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Often the resolution in QM to energy being conserved if a state is in a superposition of two energy eigenstates is 'if you consider a 'meta' Hamiltonian which accounts for more things in the Hamiltonian, then the meta system is in an energy eigenstate and thus energy is still conserved'.

(This argument can for instance be used to argue that energy is conserved when an atom is in a superposition of its ground and excited state, because the EM field is in a superposition of 1 and 0 photon states too such that the total Hamiltonian accounting for both the atom and the EM field is in an energy eigenstate.) So the point is, we must remember the naive isolated atom energy eigenstates are not actually energy eigenstates of the total system.

However, since no system can actually exist in a perfect energy eigenstate (since the lifetime is infinite), one can use this same argument on this 'meta' Hamiltonian (to an 'even more meta' Hamiltonian) over and over again until, if you want energy to be perfectly conserved, you end up with the total Hamiltonian of the universe being in a single energy eigenstate.

I know that, in reality (due to dark energy etc), the energy of the universe is not conserved. I'm asking therefore, in a model universe with constant energy, would the statement 'the universe exists in a single energy eigenstate of the Hamiltonian of the universe' be correct?

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  • $\begingroup$ We can simply argue that, since we do not have a ToE and, most likely, will not have one, we do not know its observables (even the word/notion of observable means two things: observer and measurement, anthropic principle and all the philosophical garbage), so we don't know if the Universe has a Hamiltonian. But if it did, then, simply because there's nowhere for the "leak of energy" to go to, its energy is trivially conserved, i.e. constant. $\endgroup$
    – DanielC
    Apr 17 at 23:14
  • $\begingroup$ I'm not asking anything about the actual universe, just about the framework of QM, asking if universal energy conservation within this framework would imply we are in a single energy eigenstate (or maybe a superposition of multiple degenerate energy eigenstates) of the hamiltonian containing everything $\endgroup$
    – Alex Gower
    Apr 17 at 23:44
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    $\begingroup$ The energy of a closed system is conserved, whether the system is in an eigenstate or not. That's true for absolutely any state. $\endgroup$
    – Pavlo. B.
    Apr 18 at 1:04
  • $\begingroup$ I am happy that a quantity $A$ (such as energy) can be conserved even if the wavefunction $\psi$ is in a superposition of different $A$ values, so long as $[H,A]=0$. I just always assumed that in order for this superposition $\psi$ to be produced in a unitary manner, there must be some process where 'another wavefunction is in a complimentary $A$ superposition such that the total wavefunction is in an A eigenstate'. $\endgroup$
    – Alex Gower
    Apr 18 at 20:22
  • $\begingroup$ For instance, maybe, a decay of particle X with well-defined quantity $A=1$ could decay into particles Y and Z who each have wavefunctions $\psi_X$ and $\psi_Y$ which are in a superposition of many different $A$ values, but in such a way that their combined wavefunction still has $A=1$ only. This could be a completely incorrect assumption of mine? @ChiralAnomaly $\endgroup$
    – Alex Gower
    Apr 18 at 20:24
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First, let us consider the situation described by the OP: an atom and a EM mode that are coupled and exchanging energy (as described, for example, by the Jaynes-Cummings model). As the OP says, once you take into account both the energy in the atom and in the field the total energy is conserved. However, this does not make it any closer to an eigenstate because the atom and field are coupled, and the new eigenstates are entangled states of the two. Remember, as long as there are any dynamics, the system is not in an energy eigenstate. Indeed, for any complete flopping between two states with frequency $f$ it is in an equal superposition of states separated by $E=hf$, and no amount of accounting for additional degrees of freedom will change this.

Okay, is this the end of the story? Not quite. If you are serious about asking questions about the state of the entire universe, you have to reckon with the complication that this includes you. And the whole notion of a quantum state changes meaning when the observer is part of it. For example: we know that for a normal state which does not include the observer, if it shows any dynamics it must be in a superposition of energy eigenstates. However, it is natural to ask whether we could imagine a frozen, stationary state for which an observer embedded within it might still have the perception of time. This idea is known as the Page-Wootters mechanism, and while I can't claim deep familiarity, its proponents claim that such a framework can indeed be worked out (sample paper).

