# Conservation of quantum Noether current

The Noether current for a set of scalar fields $\varphi_a$ can classically be written as:

$$j^\mu(x)=\frac{\delta \mathcal L(x)}{\partial(\partial_{\mu}\varphi_a(x))}\delta \varphi_a(x)$$

The divergence of this current can then be written as: $$\partial_\mu j^\mu(x)=\delta \mathcal L(x)-\frac{\delta S}{\delta \varphi_a(x)}\delta \varphi_a(x)$$

If the classical field equations are satisfied the second term on the right hand side vanishes. However in quantum theory the classical field equations are not satisfied. Why is the current still conserved for a symmetry in this case?

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The quantum current is preserved in a quantum average sense given by the Ward identity. It seems that OP basically answered the question(v1) himself here: physics.stackexchange.com/q/16438 –  Qmechanic Nov 4 '11 at 13:02
Indeed, I know about the Ward identity, but I am not completely certain of its physical content. The point I am not sure about is the following : in the Feynman rules for, say, QED, charge is conserved at each vertex. It would seem therefore that charge is exactly conserved, not only statistically. What am I misunderstanding? –  Whelp Nov 4 '11 at 14:57
Well, in the case of global gauge symmetry in QED, which leads to an on-shell electric charge conservation via Noether's first Theorem, there is also an underlying local gauge symmetry, which leads to a trivial off-shell conservation law via Noether's second Theorem. –  Qmechanic Nov 4 '11 at 15:25
I am not aware of Noether's second theorem. Could you explain how it comes into play, or provide a reference for me to study? –  Whelp Nov 4 '11 at 15:30
Well, let me only argue classically. Assuming that you are familiar with the proof of Noethers first Theorem, at one point in the proof one has derived $\delta S = \int d^4x ~{\cal J}^{\mu} d_{\mu} \epsilon$. If $\epsilon$ is a local gauge symmetry, one can integrate by part. Assuming that $\epsilon$ only has support in a small neighborhood, one may deduce an off-shell conservation law $d_{\mu}{\cal J}^{\mu}\equiv 0$ everywhere. Note however that the first and second Noether current may only agree on-shell. –  Qmechanic Nov 4 '11 at 16:25
In the quantum theory, the classical equations of motion are satisified as operator equations. This means that they are satisfied just as well as in the classical theory. The proof is from translation invariance of the path integral: if you integrate over $\phi(x)+h(x)$ for a fixed value h, you get the same answer as if you integrate over $\phi$, so the first order h contribution vanishes. This is covered in the Wikipedia article for path-integral formulation, and applies equally well to field theory. Note that Grassman integration is equally translation invariant, so Fermionic fields obey the classical equations too, even though they have no real classical field limit.