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Let $M$ and $N$ be mappings from $\mathcal{S}(H_A \otimes H_B)$ to itself, where $\mathcal{S}(H)$ denotes the set of density operators over the Hilbert space $H$.

If the following two conditions hold:

(i) both $M$ and $N$ are linear over $\mathcal{S}(H_A \otimes H_B)$

(ii) $M(\rho_A \otimes \rho_B) = N(\rho_A \otimes \rho_B)$ for all $\rho_A \in \mathcal{S}(H_A)$ anf $\rho_B \in \mathcal{S}(H_B)$ - i.e. they have the same behaviour on product states

Does it follow that $N = M$?

I am convinced this should work, but am struggling to find a proof (or, if it does not hold, an explicit example).

vfx01
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1 Answers1

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Yes. But there are some things to discuss first to help you understand.

When discussing a "linear mapping" one typically requires that it be defined on a full linear space, not some convex subset. For now, let $B(H)$ denote the linear space of all bounded linear operators on $H$, then $S(H)$ is a convex subset of $B(H)$.

Now we can use linearity of $M$ to allow for the equality $M(\alpha X + \beta Y) = \alpha M(X) + \beta M(Y)$ to hold for arbitrary operators $X$ and $Y$ and arbitrary complex numbers $\alpha$ and $\beta$ (i.e., not just for density operators).

Because $M$ and $N$ agree on all product density operators, by linearity of $M$ and $N$ they must agree on all product operators as well. That is, it must be the case that $$M(X\otimes Y) = N(X\otimes Y)$$ holds for all $X\in B(H_A)$ and $Y\in B(H_B)$.

If $H_A$ and $H_B$ are finite-dimensional Hilbert spaces, then $B(H_A)$, $B(H_B)$ and $B(H_A\otimes H_B)$ are all finite dimensional as vector spaces as well, where we note that $$B(H_A\otimes H_B)\simeq B(H_A)\otimes B(H_B).$$ In this case, the tensor product operator space $B(H_A\otimes H_B)$ is spanned by the product operators, $$ B(H_A\otimes H_B) = \operatorname{span}\big(\big\{X\otimes Y\, :\, X\in B(H_A), Y\in B(H_B)\big\}\big), $$ and so $M$ and $N$ must agree on the whole space.

If they are infinite dimensional, then equality of $M$ and $N$ follows from the universal property of tensor product spaces. See this answer.

Luftbahnfahrer
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  • Thank you! It is not clear to me why $N$ and $M$ need to agree on all product operators. Consider the product state $\rho_A \otimes \rho_B = (\sum_{ij} c_{ij} T_{ij}) \otimes (\sum_{kl} d_{kl} T_{kl})$, where ${T_{ij}}{ij}$ is a basis of $B(H_A)$, ${T{kl}}{kl}$ a basis of $B(H_B)$ and the coefficients chosen to satisfy positiveness and trace 1. I don't get how the condition on the sum $\sum{ijkl}c_{ij}d_{kl} M(T_{ij} \otimes T_{kl}) = \sum_{ijkl}c_{ij}d_{kl} N(T_{ij} \otimes T_{kl})$ implies the equality element-wise, $M(T_{ij} \otimes T_{kl}) = N(T_{ij} \otimes T_{kl})$. – vfx01 Aug 23 '23 at 15:14
  • You are trying to use the implication in the wrong direction. Assume that $M(X\otimes Y)=N(X\otimes Y)$ holds for all possible choices of $X$ and $Y$. (In particular, it holds when they are basis elements, i.e., if $X,Y\in{T_{i,j},:,i,j}$). Now let $c_{ij}$ and $d_{kl}$ be some complex numbers, for all possible choices of indices $ijkl$. Linearity of $M$ implies that $M(\sum_{ijkl}c_{ij}T_{ij}\otimes d_{kl}T_{kl}) = \sum_{ijkl}c_{ij}d_{kl}M(T_{ij}\otimes T_{kl})$. – Luftbahnfahrer Aug 23 '23 at 15:29
  • Because we have assumed that $M(T_{ij}\otimes T_{kl})=N(T_{ij}\otimes T_{kl})$ holds for all indices $ijkl$, we have $M(\sum_{ijkl}c_{ij}T_{ij}\otimes d_{kl}T_{kl}) = \sum_{ijkl}c_{ij}d_{kl}M(T_{ij}\otimes T_{kl}) $ $= \sum_{ijkl}c_{ij}d_{kl}N(T_{ij}\otimes T_{kl}) = N(\sum_{ijkl}c_{ij}T_{ij}\otimes d_{kl}T_{kl})$, where the last equality follows from linearity of $N$. – Luftbahnfahrer Aug 23 '23 at 15:32
  • Sorry but I am still a little bit confused. Why do you say we assume that $M(X \otimes Y) = N(X \otimes Y)$ holds for all possible choices of $X$ and $Y$, when the question simply ensures that this is true for density operators, hence only for some choices of $X$ and $Y$? – vfx01 Aug 23 '23 at 15:35
  • I understand that, if $M(X \otimes Y) = N (X \otimes Y)$ for all possible choices, then the rest follows. It's unclear to me how this is implied from the linearity of $M$ and $N$ and the restriction on density operators. Sorry if I am misunderstanding you. – vfx01 Aug 23 '23 at 15:38
  • A linear map on density operators extends uniquely to a linear map on the full linear space of all operators. See the comments heere: https://math.stackexchange.com/q/4694680/106213 – Luftbahnfahrer Aug 23 '23 at 15:48
  • Yes, which means $M(X \otimes Y)$ and $N(X \otimes Y)$ are well-defined even if $X$ and $Y$ are not density operators. However, how can we say that $M(X \otimes Y)=N(X \otimes Y)$ whenever they are not density operators? I believe that is to be derived, not assumed. – vfx01 Aug 23 '23 at 15:52
  • Let $X$ and $Y$ be arbitrary operators. They can be split uniquely into hermitian and anti-hermitian components $X=H+iK$ and $Y=Z+iW$ where $H,K,Z,W$ are Hermitian. Moreover, each can now be split into positive and negative components, i.e. $H= H_+ - H_-$, where $H_+$ and $H_-$ are positive semidefinite operators. – Luftbahnfahrer Aug 23 '23 at 15:56
  • These can now all be normalized to density operators $\rho = H_+/ (\operatorname{Tr}(H_+))$. So $M(X\otimes Y) = M((\operatorname{Tr}(H_+)(H_+)/\operatorname{Tr}(H_+) - \operatorname{Tr}(H_-)(H_-)/\operatorname{Tr}(H_-)) + i(\cdots))$ and so forth, so everything is expressed as linear combinations of density operators. – Luftbahnfahrer Aug 23 '23 at 15:56