As far as I can tell, then, it is possible either that the universe (neglecting GR) is in an energy eigenstate or is not, and it is not necessarily clear that an observer within it could ever distinguish the two.


Edit: for the sake of concreteness I will provide more detail about the Jaynes-Cummings model from the first paragraph. This model, for the case of a resonant coupling, describes a single EM mode of $|n\rangle$ photons coupled to a two-level atom with coupling $\Omega$:

$$\hat{H}_{J C}=\hbar \omega \hat{a}^{\dagger} \hat{a}+E_{e g}|e\rangle\langle e|+\hbar \Omega\left(\hat{\sigma}_{+} \hat{a}+\hat{\sigma}_{-} \hat{a}^{\dagger}\right)$$

The key point is that states of definite atom excitation and photon number such as $|e,0\rangle$ are not eigenstates of the Hamiltonian, except in the trivial case in which the atom and EM field are completely decoupled ($\Omega=0$)*. Instead, $|e,0\rangle$ is a superposition of two eigenstates, $|+\rangle$ and $|-\rangle$: $$|+\rangle=\frac{1}{\sqrt{2}}(|g,1\rangle+|e,0\rangle)$$ $$|-\rangle=\frac{1}{\sqrt{2}}(|g,1\rangle-|e,0\rangle)$$ To emphasize, these two states are eigenstates, and they are not degenerate (they are split by $\Omega$). As a result, a state initialized as $|e,0\rangle$ will Rabi flop in the familiar way to $|g,1\rangle$ and back- with frequency $\Omega$ that corresponds to the splitting between the two energy eigenstates. This is a illustration of the main general point- even if you have an 'enlarged' Hamiltonian that contains both objects that are exchanging energy, any dynamics still indicate that you are in a superposition of the full energy eigenstates, which will often be due to the interaction coupling the systems.

*and the special case of $|g,0\rangle$

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  • $\begingroup$ Thank you, I do feel closer to being happy. One thing, are you saying that a Rabi flopping state would still be an equal superposition of different energy eigenstates (separated by $E=hf$ of the total Hamiltonian? I don't quite get that bit yet. I would thought a superposition of $|g\rangle |1 \rangle$ and $|e\rangle |0\rangle$ would both have the same fixed energy of the transition (just either the atom or the photon has it). So how can the 'global/total' Hamiltonian be in a superposition of multiple energy eigenstates? $\endgroup$
    – Alex Gower
    Apr 19 at 15:32
  • $\begingroup$ Unless maybe extra energy is involved with the dynamics itself, and its this energy which is in a superposition $\endgroup$
    – Alex Gower
    Apr 19 at 15:36
  • $\begingroup$ Hi Alex, the answer to your first question is a definite 'yes.' I think this is the key point- I've added an edit to elaborate. $\endgroup$
    – Rococo
    Apr 20 at 1:47
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    $\begingroup$ Good question- it is not obvious to me either. This Hamiltonian has abstracted away all the spatial and time dependence, but I think the answer is hidden in there. If you go back to the fundamental EM-atom interaction that leads to this interaction term, it can be thought of in a classical picture as coming from a decrease of energy when the atom is oscillating in-phase with the EM field and an increase when they are out-of-phase. I think that is ultimately what is going on here, but it is pretty well buried in the typical form. $\endgroup$
    – Rococo
    Apr 22 at 2:03
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    $\begingroup$ But purely at the level of the Hamiltonian, without reference to the physical context, you can think of the sign of the interaction and the resulting splitting as an arbitrary convention, which you have to keep consistent but otherwise isn't very meaningful. $\endgroup$
    – Rococo
    Apr 22 at 2:05
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It seems like you're confused about the structure of quantum mechanics. When we say that X is conserved in quantum mechanics, we just mean that the operator X commutes with the hamiltonian. It doesn't mean that we can't have superpositions of different eigenstates of X.

It is also in general not meaningful operationally to ask whether a system is in a superposition of states. Suppose someone gives you an electron that has been prepared in some spin state, but they don't tell you anything about the state. Then it's not possible by any measurement on the electron to tell whether it was in, say, a pure state of $s_z$ or a mixture of $s_z$ states.

