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In answering another question(1) on this site, I started to consider the conservation of energy in Quantum Mechanics. Doing some research, I came across this recent paper.(2) The abstract of the paper is as follows:

We study the conservation of energy, or lack thereof, when measurements are performed in quantum mechanics. The expectation value of the Hamiltonian of a system can clearly change when wave functions collapse in accordance with the standard textbook (Copenhagen) treatment of quantum measurement, but one might imagine that the change in energy is compensated by the measuring apparatus or environment. We show that this is not true; the change in the energy of a state after measurement can be arbitrarily large, independent of the physical measurement process. In Everettian quantum theory, while the expectation value of the Hamiltonian is conserved for the wave function of the universe (including all the branches), it is not constant within individual worlds. It should therefore be possible to experimentally measure violations of conservation of energy, and we suggest an experimental protocol for doing so.

After reading the paper, I believe there are three mechanisms for which energy can appear to either be lost or gained in a system:

  1. Transfer to or from the measurement instrument

  2. Transfer to or from the surroundings

  3. Transfer to or from different worlds in the state of the universe

The third point is the main focus of the paper(2) and from responses to my answer to the aforementioned question(1) appears to be controversial. I understand the third point only holds for some interpretations for quantum mechanics, which may be one source of the controversy.

Questions

Given these points, my questions are:

    a. Is mechanism (3) accepted or controversial?

    b. In which cases will each mechanism arise, or is there an equal chance of any mechanism arising?

I have linked the original question(1) I was answering in case context is important for question (b); however, I am interested not only in this case but more generally in which situations mechanism (3) will apply.

I will link this question in my original answer and reference answers I believe answer my questions well.

Thank you for your time reading this long question and I welcome answers requiring any level of mathematics.

References

(1) What happens to an electron if given quantized energy to jump to a full orbital?

(2) Sean M. Carroll, Jackie Lodman; Energy Non-Conservation in Quantum Mechanics; 26 Jan 2021; CALT-TH-2020-40; https://arxiv.org/abs/2101.11052

Similar Questions

Below are some similar questions which do not answer my question:

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    $\begingroup$ Are you mainly talking about the expected value of the energy? The only time it truly makes sense to ask if the energy itself is conserved is when everything is in an energy eigenstate. $\endgroup$ Jul 7 at 13:40
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    $\begingroup$ Good point that I didn't think to clarify. I am intrested in the expectation value both from the perspectives of a Copenhagen type interpretation and a many worlds type interpretation (which this paper suggests are different); but, also the observed values when measured. In fact between measurements it is very clear that the expectation value of the energy is conserved for a time-independent Hamiltonian. $\endgroup$
    – Chris Long
    Jul 7 at 13:49
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    $\begingroup$ The definition of the energy of the authors as $E \equiv \langle\psi|H|\psi\rangle $ is questionable at best.. $\endgroup$
    – pp.ch.te
    Jul 7 at 15:40
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    $\begingroup$ Here's what Lubos Motl has to say on the matter: motls.blogspot.com/2021/01/… $\endgroup$ Jul 8 at 7:19
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    $\begingroup$ @NiharKarve Thank you, a very informative blog post. $\endgroup$
    – Chris Long
    Jul 8 at 7:38
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First point: the authors tend to take a very "stringent" view of "worlds" in the many worlds interpretation. I take issue with this. I think "many worlds" is a misnomer. Better names for the Everettian interpretation would be "strict unitary evolution", "universal wavefunction" or, my favorite, "church of the large Hilbert space". For the "many worlds" are "soft" boundaries rather than "hard" boundaries. This is especially because a "splitting" of worlds can always, in principle be undone by the reverse of the unitary that split them.

Now to address your question. First of all, I would say there's no point in discussing this question for a system which is not closed. Open systems do not demonstrate energy conservation almost by definition. No one expects them to. So we can leave your possibilities 1 and 2 out of the question because they are situations that are under-specified to determine the global conservation of energy.

Response to question a): So I reiterate, this insistence on the reality of worlds in MWI is naive in my opinion. It is true that a wavefunction that might be able to be expressed as a single term might evolve into one that requires multiple terms, but it can evolve backwards too. Also, sometimes what looks like many terms might only be one term if you express it in a different basis. It might be true "for all practical purposes" that you can't reverse the wavefunction in a particular case. But when we are discussing the mathematics we don't care an ounce about practicality, and the Everettian interpretation is a very mathematical one.

So do we say energy leaves one "branch" and goes to another? No.. I find that language to be horribly confusing. Let's forget the language of branches and just consider the mathematical evolution of a wavefunction under the Hamiltonian.

First, we make a much stronger statement than "The average energy of the system is conserved". Rather, more strongly, the magnitude of the universal wavefunction* to have a given energy is conserved over time. Take state $|\psi\rangle$ and express it in the energy basis:

$$\tag{1} U|\psi\rangle = e^{-i\frac{H}{\hbar}t} \sum_n c_n |n\rangle = \sum_n e^{-i \frac{E_n}{\hbar}t} c_n |n\rangle $$

The Hamiltonian cannot mix states of different energy, so the weight of each term, $|c_n|^2$ is conserved.

