Calculations with Weyl Spinor Indices in QFT

nicouh
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Homework Statement


The task is to show the invariance of a given Lagrangian (http://www.fysast.uu.se/~leupold/qft-2011/tasks.pdf" ), but my problem is just in one step (which i got from Peskin & Schröder, page 70) which i can not reproduce due to my lack of knowledge regarding spinors.

The step i am talking about is 3.147 in the attached picture or written out:


Homework Equations



C \bar \psi \psi C = (-i \gamma^0 \gamma^2 \psi)^T (-i \bar \psi \gamma^0 \gamma^2)^T
= -\gamma^0_{ab}\gamma^2_{bc} \psi_c \bar \psi_d \gamma^0_{de}<br /> \gamma^2_{ea}
= \bar \psi_d \gamma^0_{de} \gamma^2_{ea} \gamma^0_{ab} \gamma^0_{bc} \psi_c
= -\bar \psi \gamma^2 \gamma^0 \gamma^0 \gamma^2 \psi
= \bar \psi \psi

The Attempt at a Solution


Well, i browsed through Wikipedia, Google and Friends, but did not find anything.
I know how to handle the gamma matrices (like their commutation relations or (\gamma^0)^2 = 1) .
I have just no clue what these indices mean, why they are sorted the way they are ("ab bc cd de" instead of e.g. "ab cd ef gh") and how \psi^T gets converted to \psi.

Thanks!
Regards,
Nico
 

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It's been years since I've done any QFT, and I don't have any books here with me right now, so what I write might be completely wrong.

First, consider some linear algebra. If A is a square matrix, then \vec{w} = A \vec{v} becomes

w_i = \sum_j A_{ij} v_j

in component form. It looks like the matrix multiplications have been written in component form with the summation symbols omitted, which is why there are repeated indices.

Next, is it true that for a scalar that

C \bar{\psi} \psi C = \left( C \bar{\psi} \psi C \right)^T ?

Can you use this to do the calculation more simply, taking into account that fermion fields anticommute?
 
George Jones said:
First, consider some linear algebra. If A is a square matrix, then \vec{w} = A \vec{v} becomes

w_i = \sum_j A_{ij} v_j

in component form. It looks like the matrix multiplications have been written in component form with the summation symbols omitted, which is why there are repeated indices.

Thanks for your answer!
At i thought it was the standard matrix multiplication, too. But as you wrote it, it already begins with an index. So its
w_i = \sum_j A_{ij} v_j instead of
w = \sum_j A_{ij} v_j, which just confuses me.

George Jones said:
Next, is it true that for a scalar that

C \bar{\psi} \psi C = \left( C \bar{\psi} \psi C \right)^T ?

Can you use this to do the calculation more simply, taking into account that fermion fields anticommute?

Yeah, i think so, too. Since its a scalar transposing it shouldn't change anything :P
But i haven't come really far doing that task. I did others first, but ended up at the same spot with the same problem eventually. The same thing even appeared a few times more now.
 
nicouh said:
At i thought it was the standard matrix multiplication, too. But as you wrote it, it already begins with an index.

If A and B are matrices, how is the product AB written using indices?
 
ah, i think i got it.
yeah, since (as i just mentioned in my previous post :P) the end product is a scalar, it makes sense to write it this way and it really is just a index notation for a vector-many matrices-vector product. a normal scalar product of 2 vectors would have the same features when looked at it written out in components.
im kind of taking a break right now, but when i start again and still get stuck, i might will ask you further questions :P
thanks!
 
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