# Internal vector symmetry of Dirac Lagrangian

1. Nov 8, 2014

### gu1t4r5

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

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

3. 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.

2. Nov 8, 2014

### Orodruin

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

3. Nov 8, 2014

### 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)

4. Nov 8, 2014

### Orodruin

Staff Emeritus
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.