# BRS operator, ghost fields, Grassmann numbers

• christodouloum
In summary: The property you are looking for is nilpotency: s^2 = 0. This is the basic defining property of BRST operator, it's a differential operator with respect to some fermionic ghosts, which is nilpotent. Then there are other properties of it, like homotopy, etc. But I don't have the definitions/axioms at hand, so I can't tell you exactly what you need.
christodouloum
Hi, I am learning what is going on with ghost fields in string theory. Does anyone know where I can find the basics for the brs (brst) operator? Specifically I need help with the following calculation. I have for the action of the brs operator on a ghost field and on the metric
$$s \xi^\alpha = \xi^\beta \partial_\beta \xi^\alpha$$

$$s g_{\alpha\beta}=\Omega g_{\alpha\beta} + L_{\xi} g_{\alpha\beta}$$

Omega is the Weyl field and L_xi the killing derivative.

I am trying to show explicitly that s^2 is zero when acting on the fields. I cannot show it for the ghost field or for the Weyl field Omega for which being a scalar the brs operator acts as

$$s \Omega = \xi^{\alpha} \partial_{\alpha} \Omega$$

I believe it has to do with considering the ghost field as grassmann numbers and exploiting some property but I could not see what is going on only from the basic property x^2=0 and x1 x2=-x2 x1.

Last edited:
<Quantization of gauge systems> by Henneaux and Teitelboim is the best reference on BRST symmetry.

Why can't you show that $s^2 \Omega =0$ ?

bigubau please bare with me I do not want spoon feeding.

This is my first time with grassmann numbers but so far it goes like this

$$s \Omega = \xi^\alpha \Omega_{,\alpha}$$
$$s^2 \Omega = s (\xi^\alpha) \Omega_{,\alpha} + \xi^\alpha (s \Omega )_{,\alpha}$$

I am assuming here that the partial differentiation operator and s commutes (do you agree?) as with any other differential operators I know, I don't have a formal definition of s at hand so I am not sure what is going on.

$$s^2 \Omega = \xi^\beta \xi^\alpha_{,\beta} \Omega_{,\alpha} + \xi^\alpha ( \xi^\beta \Omega_{,\beta} )_{,\alpha}$$

$$s^2 \Omega = \xi^\beta \xi^\alpha_{,\beta} \Omega_{,\alpha} + \xi^\alpha \xi^\beta_{,\alpha} \Omega_{,\beta} +\xi^\alpha \xi^\beta \Omega_{,\beta \alpha}$$

I tried several manipulations to cancel out the terms but no luck. So I believe it has to do with some property of grassmann numbers.

I believe I have to use the property I came up with that since
x1x2=-x2x1

x1x2+x2x1=0

if I have something like x^a x^b A_ab with the components of A_ab real numbers and x^a grassmann numbers then x^a x^b A_ab=0 if A_ab is symmetric. In 2d I get

x^a x^b A_ab= x1x1A11 + x2x2A22 + x1x2A12 + x2x1A21.

The first two terms are zero because xx=0 but for the next two I need A12=A21. This is ok for the last term above with Omega_,ab because partial differentiation commutes. But for the other terms I am not sure how to justify things.

Omega and xi are fermionic variables, which mean that their ANTIcommutator is 0. You've made a mistake in your second equation. The BRST differential is a fermionic derivation, let's choose it acting from the left. That's why:

$$s^2 \Omega = s (\xi^\alpha) \Omega_{,\alpha} - \xi^\alpha (s \Omega )_{,\alpha}$$

That minus makes the world of difference, because it allows to establish the desired nillpotency.

Thanks, so let me see if I got it right. When I wrote

$$s^2 \Omega = \xi^\beta \xi^\alpha_{,\beta} \Omega_{,\alpha} + \xi^\alpha \xi^\beta_{,\alpha} \Omega_{,\beta} +\xi^\alpha \xi^\beta \Omega_{,\beta \alpha}$$

the two plus signs would be minus signs. So the first two terms cancel out (renaming the indices) and the third is zero because

$$\xi_\alpha \xi_\beta \Omega_{,\beta \alpha} = -\xi_\beta \xi_\alpha \Omega_{\beta \alpha}=-\xi_\beta \xi_\alpha \Omega_{,\alpha \beta}= -\xi_\alpha \xi_\beta \Omega_{,\beta \alpha}$$

hence this term is zero. would you think of it in a different way?

I don't think I understand how to derive correctly
$$s^2 \Omega = s (\xi^\alpha) \Omega_{,\alpha} - \xi^\alpha (s \Omega )_{,\alpha}$$
with a minus from the anticommutativity of omega and xi alone. Is this a matter of definition? Can you give me a defining property of fermionic derivation operators so that I can derive these things?

Thanks again

ok I am getting really confused.

A is grassmann (fermionic field). So A^2=0. For a derivation operator acting "normally"

d(A^2)=dA A + A dA =0 thus dA A=-A dA which is a property of grassmann numbers (see http://www.mathematics.thetangentbundle.net/wiki/Grassmann_number" )

but if I make the derivation using the rule you gave me for a fermionic operator

s(A^2)=sA A - A sA =0 thus sA A=A sA and so
s(A^2)=2 sA A=0

thus sA A=A sA =0

??

