Are quantum state (density) operators unique? For context, the following question is related to my question on Physics SE. Now for the question.
Consider the following requirements on an operator $\rho$ on a Hilbert space (or on a rigged Hilbert space? Obviously I am not so capable as it pertains to these subtleties):
(1) Self-adjointness: $\rho$ = $\rho^\dagger$
(2) Non-negative: For every element (again I'm not sure of exactly which space -- the Hilbert space, the nuclear space?) $u$, we have $\langle u,\rho u \rangle \geq 0.$
(3) Now for the quantity $\langle R \rangle$ (physically, the expectation value/average of an observable; mathematically, given below) defined for every self-adjoint operator $R$ on the space as
$$\langle R \rangle := \frac{\textrm{Tr}[\rho R]}{\textrm{Tr}[\rho]}, $$
consider the set over all $R$, {$\langle R \rangle_\rho$}, for the given $\rho$.
Do the three requirements above uniquely specify an operator (quantum state)? I think the answer is obviously no: in particular, consider $\rho'$ defined by
$$\rho' := \frac{\rho}{\textrm{Tr}[\rho]}.$$
Clearly, for a given $\rho$, $\rho'$ so defined gives us the same three properties. So too does $\rho'' := \rho/c$ for any $c > 0$ ($c \in \mathbb{R}$). My question is thus: are there any other classes of $\rho'''$ which also satisfy these three requirements?. I believe this question is equivalent to: do requirements (1)-(3), along with the requirement
(4) that $\rho$ be of unit trace
uniquely specify an operator? I believe this is what's tacitly argued when we say that we can conventionally take a quantum state to be represented by an operator obeying (1) - (4).
 A: Your phrasing of (3) as a "requirement" doesn't make sense because it's not a proposition that can be true or false but a collection of data. Here is what I believe is the intended question:

Question: Suppose $\rho, \rho'$ are self-adjoint non-negative trace-class unit trace operators on a Hilbert space $H$ such that $\text{tr}(\rho R) = \text{tr}(\rho' R)$ for all self-adjoint operators $R$. Does $\rho = \rho'$?

The answer is yes and the proof is simple. The last condition is equivalent to $\text{tr}((\rho - \rho') R) = 0$. Now take $R = \rho - \rho'$. By the spectral theorem $\rho - \rho'$ (which, being trace class, is in particular compact) has an orthonormal basis of eigenvectors $e_k$ with real eigenvalues $\lambda_k$, and the trace can be evaluated using this basis to give
$$\text{tr}((\rho - \rho')^2) = \sum_k \langle (\rho - \rho')^2 e_k, e_k \rangle = \sum_k \lambda_k^2 = 0$$
which is only possible if $\lambda_k = 0$ for all $k$, so $\rho - \rho' = 0$.
The conclusion is that density operators are uniquely determined by the expectation values of observables, as desired.
