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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

    Ok,

    [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?
     
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