## 6.2 Statistical mixtures

Let us start with probability distributions over state vectors.
Suppose Alice prepares a quantum system and hands it over to Bob who subsequently measures observable ^{88}
**mixture of the states** **mixed state** for short.

Remember, a mixture of states is very different from a superposition of states: a superposition *always* yields a definite state vector, whereas a mixture does *not*, and so must be described by a density operator.

Bob knows the ensemble of states ^{89}
**density operator**, since it has all the defining properties of the density operator (the convex sum of rank one projectors).
It depends on the constituent states

Once we have

Alice flips a fair coin. If the result is

\texttt{Heads} then she prepares the qubit in the state|0\rangle , and if the result is\texttt{Tails} then she prepares the qubit in the state|1\rangle . She gives Bob the qubit without revealing the result of the coin-flip. Bob’s knowledge of the qubit is described by the density matrix\frac12|0\rangle\langle 0| + \frac12|1\rangle\langle 1| = \begin{bmatrix} \frac12 & 0 \\0 & \frac12 \end{bmatrix}. Suppose Alice flips a fair coin, as before, but now if the result is

\texttt{Heads} then she prepares the qubit in the state|\bar{0}\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) , and if the result is\texttt{Tails} then she prepares the qubit in the state|\bar{1}\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle) . Bob’s knowledge of the qubit is now described by the density matrix\begin{aligned} \frac12|\bar{0}\rangle\langle\bar{0}| + \frac12|\bar{1}\rangle\langle\bar{1}| &= \frac12 \begin{bmatrix} \frac12 & \frac12 \\\frac12 & \frac12 \end{bmatrix} + \frac12 \begin{bmatrix} \frac12 & -\frac12 \\-\frac12 & \frac12 \end{bmatrix} \\&= \begin{bmatrix} \frac12 & 0 \\0 & \frac12 \end{bmatrix}. \end{aligned} Suppose Alice picks up any pair of orthogonal states of a qubit and then flips the coin to chose one of them. Any two orthonormal states of a qubit,

|u_1\rangle ,|u_2\rangle , form a complete basis, so the mixture\frac12|u_1\rangle\langle u_1|+\frac12|u_2\rangle\langle u_2| gives\frac12\mathbf{1} .

As you can see, these three different preparations yield precisely the same density matrix and are hence statistically indistinguishable. In general, two different mixtures can be distinguished (in a statistical sense) if and only if they yield different density matrices. In fact, the optimal way of distinguishing quantum states with different density operators is still an active area of research.