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)? $\endgroup$ Commented Oct 19, 2019 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. $\endgroup$ Commented Oct 19, 2019 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. $\endgroup$ Commented Oct 20, 2019 at 4:31
  • $\begingroup$ If we don't even know about the existence of the Schroedinger picture, then where does operator $U(t)$ come from? Unitarity is a mathematical reflection of time-reversal symmetry of the laws of motion. So, if we do not want to attach ourselves to Hilbert space, etc., we should not talk about unitarity, but only about the equations-of-motion and their symmetry in respect to the time reversal. $\endgroup$
    – Roger V.
    Commented Oct 20, 2021 at 11:48

2 Answers 2


It depends on the hypotheses, in particular on the set of observables, you assume.

If assuming that elementary YES-NO observables (also known as tests) are the orthogonal projectors on a Hilbert space a proof naturally arises.

In this view, the lattice structure is expected to be preserved by time evolution in view of its logic interpretation.

In other words, if $P\in {\cal L}(H)$ is an elementary observable, where ${\cal L}(H)$ is the lattice of orthogonal projectors on the Hilbert space $H$, the time evolution is a family of maps $$s_t : {\cal L}(H) \ni P \mapsto s_t(P) \in {\cal L}(H)\:,$$ for $t\in \mathbb R$ satisfying some properties. These properties are natural for isolated (closed) systems or systems evolving in a stationary environment:

(1) It should be bijective (time evolution is expected to be bijective for closed systems in a stationary environment).

(2) It should preserve the orthocomplemented lattice structure (time evolution preserves logic relations as already said).

(3) It should be additive $s_t s_u = s_{t+u}$ (stationary environment).

Assuming the technical hypothesis that $H$ is

(4) separable with dimension $>2$,

then the Gleason and Kadison theorems imply that, for every $t\in \mathbb R$ there is a unitary or antiunitary map $U_t : H \to H$, defined up to multiplicative phases depending on $t$, such that $$s_t(P) = U_t P U_t^{-1}\:.$$ Furthermore it is not difficult to prove that (3) yields $U_tU_u = \chi(t,u) U_{t+u}$ where $\chi(t,u) \in \mathbb{C}$ with $|\chi(t,u)|=1$. Hence $\chi(t/2,t/2)^{-1} U_{t/2}U_{t/2} = U_{t}$. From that we have that $U_t$ must be unitary.

Up to now we have obtained that $\mathbb{R} \ni t \mapsto U_t$ is a unitary-projective representation of the Abelian Lie group $\mathbb{R}$.

Let us assume a further continuity hypotesesis (it can be justified by stating that expectation values of observables evolve continuously in time for any quantum state [=a probability measure on the lattice of elementary observables])

(5) $\mathbb{R} \ni t \mapsto |\langle \psi, U_t \phi\rangle|$ is continuous for every $\psi, \phi \in H$.

A theorem due to Bargmann (using the fact that $\mathbb R$ has trivial cohomology as a Lie group) implies that, it is possible to change the phases of $U_t$, i.e. replacing $U_t$ for $V_t := \chi(t) U_t$ and suitable phases $\chi(t)$, in order that $$\mathbb{R} \ni t \mapsto V_t$$ is a unitary, strongly-continuous representation of $\mathbb{R}$.

In other words $\mathbb{R} \ni t \mapsto V_t\psi$ is continuous for every $\psi \in H$, $V_tV_u = V_{t+u}$, $V_0=I$ and every $V_t : H \to H$ is unitary.

By construction, $$s_t(P) = V_t P V_t^{-1}$$ By spectral decomposition, if $A^* = A = \int_\mathbb{R} \lambda dP^{(A)}_\lambda$ the above action extends to general observables: $$s_t(A) := \int_\mathbb{R} \lambda dP^{(V_tAV_t^{-1})}_\lambda = V_t A V_t^{-1}\:.$$

Strong continuity implies that there is a unique observable $H$, such that $V_t = e^{itH}$ "discovering" the Heisenberg picture.

The crucial assumption here is that the set of elementary propositions consists of the whole family of orthogonal projectors. That is equivalent to requiring that the von Neumann algebra of observables is made of all (bounded) selfadjoint operators on $H$. We know that this hypothesis is physically untenable (in presence of superselection rules or a gauge group for instance). In that case the proof above does not hold. However the result may be true in any case depending on the further hypotheses one assumes.

  • $\begingroup$ How would a gauge group play in here? Is it because two gauge equivalent projectors would yield the same outcome for any observable? And how would one safe the unitarity here? $\endgroup$ Commented Jun 29, 2022 at 9:37

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. $\endgroup$ Commented Oct 16, 2019 at 10:36

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