The Bloch Vector of ρ

For a single qubit, the density matrix has an especially vivid geometric form: a point in or on a ball of radius one — the Bloch ball.

The Bloch decomposition

Any $2\times2$ density matrix can be written using the identity and the Pauli matrices $X, Y, Z$ as

\rho = \frac{1}{2}\big(I + r_x X + r_y Y + r_z Z\big) = \frac{1}{2}\big(I + \vec r\cdot\vec\sigma\big),

where $\vec r = (r_x, r_y, r_z)$ is the Bloch vector. This is exact and unique: there is a perfect one-to-one correspondence between qubit density matrices and Bloch vectors.

Extracting the Bloch vector

Because the Paulis are traceless and $\operatorname{Tr}(\sigma_a \sigma_b) = 2\delta_{ab}$ , each component is an expectation value:

r_x = \langle X\rangle = \operatorname{Tr}(\rho X), \quad r_y = \langle Y\rangle = \operatorname{Tr}(\rho Y), \quad r_z = \langle Z\rangle = \operatorname{Tr}(\rho Z).

Carrying out the traces in terms of matrix entries gives the convenient formulas

r_x = 2\,\operatorname{Re}(\rho_{01}), \qquad r_y = -2\,\operatorname{Im}(\rho_{01}), \qquad r_z = \rho_{00} - \rho_{11}.

Length encodes purity

The length of the Bloch vector measures how pure the state is:

\operatorname{Tr}(\rho^2) = \frac{1 + |\vec r|^2}{2}.

So:

$|\vec r| = 1$ — the surface of the ball — corresponds to a pure state ( $\operatorname{Tr}(\rho^2)=1$ ).
$|\vec r| = 0$ — the center — is the maximally mixed state $\rho = I/2$ .
$0 < |\vec r| < 1$ — the interior — is a partially mixed state.

The pure states from earlier modules sat on the surface of the Bloch sphere; mixed states fill the inside. Decoherence shrinks $|\vec r|$ , pulling the state toward the center.

Try it

The mixed state $\rho = \begin{pmatrix}3/4 & 1/4\\ 1/4 & 1/4\end{pmatrix}$ is real, so $r_y = 0$ . Compute its Bloch vector and return the length $|\vec r|$ . Since $0 < |\vec r| < 1$ , the point lies strictly inside the Bloch ball — the geometric signature of a mixed state.

Run your code to see the quantum state.