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Derivative of a Noether current from Dirac Equation

  1. Nov 25, 2014 #1
    1. The problem statement, all variables and given/known data
    Hey guys,
    Consider the U(1) transformations
    [itex] \psi'=e^{i\alpha\gamma^{5}}\psi [/itex] and [itex] \bar{\psi}'=\bar{\psi}e^{i\alpha\gamma^{5}} [/itex] of the Lagrangian [itex]\mathcal{L}=\bar{\psi}(i\partial_{\mu}\gamma^{\mu}-m)\psi[/itex].

    I am meant to find the expression for [itex]\partial_{\mu}J^{\mu}[/itex].

    2. Relevant equations
    Gamma matrices anticommute
    Noether current is [itex] \delta J^{\mu}=\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\psi)}\delta\psi+\delta x \mathcal{L} [/itex]
    not sure of anything else...

    3. The attempt at a solution
    So here's what I've done so far. Since its a U(1) transformation, the coordinates arent changing, so the Noether current is given by [itex] \delta J^{\mu}=\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\psi)}\delta\psi [/itex]. I found that [itex]\delta\psi=i\alpha\gamma^{5}\psi[/itex], so that

    [itex]\delta J^{\mu}=-\bar{\psi}\alpha\gamma^{\mu}\gamma^{5}\psi[/itex], then I drop the infinitesimal parameter to get
    [itex]J^{\mu}=-\bar{\psi}\gamma^{\mu}\gamma^{5}\psi[/itex].

    So the next step is to calculate the derivative of this. Doing so, I get

    [itex]\partial_{\mu}J^{\mu}=-(\partial_{\mu}\bar{\psi}\gamma^{\mu}\gamma^{5}\psi+\bar{\psi}\gamma^{\mu}\gamma^{5}\partial_{\mu}\psi)[/itex]

    And at this point im stuck...im not sure if this is right and/or if I can simplify this or do something neat with it? because I think I'm meant to be using the transformed Lagrangian
    [itex]\mathcal{L}'=i\bar{\psi}\partial_{\mu}\gamma^{\mu}\psi-m\bar{\psi}e^{2i\alpha\gamma^{5}}\psi[/itex]
    for something but I dont really know.

    Thanks guys..
     
  2. jcsd
  3. Nov 25, 2014 #2

    Orodruin

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    What equations of motion do you have?
     
  4. Nov 25, 2014 #3
    I think its just the equations of motion from the untransformed Lagrangian, which are:
    [itex](i\partial_{\mu}\gamma^{\mu}-m)\psi=0[/itex]
    and
    [itex]i(\partial_{\mu}\bar{\psi})\gamma^{\mu}+m\bar{\psi}=0[/itex]
     
  5. Nov 25, 2014 #4

    Orodruin

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    So I suggest using those. :)
     
  6. Nov 25, 2014 #5
    I thought of that but I'm not sure if I can use these because I'm considering the transformed Lagrangian...and the above transformations arent a symmetry unless m = 0. So how can I use these equations of motion?
     
  7. Nov 25, 2014 #6

    Orodruin

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    The fields will follow their equations of motion. Had the transformation of the fields been a symmetry of the Lagrangian, the divergence of the current would be zero. Since it is not, you will simply get a non-zero expression if you anyway chose to write down the current that would be conserved if the symmetry breaking parameter was zero.
     
  8. Nov 25, 2014 #7
    I see...so I guess Im doing it wrong? I mean I've found the equations of motion from the transformed Lagrangian
    [itex](i\partial_{\mu}\gamma^{\mu}- m e^{2i\alpha\gamma^{5}})\psi=0[/itex]
    and
    [itex]i \partial_{\mu}\bar{\psi}\gamma^{\mu}+m\bar{\psi}e^{2i\alpha\gamma^{5}}=0[/itex]

    and now I'm trying to use THESE inside my expression for [itex]\partial_{\mu}J^{\mu}[/itex]. I guess you're saying to just use the orignals I posted in post #3?
     
  9. Nov 25, 2014 #8
    I get a nonzero answer either way but I'm not sure which one is right...:(
     
  10. Nov 25, 2014 #9

    Orodruin

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    Yes, this willbe the divergence of the current.
     
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