1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Mandl and Shaw 4.3

  1. May 11, 2008 #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.
  2. jcsd
  3. May 12, 2008 #2
    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.
  4. Jun 18, 2009 #3
    Re: [SOLVED] Mandl and Shaw 4.3


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

    where j[tex]^{\mu}[/tex] =[tex]\overline{\psi}(x)[/tex][tex]\gamma[/tex][tex]^{\mu}[/tex][tex]\psi (x)[/tex]

    Is this true?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook