Simultaneous transformations on states and observables which leave predictions invariant

Question

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?

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$^\ddagger$ 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.

Answer 1

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 $\mathcal{A}$.

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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.

Answer 2

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.

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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.)