Mixed States and Density Matrices

A pure state $|\psi\rangle$ carries all the information about a quantum system; you know the exact state vector. Real hardware, however, suffers noise: a qubit may have been prepared incorrectly, interacted with its environment, or relaxed before you could use it. In those cases you only know that the qubit is in state $|\psi_i\rangle$ with probability $p_i$ . This is a mixed state, and the right tool to describe it is the density matrix.

From state vectors to density matrices

For a pure state $|\psi\rangle$ the density matrix is the outer product

$\rho = |\psi\rangle\langle\psi|.$

For a mixed state — a classical probability distribution over pure states — it is the weighted average

$\rho = \sum_i p_i\, |\psi_i\rangle\langle\psi_i|, \qquad \sum_i p_i = 1,\; p_i \ge 0.$

Every valid density matrix is Hermitian ( $\rho^\dagger = \rho$ ), positive semi-definite ( $\langle\phi|\rho|\phi\rangle \ge 0$ for all $|\phi\rangle$ ), and has unit trace ( $\text{tr}(\rho) = 1$ ).

A one-qubit example

Suppose a noisy source emits $|0\rangle$ with probability $p_0 = 0.7$ and $|1\rangle$ with probability $p_1 = 0.3$ . The density matrix is

$\rho = 0.7\,|0\rangle\langle 0| + 0.3\,|1\rangle\langle 1| = \begin{pmatrix} 0.7 & 0 \\ 0 & 0.3 \end{pmatrix}.$

The diagonal entries are the probabilities of each basis state. There are no off-diagonal terms because this is a classical mixture — no quantum coherence between $|0\rangle$ and $|1\rangle$ . Compare this with the equal-superposition pure state $|{+}\rangle = (|0\rangle + |1\rangle)/\sqrt{2}$ , whose density matrix is

$\rho_{+} = |{+}\rangle\langle{+}| = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}.$

The off-diagonal entries (coherences) are a hallmark of genuine quantum superposition. Decoherence is the process by which those off-diagonal entries decay toward zero, turning the pure $\rho_{+}$ into a diagonal mixed state.

Purity: how mixed is the state?

The purity of a state is

$\gamma = \text{tr}(\rho^2), \qquad \frac{1}{d} \le \gamma \le 1,$

where $d$ is the dimension ( $d = 2$ for a qubit). $\gamma = 1$ if and only if $\rho$ is pure. The maximally mixed state $\rho = I/d$ has the minimum purity $\gamma = 1/d$ .

For a diagonal density matrix with entries $p_0$ and $p_1$ (so $p_0 + p_1 = 1$ ),

$\text{tr}(\rho^2) = p_0^2 + p_1^2.$

Substituting $p_0 = 0.7$ , $p_1 = 0.3$ :

$\gamma = 0.49 + 0.09 = 0.58.$

This is strictly less than 1 (not pure) and greater than $1/2$ (not maximally mixed) — exactly what we expect for a moderately noisy qubit.

Measurement from the density matrix

The probability of measuring outcome $b$ is

$P(b) = \text{tr}\!\left(\,|b\rangle\langle b|\,\rho\right) = \langle b|\rho|b\rangle = \rho_{bb},$

just the corresponding diagonal entry. For the noisy qubit above, $P(0) = 0.7$ and $P(1) = 0.3$ .

Try it

A qubit is prepared as a classical mixture: $|0\rangle$ with probability 0.7 and $|1\rangle$ with probability 0.3. Compute and return the purity $\text{tr}(\rho^2)$ of this mixed state.

Run your code to see the quantum state.