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My intuition says that $M$ has to be the identity matrix, but I am not able to show it rigorously. I tried playing around using spectral decomposition. If $$ \rho = \sum_i \lambda_i |\lambda_i \rangle \langle \lambda_i|\,, $$ then we get the condition $\sum_i \lambda_i M_{ii} = 1$, where $\lambda_i$s are the eigenvalues of the density matrix and $M_{ii}$ are the diagonal matrix elements of $M$ in the eigenbasis of $\rho$.

This should be true for any $\{ \lambda_i \}$ which sums to 1 with diagonal matrix elements $M_{ii}$ in the corresponding eigenbasis.

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TL;DR: Yes, if $\mathrm{tr}(M\rho)=1$ for every density matrix $\rho$, then $M=I$.

We need three facts. First, the inner products of any element $A$ of a Hilbert space with the elements of a basis $\mathcal{B}$ determine $A$ uniquely$^1$. Second, $\langle A,B\rangle_{HS}:=\mathrm{tr}(A^\dagger B)$ is an inner product$^2$. Third, the set of density matrices $\mathcal{D}$ contains$^3$ a basis$^4$ $\mathcal{Q}$. Thus, just the finite set of equations $\mathrm{tr}(M\rho)=1$ for $\rho\in\mathcal{Q}\subset\mathcal{D}$ fixes $M$ uniquely, let alone the infinite set $\mathrm{tr}(M\rho)=1$ for $\rho\in\mathcal{D}$.


$^1$ This follows from the fact that these inner products are coefficients in the representation of $A$ as a linear combination of the elements of the basis dual to $\mathcal{B}$.
$^2$ Also known as the Hilbert-Schmidt inner product.
$^3$ Exercise. Hint: Construct the one-hot basis from projectors.
$^4$ Unfortunately, none of the bases in $\mathcal{D}$ are orthogonal with respect to $\langle.,.\rangle_{HS}$, which is why we don't assume it in $^1$ and instead resort to the dual basis.

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  • $\begingroup$ Can you please explain the third fact again? It's not clear to me. $\endgroup$
    – FDGod
    Sep 21, 2023 at 18:03
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    $\begingroup$ There exist a basis $\mathcal{Q}$ of the complex vector space of all linear operators on the Hilbert space in which every element is a valid density matrix. For example, in the single-qubit case we can take $\mathcal{Q}=\{|0\rangle\langle 0|, |1\rangle\langle 1|, |+\rangle\langle +|, |{+i}\rangle\langle{+i}|\}$. $\endgroup$ Sep 21, 2023 at 23:55

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