6

Generally, we say that conservation of energy is a local law; the change in energy in some small region of space is equal to the energy flux out of that region. However, in quantum mechanics, we can have superpositions of energy states. Then, when we measure them, they "instantly" achieve a certain energy. I'm not sure how to reconcile this with local energy conservation.

To be specific, let's consider the following case: we have two identical copies of some two-state system with energy levels $0$ and $E$, and we prepare them in an entangled state given by

$$ |0E\rangle +|E0\rangle $$

Let's assume one atom is in our lab, the other is across the hall. Then locally, their (expected) energies before measurement are each $E/2$. If we measure the electron in our lab, it instantaneously has energy $0$ or $E$--and the same thing happens across the hall! It seems like if we replace "energy density" with "expected energy density", we can have discontinuous jumps in the energy.

Is there any way to formulate local energy conservation in quantum mechanics? Especially if we assume nothing has interacted with the electron across the hall?

Jahan Claes
  • 8,080
  • 2
    If you won't consider wavefunction collapse to happen in the middle of your experiment, the energy is conserved (meaning that each eigenstate of the Hamiltonian is living its own life, which is pretty much what stationary Schrodinger's equation tells us). Nobody knows for sure what is wavefunction collapse, and it might very well be that there is no such thing at all. This is closely related to interpretations of QM which is a very very very very murky topic and is best left untouched :) – Prof. Legolasov Sep 01 '15 at 21:52
  • @Hindsight I realize that, but it still seems odd that even if we assume some interpretation of quantum mechanics (let's say my favorite: decoherence+spontaneous wavefunction collapse), the energy of the atom across the hall seems to discontinuously gain energy. – Jahan Claes Sep 01 '15 at 22:38
  • if you consider a measurement device and an atom to be a single quantum system (like most people who preach decoherence would do), than your energy is transfered from the atom to the measurement device and the other way around in the process of measurement. How can one see this? Consider the overall Hamiltonian dynamics. The total energy has to be conserved. Spontaneous collapse, however, obscures this conclusion drastically. But one can not have an objective collapse without a theory of such collapse, so you implicitly consider QM incomplete in the first place. – Prof. Legolasov Sep 02 '15 at 01:07
  • @Hindsight, that explains why the particle in the LAB can gain/lose energy. But it doesn't do much to explain how the particle across the hall does! – Jahan Claes Sep 02 '15 at 02:16
  • I don't get it. Why not? – Prof. Legolasov Sep 02 '15 at 02:44
  • @Hindsight The particle across the hall has energy $E/2$. After observation, the particle across the hall has either energy $E$ or $0$. But nothing was touching the particle across the hall. The only way for energy to get across the hall from the measurement device would be in some non-local manner – Jahan Claes Sep 02 '15 at 03:17
  • @Hindsight I guess I still am assuming things about wavefunction collapse when I say that. But wavefunctions DO collapse, right? – Jahan Claes Sep 02 '15 at 03:18
  • The energy difference of $E/2$ was given to (or taken from) your particle by the measurement device. If nothing touched the particle, then there could be no observation. I guess they don't, or if they do (which spontaneous collapse is all about), then energy is conserved only probabilistically (its like saying that you don't win or loose in the long run when you constantly flip the coin for 1/1000000 of your bankroll). – Prof. Legolasov Sep 02 '15 at 13:04
  • The point is that your problem only arises when considering wavefunction collapse. And nobody understands completely what this collapse actually is. – Prof. Legolasov Sep 02 '15 at 13:06

3 Answers3

3

As for all EPR-type situations, the answer is in the local statistics over an ensemble of identical copies. Say you do measure one pair of systems and find that system 1 is state $|0\rangle$ while its counterpart across the hall, system 2, is found in state $|E\rangle$. You can say that there is some "spooky-energy-transfer-at-a-distance" by means of wave-function collapse. Great. Now try to get the next pair of systems to do exactly the same thing and cement the result.

You'll find out that it is impossible: there is no way to predict in what state you will measure the 1st system, and this means, unfortunately, no way to predict the energy of the 2nd system. The only thing you can do is average your results over as many attempts as possible. When you do that, you simply find that the average energy both for the measured systems and for their counterparts across the hall is E/2. The only conclusion available is that "on average energy is conserved", although in individual pairs it is redistributed.

