# Ergodocity and unitary evolution

1. Jan 30, 2016

### A. Neumaier

I don't know a reference but I can indicate a proof. Ergodicity of unitary evolution of the wave function says that for given $\psi_0$ of norm 1, $\psi(t)=e^{-itH}\psi_0$ comes for some $t$ arbitrarily close to any $\psi$ of norm 1. This is equivalent to asserting that the inner product $\langle\psi_0|e^{itH}|\psi\rangle$ gets arbitrarily close to 1. Now represent $\psi_0$ and $\psi$ in a basis where $H$ is diagonal, to verify that when $H$ is unbounded, this is for given $\psi_0$ the case only for a small minority of states $\psi$.

Last edited by a moderator: May 7, 2017
2. Jan 30, 2016

### akhmeteli

I have some doubts about your definition of ergodicity. For example, it does not seem to take into account energy conservation. A similar definition for a classical system would also forbid any ergodicity, as standard evolution of a classical system conserves energy.

3. Jan 31, 2016

### A. Neumaier

According to most interpretations, energy doesn't have sharp values in QM. Almost no system is in an energy eigenstate - hence how can energy be conserved in quantum mechanics? Only $\langle H\rangle$ is conserved.

On the other hand, even if your require $\psi_0$ and $\psi$ to be restricted by having the same expectation for every conserved quantity, the same argument still applies - the closure of the trajectory $\psi(t)$ is far too small to cover any reasonable domain of state vectors.

Last edited: Jan 31, 2016
4. Jan 31, 2016

### akhmeteli

In my book, energy is conserved in quantum mechanics as Hamiltonian commutes with itself:-) But even if you deny energy conservation in quantum mechanics, your definition of ergodicity is still unreasonable, as it requires that evolution of an eigenstate of energy bring it arbitrarily close to another eigenstate with a different eigenvalue. A similar definition in classical mechanics would require that evolution of a state with a certain energy bring it arbitrarily close to a state with a different energy, so there would be no ergodicity in classical mechanics, not even for Sinai's billiards:-)
Sorry, this does not look obvious to me, and you don't have any reference.

If you give a proof here, that would be interesting, and I would appreciate such a proof, but such a proof should start with a generally accepted definition of ergodicity, not some definition introduced by you here for the first time.

Let me note that I had no intention to ask a "gotcha" question. Your statement seemed 1) interesting; 2) not obvious; 3) not obviously wrong, so I wanted to know its status. Right now it looks like this is your personal opinion.

I should also note that I have my doubts about this statement, as, for example, there is a quantum recurrence theorem (Phys. Rev. 107, 337, 1957). This is certainly not the same as ergodicity, but it makes me wonder if quantum mechanics is that much different from classical mechanics in this respect.

5. Jan 31, 2016

### wle

So the ergodic theorem isn't generally valid then. (Or, more reasonably, you would have to carefully qualify what situations it is meant to apply to before claiming it should hold.) Why should this be a problem for MWI?

6. Jan 31, 2016

### A. Neumaier

Because the observable probabilities agree only in a tiny fraction of the possible worlds (namely in those where ergodicity holds - if there is one at all) with the ensemble probabilities. Now MWI is made such the ensemble probabilities agree with the quantum mechanical predictions, while in our particular world we can measure only observable probabilities (that come from repeating experiments in this particular world). Thus MWI fails to explain why our world is so special that it belongs to the ergodic minority.

7. Jan 31, 2016

### wle

I don't see how you've done anything to justify that (at least, not for any version of the ergodic theorem that we actually believe is true). Like akhmeteli pointed out, essentially all you're saying is that if you write $\lvert \psi(t) \rangle = \sum_{n} a_{n}(t) \lvert n \rangle$ for a (pure) quantum state in a basis $\{\lvert n \rangle\}$ that diagonalises the Hamiltonian, then the absolute values $\lvert a_{n}(t) \rvert$ of the coefficients must be independent of time for all $n$. This is just QM's version of conservation of energy. There's nothing remarkable about this. Most serious theories and even toy models in physics obey conservation laws that imply constraints on the future evolution of an initial state. This is not something unique to QM.

8. Jan 31, 2016

### A. Neumaier

I had argued this under the assumption (temporarily granted by you) that the ergodic theorem doesn't hold.
Now you retracted this temporary assumption to make a different point. I'll make my argument more precise once I have more time to do so.

9. Jan 31, 2016

### wle

I don't see where you get that idea. There are two ideas called "ergodic theorem" in physics. One of them, the one you contradict in post #63, roughly says that a physical system will eventually visit all of the available state/configuration space (generally after some astronomically long time). The way you stated your argument was as if you assumed the ergodic theorem had unrestricted validity, i.e., it applies to any or all degrees of freedom for any physical system. You then showed an example from quantum physics showing that isn't true and concluded this was somehow a problem for MWI, rather than a more mundane example showing that, in general, not all physical systems are ergodic. I don't know any a priori reason to believe that the entire universe as a whole must be an ergodic system for instance, so I don't see any reason it should be a problem if MWI implies it is not.