So the question of whether the universe is in an eigenstate of energy is meaningless, for the same reason.

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  • $\begingroup$ @Chiral Anomoly Is measuring an observable of the "universe's wavefunction" a realistic concept - considering that we, the observers, are part of that wavefunction? $\endgroup$
    – Penguino
    Apr 18 at 21:56
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    $\begingroup$ @Chiral Anomaly With your emphasis on "for the same reason" I now understand and agree with your comment (although still feel uncertain about whether we can say anything sensible about observation of the universe's wavefunction). $\endgroup$
    – Penguino
    Apr 18 at 23:03
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I have a distinct feeling you chose a huge number of interacting particles so you won't have to model them in your mind. Choose just two states, and consider a quantum flipflop, describable by a two-vector, $|\psi(t)\rangle$.

In the most general unitary evolution, your energy is time-independent, $$ \langle \psi(t)|H|\psi(t)\rangle = \langle \psi(0)| e^{itH/\hbar} H e^{-itH/\hbar}|\psi(0)\rangle = \langle \psi(0)|H|\psi(0)\rangle. $$ Now generalize to a large number of particles. Huge wave function and Hilbert space. The answer sure looks identical. Is this your question? Why are you talking about stationary states at all? Why should the universe be a stationary state?

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  • $\begingroup$ I probably have faulty assumptions, but I was assuming that your result implied that the average energy was conserved in time (which obviously should be true too). However I was also under the impression that if we follow quantum systems that only evolve unitarily in time, we could get further results. $\endgroup$
    – Alex Gower
    Apr 18 at 20:31
  • $\begingroup$ E.g. when I read (about energy conservation in Rabi flopping) that eventhough the electron is in a superposition of $|g \rangle$ and $| e \rangle$ states, the EM field is also in a superposition of $|0\rangle$ and $|1 \rangle$ photon states such that, eventhough the energy of the atom and EM fields individually are not well-defined, the energy of their combined system is. I assumed this meant that a single energy eigenstate of the combined system could be decomposed into a product of atom and EM field states which each had ill-defined energy. My OP asked if this logic could apply globally. $\endgroup$
    – Alex Gower
    Apr 18 at 20:34
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    $\begingroup$ I don't understand your picture. If you are dealing with a stationary system, what is it that is driving the flip-flopping? You are excluding the possibility in principle that you can have an isolated system with off-diagonal terms in its hamiltonian. Switch EM off. What you see is what you get. The energy is $\langle H \rangle$. $\endgroup$ Apr 18 at 20:54
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    $\begingroup$ There have, of course, been contrarian speculations. $\endgroup$ Apr 18 at 21:14
  • $\begingroup$ I'm not sure if I can agree with your answer. If the universe is initially in an energy eigenstate, the evolution from $|\psi(0)>$ to $|\psi(t)>$ just adds a phasefactor that can be eliminated if you move to a rotating frame of reference. If the system is not initially in an energy eigenstate, there will be a huge amount of dephasing outside of a vanishingly small energy shell. $\endgroup$
    – Wouter
    Apr 20 at 7:01
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For a large interacting system, the energy eigenstates will be extremely bizarre states with lots of non local features. In real life, all the complexity of time evolution comes from the fact that the state itself will be a superposition of energy eigenstates.

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I actually believe that you are right. If the universe, described by a density matrix $\rho$ is stationary (time-translation is a symmetry of the system), then we know that $[H,\rho]=0$ or $H$ and $\rho$ are diagonal in the same basis. We know have that $\rho$ is a classical mixture of eigenstates. This can easily considered a Bolzmann-Gibbs distribution or similar according to equilibrium thermodynamics (this relates to the so-called eigenstate thermalization hypothesis), but in principle we could say that we are and remain in one particular pure eigenstate $|\psi>$.

Note that in principle, a catch of my answer would be if the time-translation symmetry is broken by the initial state. But because you say that energy is conserved, this means the universe is in a microcanonical ensemble (only eigenstates with energy in a vanishingly small energy window are allowed), and dephasing should kill superpositions over time anyway.

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