Are quantum state (density) operators unique?

Question

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).

Answer 1

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.