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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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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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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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