In Zinn-Justin's Quantum Field Theory and Critical Phenomena they start with an action for a free massless scalar field:

$$S(\varphi) = \frac{1}{2}\int d^{2}x\left[\partial_{\mu}\varphi(x)\right]^{2}$$

And say that the action is invariant under constant translations of the field

$$\varphi(x) = \varphi^{\prime}(x) + \theta$$

This is all fine, the problem I'm having is with the next part. They say a conserved current $J^{V}_{\mu}$ corresponds to this symmetry:

$$J^{V}_{\mu} = \partial_{\mu}\varphi(x)$$

I just can't see how to get that out with $\theta$ being a constant - unless I'm missing something in the notation.


For each continuous symmetry, infinitesimal transformations may be expressed, by a bracket involving the conserved charge operator associated to the symmetry :

$$\delta_\epsilon \phi(x) = i\epsilon [Q, \phi(x)] \tag{1}$$

In our case, we must have :

$$\epsilon \theta = i\epsilon [Q_\theta, \phi(x)] \tag{2}$$

A solution is then :

$$Q_\theta = \theta \int d^3 x' \,\partial_0\phi(x',t) \tag{3}$$

This is because the quantization of fields is expressed as :

$$[ \phi(x,t),\partial_0\phi(x',t)]= i \delta^3(x-x') \tag{4}$$

From $(3)$, we see that the zero component $J_0$ of the current associated to the conserved quantity $Q_\theta$, is $\theta \, \partial_0\phi$, so the current $J_\mu$ is simply $\theta \,\partial_\mu \phi$, and the fact that $Q_\theta$ is a conserved quantity, implies that the associated current is conserved : $\partial^\mu J_\mu=0$, that is in fact $ \theta \,\partial^\mu \partial_\mu \phi=0$, which is nothing, dividing by a non-zero $\theta$, that the equations of movement for the field $\phi$, obtained from the Lagrange-Euler equations.


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