Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I need to know the mathematical argument that how the relation is true $(C^{-1})^T\gamma ^ \mu C^T = - \gamma ^{\mu T} $ . Where $C$ is defined by $U=C \gamma^0$ ; $U$= non singular matrix , $T$= transposition, $\gamma^0= $Dirac gamma matricx = $\beta$

I need to know the significance of these equation in charge conjuration .

share|improve this question
    
This can be taken as a definition of $C$ if you like, although it does not uniquely specify a $C$. You have to normalize. Let $\gamma ^\mu$ be an irreducible representation of the Clifford algebra $\mathcal{C}\ell (1,3)$. You can check that the matrices $\gamma '^\mu :=-(\gamma ^\mu )^T$ also define a representation of this algebra. There is a theorem about Clifford algebras that says there is essentially only one (faithful) irreducible representation, and hence these two representations must be equivalent, i.e. there is a unitary $C$ such that $C\gamma ^\mu C^{-1}=-(\gamma ^\mu )^T$. –  Jonathan Gleason Feb 12 '13 at 22:29
add comment

1 Answer

up vote 0 down vote accepted

First, $U$ is surely not "any non-singular matrix". For a given basis, $U$ is almost completely determined i.e. unique. It contains $\gamma_2$ because it's derived from the only imaginary Pauli matrix.

Because of the basic Dirac algebra $$ \{ \gamma_\mu,\gamma_\nu \} = 2\cdot 1_{2\times 2} \cdot g_{\mu\nu} $$ one may see that $\gamma_0$ is Hermitian, $\gamma_0=\gamma_0^\dagger$, while the spatial ones are anti-Hermitian, $\gamma_i=-\gamma_i^\dagger$.

In your identity, you want to relate $\gamma^\mu$ to its transposition $\gamma^{\mu T}$. Up to the sign that depends on the spatial or temporal character of $\mu$, the transposition is the same thing as complex conjugation.

So a related problem is whether the complex conjugate matrices $\gamma^{\mu*}$ can be related to $\gamma^\mu$ by something like a conjugation. And the answer is Yes. The main fact behind the exercise is that $\sigma^2$ is the only imaginary Pauli matrix, so complex conjugation of Pauli matrices is equivalent to the conjugation by $\sigma^2$ with an extra sign. This may be easily generalized if you also include the temporal 0th component and if you use the normal basis.

You should check the identity you want to verify in a particular convenient basis, i.e. with an explicit form of the gamma matrices. The verification is most convenient if you write the gamma matrices in block form, with $2\times 2$ blocks being either multiples of Pauli matrices or the unit matrix.

In a more general representation, the Dirac gamma matrices differ from those in the particular basis you will have verified by a conjugation only, and this may only mean that $U$ is changed in the formula, but the essence of the conjugation is unchanged.

These equations are important because $C$ is related to the charge conjugation – the replacement of particles by antiparticles (e.g. exchange of electrons and positrons). Mathematically, the most important part of the charge conjugation is complex conjugation which is why we needed to express the "complex conjugate gamma matrices as some conjugations of the normal ones".

Theories with a symmetry between matter and antimatter are symmetric under C - the charge conjugation symmetry. Spinors are mapped to $\psi\to C\psi$ etc. and the only hard part of the symmetry of the Lagrangian is a step that requires you to conjugate the gamma matrices by $C$ which is why it's good that we have a way to simplify $C^{-1T} \gamma^\mu C^T$.

share|improve this answer
    
But I found in "relativistic quatum mechanics by Greiner " as non singular matrix. –  Unlimited Dreamer Nov 5 '12 at 13:47
    
I need mathematical proof . –  Unlimited Dreamer Nov 17 '12 at 15:43
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.