Avodyne]

Thus I say:

$$\delta _{ \theta } ( s \psi ) = ... = i t_{ \alpha } \omega _{ \alpha } \delta _{ \theta } \psi$$

This is wrong. In accord with the product rule, it should be

$$\delta _{ \theta } ( s \psi ) = i t_{ \alpha }[(\delta_\theta \omega _{ \alpha })\psi+ \omega _{ \alpha }( \delta _{ \theta } \psi)]$$

 

[topsquark]

This is wrong. In accord with the product rule, it should be

$$\delta _{ \theta } ( s \psi ) = i t_{ \alpha }[(\delta_\theta \omega _{ \alpha })\psi+ \omega _{ \alpha }( \delta _{ \theta } \psi)]$$

Okay, I see where you are. Sorry. This brings up two questions.

First, applying the derivative rule directly
$$\delta _{ \theta } (s \psi) = ( \delta _{ \theta } s ) \psi + s ( \delta _{ \theta } \psi )$$
what I am I to do with the operator $$\delta _{ \theta } s$$ ? I don't know how to transform s?

Second, your comment (and the fact that my derivation gives the wrong answer) implies that $$s \psi$$ is not a fermionic matter field and thus will not transform as one?

Sorry if I seem really muddled about this. There's something that simply isn't clicking here for me.

-Dan Boyce

 

[Avodyne]

First, applying the derivative rule directly

It doesn't work like this. To compute $\delta _{ \theta } (s \psi)$, you first apply $s$, and then $\delta _{ \theta }$.

Second, your comment (and the fact that my derivation gives the wrong answer) implies that $$s \psi$$ is not a fermionic matter field and thus will not transform as one?

Correct. It is (and transforms like) a ghost field times a matter field.

 

[topsquark]

It doesn't work like this. To compute $\delta _{ \theta } (s \psi)$, you first apply $s$, and then $\delta _{ \theta }$.


Correct. It is (and transforms like) a ghost field times a matter field.

Thanks. I'll work with that for a bit.

-Dan Boyce

 

[topsquark]

Oh Heavens. That was almost ridiculously simple. I've got it now. Thanks for your time and patience.

(There's a way to mark this thread as "solved." How do you do that?)