# Simultaneous transformations on states and observables which leave predictions invariant

Consider (for the sake of simplicity) a finite-dimensional complex Hilbert space $$H$$. Let $$\mathcal O(H)$$ and $$\mathcal S(H)$$ denote the set of all hermitian operators and the set of all density operators on $$H$$, respectively. Define $$\varphi^U_\mathcal{O}: \mathcal O(H) \longrightarrow \mathcal O(H)$$ as $$\varphi^U_{\mathcal {O}}(A):=U\,A\,U^\dagger$$ and $$\varphi^U_{\mathcal{S}}: \mathcal S(H) \longrightarrow \mathcal S(H)$$ as $$\varphi^U_{\mathcal{S}}(\rho):=U\,\rho \,U^\dagger$$, for $$U$$ unitary.

We immediately find

$$\mathrm{Tr}\,\rho\, A=\mathrm{Tr}\,\varphi_{\mathcal{S}}^U(\rho)\,\varphi_{\mathcal{O}}^U(A) \tag{1} \quad ,$$

for all $$A\in \mathcal O(H)$$ and $$\rho \in \mathcal S(H)$$. Thus, the joint application of $$\varphi_{\mathcal{S}}^U$$ and $$\varphi_{\mathcal{O}}^U$$ to $$\mathcal S(H)$$ and $$\mathcal O(H)$$, respectively, leaves the predictions of our model invariant.$$^\ddagger$$

My question is whether all maps $$\varphi_S$$ and $$\varphi_O$$ which fulfill $$(1)$$ are necessarily of the form as above, i.e. defined via some unitary operator. If not, under which conditions are these maps necessarily defined through some unitary transformation?

$$^\ddagger$$ Two (trivial) examples: If we have two maps defined as $$\varphi_\mathcal S(\rho):=\rho^\prime$$ for some $$\rho^\prime \in\mathcal S(H)$$ and $$\varphi_\mathcal O(A):=A$$ we find that both maps fulfill equation $$(1)$$ for all $$\rho \in \mathcal S(H)$$ and $$A\in \mathcal O(H)$$ only if $$\rho^\prime=\rho$$, i.e. $$\varphi_\mathcal S$$ is the identity map. Similarly, if we define $$\varphi_\mathcal O(A):=A^\prime$$ and $$\varphi_\mathcal S(\rho)=\rho$$ and assume that these functions fulfill $$(1)$$, then $$A^\prime=A$$. Thus, if there are two functions fulfilling $$(1)$$ and one of the maps is the identity function on its domain, then the other must be the identity function of their respective domain, too.

Let us introduce a bit of abstract machinery first: The usual way to formalize thinking about quantum mechanical observables and state without tying ourselves to a specific Hilbert space or a specific operator representation of an observable is by thinking about the observables as being given as the data of an abstract $$C^\ast$$-algebra $$\mathcal{A}$$. The assignment of concrete operators on a Hilbert space is then given by a $$C^\ast$$-algebra representation $$(H,\pi)$$, i.e. a $$C^\ast$$-algebra isomorphism $$\pi : \mathcal{A}\to L(H)$$ from the abstract algebra to the algebra of linear operators on a Hilbert space $$H$$.

States on the abstract algebra are positive linear functionals on $$\mathcal{A}$$ (as motivation, note that $$A\mapsto \mathrm{Tr}(\rho A)$$ is a positive linear functional on linear operators), denoted by $$\mathcal{E}(\mathcal{A})$$.

Given a representation $$(H,\pi)$$, a state $$\omega\in\mathcal{E}(\mathcal{A})$$ is called normal with respect to $$\pi$$ if there is a density matrix $$\rho_\omega$$ on $$H$$ such that $$\mathrm{Tr}(\rho_\omega \pi(A)) = \omega(A)$$ for all $$A\in\mathcal{A}$$. Two representations $$(H_1,\pi_1)$$ and $$(H_2,\pi_2)$$ are called quasi-equivalent if every state that is normal w.r.t. $$\pi_1$$ is also normal w.r.t. $$\pi_2$$ and vice versa.

Your question in this framework is now simply: Are two quasi-equivalent representations necessarily unitarily equivalent?

As mentioned in this MO answer, this question has a straightforward answer in the usual case where we know that both representations are GNS representations associated to a pure state:

In this case, quasi-equivalence implies unitary equivalence, i.e. indeed that we have $$\mathrm{Tr}(\rho_\omega\pi(A)) = \omega(A) = \mathrm{Tr}(\rho_\omega'\pi'(A))$$ by quasi-equivalence for two representations $$(H,\pi)$$ and $$(H',\pi')$$ and all $$\omega$$ that are normal and all $$A\in\mathcal{A}$$ implies that there is a unitary equivalence between $$H$$ and $$H'$$ that carries $$\pi(A)$$ to $$\pi'(A)$$ and $$\rho_\omega$$ to $$\rho_\omega'$$.

