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Is the reason why the time evolution operator is unitary based on purely physical arguments, i.e. that the physical processes that an isolated system undergoes shouldn't depend on any particular instant in time (homogeneity of time); thus two experimenters who conduct the same experiment starting from the same initial state, but at different times, should have the same probability amplitude for that state?! Or is there some mathematical argument as well?

Also, is the reason why the time evolution operator is linear implied by the superposition principle (as an arbitrary state can be expressed as a linear combination of basis states the operator should act linearly as otherwise the state as a whole would evolve differently to the superposition of states that it was initially represented by)?!

Will
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  • While there is a standard physical argument see the answer given by ACuriousMind, I think that the notion of time evolution is fundamentally flawed. Suppose that there exists a multiverse of unverses described by the same QM laws such that only the initial conditions are different. Then you can always consider an alternative time evolution that maps initial states from one universe to final states taken from some other universe, or arbitrary superpositions of these. So, the notion of time evolution is ambiguous. That time evolution is unitary is a tautology. – Count Iblis Mar 12 '15 at 16:25
  • It's because information should be conserved,i.e. Conservation of Information, you can see it by noting that the unitarity of time evolution results in constancy of fine grained entropy, and constancy of fine grained entropy means the amount of information we have about the system doesn't change by time, you could call the conservation of information the zeroth law of physics, –  Jun 27 '15 at 17:22

2 Answers2

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Time evolution is the exponential of the Hamiltonian, since the Hamiltonian is the generator of time-translation (equivalently: Energy is the charge of time translation).

As a physical observable corresponding to energy, the Hamiltonian has to be self-adjoint.

The exponential of a self-adjoint operator is unitary by Stone's theorem.

A "physical" argument is that time evolution should preserve whatever normalization we have chosen for our states (because the probability to find the state $\psi$ in $\phi$ at $t_0$ should be the same as finding the evolved state $\psi$ in the evolved state $\phi$ at $t_1$), i.e. it should preserve the inner product, i.e. it should be unitary.

ACuriousMind
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    Note that the physical argument $$\mathrm{normalisation} \Rightarrow \mathrm{unitarity} $$ presumes linearity. One could imagine more complicated non-linear evolutions which preserve normalisation. However this would violate the superposition principle for probability amplitudes, as noted in the OP. – Mark Mitchison Mar 12 '15 at 16:11
  • ``As a physical observable corresponding to energy, the Hamiltonian has to be self-adjoint." is this because as a physical observable it should have real eigenvalues?! Is it valid to argue that linearity follows for the reason I said. Also, isn't the exponential expression only valid for a time independent Hamiltonian?! Finally, does the identification of the Hamiltonian as the generator of time translation follow from applying Noether's theorem to a time translation of a quantum state?! – Will Mar 12 '15 at 16:27
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    @MarkMitchison (normal) States are density matrices, i.e. trace class positive operators with trace one (in some Hilbert space). There are positivity-preserving, trace-preserving linear semigroups on the density matrices that are not unitarily implemented (e.g. the Lindbladians). So on purely mathematical grounds there is no need of unitarity for the normalisation of the state to be preserved. – yuggib Mar 12 '15 at 16:30
  • @Will: Observables need to have an eigenbasis on the space and should have real eigenvalues - this is indeed what self-adjointness does. That the Hamiltonian is the generator of time translation is inherent in the formulation of all quantization procedures, of which the simplest is the heuristic replacement of the classical Poisson bracket by the commutator. You can't apply "Noether's theorem" to normal QM, the quantum version of conservation laws are the Ward-Takahashi identities. – ACuriousMind Mar 12 '15 at 16:42
  • So, for example, can one infer from the Schrödinger equation that the Hamiltonian generates time translation as its action on a state vector is proportional to the time derivative of the state vector? Sorry to go on about this, but is it valid to claim that the superposition principle of quantum states implies that the evolution operator should be linear, as it should evolve each state in a uniform fashion (if it were non-linear then this would imply that time is inhomogeneous)?! – Will Mar 12 '15 at 17:07
  • @Will: Yes, the Schrödinger equation is one way of stating that the Hamiltonian generates time translations - it is more obvious though in the equivalent Heisenberg equation of motion. And, indeed, time evolution has to be a linear operator because it should not matter whether we evolve a state as a whole or choose a certain basis, decompose every state into the basis, evolve the basis, and reassemble the states. – ACuriousMind Mar 12 '15 at 17:18
  • Ok, thanks for your help. Would it also be correct to say that the reason why $U(t_{2},t_{0})=U(t_{2},t_{1})U(t_{1},t_{0})$ is because it shouldn't matter if one starts in a particular state at $t=t_{0}$, $\vert\psi (t_{0})\rangle$ evolve it to $t=t_{2}$, $\vert\psi (t_{2})\rangle$ and then make a measurement, – Will Mar 12 '15 at 18:37
  • or if one starts off with a state at $t=t_{1}$, $\vert\psi (t_{1})\rangle$ which is equal to the evolved original state at $t=t_{1}$, i.e. $\vert\psi (t_{1})\rangle =U(t_{1},t_{0})\vert\psi (t_{0})\rangle$, and then evolve this to $t=t_{2}$ and make the same measurement. In both cases we should get the same result. Is this because we consider the system to be isolated and thus as there is no external influence, it should evolve according to any internal interactions, and the way in which such interactions affect the evolution of the system should not depend on what time we start the process?! – Will Mar 12 '15 at 18:38
  • @Will: This is indeed the physical intuition behind it, but we don't need to postulate that property - it follows directly from the way the exponential works, and that it is the exponential follows directly from saying it is generated as a linear operator by the Hamiltonian. – ACuriousMind Mar 12 '15 at 18:41
  • Ah ok, I was just wondering if one could motivate it before determining it's functional form. – Will Mar 12 '15 at 19:38
  • This assumes that a hamiltonian exists, which needs not be the case.

