I am trying to solve the following problem
Consider the ensemble consisting of the qubit states $|0⟩⟨0|$ and $|1⟩⟨1|$ occuring with probabilities 2/3 and 1/3, respectively. Compute the quantum state $\rho$ corresponding to this ensemble. Is $ρ $ pure or mixed?
My try:
I think there is nothing to compute. The quantum state is $\rho=\frac{2}{3}|0⟩⟨0|+\frac{1}{3}|1⟩⟨1|= \pmatrix{2/3 & 0 \\ 0 & 1/3} $. ...,is there? Is this how the quatum state is always computed given an ensamble?
To see if it is pure: If it where pure there would exist a unit vector $|\psi⟩=(a,b)^T$ such that $\rho=|\psi⟩⟨\psi|=\pmatrix{a^2 & ab \\ ab & b^2}$, so $a^2=2/3$, $b^2=1/3$,$ab=0$ where a,b are complex numbers. The system is incompatible because $a=0 $ or $b=0$ contradicts that$a^2=2/3$ and $b^2=1/3$. Therefore since such $|\psi⟩$ does not exist, the state is mixed
Any feedback would be appreciated. Is this correct? Is there something wrong?
