What is the underlying reason that the same pairs of conjugate variables (e.g. energy & time, momentum & position) are related in Noether's theorem (e.g. time symmetry implies energy conservation) and likewise in QM (e.g. $\Delta E \Delta t \ge \hbar$)?
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$\begingroup$ Related: physics.stackexchange.com/q/28957/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Dec 4, 2012 at 12:05
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$\begingroup$ In QM momentum and coordinate are not "related in Noether's theorem" except for a free motion. $\endgroup$– Vladimir KalitvianskiCommented Dec 4, 2012 at 13:01
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$\begingroup$ This is an interesting question. $\endgroup$– Terry BollingerCommented Dec 4, 2012 at 17:29
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2 Answers
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Good observation.
The pairs of variables in Noether's theorem are conjugate momenta. Basically they are generalized version of mechanical momentum. Interesting note is that even without relativity, energy is recognized as the momentum of time. Generalized momenta are extensively used in the Hamiltonian formulation of classical physics. In quantization, basically the Hamiltonian and most related structures are unchanged, so the conjugate pairs remain the same.
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1$\begingroup$ This leaves the question why quantization has to make conjugate pairs non-commutative. If you just look at the geometric meaning of conjugate pairs you can avoid that question. $\endgroup$– A.O.TellCommented Dec 4, 2012 at 12:32
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$\begingroup$ Were they ever commutative? Also, I don't think that was what the O.P. asked, and probably wasn't in his interests. $\endgroup$– resghCommented Dec 4, 2012 at 12:48
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1$\begingroup$ What do you mean, "were they ever commutative?"?. The observable algebra on the classical phase space IS commutative. My point is merely that you point to another question, namely why quantization works like that, which is in no way obvious. $\endgroup$– A.O.TellCommented Dec 4, 2012 at 13:29
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$\begingroup$ Conjugate pairs? ${x,p} = 1$ holds in classical physics. I didn't think you meant the observable algebra. Sorry about that. However, I think this is (mostly)irrelevant to the current matter so I omitted it. $\endgroup$– resghCommented Dec 4, 2012 at 14:33
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$\begingroup$ Oops. Just noticed. Should be $\{x,p\}=1$. Bad formatting. $\endgroup$– resghCommented Dec 4, 2012 at 17:26
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Both commutation and conservation work in the context of geometry, specifically the generation of transformations. It is important to understand that conjugate pairs like position and momentum relate as generators of translations. If these translations leave the system unchanged, i.e. you have a symmetry generated by the conjugate momentum, then the conjugate momentum must be conserved.
In quantum theory the conjugate pairs are not independent. This is because on a hilbert space the generators of translations and the coordinate they act on relate like the derivative and the coordinate, which don't commute.
So it really comes down to the geometry of the phase space of a system. In classical physics the phase space is just a normal manifold with the usual geometric structure. But in quantum theory the construction using translations on the hilbert space result in a non-commutative geometry. There is no way to avoid this.
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$\begingroup$ Conservation means commutativity with the Hamiltonian, not with a conjugated variable. $\endgroup$ Commented Dec 4, 2012 at 13:03
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$\begingroup$ I never claimed anything else? You must have misunderstood my argument. I never relate conservation to commutativity, instead I relate non-commutativity to unitary representations of translations. The part about conservation is solely about translations being generated by their conjugate momenta. Two different things. $\endgroup$– A.O.TellCommented Dec 4, 2012 at 13:31