    For example, in (classical and most likely quantum) GR the lagrangian is not a convex function of the first derivatives, so there is no legendre transform to a hamiltonian formulation, and time translation is not well-defined. QFT's in general are given by lagrangians and a similar situation may occur in general.

     – saolof Apr 16 '18 at 00:09
  • One could also consider a theory in which the state vector gets "continuously renormalized" over time, so that the probabilistic interpretation still makes sense even with non-unitary (but still linear) time evolution. As discussed here, this theory has some "nasty" physical consequences, but is still perfectly logically consistent. – tparker Jul 06 '18 at 18:42
  • Isn't the reasoning the other way around? We simply postulate the time-translational symmetry and Wigner's theorem then dictates that the evolution of a state must be described through a unitary operator. This ensures that the generator of this evolution would be Hermitian and makes it an observable. –  May 28 '19 at 19:36
  • @FeynmansOutforGrumpyCat There is no unique set of axioms, no "the reasoning". You can indeed just as well postulate that time translation is a symmetry instead of that the Hamiltonian generates time translations. – ACuriousMind May 28 '19 at 20:02
  • @ACuriousMind Ok, I see. Thanks for your response. –  May 28 '19 at 20:16
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It is a consequence of the conservation of the total probability, that is, that $1=\langle a | a \rangle,$ being $|a\rangle$ the state in which your system is. As time makes the state evolve, the final state must also be normalized that way, so that the probability of finding it in the state it will be is one. An easy mathematical calculation leads to the fact that the adjoint of $U$ times $U$ ( $U$ is the time evolution operator ) must conserve distances. From that, that it must conserve any scalar product, and from that the unitarity. You can see this detailed in Leonard Susskind's freely available video of his lecture 9 on quantum entanglements.

César Tejeda Hernández
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  • Hi! Welcome to Physics Se. Please note that this is a MathJax-enabled site. Do use that to format your equations. For a quick reference, please check this meta post:http://meta.math.stackexchange.com/q/5020 –  Mar 02 '16 at 16:10