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If we can assume that quantum mechanics does not have a bound on its applicability, i.e. there are no inherently classical properties of the universe, we can represent the physical state of the entire universe with a state vector $|\psi\rangle$. From normal quantum mechanics we know that the operator governing the evolution of this system, in natural units, is $\hat{U}(t)=\mathrm{exp}(-i\hat{H}t)$, assuming that there are no forces changing $\hat{H}$ at any time.

Here are my first questions. Can we assume that $\hat{H}$ of the universe will stay constant, and whatever model that can bring together gravity and all of the other quantum phenomena will be in agreement with the assumptions I made in the first paragraph?

If we can assume these things to be true, here is my next assumption. The $\hat{H}$ will have a $\hat{V}$ term, of course. In any Hamiltionian with $\hat{V}$ I have seen, the energy eigenstates form a discrete basis. I will assume now that the universe will have a discrete energy basis as well. Is this a good assumption?

Given all of these assumptions, we can express $\hat{H}$ in the energy eigenstate basis as a diagonal matrix with the energy eigenvalues as diagonal terms, denoted here as $\hat{H}_{ii}=\Lambda_{i}$. If one calculates the matrix exponential for a diagonal matrix, one will get a diagonal matrix with values on the diagonal $\hat{U}_{ii}=\mathrm{exp}(-i\Lambda_i t)$.

If one wants to show that the time evolution is cyclic, one must show that for some $t_c\neq0$, $\hat{U}(t_c)=\hat{I}$ which corresponds to all of the exponential terms being equal to one. Further one must show that this condition is fulfilled periodically. From fourier series analysis we can say that this is true, as the diagonal values are just complex exponentials spinning at different frequencies, corresponding to a truncated fourier series.

Qmechanic
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Joel Järnefelt
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    In QM the time $t$ is not an observable i.e. there is no time operator that returns the time. By contrast the positions $x$, $y$ and $z$ are observables with operators $\hat{x}$ etc. In quantum gravity this is no longer the case since time has to be treated in the same way as space, and the surprising result of this is that in quantum gravity the Hamiltonian is zero i.e. it becomes a constraint $H = 0$ and not a time evolution operator. This is the notorious problem of time. – John Rennie Nov 05 '23 at 08:17
  • Wow, I didn't know that the Hamiltonian goes to zero! That is interesting. But what do you mean by time operator? I believe I talked about the time evolution operator, which takes in time and gives an operator that changes $|\psi_0\rangle$ to some $|\psi_t\rangle$. – Joel Järnefelt Nov 05 '23 at 08:31
  • Ah do you mean that in QM gravity there is a time operator and in normal QM there is not? – Joel Järnefelt Nov 05 '23 at 08:33
  • @Joeel Järnefelt: Yes. – John Rennie Nov 05 '23 at 16:07
  • But what is a QM state, which is not a time eigenstate? And even if you have a time eigenstate, does it mean that the state exists only for one instant? – Joel Järnefelt Nov 05 '23 at 18:20

2 Answers2

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What you are referring to is called a "quantum revival" via a recurrence theorem (although your math/logic I can't completely follow). Any isolated system experiences such a revivial.

But since you want to know if this occurs for the whole universe, then this is different. The time evolution of the universe is cyclic if you believe in the many world's interpretation.

The unitary operator you write, when talking about the entire universe, describes the superposition state of the universe to an outside observer who has been disconnected from the universe from the beginning of time. To such an observer, the universe would appear to oscillate infinitely due to the cyclic nature of the operator.

But if you reject the MWI, and only consider the universe in your own reference frame, wavefunction collapse occurs and what you have written is no longer true. This is no longer cyclic evolution of a unitary operator, but indeterministic jumps occur. From this, revival doesn't necessarily happen.

Steven Sagona
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  • Oh, I did not know that this problem had a name. Just thought about this when I was walking. Thanks for the answer, I got it now! – Joel Järnefelt Nov 05 '23 at 18:05
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There are many misconceptions in this question but let's address the question as written:

The question claims that for a self-adjoint operator $H$ the one-parameter group $\mathrm{e}^{\mathrm{i}Ht}$ is "cyclic", i.e. isomorphic to $\mathrm{U}(1)$ not $\mathbb{R}$. This is incorrect, as the counterexample of the momentum operator shows: The momentum operator $p = \mathrm{i}\partial_x$ is self-adjoint on $L^2(\mathbb{R})$ and it's corresponding one-parameter group $T(x_0) = \mathrm{e}^{\mathrm{i}px_0}$ acts on functions as $T(x_0)\psi(x) = \psi(x-x_0)$. This never becomes cyclic in the sense of the question - for any $x_0\neq 0$, $T(x_0)\neq 1$.

ACuriousMind
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  • Can you put this in a more simple way? The question does not ask for an answer in terms of group theory, and it's at least not obvious to me why his question must be analgous to your framing of it. – Steven Sagona Nov 05 '23 at 15:22
  • @StevenSagona I assumed this would be understandable even if you ignore the group theory language. To make contact with your answer: The "quantum revival" from your answer is only guaranteed on finite-dimensional space of states (such as the truncation of the hydrogen atom's state space to the first 100 orbitals), and the momentum operator is a simple example of this failing on an infinite-dimensional space of states. So there is no reason to expect this kind of recurrence in general in quantum mechanics, and in particular not for the whole universe. – ACuriousMind Nov 05 '23 at 15:27
  • This was part of my question. I assumed that the energy spectrum of the universe is discrete, since I figured that $\hat{H}$ of the universe had a $\hat{V}$ term in it. All of the Hamiltonians with a potential term I've encountered, have had discrete energy eigenstates. If this is not true, can you show a counterexample, i.e. a physical Hamiltonian with a potential term with continous energy eigenstates? Of course a proof works too but a counterexample is more simple. – Joel Järnefelt Nov 05 '23 at 18:13
  • @JoelJärnefelt A discrete spectrum is not enough, and Hamiltonians with potentials don't have discrete spectra, they typically have a discrete and continuous part of their spectra - like for hydrogen, where you have a bunch of discrete bound states and then a continuum of unbound states. – ACuriousMind Nov 05 '23 at 19:29
  • Ah okay so, for example, a free particles wavefunction will dissipate out into space for eternity if there's no end boundary. But is this true even for a system with a closed boundary? – Steven Sagona Nov 05 '23 at 21:51
  • @ACuriousMind Oh right, I didn't remember that! Well I guess that answers my question. If universe has $\hat{V}$ that is either continous or has a continous part, and the state vector of the universe is in that potential regime, the quantum revival does not happen, right? – Joel Järnefelt Nov 06 '23 at 10:30