Is chirality dependent on the representation of the gamma matrices?

1. Nov 3, 2013

AlbertEi

Hi,

In QFT we define the projection operators:

P_{\pm} = \frac{1}{2} ( 1 \pm \gamma^5)

and define the left- and right-handed parts of the Dirac spinor as:
\begin{align}
\psi_R & = P_+ \psi \\
\psi_L & = P_- \psi
\end{align}
I was wondering if the left- and right-handed parts of the Dirac spinor are dependent on the representation of the gamma matrices (for instant Dirac representation, Weyl representation or Majorana representation)?

For instance, a general solution of the Dirac equation (following the book "Symmetry and the Standard Model" by Matthew Robinson who works with the signature (-,+,+,+)) is given by:

\psi(x) = a v(\mathbf{p}) e^{i p_\mu x^\mu} + b u(\mathbf{p}) e^{- i p_\mu x^\mu} \label{15.1}

Then, it is not too difficult to show that for a particle at rest using the Weyl representation this results in:

v_L=-v_R

and:

u_L=u_R

Subsequently, he derives that in the Dirac representation the above two equations also hold. However, I do not agree with his derivation and his result, and according to my derivation the above two equations do not hold in the Dirac representation. So my question is: do the above two equations must hold for any representation (i.e. is my derivation definitely wrong)?

Last edited: Nov 3, 2013
2. Nov 3, 2013

fzero

Since the gamma matrices anticommute, if we were to swap two of them (say $\gamma^2\leftrightarrow \gamma^3$), the sign of $\gamma^5$ can change. I think you can show that chirality is fixed under a unitary transformation $\gamma^\mu \rightarrow U \gamma^\mu U^\dagger$, but this requires that you also perform the unitary transformation on the spinor $\psi \rightarrow U \psi$. Also, you can see that if $\gamma^5$ is different between representations, an eigenvector of one $\gamma^5$ matrix need not be an eigenvector of the other choice $\gamma^{'5}$. So in the Dirac basis, the chiral projections of a generic spinor do not take the same form as the projections in the Weyl basis.

3. Nov 5, 2013

andrien

you are right,the left handed and right handed parts will only come out when you use the chiral representation of gamma matrices i.e. weyl representation.Just put the γ5 in chiral representation and see the projection operator.

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