7
$\begingroup$

In non-relativistic Quantum Mechanics, energy can be interpreted as a frequency in time, and momentum as a frequency in space.

In Special Relativity, Energy turns out to be the temporal component, and momentum, the spatial component of the four-momentum vector.

These two are very unrelated theories. I mean, neither relies on the other. Both are independent extensions of Newtonian Mechanics. Is there any deeper reason energy tends to get paired with time, and momentum with space?

$\endgroup$

2 Answers 2

6
$\begingroup$

Yes, because via Noether's theorem, conservation of energy follows directly from time symmetry (if you do an experiment today, and repeat the same experiment in the same place tomorrow or 100 yrs from now, the result will be the same), and likewise conservation of momentum follows directly from spatial symmetry (if you perform an experiment at one location, then 100 yards away, then 1 light year away, the result will be the same).
A consequence of this is that taking our universe as a whole, energy is NOT conserved in some respects. If you performed an experiment 1 sec after the Big Bang, and then did it now at the same location, the results would be different because of the expansion of the universe. One way this is manifest is the cosmic microwave background has lost energy over time. This energy hasn't "gone anywhere;" it's just off the balance sheet.

This is a great insight you've had from studying the relationships of these quantities in disparate equations, and points to deeper realities. That's how science discoveries are made.

$\endgroup$
7
  • $\begingroup$ I was wondering if something like a four-frequency exists in relativistic QM. A four-frequency associated with particles, whose components transform in a Lorentz way. $\endgroup$
    – Egg Man
    Dec 7, 2021 at 6:42
  • $\begingroup$ Also, maybe the probability of finding a particle would become frame-dependent. Because the four-frequency of the particle, and hence the wave-function, would be frame-dependent. $\endgroup$
    – Egg Man
    Dec 7, 2021 at 6:46
  • $\begingroup$ If this question was misinterpreted, you should edit it so it correctly reflects your intentions. Otherwise, you should ask these as new questions. You can always link to this question or this answer in the new questions to give potential answerers context. $\endgroup$
    – Paul T.
    Dec 7, 2021 at 14:17
  • 1
    $\begingroup$ The lack of energy conservation in GR is true. But it's not a consequence of the fact that you can use Noether's theorem on systems with more global symmetry. $\endgroup$ Dec 7, 2021 at 18:45
  • 1
    $\begingroup$ @EggMan Yes! Such a "four-frequency" does exist, usually called the wave four-vector. Its components are the frequency $\omega$ and spatial wave-vector $\vec k$. $\endgroup$
    – Carmeister
    Dec 7, 2021 at 19:01
3
$\begingroup$

This is a great question, and unfortunately you've gotten a completely incorrect answer from RC_23.

The first thing to realize is that the analogy is not quite as close as it seems. The relevant quantity in quantum mechanics is the Hamiltonian, and this is actually not always the same thing as the energy of a system. For example, in a rotating frame, they are different things.

Further breaking the analogy is the fact that although in relativity time is just another coordinate, this is not the case in quantum mechanics. In QM, time is a parameter, and there is no such thing as a time operator or a time observable. (For example, if you were to apply a time operator to an electron in the ground state of the hydrogen atom, there is clearly nothing sensible that could come out as a result, and similarly for a time measurement of this system, which is too simple to act as a clock.)

If we look at quantum mechanics from a modern perspective, then foundationally it really has nothing to do with things like momentum or energy. If you look at one of these formulations, e.g., https://arxiv.org/abs/quant-ph/0101012, you will see that the basic building blocks are more like qubits than particles. It's possible to have rich, interesting quantum systems such as quantum computers for which there is a Hamiltonian and a time parameter, but no such thing as momentum or position.

So foundationally, we just have wavefunctions, and the Hamiltonian tells us how the wavefunction evolves with time. There's nothing about position or momentum that's inherent to QM. However, if you are going to have building blocks such as particles living in space, then you are going to have to obey the correspondence principle, which means that you have to be able to recover the classical equations of motion in the appropriate limit. The way we normally do this is by associating the QM Hamiltonian with the Hamiltonian from classical mechanics. If you look at the Hamiltonian equations of motion, then you'll see that there is the kind of symmetrical treatment of Hamiltonian, time, position, and momentum that is described in your question.

