I Lorentz transformation of the "bilinear spinor matrixelement"

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The discussion centers on the Lorentz transformation of a bilinear spinor matrix element, specifically addressing confusion around the transformation of the matrix elements and the role of the gamma matrices. Key questions include the transition from Γ12T to ΓμνT and the origin of the Γ02T element. Participants explore the transformation process using Lorentz matrices and the implications of index ordering. A breakthrough occurs when one participant realizes the correct application of the transformation, leading to a clearer understanding of the relationships between the matrices. The conversation concludes with a sense of accomplishment in resolving the complexities of the problem.
Garlic
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Dear reader,

there is a physics problem where I couldn't understand what the solutions.
It is about the lorentz transformation of a bilinear spinor matrix element thing.

So the blue colored equation signs are the parts which I couldn't figure out how.
There must be some steps in between which were skipped on the solution.

For example, we know that the original matrixelement contains Γ12T. Why does it turn back to ΓμνT? Where does the Γ02T element come from?

And lastly, why did we have to transpose the Γ's in the end, so that the indices get to the top?

Thank you very much for your time :)

-Garlic
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Garlic said:
For example, we know that the original matrixelement contains Γ12T. Why does it turn back to ΓμνT?

First question first. Suppose
$$A_{\mu \nu} =\bar{\psi} \Gamma^{T}_{\mu \nu} \psi.$$
In the primed frame, what is ##A'_{\alpha \beta}##?
 
George Jones said:
First question first. Suppose
$$A_{\mu \nu} =\bar{\psi} \Gamma^{T}_{\mu \nu} \psi.$$
In the primed frame, what is ##A'_{\alpha \beta}##?

I suppose:
$$
A'_{\alpha \beta}= \Lambda_{\alpha}^{\: \mu} \Lambda_{\beta}^{\: \nu} A'_{\mu \nu} = \Lambda_{ \alpha}^{\: \mu} \Lambda_{\beta}^{\: \nu} \bar{\psi} \Gamma^{T}_{\mu \nu} \psi
$$

So M should transform like this:
$$
M_{\alpha \beta} \rightarrow M'_{\alpha \beta}= \Lambda_{\alpha}^{\: 1} \Lambda_{ \beta}^{\: 2} \bar{\psi} \Gamma^{T}_{1 2} \psi
$$
My intuition tells me the opposite of the solution...

Edit: I have an idea. ##\Gamma_{12}^T =\gamma^1 \gamma^2## so the indices on lorentz transformations should be inverted?

$$
M_{\alpha \beta} \rightarrow M'_{\alpha \beta}= \Lambda_{1}^{\: \alpha} \Lambda_{ 2}^{\: \beta} \bar{\psi} \Gamma^{T}_{1 2} \psi
$$
But still the \Gamma matrix has non specific indices.

Also I probably did the index ordering wrong. ##\Lambda^{ 1}_{\: \alpha}## or ##\Lambda_{ \alpha}^{\: 1}##
 
Last edited:
Garlic said:
So M should transform like
Sorry, my hint wasn't very good. Try this. Write
$$M' =\bar{\psi'} \left(x'\right) \Gamma^{T}_{12} \psi' \left(x'\right).$$
Now write each ##\psi' \left(x'\right)## in terms of ##\psi \left(x\right)##.
 
George Jones said:
Sorry, my hint wasn't very good. Try this. Write
$$M' =\bar{\psi'} \left(x'\right) \Gamma^{T}_{12} \psi' \left(x'\right).$$
Now write each ##\psi' \left(x'\right)## in terms of ##\psi \left(x\right)##.

Maybe like this?

$$
M' = \bar{\psi} \: (x) S(\Lambda^{-1}) \Gamma^{T}_{12} S(\Lambda) \psi \: (x)
$$
 
Garlic said:
Maybe like this?

$$
M' = \bar{\psi} \: (x) S(\Lambda^{-1}) \Gamma^{T}_{12} S(\Lambda) \psi \: (x)
$$

Now use ##I = S\left(\Lambda\right) S\left(\Lambda^{-1}\right)## between the ##\gamma## s.
 
So I just figured it out. :)
Although the indices are upside down, but I have a feeling they are the same as in the solution.

Now I'm stuck at the second part:
what am I doing wrong here? Am I using a wrong lorentz matrix?

IMG_0194.JPG
 
I understand it! I can't believe I did it! I feel intelligent :cool:

IMG_0195.JPG


Thank you so much for helping me!
 
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