is a mixed states really a statistical mixture?

Question

$\DeclareMathOperator{\Tr}{Tr}\newcommand{\ket}[1]{|#1\rangle}\newcommand{\bra}[1]{\langle#1|}$Let consider the pure state $$ \ket{\psi_{1}}=\dfrac{\ket{0}+\ket{1}}{\sqrt{2}} $$ whose density matrix is $$ \rho_{1} = \frac{1}{2} \mathbb{1}_{2} \text{ (matrix of ones)}$$ and the mixed state $$ \begin{equation*} \rho_{2} = \dfrac{1}{2}(\ket{0} \bra{0} + \ket{1}\bra{1}) \end{equation*} $$ in the first case it "means" that the state is both in $\ket{0}$ and $\ket{1}$ while in the second it "means" that we lack information on the system and that it is either in $\ket{0}$ or $\ket{1}$. however if I compute the probability that $\rho_{2}$ is in $\psi_1$ which is $$\Tr(\Pi_{\psi_1}\rho_{2}) = \Tr(\rho_{1}\rho_{2}) = \frac{1}{2}$$. Can someone explain to me this " contradiction " ? perhaps I misunderstood the meaning of a mixed state.

Answer 1

$\DeclareMathOperator{\Tr}{Tr}\newcommand{\ket}[1]{|#1\rangle}\newcommand{\bra}[1]{\langle#1|}$The math that you have done is correct, and so is your general understanding of mixtures and density matrices. The problem is that your interpretation of your measurement math is wrong.

Measurements are with respect to some basis (such as the computational basis $\ket{0}$ and $\ket{1})$. When you take the measurement here, you are measuring in a basis with $\ket{\psi_1}$ (with some other basis vector, such as $\ket{-}$). When you say,

however if I compute the probability that $\rho_2$ is in $\psi_1$ which is $$\Tr{(\Pi_{\psi_1}\rho_2)} = \Tr(\rho_1\rho_2) = \frac{1}{2}$$

The actual math is that you are measuring in the basis of $\ket{\psi_1}$, i.e. the likelihood that this statistical mixture of $\ket{0}$ and $\ket{1}$ will be measured to be in the $\ket{\psi_1}$ state when measured in $\ket{\psi_1}$'s basis. Note that both $\ket{0}$ and $\ket{1}$ are superpositions of $\ket{\psi}$, both with a 50% chance of being $\ket{\psi_1}$ when measured in $\ket{\psi_1}$'s basis. The probability computes to: $$ P(\rho_2\text{ is }\ket{0}) \cdot P(\ket{0}\text{ measures as }\ket{\psi_1}) + P(\rho_2\text{ is }\ket{1}) \cdot P(\ket{1}\text{ measures as }\ket{\psi_1})\\ = \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) + \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) = \frac{1}{2} $$ In summary, the measurement you calculated does not mean the probability that "$\rho_2$ is in $\psi_1$", but rather the probability that its ensemble states $\ket{0}$ and $\ket{1}$ would measure to be $\ket{\psi_1}$.