This is no different than any attempt to use entanglement and projection-at-a-distance for faster-than-light communication. Energy transfer is subject to the same rules.

udrv
  • 10,371
3

This is not a problem for mainstream interpretations of quantum mechanics where there is no physical collapse of wavefunctions.

It's only a problem for fringe theories, such as spontaneous localization (aka objective collapse), where quantum mechanics is modified to induce a real collapse. From what I understand, violating conservation of energy has always been the biggest problem with those set of interpretations (which, strictly speaking should not be considered interpretations but rather, extensions of quantum mechanics).

If the wavefunction is considered to be real, as in Everett, then it never collapses at all. So energy and information flow is always local.

If, on the other hand, the wavefunction is assumed instead to be epistemic, as in Copenhagen or QBism, then the situation is analogous to classical mechanics when you have some uncertainty about a state. For example, if you don't know whether a coin is heads or tails, then finding out instantly from a friend who looked at it gives you information and "collapses" the 2 states you were imagining each with 50% probability into a single known state. The information flow associated with that collapse is nonlocal in a sense, but not one that's relevant to causality or physics. In your example, the expected value of energy E/2 is nothing more than your expectation that it might be 0 and it might be E. You don't know until you make an observation.

It's only the people who try to modify quantum mechanics (usually, by adding non-linear terms to the Schrodinger Equation) to make this collapse physical who run into a conflict with locality and energy conservation.

reductionista
  • 1,801
  • And this is exactly my point (in the comments to OP's question). – Prof. Legolasov Sep 02 '15 at 13:11
  • So you're saying there's no (as far as you know) good explanation of wavefunction collapse that doesn't violate local conservation of energy? – Jahan Claes Sep 02 '15 at 17:05
  • @JahanClaes wavefunction collapse is not something that has to be explained. We are ought to explain experimentally measurable phenomena, like anomalous magnetic moment of the electron, or the existence of gravity. True collapse (which could not be explained by unitary evolution of an extended system + measurement device) have never been experimentally detected ever. The best evidence for its existence we possess is based on metaphysical expectations of the minority of people. Even if collapse is real, it might turn out to require something much stranger and fundamental than quantum mechanics. – Prof. Legolasov Sep 02 '15 at 18:11
  • @Hindsight but it's undeniable that before measurement, the wavefunction is different than afterwards--in a way that doesn't seem to obey local energy conservation. – Jahan Claes Sep 02 '15 at 18:25
  • 1
    @JahanClaes you are missing the point. You should include your measurement device into the system. The total wavefunction now is subject to the unitary evolution, defined by the Hamiltonian, which represents the interaction between the system and the measurement device - "measurement". – Prof. Legolasov Sep 02 '15 at 18:28
  • @Hindsight I realize that, but your measurement device is completely separate from one of the two atoms, thus can't transfer energy locally. – Jahan Claes Sep 02 '15 at 23:05
  • @JahanClaes thus can't measure anything :) – Prof. Legolasov Sep 03 '15 at 01:17
  • @Hindsight the particles are entangled. Measuring one automatically measures the other, even if it's across the room. At least in standard Copenhagen QM. I'm just trying to see if standard Copenhagen QM (or any other similar interpretation) has any kind of local energy conservation. It looks like the answer is no? – Jahan Claes Sep 03 '15 at 03:47
0

Conservation of Energy is problematic in the Quantum world. Noether's approach applied to the quantum world only gives energy conservation on average, which falls short of what we want. Further suppose |E1> and |E2> are energy eigenvectors of a system that starts off in state |E1>. Measurement are then made of a non-commuting observable quantity A followed by a second measurement of energy. In general, energy will not be conserved. The usual explanation is that the measuring device imparted/absorbed energy - the system was not closed. But it is difficult to make the conservation of energy in closed systems rigorous. I believe it can be done is specific cases. I don't think interpretations have much to do with it.

Energy is not necessarily conserved in General Relativity (GTR). If Quantum Mechanics is a more fundamental theory that GTR, and is essentially correct, then it should not be possible to "prove" conservation of energy in Quantum Mechanics for the most general case.

shaunokane001
  • 538
  • 2
  • 8