There are various versions of this question already on this site, which attempt to justify / make plausible that the time evolution of quantum mechanical observables is unitary. Most of these questions already employ the assumptions of states being elements of a complex Hilbert space, linearity in the time-evolution, hermitian operator's eigenvectors spanning this Hilbert space and that the squared 2-norm of the projection of a state to the eigenspace of an operator gives the probabillity to measure the eigenvalue of the operator (See for example this or this question).

The answer to the question usually is that time-evolution should preserve the norm of a given state, because we interpret this norm as the overall-probabillity. An Answer also suggests that even if we drop the probabillity interpretation with the squared 2-norm, this paper shows that non-unitary comes with nasty properties like superliminal signalling or distinguishabillity of non-orthogonal states.

However - every of these questions (and even the paper given) answers from the perspective of the Schroedinger-picture only:

I'd like to know if similar reasonings can be given if we are in the Heisenberg-picture from the start (and don't even know about the existence of the Schroedinger-picture). Has something like that ever been formulated? Or is this the Schroedinger-picture more general in that sense?

What I came up with: If I assume (for whatever reason) that $$\hat{A}(\delta t) = \hat{U}^{-1} A(0) \hat{U}$$ and $\hat{A}(0)$ was a normal operator, then $\hat{A}(\delta t)$ should be normal as well (because I at least want $\hat{A}(0)$ and $\hat{A}(\delta t)$ to span the complete hilbert-space with it's eigenvectors. This something that I consider a plausible requirement for operators that correspond to observables. If I choose $\hat{U}$ to be unitary, then $\hat{A}(\delta t)$ is a normal operator - I however don't know how to show that this is the only way for $\hat{A}(\delta t)$ to be normal.

Edit: Since all arguments that have been stated in answers dealing with the Schroedinger-picture only talk about closed (contrary to open) quantum systems, I don't see the need to talk about cases in open quantum systems (or subsystems), where time evolution isn't necessarily unitary.

  • $\begingroup$ If my interpretation of Proposition 2 in Unitary implementation of automorphism groups on von Neumann algebras is correct, then a 1-parameter group of automorphisms of a von Neumann algebra on a given Hilbert space is always implemented by a 1-parameter group of unitary transformations. With that, the question becomes: why should time-evolution be given by automorphisms of the operator algebra (in the Heisenberg picture)? $\endgroup$ – Chiral Anomaly Oct 19 '19 at 3:23
  • $\begingroup$ @ChiralAnomaly I agree with your conclusion, although I don't have a clue at all about Von-Neumann Algebras, and what is encoded in them. $\endgroup$ – Quantumwhisp Oct 19 '19 at 7:44
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    $\begingroup$ A vN algebra is a natural way of "completing" a given set of observables with respect to sums, products, and limits. If $\Omega$ is a set of observables and $\Omega'$ is the set of all operators that commute with everything in $\Omega$, then $(\Omega')'$ is the smallest vN algebra containing $\Omega$. In general, $(\Omega')'$ doesn't contain all of the operators on the Hilbert space (because some of them may commute with all of the theory's observables), and I wondered if it might then admit non-unitary 1-parameter automorphism groups. The theorem I cited seems to say that doesn't happen. $\endgroup$ – Chiral Anomaly Oct 20 '19 at 4:31

Yes, the non-unitary Lindbladian time-evolution of an open quantum system can be formulated in the Heisenberg picture: https://en.wikipedia.org/wiki/Lindbladian#Heisenberg_picture.

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    $\begingroup$ To be honest I don't see how this answers my question. $\endgroup$ – Quantumwhisp Oct 16 '19 at 10:36

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