# Mandl and Shaw 4.3

jdstokes
[SOLVED] Mandl and Shaw 4.3

The question is to show that the charge current density operator $s^\mu = - ec \bar{\psi}\gamma^\mu\psi$ for the Dirac Lagrangian commutes at spacelike separated points. Ie

$[s^\mu(x),s^\nu(y)] = 0$ for $(x-y)^2 < 0$.

By microcauality we have $\{ \psi(x), \bar{\psi}(y) \} = 0$.

The commutator is

$e^2c^2( \bar{\psi}(x)\gamma^\mu\psi (x) \bar{\psi}(y)\gamma^\nu\psi(y)-\bar{\psi}(y)\gamma^\nu\psi(y) \bar{\psi}(x)\gamma^\mu\psi (x) )$

I tried to evaluate this in index notation. The first term is

$\left(\bar{\psi}(x)\gamma^\mu\psi (x) \bar{\psi}(y)\gamma^\nu\psi(y)\right)_{\alpha\beta} = \left(\bar{\psi}(x)\gamma^\mu\psi (x) \right)_{\alpha\epsilon}\left( \bar{\psi}(y)\gamma^\nu\psi(y)\right)_{\epsilon\beta} = \bar{\psi}_\alpha (x) (\gamma^\mu)_{\epsilon\gamma} \psi_\gamma (x) \bar{\psi}_\epsilon (y)(\gamma^\nu)_{\beta\delta}\psi_\delta(y)$

$=\bar{\psi}_\alpha(x) \psi_\gamma (x) \bar{\psi}_\epsilon(y)\psi_\delta (y)(\gamma^\mu)_{\epsilon\gamma} (\gamma^\nu)_{\beta\delta}$.

Minus the second term is

$\left(\bar{\psi}(y)\gamma^\nu\psi (y) \bar{\psi}(x)\gamma^\mu\psi(x)\right)_{\alpha\beta}$.

If I simply expand this as $\left(\bar{\psi}(y)\gamma^\nu\psi (y)\right)_{\alpha\epsilon}\left( \bar{\psi}(x)\gamma^\mu\psi(x)\right)_{\epsilon\beta}$ I get a different answer to the first term. What I would like to do is to equate this to

$\left(\bar{\psi}(y)\gamma^\nu\psi (y)\right)_{\epsilon\beta}\left( \bar{\psi}(x)\gamma^\mu\psi(x)\right)_{\alpha\epsilon}$ and then use the anti-commutation relations to show this is the same as the first term.

If A and B are Hermitian and so is AB then $(AB)_{\alpha\beta} = (AB)^\ast_{\beta\alpha} = a_{\beta\epsilon}^\ast b_{\epsilon\alpha}^\ast = a_{\epsilon\beta}b_{\alpha\epsilon}$. But in my case the product of the two matrices is not Hermitian so I can't do that.

jdstokes
Turned out to be something totally stupid. I was interpreting the current as quadruple of matrices when it is in fact a quadruple of complex numbers.

Love*Physics

Ok,

[j$$^{\mu}$$, j$$^{\nu}$$] =0

where j$$^{\mu}$$ =$$\overline{\psi}(x)$$$$\gamma$$$$^{\mu}$$$$\psi (x)$$

Is this true?