Is Feynman correct when he suggests that Noether's Theorem requires quantum mechanics?

Question

I'm reading the Feynman lectures on physics. In 52-3 he discusses how for each of the rules of symmetry there is a corresponding conservation law. I'm assuming he is referring to Noether's Theorem, though he doesn't say so explicitly. I was a little surprised to see that Feynman seems to suggest that Noether's Theorem only works when we consider quantum mechanics. Feynman says:

"..in quantum mechanics, for each of the rules of symmetry there is a corresponding conservation law."

Further on he says:

"The fact, for example, that the laws are symmetrical for translation in space when we add the principle of quantum mechanics, turns out to mean that momentum is conserved. That the laws are symmetrical under translation in time means, in quantum mechanics, that energy is conserved."

Does Noether's theorem hold when working within classical physics or do we need to invoke quantum mechanics?

Answer 1

If you read carefully, Feynman never says that QM is necessary for NT. The only thing he says is that already in QM you get a statement that coincides with Noether's theorem in classical mechanics (and in a much more simple way). Coming to the question in the last line, one does not need to know QM to prove NT, as Emmy Noether herself famously did.

Answer 2

Noether's theorem holds in quantum mechanics and in classical mechanics. Noether's theorem sounds like magic when you first hear of it: to help put it on solid ground for you, I'll run through the logic of Noether's theorem (from a physicist's perspective) for a single point particle in a potential, so you can get a feel for how it works and what it means.

---

Take a simple Lagrangian $$L=\frac{1}{2}m\dot{q}^2+V(q).$$ We will show that if the potential $V$ is explicitly time-independent (that is, $\partial_tV=0$) that there is an associated conserved quantity (this will be the energy). The action is defined as $$S=\int\text{d}t\,L=\int\text{d}t\left[\frac{1}{2}m\dot{q}^2+V(q)\right]$$ Let's assume that the action is invariant under a time translation $t'=t+\delta t$. Then, we may write $\delta S=0$, because $\partial_t L=0;$ if the Lagrangian isn't dependent on time, then the action shouldn't care about a redefinition of the time coordinate by an infinitesimal constant. We may then write \begin{align*} \delta S&=0 \\ &=\int\delta (\mathrm{d}t\,L)\\ &=\int \text{d}t\left(\partial_t\delta t+\text{t} \delta L\right)\\ &=\int \text{d}t\, \delta L, \end{align*} We do a standard change of variables to compute the change in $\text{d}t$, but since we are varying by a constant, that change is zero. The general variation of the Lagrangian follows from the usual chain rule: $$\delta L=L'\delta t+ \frac{\partial L}{\partial q}\delta q+ \frac{\partial L}{\partial q'}\delta q',\,f':=\frac{\partial L}{\partial t}.$$ Don't get tripped up! The dot on $q$ in the Lagrangian is a total time derivative, while our prime notation denotes a partial one. Using the Euler-Lagrange equations, we may write \begin{align*} \delta L &= \frac{\partial L}{\partial q}\delta q + \frac{\partial}{\partial t}\left(\frac{\partial L}{\partial q'}\delta q\right)-\frac{\partial}{\partial t}\left(\frac{\partial L}{\partial q'}\right)\delta q\\ &=_\text{on-shell}\underbrace{\left[\frac{\partial L}{\partial q}-\frac{\partial}{\partial t}\left(\frac{\partial L}{\partial q'}\right)\right]}_{=0}\delta q+ \frac{\partial}{\partial t}\left(\frac{\partial L}{\partial q'}\delta q\right)\\ &=\frac{\partial}{\partial t}\left(\frac{\partial{L}}{\partial \dot{q}}\delta q\right) \end{align*}

Putting this into the action, we get $$\delta S=\int\text{d}t\frac{\partial}{\partial t}\left(L+\frac{\partial L}{\partial \dot{q}}\delta q\right).$$ Now, the total variation of the position is $$\delta q_T=\delta q+q'\delta t.$$ Under an infinitesimal, constant time translation, $\delta q_T=0$, so we have $\delta q=-\dot{q}\delta t.$ Plugging this in (and remembering that $\delta q$ is a constant, so we can pull it out of the derivative) we get $$0=\delta S=\int\text{d}t\frac{\partial}{\partial t}\left(L- \frac{\partial L}{\partial \dot{q}}\dot{q}\right)\delta t.$$ Since this must hold for arbitrary constant $\delta t$, we have $$\boxed{E=\frac{\partial L}{\partial \dot{q}}\dot{q}-L,}$$ where $E$ is some constant. For our Lagrangian, we have \begin{align*} E&=m\dot{q}^2-\frac{1}{2}m\dot{q}^2+V(q)\\ &=\frac{1}{2}m\dot{q}^2+V(q) ,\end{align*} the total energy.

---

This is the substance of Noether's theorem: a symmetry of the action leads to a conserved quantity. The generalization to classical field theory is easy, because classical mechanics is just a one-dimensional classical field theory. Instead of $q(t)$, you have $\phi(t,\mathbf{x})$, and you follow a similar set of steps above to find that energy-momentum is conserved for spacetime translations. This can be carried over to quantum mechanics/field theory fairly easily; the technical details are beyond the scope of this answer.

---

Note: For internal symmetries of the fields, you can also derive conserved quantities. This is how, for example, electric charge is derived. In our example from classical mechanics, $q$ is the field, and if the system is translationally invariant, you get the momentum. This is peculiar to mechanics; in field theory, position is not a field, but a label on a field, like time for $q(t)$, and momentum conservation is derived in an analogous way as the above.

Answer 3

I don't have Feynman's exact words in front of me, except the few you quoted. But there are a couple of things going on here:

• In classical mechanics, Noether's theorem pairs action-preserving continuous transformations with conservation laws. In quantum mechanics, where a Hamiltonian formulation is much more convenient than a Lagrangian one, the symmetries can be characterized in terms of the Hamiltonian operator, state kets per picture etc., with conserved quantities commuting with the Hamiltonian, but it's the same basic idea. (Note that Noether's theorem looks different again in quantum field theory.)
• Noether's theorem follows from the stationary action principle. But why is that principle true? Feynman contributed to a quantum-mechanical motivation for it (again, the QFT details add a further perspective).

I'm pretty sure, however, Feynman's point in what you've read is the first of the above bullet points, not the second.

Answer 4

I do not know what Feynman intended to communicate, but a world where classical mechanics is valid and quantum mechanics is false, from a theoretical point of view, is logically admissible (it would not be our universe). There, the Noether theorem would be true without quantum mechanics. So they are logically independent.

After all the Noether theorem (in its Lagrangian version) was proved around ten years before the Schroedinger equation was written.

It is true that the stationary action principle, that implies the Noether theorem, can be derived from a suitable interpretation of the Feynman integral in a classical limit. But, again, from a purely logical perspective we could have a world where the stationary action principle is valid but the double slit experiment does not show the interference patterns.

Finally, it seems to me from the quoted text that Feynman referred to the Noether theorem in Hamiltonian formulation. What he stated in the quoted paragraphs exactly holds for Hamiltonian mechanics where dynamical symmetries are equivalent to conservation laws, exactly as in QM. Within the Lagrangian formalism, this equivalence is not perfect. But again, logical independence holds also in this case.