Internal vector symmetry of Dirac Lagrangian

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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.
 
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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)