Can changing representation change the meaning of density operator?

Question

I had posted a question What is the actual meaning of the density operator?. After that I understood that if I have the expression of a density operator $$\rho=\sum_{i=1}^{i=k}p_i|\psi_i\rangle \langle \psi_i| ...(0)$$ then it means that the system of qubits defined by $\rho$ is in one of the states ( state vector ) $|\psi_i\rangle$ but I don't in which but I know the probability with which it can be in a state. But then I came across this $$\rho_1 = \frac{1}{2}|a\rangle \langle a|+\frac{1}{2}|b\rangle \langle b |...(1)$$ $$|a\rangle=\sqrt{\frac{3}{4}}|0\rangle+\sqrt{\frac{1}{4}}|1\rangle$$ $$|b\rangle=\sqrt{\frac{3}{4}}|0\rangle-\sqrt{\frac{1}{4}}|1\rangle$$ $$\rho_2=\frac{3}{4}|0\rangle \langle 0| + \frac{1}{4}|1\rangle \langle1|...(2)$$ Now $\rho_1..(1)$ and $\rho_2..(2)$ are mathematically the same but by definition I gave alongside equation $(0)$, equation $(1)$ means qubit is in either state $|a\rangle$ or $|b\rangle$ whereas equation $2$ tells it is in either $|0\rangle$ or $|1\rangle$.Thus being mathematically equal they don't seem equal to me when I see them by this basic definition of density operator. What am I missing?

PS : Sorry for vague terms like state or system. Basically my question is just restricted to one qubit system where state is the vector state of the qubit in $|0\rangle,|1\rangle$ basis. And this example is from Nielsen and Chuang.

Answer 1

The two density matrices are expressed in a different basis of the same Hilbert space. If you compute the expectation values on any state you like you will find the same result with both $\rho_1$ and $\rho_2$. Hence by polarization you can conclude they are indeed the same operator, i.e. they're acting in the same way on the Hilbert space in question.