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Internal vector symmetry of Dirac Lagrangian

  1. Nov 8, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the conserved Noether current [itex] j^\mu [/itex] of the Dirac Lagrangian
    [tex] L = \bar{\psi} ( i \partial_\mu \gamma^\mu - m ) \psi [/tex]
    under the transformation:
    [tex] \psi \rightarrow e^{i \alpha} \psi \,\,\,\,\,\,\,\,\,\, \bar{\psi} \rightarrow e^{-i \alpha} \bar{\psi} [/tex]

    2. Relevant equations
    [tex] j^\mu = \frac{\partial L}{\partial(\partial_\mu \psi)} \Delta \psi + \frac{\partial L}{\partial(\partial_\mu \bar{\psi})} \Delta \bar{\psi} - J^\mu [/tex]

    3. The attempt at a solution
    Substituting the transformations into the langrangian shows it's invariant, so [itex] J^\mu = 0[/itex].
    For infinitesimal [itex] \alpha [/itex] , [itex] \Delta \psi = i \alpha \psi \,\,\,\,\,\,\,\, \Delta \bar{\psi} = -i \alpha \bar{\psi} [/itex] .

    The conserved current then becomes:
    [tex] j^\mu = \bar{\psi} i \gamma^\mu . i \alpha \psi = - \alpha \bar{\psi} \gamma^\mu \psi [/tex]

    Whenever I have seen this result states however, the [itex] - \alpha [/itex] seems to have been dropped. The derivative of my result will still be zero (so my derived current is conserved as it should be) but I cannot see why the result is usually quoted as
    [tex] j^\mu = \bar{\psi} \gamma^\mu \psi [/tex]

    Has this multiplicative constant simply been dropped as it is irrelevant to the conservation, or am I missing something else?

    Thanks.
     
  2. jcsd
  3. Nov 8, 2014 #2

    Orodruin

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    Yes. The normal thing is to define the current as the derivative of your expression with respect to ##\alpha##.
     
  4. Nov 8, 2014 #3
    Thanks for the reply. I'm a little confused though sorry, do you mean [itex] j^\mu [/itex] is usually defined as it's derivative w.r.t the multiplicative constant ( ie, w.r.t [itex] - \alpha [/itex] here ), as the derivative w.r.t. [itex] \alpha [/itex] would make make [itex] j^\mu = - \bar{\psi} \gamma^\mu \psi [/itex] (ie, still out by a negative)
     
  5. Nov 8, 2014 #4

    Orodruin

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    You can always multiply by a constant and still have a conserved current. I could also define a transformation which is using ##e^{2\alpha}##, but of course this does not correspond to any other physics.
     
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