gu1t4r5
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Homework Statement
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]
Homework 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]
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.