Last edited by a moderator:
christodouloum said:
Thanks, so let me see if I got it right. When I wrote

$$s^2 \Omega = \xi^\beta \xi^\alpha_{,\beta} \Omega_{,\alpha} + \xi^\alpha \xi^\beta_{,\alpha} \Omega_{,\beta} +\xi^\alpha \xi^\beta \Omega_{,\beta \alpha}$$

the two plus signs would be minus signs. So the first two terms cancel out (renaming the indices) and the third is zero because

$$\xi_\alpha \xi_\beta \Omega_{,\beta \alpha} = -\xi_\beta \xi_\alpha \Omega_{\beta \alpha}=-\xi_\beta \xi_\alpha \Omega_{,\alpha \beta}= -\xi_\alpha \xi_\beta \Omega_{,\beta \alpha}$$

hence this term is zero. would you think of it in a different way?

No, excellent.

christodouloum said:
I don't think I understand how to derive correctly
$$s^2 \Omega = s (\xi^\alpha) \Omega_{,\alpha} - \xi^\alpha (s \Omega )_{,\alpha}$$
with a minus from the anticommutativity of omega and xi alone. Is this a matter of definition? Can you give me a defining property of fermionic derivation operators so that I can derive these things?

I don't have Henneaux's book, nor the notes from the courses I took on BRST, so I can't give you the full definitions/axioms. So take them like that and I will update you on this at a certain point in the future.

christodouloum said:
ok I am getting really confused.

A is grassmann (fermionic field). So A^2=0. For a derivation operator acting "normally"

d(A^2)=dA A + A dA =0 thus dA A=-A dA which is a property of grassmann numbers (see http://www.mathematics.thetangentbundle.net/wiki/Grassmann_number" )

but if I make the derivation using the rule you gave me for a fermionic operator

s(A^2)=sA A - A sA =0 thus sA A=A sA and so
s(A^2)=2 sA A=0

No, it's wrong, where did you get that ? s(A^2) = s(0) = 0 = (sA)A-A(sA), because both s and A are fermionic => (sA)A=A(sA) which brings nothing new, because it's obvious from sA being a bosonic variable.

Last edited by a moderator:
Forget about my last post I was getting tired... thanks for all the help I am much better off now. I will look for the book you proposed

Firstly after some studying the fermionic operators are Z2-graded (even-odd rank for a differential form) differential operators, and s here was a left such derivative. That means that depending on the grade of the two forms a,b in s(ab) the sign is decided as + if a is of even rank and - if odd.

Second, I noticed something interesting related to our discussion before. A p-form is defined as fully antisymmetric in its arguments, which for a 2-form translates as antisymmetry in the two indices of the components:

$$/omega_{/alpha /beta}=-/omega_{/beta /alpha}[\tex] and so I noticed that the contraction of a 2-form with any two vector fields would vanish because [tex]/omega_{a b} u^a v^b=-/omega_{ b a} u^a v^b=-/omega_{b a} u^b v^a=-/omega_{a b} u^a v^b[\tex] which is the same think tha happened two posts above it doesn't vanish, the 2 form picks up the antisymmetrized product of the 2 vectors. Please, use$$

## 1. What is a BRS operator?

A BRS operator is a mathematical operator used in quantum field theory to describe symmetries of the theory. It stands for Becchi-Rouet-Stora (BRS) operator, named after the physicists who first proposed it. It is a tool for studying the symmetries of gauge theories, which are fundamental to our understanding of the fundamental forces of nature.

## 2. What are ghost fields?

Ghost fields are mathematical constructs used in quantum field theory to help maintain the symmetry of the theory. They are associated with gauge symmetries and do not represent physical particles, but rather serve to maintain the consistency of the mathematical formalism.

## 3. How do Grassmann numbers relate to quantum field theory?

Grassmann numbers are mathematical objects that are used to describe fermionic fields in quantum field theory. They are anticommuting numbers, meaning that they do not follow the usual rules of multiplication, and are essential for describing the behavior of quantum particles with half-integer spin, such as electrons and quarks.

## 4. What is the significance of the BRS operator, ghost fields, and Grassmann numbers in quantum field theory?

The BRS operator, ghost fields, and Grassmann numbers are all important mathematical tools in quantum field theory. Together, they help us understand the symmetries and interactions of fundamental particles and fields, and allow us to make precise predictions about their behavior. They are crucial for the development of theories such as the Standard Model, which describes the fundamental particles and forces of nature.

## 5. Are the BRS operator, ghost fields, and Grassmann numbers relevant in other fields of science?

While these mathematical concepts were originally developed for quantum field theory, they have found applications in other areas of physics, such as condensed matter physics and string theory. They have also been used in other fields such as mathematics and computer science. Their significance extends beyond just quantum field theory and demonstrates the power and versatility of these mathematical tools.

• Quantum Physics
Replies
16
Views
1K
• Quantum Physics
Replies
4
Views
2K
Replies
0
Views
538
• Quantum Physics
Replies
5
Views
1K
• Quantum Physics
Replies
9
Views
2K
• Quantum Physics
Replies
7
Views
3K
• Special and General Relativity
Replies
1
Views
651
• Differential Geometry
Replies
3
Views
1K
• Quantum Physics
Replies
1
Views
1K
• Quantum Physics
Replies
3
Views
3K