$\DeclareMathOperator{\Tr}{Tr}\newcommand{\ket}[1]{|#1\rangle}\newcommand{\bra}[1]{\langle#1|}$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.

  • $\begingroup$ What would happen if you apply a Hadamard gate to your first qubit? What is its density matrix then? How about doing the same for your second qubit? $\endgroup$ Dec 10, 2023 at 13:06
  • $\begingroup$ for the first qubit we get $\ket{0}\bra{0}$ (seems logical to me because we go form 0 to + with hadamard and vice versa) and the second we get $\rho_{2}$ . Sorry but I don't exactly see what you are trying to show me. $\endgroup$
    – yosh
    Dec 10, 2023 at 13:47
  • $\begingroup$ Thanks. Your question seems to be what’s the difference between a single (pure) qubit in superposition and one in a mixed state. I was pointing out that you can tell the difference between the two with a Hadamard gate. Otherwise I don’t understand your question. $\endgroup$ Dec 10, 2023 at 14:25
  • $\begingroup$ Sorry if I was not clear but my question is : If mixed states are "physically" different from superposition, how come the probabilty that $\rho_2$ is in the state $\psi_1$ is not 0 since it is supposed to be either in the state $\ket{0}$ or $\ket{1}$ if it is really a statistical mixture $\endgroup$
    – yosh
    Dec 10, 2023 at 14:37
  • $\begingroup$ If I flip a coin and hide the result from you what is the outcome probability that it is heads? It’s already landed, you can see that I’d never flip it myself. Otherwise I’m sorry I still don’t understand your question. $\endgroup$ Dec 10, 2023 at 17:55

1 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 $\ket{\psi_1}$.

  • $\begingroup$ Thank you a lot, that explains everything :) $\endgroup$
    – yosh
    Dec 10, 2023 at 18:17

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