You might object that this doesn't solve the question as written: The question actually just poses the existence of a representation $$(H,\pi)$$ and some vague map $$\varphi : L(H)\to L(H')$$ such that $$\mathrm{Tr}(\rho_\omega \pi(A)) = \mathrm{Tr}(\varphi(\rho_\omega)\varphi(\pi(A)))$$. In order for the statement about two quasi-equivalent representations to apply, we need that $$\varphi\circ\pi$$ is actually a representation:

For additivity we have that $$\mathrm{Tr}(\rho_\omega \pi(A+B)) = \mathrm{Tr}(\rho_\omega \pi(A)) + \mathrm{Tr}(\rho_\omega \pi(B))$$ implies $$\mathrm{Tr}(\varphi(\rho_\omega)\varphi(\pi(A+B)) = \mathrm{Tr}(\varphi(\rho_\omega)(\varphi(\pi(A)) + \varphi(\pi(B)))$$. Now we just need that $$\mathrm{Tr}(\rho X) = \mathrm{Tr}(\rho Y)$$ for all $$\rho$$ implies $$X=Y$$, which is already true when the equation holds only for pure $$\rho$$, see this answer of mine - the only operator with all expectation values zero is the zero vector, so equality of all expectation values implies equality of operators.

Therefore, $$\varphi(\pi(A+B)) = \varphi(\pi(A)) + \varphi(\pi(B))$$, i.e. $$\varphi\circ\pi$$ is linear.

What does not follow is that $$\varphi(\pi(AB)) = \varphi(\pi(A))\varphi(\pi(B))$$. Indeed, due to cyclicity of the trace, we get $$\varphi(\pi(AB)) = \varphi(\pi(B))\varphi(\pi(A))$$ as an alternative solution. So $$\varphi$$ is either an algebra homomorphism or an algebra homomorphism composed with an involution. This involution preserves the $$C^\ast$$-algebra norm and is therefore unique (see this math.SE question), namely the adjoint operation. So either $$A\mapsto \varphi(\pi(A))$$ or $$A\mapsto \varphi(\pi(A))^\dagger$$ is a representation of $$\mathcal{A}$$.

Finally, note that for the most common case of quantum mechanics, this result is much simpler to obtain: When $$\mathcal{A}$$ is the algebra of the canonical commutation relations, then the Stone-von Neumann theorem implies directly that two irreducible representations must be isomorphic without even using the hypothesis of quasi-equivalence.

• @ACuriousMind Ok, then I'm not sure I understand your answer. Do you claim it answers the question as it is stated, without using anything else not stated in the question? (And if yes, how does it rule out dual representations -- because these are included in the question, as is complex conjugation.) My feeling is that the representation structure of $(H,\pi)$ is not implied by the question as stated. Commented Nov 23, 2022 at 22:24
• Comments are not for extended discussion; this conversation has been moved to chat. Commented Nov 23, 2022 at 23:25
• @NorbertSchuch I've finally understood your point about the vagueness of $\varphi$ in the question and I've added some paragraphs discussing that. Commented Nov 23, 2022 at 23:43
• Thanks, that indeed helps. Showing that $\varphi\circ\pi$ is a representation is indeed the key point. Yet (sorry!), you are assuming that $\varphi$ is surjective (on the density operators, or a suitable subspace). (It was a bit along the lines of what I had in mind, but I indeed did (and do) not see how to get rid of the trace, without knowing that $\varphi$ is surjective.) Furthermore, is it clear that a solution has to be either a homomorphism or antihomomorphism -- all you show (leaving the surjectivity aside) is that it is linear on the vector space. (P.S.: Happy you see the point, ... Commented Nov 24, 2022 at 0:15
• ... I was already doubting my mind.) Commented Nov 24, 2022 at 0:17

No -- choosing both maps to be the transpose map, $$\varphi:X\mapsto X^T$$, is a clear counterexample, since $$\mathrm{tr}[\rho^T A^T] =\mathrm{tr}[\rho A]\ .$$

That $$\varphi:X\mapsto X^T$$ cannot be written as $$X\mapsto UXU^\dagger$$ can be seen by noting that the transpose map is not completely positive, while conjugation by a unitary is.

Note that the transpose map has almost all nice properties you want -- in particular, it is linear (and of course, it sends positive operators to positive operators, as required by the question) -- but it is an antihomomorphism of the algebra.

My conjecture would be that under the conditions stated in the question (in its current version), there are only two types of solutions: Conjugation by a unitary, and conjugation by a unitary composed with transposition.

(I seem to remember a result in quantum information that any positive map can be decomposed into a convex combination of conjugation and conjugation + transpose -- maybe this helps, though I am not sure how: But it suggests it might be sufficient to show that the maps $$\varphi$$ are positive maps (i.e. map positive operators to positive operators), together with some extremality property.)

• Both. (This will conjugate the expectation value: But it is real.) -- Should I write this out? Commented Nov 23, 2022 at 18:06
• It is all fine. The only thing which is not linear is the map from the observable algebra to their representation. But you did not ask about this (or for this) at all. (Did I mention I feel the question is underspecified for what you actually would like to know? It seems you would like to know under which conditions the unitary equivalence holds; then you should ask for that!) Commented Nov 23, 2022 at 18:24
• @TobiasFünke No -- not yet, at least ;) -- otherwise I would have written another answer. -- Mostly, I think it would be aesthetically pleasing, as it would remove the annoying feature of the complex conjugation that it is a basis dependent notion. -- But it seems like quite some work, basically one has to start to even derive linearity over R etc. from the property you demand. Commented Nov 23, 2022 at 18:55
• @TobiasFünke Now with an improved counterexample. (Linear!) Commented Nov 23, 2022 at 20:43
• Regarding the idea, it is not that simple -- I have to check the QI statement, but in particular, IIRC it says that any positive map can be written as a CP map and a CP composed with transposition -- but of course, CP maps are not of the form $U\rho U^\dagger$ (but $\sum K_i\rho K_i^\dagger$) unless they satisfy some additional (extremality?) property. Commented Nov 23, 2022 at 21:13