Internal vector symmetry of Dirac Lagrangian

[gu1t4r5]

Homework Statement

Find the conserved Noether current j^\mu of the Dirac Lagrangian
L = \bar{\psi} ( i \partial_\mu \gamma^\mu - m ) \psi
under the transformation:
\psi \rightarrow e^{i \alpha} \psi \,\,\,\,\,\,\,\,\,\, \bar{\psi} \rightarrow e^{-i \alpha} \bar{\psi}

Homework Equations

j^\mu = \frac{\partial L}{\partial(\partial_\mu \psi)} \Delta \psi + \frac{\partial L}{\partial(\partial_\mu \bar{\psi})} \Delta \bar{\psi} - J^\mu

The Attempt at a Solution

Substituting the transformations into the langrangian shows it's invariant, so J^\mu = 0.
For infinitesimal \alpha , \Delta \psi = i \alpha \psi \,\,\,\,\,\,\,\, \Delta \bar{\psi} = -i \alpha \bar{\psi} .

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

Whenever I have seen this result states however, the - \alpha 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
j^\mu = \bar{\psi} \gamma^\mu \psi

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

Thanks.

 

[Orodruin]

Has this multiplicative constant simply been dropped as it is irrelevant to the conservation?

Yes. The normal thing is to define the current as the derivative of your expression with respect to $\alpha$.

 

[gu1t4r5]

Thanks for the reply. I'm a little confused though sorry, do you mean j^\mu is usually defined as it's derivative w.r.t the multiplicative constant ( ie, w.r.t - \alpha here ), as the derivative w.r.t. \alpha would make make j^\mu = - \bar{\psi} \gamma^\mu \psi (ie, still out by a negative)

 

[Orodruin]

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.