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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?

Whelp
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1 Answers1

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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.

Ron Maimon
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