What is the proof of Davies theorem: if a map on pure quantum states transforms equivalent ensembles to equivalent ones, then the map is unitary?

Question

In the following paper (Dynamical Reduction Models by Bassi and Ghirardi), at the end of section 5.3, the following claim is made.

Consider a bijective(*) map on pure states (not necessarily unitary or even linear),

$$S_t |\psi \rangle \rightarrow |\psi (t) \rangle$$

such that a mixture of states $|\psi_i\rangle$ with weights $x_i$ evolves into a mixture of states $S_t|\psi_i\rangle$ with the same weights $x_i$. In particular, $$\sum_i x_i |\psi_i\rangle \langle \psi_i| \rightarrow \sum_i x_i S_t |\psi_i\rangle \langle \psi_i| S_t ^\dagger $$

Let us consider now, at the initial time $t=0$, two physically different ensembles $E(0)$ with states $|\psi_i \rangle$ and weights $x_i$ and $E'(0)$ with states $|\chi_i\rangle$ and weights $y_i$, which are equivalent, i.e. $\rho(0)= \sum_i x_i |\psi_i\rangle \langle \psi_i| = \sum_j y_j |\chi_j\rangle \langle \chi_j|$.

Then, if we assume that

$$\sum_i x_i S_t |\psi_i\rangle \langle \psi_i| S_t ^\dagger = \sum_j y_j S_t |\chi_j\rangle \langle \chi_j| S_t ^\dagger$$

That is, the map $S_t$ has a special property that it maps two ensembles with the equivalent corresponding density matrices to final two ensembles with equivalent corresponding density matrices.

Then according to an elusive theorem of Davies, $S_t$ is a Unitary map.

The problem is, the citation to this thoerem of Davies points to a 180 pages long textbook- E.B. Davies, Quantum Theory of Open Systems, Academic Press, London (1976). There is not even a mention of the page number or section of the book the theorem belongs to.

Does anyone know of a proof of this theorem?

The proof of this theorem will establish why nonlinear maps on states will ALWAYS lead to superluminal signalling.

EDIT: I found another paper, published in 2015, which has one co-author in common with the 2003 paper, which claims to prove the same result. But unfortunately, again, the crucial part of the proof is not provided, and is mentioned to be present in a 1991 reference which is inaccessble this time. A softcopy of this reference is not even behind a paywall, anywhere.

This is the reference that is said to contain the proof- "Ghirardi, G.C. and Grassi, R., Dynamical Reduction Models: some General Remarks, in Nuovi Problemi della Logica e della Filosofia della Scienza, D. Costantini et al. (eds), Editrice Clueb, Bologna. (1991)"

(*) See answer of @quantummechanic to see the necessity of assuming bijectivity.

Answer 1

I do not know this Davies theorem. But your statement is a straightforward reformulation of one of the many ways (I think 6) to define a quantum symmetry, when you assume that the map is bijective on the convex set of generally mixed states. Your type of symmetry is mastered by the Kadison theorem.

The point is that the final requirement is just a coherence requirement. When you extend $S$ from pure states to mixed states, you have to consider the fact that there may be several possible but equivalent decompositions of a given mixed state into a convex combination of pure states. The final requirement just imposes that all these equivalent decompositions of a given mixed state produce the same final mixed state.

At this juncture, you have coherently constructed a map from mixed states to mixed states which leaves fixed the coefficients of every convex decomposition.

The Kadison theorem (equivalent to the more known Wigner theorem) proves that this map can be implemented by a unitary or anti unitary operator (depending on the map), defined up to a phase. A proof of this theorem appears for instance in my book on Spectral Theory and Quantum Mechanics, Springer (2018 I think).

As a final comment, I stress that your notation is a bit misleading (*). The map $S$ may act on pure states as a whole: $S(|\psi\rangle\langle \psi|)$, since it is not a map on vectors. I mean, the theorem holds with this generality.

The general statement is like this.

THEOREM. If the map from generally mixed states to generally mixed states of a Hilbert space $$S: \rho =\sum_k p_k P_k \mapsto \sum_k p_k S(P_k) =: S(\rho)$$ is well-posed and bijective, then there is a unitary or anti unitary map $U$ such that $$S(\rho) = U_S\rho U_S^{-1}$$ for every generally mixed state $\rho$. $U_S$ is defined up to a phase.

Above $$\rho =\sum_k p_k P_k$$ is a convex decomposition into pure states of mixed state so that: $P_k$ is any pure state and the $p_k$ are a numerable family of non-negative reals whose sum is $1$, and the the linear combination converges in the strong operator topology. There may be several different such decompositions of the same $\rho$.

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(*) In particular, if your $S$ defned on vectors is not linear, the map $S^\dagger$ you use on the "bra" vectors is not defined.

Answer 2

This seems too stringent to be true. An arbitrary map includes $$S_t|\phi\rangle\langle \phi|S_t^\dagger\to|0\rangle\langle 0|\quad \forall|\phi\rangle.$$ This is linear but not unitary and would trivially satisfy $$\sum_i x_i S_t|\psi_i\rangle\langle\psi_i|S_t^\dagger=|0\rangle\langle 0|=\sum_i y_i S_t|\chi_i\rangle\langle\chi_i|S_t^\dagger,$$ thereby disobeying the consequences of the quoted theorem.