Proving an Identity Involving Gamma Matrices: Help Needed

[LayMuon]

Can anyone help me in proving the following identity:

<br /> <br /> (\gamma ^{\mu} )^T = \gamma ^0 \gamma ^{\mu} \gamma ^0<br /> <br />

I understand that one can proceed by proving it say in standard representation and then proving that it's invariant under unitary transformations. this last thing is the one that I am not able to prove.

Are there other ways to go?

Thanks.

 

[vanhees71]

I think you rather want to prove
(\gamma^{\mu})^{\dagger}=\gamma^0 \gamma^{\mu} \gamma^0.
For an intro to the Dirac equation and all that, see

http://fias.uni-frankfurt.de/~hees/publ/off-eq-qft.pdf

 

[Bill_K]

LayMuon, There are several matrices like the one you're seeking:

Aγ^μA^-1 = - (γ^μ)^†
Bγ^μB^-1 = (γ^μ)^~
Cγ^μC^-1 = (γ^μ)^*

and so on. You can express each of them as a specific product of gamma matrices in the standard representation, or any other given representation, but it will not hold when you go from one representation to another. Generally we just treat them as an independent matrix and let it go at that without trying to write them in terms of the gammas.

 

[LayMuon]

I think you rather want to prove
(\gamma^{\mu})^{\dagger}=\gamma^0 \gamma^{\mu} \gamma^0.
For an intro to the Dirac equation and all that, see

http://fias.uni-frankfurt.de/~hees/publ/off-eq-qft.pdf

No, I can prove the one with dagger, I need the proof for the transposed version. Bill is saying it's representation dependent. Dagger version is not, I am not sure about the transposed version. Judging from my textbook it should be representation independent too.

 

[LayMuon]

Bill, this post is a follow-up of my previous one, I need this identity to prove the charge conjugation relation in my way. I had a look at the equations you brought in Bjorken and Drell.

 

[andrien]

you should better start with the anticommutation relation followed by gamma matrices and treat γ₀ as a single case when μ=0 and how the hermitian conjugate of different gamma matrices is defined.Like γ₀ is hermitian in one representation then it's hermitian conjugate is γ₀ itself but in other representation it may not be but the relation will prove the consequence.