Are two states with the same measurement probabilities necessarily equal up to unitary equivalence? Let $\rho$ and $\rho'$ be $n\times n$ density matrices, and suppose that for every observable $A$ and every $\lambda$ in the spectrum of $A$ we have
$$
\text{tr}(\rho P_{\lambda})=\text{tr}(\rho' P_{\lambda}),
$$
where $P_{\lambda}$ is the orthogonal projection onto the $\lambda$-eigenspace of $A$. Does it then necessarily follow that $\rho'=U\rho U^{\dagger}$ for some unitary $U$?
 A: The answers from Jason Funderberker and Valter Moretti are perfect at explaining mathematically how the equivalence of all possible measurements requires the two density matrices to be the same.
However, something I think is important to highlight is that this is not some kind of mathematical accident, it is an essential part of a good theroy. If two things are completely impossible to distinguish from one another in real life (with experimental measurements), then it is really bad if the theory says the two are in fact distinct. If the two could not be distinguished by the experimentalists, then why did the theorist trying to fit a model to there results insist that the two were in fact different? Its is a big no-no.
If density matrices did not have that property they would not be a good representation of the system (and our ignorance of the system), and they would need to be replaced with a different approach.
A: Let $H$ denote a finite-dimensional complex Hilbert space, $\rho$ and $\rho^\prime$ be two density matrices, i.e. positive semi-definite operators with unit trace and $A$ an arbitrary hermitian operator. Its spectral representation reads
$$ A=\sum\limits_\lambda \lambda(A)\, P_\lambda (A) \quad .$$
If the equality in the question holds for all $A$, then $\rho^\prime = \rho$. To see this, multiply the equality with $\lambda(A)$ and sum over all $\lambda$. We then find
$$\mathrm{Tr}\rho A = \mathrm{Tr}\rho^\prime A \quad, $$
which we can write as
$$\mathrm{Tr} (\rho-\rho^\prime)A = 0 \quad  \forall A  \quad .$$
Now there are several ways to see that this implies $\rho-\rho^\prime =0$. For example, pick $A:=\rho-\rho^\prime$. Then $\mathrm{Tr} A^2 = 0$. But $A^2 \geq 0$ and thus $A=0$.
A: Actually $\rho=\rho'$. Indeed, specializing $A= P= |\psi\rangle \langle \psi|$ for $||\psi||=1$, the hypothesis implies $\langle\psi| (\rho-\rho')\psi\rangle =0$. Linearity permits to relax the requirment $||\psi||=1$. By polarization, in turn,  it implies $\langle\psi| (\rho-\rho')\phi\rangle =0$ for every pair of vectors $\psi,\phi$. Fixing $\phi$ and choosing $\psi:= (\rho-\rho')\phi$, we conclude that, for every vector $\phi$
$$|| (\rho-\rho')\phi||^2 =0 $$
which means $\rho-\rho'=0$.
The result is valid also for infinite dimensional Hilbert spaces.
