- #1
jdstokes
- 523
- 1
[SOLVED] Mandl and Shaw 4.3
The question is to show that the charge current density operator [itex]s^\mu = - ec \bar{\psi}\gamma^\mu\psi[/itex] for the Dirac Lagrangian commutes at spacelike separated points. Ie
[itex][s^\mu(x),s^\nu(y)] = 0 [/itex] for [itex](x-y)^2 < 0[/itex].
By microcauality we have [itex]\{ \psi(x), \bar{\psi}(y) \} = 0[/itex].
The commutator is
[itex]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) )[/itex]
I tried to evaluate this in index notation. The first term is
[itex]\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) [/itex]
[itex]=\bar{\psi}_\alpha(x) \psi_\gamma (x) \bar{\psi}_\epsilon(y)\psi_\delta (y)(\gamma^\mu)_{\epsilon\gamma} (\gamma^\nu)_{\beta\delta}[/itex].
Minus the second term is
[itex]\left(\bar{\psi}(y)\gamma^\nu\psi (y) \bar{\psi}(x)\gamma^\mu\psi(x)\right)_{\alpha\beta}[/itex].
If I simply expand this as [itex]\left(\bar{\psi}(y)\gamma^\nu\psi (y)\right)_{\alpha\epsilon}\left( \bar{\psi}(x)\gamma^\mu\psi(x)\right)_{\epsilon\beta}[/itex] I get a different answer to the first term. What I would like to do is to equate this to
[itex]\left(\bar{\psi}(y)\gamma^\nu\psi (y)\right)_{\epsilon\beta}\left( \bar{\psi}(x)\gamma^\mu\psi(x)\right)_{\alpha\epsilon}[/itex] 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 [itex](AB)_{\alpha\beta} = (AB)^\ast_{\beta\alpha} = a_{\beta\epsilon}^\ast b_{\epsilon\alpha}^\ast = a_{\epsilon\beta}b_{\alpha\epsilon}[/itex]. 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 [itex]s^\mu = - ec \bar{\psi}\gamma^\mu\psi[/itex] for the Dirac Lagrangian commutes at spacelike separated points. Ie
[itex][s^\mu(x),s^\nu(y)] = 0 [/itex] for [itex](x-y)^2 < 0[/itex].
By microcauality we have [itex]\{ \psi(x), \bar{\psi}(y) \} = 0[/itex].
The commutator is
[itex]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) )[/itex]
I tried to evaluate this in index notation. The first term is
[itex]\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) [/itex]
[itex]=\bar{\psi}_\alpha(x) \psi_\gamma (x) \bar{\psi}_\epsilon(y)\psi_\delta (y)(\gamma^\mu)_{\epsilon\gamma} (\gamma^\nu)_{\beta\delta}[/itex].
Minus the second term is
[itex]\left(\bar{\psi}(y)\gamma^\nu\psi (y) \bar{\psi}(x)\gamma^\mu\psi(x)\right)_{\alpha\beta}[/itex].
If I simply expand this as [itex]\left(\bar{\psi}(y)\gamma^\nu\psi (y)\right)_{\alpha\epsilon}\left( \bar{\psi}(x)\gamma^\mu\psi(x)\right)_{\epsilon\beta}[/itex] I get a different answer to the first term. What I would like to do is to equate this to
[itex]\left(\bar{\psi}(y)\gamma^\nu\psi (y)\right)_{\epsilon\beta}\left( \bar{\psi}(x)\gamma^\mu\psi(x)\right)_{\alpha\epsilon}[/itex] 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 [itex](AB)_{\alpha\beta} = (AB)^\ast_{\beta\alpha} = a_{\beta\epsilon}^\ast b_{\epsilon\alpha}^\ast = a_{\epsilon\beta}b_{\alpha\epsilon}[/itex]. But in my case the product of the two matrices is not Hermitian so I can't do that.