Lorentz transformation of the "bilinear spinor matrixelement"

  • #1
Garlic
Gold Member
176
68

Main Question or Discussion Point

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


IMG_0192.JPG
 

Answers and Replies

  • #2
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,360
942
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}##?
 
  • #3
Garlic
Gold Member
176
68
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:
  • #4
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,360
942
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)##.
 
  • #5
Garlic
Gold Member
176
68
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?

$$
M' = \bar{\psi} \: (x) S(\Lambda^{-1}) \Gamma^{T}_{12} S(\Lambda) \psi \: (x)
$$
 
  • #6
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,360
942
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.
 
  • #7
Garlic
Gold Member
176
68
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
 
  • #8
Garlic
Gold Member
176
68
I understand it! I can't believe I did it!! I feel intelligent :cool:

IMG_0195.JPG


Thank you so much for helping me!!!
 
  • Like
Likes vanhees71

Related Threads on Lorentz transformation of the "bilinear spinor matrixelement"

Replies
6
Views
729
  • Last Post
Replies
5
Views
758
Replies
1
Views
5K
Replies
80
Views
7K
  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
6
Views
3K
Replies
1
Views
6K
Top