I'm digressing a little. I want to look more at the example in the linked question of photon absorption. The atom-photon Hamiltonian in a closed system (such as a high quality optical cavity) is $(\hbar=1)$, in the rotating frame (the rotating frame makes the Hamiltonian time independent, alternatively we could consider a system that just has this Hamiltonian out the door):

$$ H = -\Delta a^{\dagger} a + \frac{\Omega}{2}(a^{\dagger}\sigma^+ + a\sigma^-) $$

$\Delta = \omega_{\text{photon}} - \omega_{\text{atom}}$. In this rotating frame, if $\Omega = 0$ (no coupling), then we see that the state $|0, g\rangle$ has 0 energy, the state $|0, e\rangle$ also has 0 energy (this is because the choice of rotating frame). The states $|1, g\rangle$ and $|1, e \rangle$ each have $-\Delta$ energy.

An absorption transition from $|1, g\rangle \rightarrow |0, e\rangle$, naively, looks like it does not conserve energy by an amount $\Delta$.

However, such a transition does not happen unless $\Omega$ is non-zero. But, when $\Omega$ is non-zero we can see that $|1, g\rangle$ is no longer actually an eigenstate! The actual eigenstates are superpositions of the $|1, g\rangle$ and $|0, e\rangle$ states: $|+\rangle$ and $|-\rangle$, the so-called dressed states!

\begin{align} |+\rangle =& \sin(\theta)|g\rangle + \cos(\theta)|e\rangle\\ |-\rangle =& \cos(\theta)|g\rangle - \sin(\theta)|e\rangle \end{align}

Where $\theta$ is defined by $\tan(2\theta) = -\Omega/\Delta$. Inverting this transformation we have:

\begin{align} |g\rangle =& \sin(\theta)|+\rangle + \cos(\theta)|-\rangle\\ |e\rangle =& \cos(\theta)|+\rangle - \sin(\theta)|-\rangle \end{align}

So if the system starts in $|1, g\rangle$ it is actually already a superposition of $|+\rangle$ and $|-\rangle$, so it ALREADY contains a superposition of energy states. The energies are given by

$$ E_{\pm} = -\frac{\Delta}{2} \pm \frac{\sqrt{\Omega^2 + \Delta^2}}{2} $$

For $\Omega \rightarrow 0$ we have $E_+\rightarrow 0$ and $E_-\rightarrow -\Delta$. Because $|1, g\rangle$ is not an eigenstate of the system it can change, under Hamiltonian evolution, into a different state, for example $|0, e\rangle$. Under our naive thoughts about energy it looks like the system has changed in energy by amount $\Delta$. But, if you decompose the system into the actual energy eigenbasis, you will see that the weightings of the dressed states, $|\pm\rangle$ remain unchanged other than their relative phase.

Not sure what more to say here.. Eq. (1) shows us that a system that starts in a superposition of energy states always remains in the same-weighted superposition of those energy states. This is the best quantum mechanics can do. This is what I mean when I say quantum mechanics conserves energy. This of course implies the weaker statement that the average energy of a system is conserved.

The second example reminds us that when systems are coupled their energies change and we must think in the "dressed" basis if we want to be very careful about tracking energy in the system. We must recall that the undressed states are not eigenstates of the system. This example is very relevant when thinking about absorption of off-resonant photons which is what spurred the discussion in the first place.

*Commonly called the probability of that component, but of course in MWI probability is a little controversial.

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  • $\begingroup$ This answer meanders a bit. It maybe should have been written in a reverse order, first looking at the practical and relevant atom/photon Hamiltonian, then energy conservation, then digressing into philosophical MWI stuff. My apologies for this. $\endgroup$
    – Jagerber48
    Jul 7 at 15:20
  • $\begingroup$ Thank you, I am very grateful for your help both here and on the original question in advancing my understanding. I will need some time to digest and think about you answer! $\endgroup$
    – Chris Long
    Jul 7 at 17:13
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    $\begingroup$ I often call the Everett interpretation "quantum mechanics at face value". $\endgroup$ Jul 7 at 17:51
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    $\begingroup$ Shouldn't $c_n$ be inside the sum after the first equality sign in $(1)$? $\endgroup$
    – Ruslan
    Jul 7 at 19:42
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It appears to me that the report (I don't think it's been peer reviewed) by Carroll and Lodman is just wrong.

You could imagine a pair of dice that were thrown a long time ago (unknown initial conditions) and then remain stationary forever (time evolution conserving the value of each die). Before you look at either die, your expectation for the total value is 7. You look at one of them and find it shows 2. Your expectation for the total value jumps from 7 to 5½. You look at the other and find that it shows 6. Your expectation for the total value jumps from 5½ to 8. The fact that the expectation value keeps changing does not mean that the sum isn't conserved. You have no reason to believe that the actual sum wasn't always 8.

I skimmed the report and I see nothing in it that's more sophisticated than this example. They only consider measurements that commute with the Hamiltonian, so their whole argument is essentially classical.

They suggest a way to experimentally observe violation of energy conservation:

  1. Put a primary system 1 into a known quantum state that is a superposition of energy eigenstates.
  2. Entangle that system with a probe system 2, in a way that does not involve substantial transfers of energy.
  3. Measure the state of the probe system 2, again in a way that does not involve substantial transfers of energy.
  4. Finish with the primary system 1 in an (at least approximate) energy eigenstate, with a substantially different value of the energy than the system started with.

What's the energy cost of performing step 1? You don't know, because it depends on the energy of the primary system after step 1, which by assumption you don't know. If you could measure how much energy you spent in step 1, then the primary system would be in an energy eigenstate after that measurement, and you couldn't perform steps 2–4. But if you can't measure how much energy you spent, then you have no evidence that energy wasn't conserved throughout the experiment.

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