In special relativity, the situation is completely different except that we still have to obey the correspondence principle so that we can recover Newton's laws in the nonrelativistic limit. The energy-momentum/time-position analogy is a fully correct one in SR (unlike in QM), and it's not difficult to see where it comes from. Basically relativity is a geometrical theory, so all of our observables basically need to be scalars or four-vectors. (They could also be higher-rank tensors or pieces thereof, but basically they need to be things that use a consistent inner product and parallel transport in a consistent way.) If a particle's world-line is inertial and connects points in spacetime differing by a displacement $(\Delta t,\Delta x)$, then that's some four-vector.

We could ask about this particle's energy and momentum as separate objects, but that won't work, because energy isn't a relativistic scalar, and momentum isn't a four-vector. If we're going to have a quantity that makes sense relativistically and connects in some way to the nonrelativistic E and p, as required by the correspondence principle, then we end up having to have something like an energy-momentum four-vector. Now if we ask about its energy-momentum four-vector, there is no rule we can use to give this four-vector a direction, unless we make it point in the same direction as the displacement vector. (If they pointed in different directions, then in the particle's rest frame, picking a spatial direction for the nonzero momentum would violate rotational symmetry.)

RC_23 says this can be explained by Noether's theorem. This is wrong. Noether's theorem is a consequence of Hamiltonian mechanics, and only in the special case where there is a symmetry. A system does not even need to have such a symmetry, and yet in such a situation all of these relationships involving conjugate variables still have to hold.

RC_23's answer further compounds the error by trying to make some kind of unclear but erroneous claim that this has to do with nonconservation of energy in relativity.

A consequence of this is that taking our universe as a whole, energy is NOT conserved in some respects. If you performed an experiment 1 sec after the Big Bang, and then did it now at the same location, the results would be different because of the expansion of the universe. One way this is manifest is the cosmic microwave background has lost energy over time. This energy hasn't "gone anywhere;" it's just off the balance sheet.

This has nothing to do with your question, which was about QM and SR, not QM and GR. Energy is locally conserved in GR (as expressed by the zero divergence of the stress-energy tensor), but is not globally conserved. This lack of global conservation has nothing to do with the fact that spacetime lacks some symmetry. For example, energy-momentum is conserved for asymptotically flat spacetimes, even though they lack the relevant symmetries. Noether's theorem simply doesn't work as a tool for this purpose in GR, for technical reasons. The relevant symmetry for GR would be diffeomorphism invariance, but Noether's theorem doesn't provide a conserved quantity relating to this symmetry.

$\endgroup$
7
  • 5
    $\begingroup$ Directly attacking another user doesn't make a good impression. You might want to consider writing a great answer that doesn't concentrate on mistakes in other answers. $\endgroup$ Dec 7, 2021 at 17:22
  • 1
    $\begingroup$ It's reasonable to assume the OP is asking about connections between QM and SR when they are applied to systems that have those symmetries. $\endgroup$ Dec 7, 2021 at 18:41
  • 1
    $\begingroup$ @JohnRennie On the other hand, it is good to be clear when you think the most highly voted answer is wrong. It would be good to concentrate on the content of the answer rather than the author. This could be done by citing the answer (hit the cite button below the answer to get a link!), rather than the author. $\endgroup$ Dec 7, 2021 at 19:18
  • $\begingroup$ -1 for two reasons: (1) You mention that Noether's theorem only applies in Hamiltonian dynamics. I think your point is that since QM is not Hamiltonian dynamics, Noether's theorem doesn't apply there. But this is not true! I'm not a symplecticist, but if nothing else, the proof goes through: just replace the Poisson brackets with commutators. (2) Yes, QM describes systems without positions. So what? Hamiltonian dynamics does too, and that's not to say that there isn't a commonality that appears when we apply them to systems that do have position coordinates. $\endgroup$ Dec 7, 2021 at 23:24
  • 1
    $\begingroup$ I believe $[X,P]=i$ is at the heart of quantum mechanics. Classical probability theory is applicable both to a deck of cards and to statistical mechanics. Just like that, quantum probability theory, that you linked, is applicable both to quantum computers as well as quantum mechanics. Quantum mechanics is still fundamentally about $[X,P]=i$, even if quantum computers aren't. $\endgroup$
    – Egg Man
    Dec 8, 2021 at 10:35

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.