- #1
jdstokes
- 520
- 1
[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.
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.