10. Feb 4, 2016

### A. Neumaier

Ok, let me take energy conservation into account. The argument becomes even simpler. Suppose we consider only (improper) states with a fixed energy $E$ in the continuous spectrum of a multiparticle system (and, in case you want, with all other conserved quantities required by system symmetries fixed, too) . Then $H\psi_0=E\psi_0$. Thus $\psi(t)=e^{-itH}\psi_0=e^{-itE}\psi_0$ is parallel to $\psi_0$, hence the subspace accessible to the trajectory is only a 1-dimensional subspace of the (for a system that can separate into more than two asymptotic parts) infinite-dimensional space of possible eigenfunctions for the eigenvalue $E$. This is as nonergodic as one can have it!

11. Feb 4, 2016

### andresB

Classical evolution of states is unitary as evidenced in the Koopman-Von Neumann version of classical mechanics. If my memory serves well, it was this version of CM (with its unitary evolution of vectors in a Hilbert space) the one Von Neumann used to proves his ergodicity theorem in CM.

12. Feb 4, 2016

### A. Neumaier

Yes, one can use an operator formalism (using diagonal operators on a Hilbert space of phase space functions) to prove ergodic theorems in classical mechanics. But the wave function dynamics considered as a deterministic dynamical system is definitely nonergodic.

13. Feb 5, 2016

### akhmeteli

Thank you. However, if the volume is finite, I would not expect a continuous spectrum? Or am I wrong? And one should not expect ergodicity for an infinite volume, be it classical mechanics or quantum mechanics. Or am I wrong again?

I still have two other questions.

1. So what is your definition of ergodicity for quantum theory?
2. Is it a generally accepted definition?

For example, my understanding is that, say, in classical mechanics, ergodicity means something like "the average over time equals the average over the ensemble". It does not look like whatever definition you use has much to do with that. However, if you use some generally accepted definition, that should be fine.

14. Feb 5, 2016

### A. Neumaier

Yes, that's the generally accepted meaning, and that's what I use. If a mixed quantum mechanical state is considered as an ensemble of pure states, the same definition makes sense in the quantum case.
The electrons in a finite piece of metal form conduction bands, which are continuous spectra. In general, it is the ability to move rather than only vibrate or rotate that makes a spectrum continuous.

The context of the original question was that MWI considers our universe [a system presumably of infinite volume; in any case it has clearly moving parts and hence a continuous energy spectrum] as a particular realization of an ensemble of universes, each of them described by a wave function. My complaint was that unless one can interpret the ensemble average as a time average (and hence is ergodic), one can infer nothing about observable probabilities (which typically come from temporal repetitions in one particular realization - the only one we have) from the ensemble probabilities postulated in MWI. Since quantum mechanics is indeed not ergodic, it means that MWI is mute about observable probabilities. It explains nothing.

15. Feb 5, 2016

### akhmeteli

But what ensemble do you consider (one cannot reasonably consider an arbitrary ensemble) and what observables do you average?
I am not sure the spectrum is continuous for a finite number of electrons in a finite volume. And an infinite number of electrons in a finite volume does not make much sense.
Again, not sure the spectrum can be continuous for a finite volume anyway.
Again, I doubt about a continuous spectrum for a finite volume. However, I don't mind the presumption that the Universe is infinite, so I don't mind your argument against MWI. On the other hand, I doubt there is ergodicity for an infinite volume in the classical case either.

16. Feb 5, 2016

### A. Neumaier

Indeed, you are probably right. But every metal reflects light, which makes the system consisting of metal and light unbounded (unless one assumes that evrey photon is absorbed somewhere in the universe....)

On the other hand, the question whether a classical unbounded system is ergodic is not relevant for the application to MWI.

17. Feb 5, 2016

### akhmeteli

I am not sure I understand that: if the Universe (or any other system) has a finite volume, it has a finite volume for photons as well, and the spectrum is not continuous. If, however, you consider photons as part of an external thermostat (or is it called bath?), there is no unitary evolution anyway.
MWI's problems are not my problems, but maybe an MWI's fan would say that, as classical mechanics is (arguably) a limit case of quantum mechanics, classical ergodicity can serve as an approximate justification of the Born rule in MWI. But again, this is not my problem, and again, I am not sure there is classical ergodicity in an infinite volume.

18. Feb 5, 2016

### A. Neumaier

Yes, but my intention was to talk about an ordinary finite piece of metal in our actual universe, which seems to have infinite volume.

19. Feb 5, 2016

### andresB

But then, it seems like the important point is if the spectrum is continuum or not, and not unitary evolution in general.

20. Feb 5, 2016

### A. Neumaier

If the spectrum is highly degenerate then my argument still applies. If the spectrum is nondegenerate then the assumption of fixed energy makes the eigenspaces 1-dimensional, and fixed energy orbits are of course ergodic in these 1D spaces. However, quantum mechanics is trivial in 1-dimensional spaces.

Then the question is what one means by ergodicity when one evolves a state where energy is not constant. In this case one cannot argue that energy is constant, and restricting to constant expected energy yields a huge space (infinite-dimensional already for an anharmonic oscillator) that cannot be filled densely.