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?

  • $\begingroup$ 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 $\endgroup$ – Qmechanic Nov 4 '11 at 13:02
  • $\begingroup$ 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? $\endgroup$ – Whelp Nov 4 '11 at 14:57
  • $\begingroup$ 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. $\endgroup$ – Qmechanic Nov 4 '11 at 15:25
  • $\begingroup$ 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? $\endgroup$ – Whelp Nov 4 '11 at 15:30
  • 1
    $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ This isn't entirely accurate. The Ehrenfest theorem says that the expectation values obey approximately the equations of motion of the classical system. I don't know what you mean by 'classical equations of motion are satisfied as operator equations'. The point is there are no classical equations of motion in QM, just the Schrodinger equation. That's part of the reason for the contact terms in Ward-Takahashi identities. $\endgroup$ – Edward Hughes Aug 22 '13 at 9:23
  • 1
    $\begingroup$ @EdwardHughes: This is entirely accurate. I am not exactly talking about the Ehrenfest theorem, rather the Heisenberg equation of motion (they are actually equivalent). The classical equations of motion are satisfied as operator equations, meaning that the quantum mechanical operators in the Heisenberg representation obey the classical equations of motion. The contact terms are there, but they only modify this at coinciding insertions, it doesn't change the result very much. I have shown this on a recent answer regarding supergravity. $\endgroup$ – Ron Maimon Aug 22 '13 at 19:23
  • $\begingroup$ I'm afraid I still disagree with you. There is not even the right notion of product or derivative on the space of operators to yield your claim in general, I think. If you can give me a simple mathematical argument illustrating your point then I might be able to see where we are misunderstanding each other. Cheers! $\endgroup$ – Edward Hughes Aug 22 '13 at 22:24
  • 2
    $\begingroup$ @EdwardHughes: We are not misunderstanding anything, you are just wrong. The Heisenberg equation are the classical equations, and I derived them in some generality from the path integral here (it's standard): physics.stackexchange.com/questions/26888/… . There is a proper notion of product (product of operators) and a right notion of derivative (take the derivative). $\endgroup$ – Ron Maimon Aug 22 '13 at 23:16

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.