I'm not sure if I understand the concept correctly. Given an infinitesimal transformation

$$\phi \rightarrow \phi + \alpha \Delta\phi$$

the change in the Lagrangian density $\mathcal{L}(\phi,\partial_\mu \phi)$ is

$$\mathcal{L} \rightarrow \mathcal{L} + \alpha \Delta\mathcal{L}$$

For the transformation to be a symmetry, the new Lagrangian can differ only by a four-divergence so that

$$\Delta\mathcal{L} = \partial_\mu J^\mu$$

for some four-vector $J^\mu$.

Now, we have, using E.L. equations, the identity

$$\partial_\mu J^\mu = \partial_\mu\left( \frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)}\Delta\phi\right)$$

From which finally the "conserved current" is:

$$j^\mu \equiv J^\mu-\frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)}\Delta\phi$$

Anyway, I'm trying to do a calculation for a concrete example of the Lagrangian $\mathcal{L} = \frac{1}{2} \partial_\mu \phi \partial^\mu \phi$ and transformation $\phi \rightarrow \phi + \alpha$ for constant $\alpha$.

For this, $\Delta\mathcal{L} = 0 = \partial_\mu J^\mu$. Also $\Delta\phi = 1$. So

$$\frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)}\Delta\phi = \partial^\mu \phi$$


$$j^\mu = J^\mu - \partial^\mu \phi$$

Peskin & Schroeder say that the conserved current is just $j^\mu = \partial^\mu \phi$. I suppose this is because it's defined up to a 4-divergence. So in this case $J^\mu$ can be omitted and also the minus sign doesn't matter since it's defined up to a multiplicative constant as well.

Please correct my understanding of this. What I'm having most trouble understanding here is how the different objects are defined 'up to' something.

  • 1
    $\begingroup$ As $\Delta\mathcal L=0$, you have $J^\mu=0$. Therefore, $j^\mu\propto\partial^\mu\phi$ (you set to $-1$ the constant of proportionality, P&S to $+1$: the actual value is irrelevant. If $j^\mu$ is conserved, so is $\tilde j^\mu=A j^\mu$ for any $A\in\mathbb C$). You found the correct answer, but the thing is, the answer is not unique! that's why yours and P&S's differ. (note that $\partial_\mu J^\mu=0$ doesnt imply that $J^\mu=0$, but as $j^\mu$ is defined modulo a closed field, you can set $J^\mu=0$ WLOG). $\endgroup$ Dec 25, 2015 at 12:41
  • $\begingroup$ Please elaborate what you mean by 'defined modulo a closed field'. I suspect this is the crux of the thing. $\endgroup$ Dec 25, 2015 at 12:47
  • 1
    $\begingroup$ I would recommend to first derive the Noether theorem without the $J^{\mu}$ and only then realise than if one adds an additional current nothing really changes very much. $\endgroup$
    – gented
    Dec 25, 2015 at 12:54
  • 1
    $\begingroup$ @DepeHb A closed field $f^\mu$ is any field such that $\partial_\mu f^\mu=0$. For example, $j^\mu$ is a closed field. If you add two closed fields, the result is also closed ($\partial_\mu(f^\mu+j^\mu)=\partial_\mu f^\mu+\partial_\mu j^\mu=0$). This means that the Noether current $j^\mu$ is not unique: if you add any closed field to $j^\mu$, the resulting current is also conserved, and so equally valid as the original current. In your example, you dont know what $J^\mu$ is, but you are certain that it is closed. Thus, you can set $J^\mu=0$ WLOG. $\endgroup$ Dec 25, 2015 at 12:55
  • $\begingroup$ What if there's a complex scalar field? Is there two conserved currents? $\endgroup$ Dec 25, 2015 at 17:35

1 Answer 1


For the given case since $\partial _\mu J^{\mu} = 0$, there is no need of adding this boundary term to the conserved current $j^{\mu}$ as the addition of it is meaningless as it won't play any part except adding up to a meaningless factor. We already have for the equation of the conserved current $\partial_{\mu} j^{\mu} = 0$ and we can choose to add any term $a^{\mu}$ to $j^{\mu}$ as long as it satisfies $\partial _{\mu} a^{\mu} = 0$. Hence it is really irrelevant to have the $J^{\mu}$ part.

As for the second part of the question, you are right in saying that the minus sign can be omitted as it is defined upto a multiplicative constant.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.