2
$\begingroup$

The usual Weyl representation of the Dirac matrices is defined like this: $$\tag{1}\gamma_W^a = T_W \, \gamma^a \, T_W^{-1},$$ where \begin{align}\tag{2} T_W &= \frac{1}{\sqrt{2}} (1 + \gamma^5 \, \gamma^0), & T_W^{-1} &= \frac{1}{\sqrt{2}} (1 - \gamma^5 \, \gamma^0) \equiv T_W^{\dagger}. \end{align} We then get a kind of rotation in the Dirac matrices space (notice the sign in $\gamma_W^5$): \begin{align}\tag{3} \gamma_W^0 &= \gamma^5, &\gamma_W^i &= \gamma^i, &\gamma_W^5 &= -\, \gamma^0. \end{align} This is the Weyl representation of the Dirac matrices.

Now, I wonder if there's a similar transformation that would perform a flipping of $\gamma^0$ and $\gamma^5$, instead of a rotation in the $(\gamma^0, \, \gamma^5)$ "plane". I'm looking for a matrix $V$ (probably unitary) such that \begin{align} \gamma_V^0 &= V \, \gamma^0 \, V^{-1} = \gamma^5, \tag{4} \\[1ex] \gamma_V^i &= V \, \gamma^i \, V^{-1} = \gamma^i, \tag{5} \\[1ex] \gamma_V^5 &= V \, \gamma^5 \, V^{-1} = \gamma^0. \tag{6} \end{align} Is such a transformation possible, using some unitary matrix $V$? How can we find it explicitely?

Transformations (4) and (6) imply that both $\gamma^0$ and $\gamma^5$ commute with the matrix $V^2 \equiv V \, V$: \begin{align}\tag{7} V^2 \, \gamma^0 &= \gamma^0 \, V^2, & V^2 \, \gamma^5 &= \gamma^5 \, V^2. \end{align} My intuition tells me that there's no unitary matrix $V$ satisfying (4)-(6), but I'm probably wrong. The sign in (3) pisses me off!

$\endgroup$
2
$\begingroup$

Indeed, you are doomed. There is no such V.

Suppose there were an equivalence (4,5,6).

Then consider $\gamma_5 = i \gamma_0 \gamma_1 \gamma_2 \gamma_3$, as well as its transform, independently of basis or representation, $$ V\gamma_5 V^{-1} = i V\gamma_0 \gamma_1 \gamma_2 \gamma_3 V^{-1} \implies \\ \gamma_0 = i \gamma_5 \gamma_1 \gamma_2 \gamma_3 \\ = i \gamma_5 \gamma_0\gamma_0\gamma_1 \gamma_2 \gamma_3= \gamma_5\gamma_0\gamma_5= - \gamma_0. $$

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Wow! Excellent demonstration! Thank you very much, it's a nice one! $\endgroup$ – Cham Mar 6 at 17:41
  • $\begingroup$ Thanks. The tensor product rep of the Dirac rep all but leads you to it. $\endgroup$ – Cosmas Zachos Mar 6 at 17:43
  • 1
    $\begingroup$ I think that this demonstration is well worth a notice in a book. Maybe as an exercice. $\endgroup$ – Cham Mar 6 at 18:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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