6
$\begingroup$

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)?!

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Count Iblis Mar 12 '15 at 16:25
  • $\begingroup$ 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, $\endgroup$ – user55867 Jun 27 '15 at 17:22
6
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – Mark Mitchison Mar 12 '15 at 16:11
  • $\begingroup$ ``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?! $\endgroup$ – Will Mar 12 '15 at 16:27
  • 1
    $\begingroup$ @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. $\endgroup$ – yuggib Mar 12 '15 at 16:30
  • $\begingroup$ @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. $\endgroup$ – ACuriousMind Mar 12 '15 at 16:42
  • $\begingroup$ 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)?! $\endgroup$ – Will Mar 12 '15 at 17:07
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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:meta.math.stackexchange.com/q/5020 $\endgroup$ – user36790 Mar 2 '16 at 16:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.