## The Bloch ball

We have already talked in some depth about the Bloch sphere in Chapter 2 and Chapter 3, but now that we are considering density operators (which are strictly more general than state vectors), we are actually interested in the Bloch *ball*, i.e. not just the sphere of vectors of magnitude 1, but instead the ball of vectors of magnitude *less than or equal to* 1.

The most general Hermitian (2\times 2) matrix has four real parameters and can be expanded in the basis composed of the identity and the three Pauli matrices: \{\mathbf{1}, \sigma_x, \sigma_y, \sigma_z\}.
Since the Pauli matrices are traceless, the coefficient of \mathbf{1} in the expansion of a density matrix \rho must be \frac12, so that \operatorname{tr}\rho=1.
Thus \rho may be expressed as
\begin{aligned}
\rho
&= \frac12\left( \mathbf{1}+\vec{s}\cdot\vec{\sigma} \right)
\\&= \frac12\left( \mathbf{1}+ s_x\sigma_x + s_y\sigma_y + s_z\sigma_z \right)
\\&= \frac12
\begin{bmatrix}
1 + s_z & s_x - is_y
\\s_x + is_y & 1 - s_z
\end{bmatrix}.
\end{aligned}
The vector \vec{s} is called the **Bloch vector** for the density operator \rho.
Any real Bloch vector \vec{s} defines a trace one Hermitian operator \rho, but in order for \rho to be a density operator it must also be non-negative.
Which Bloch vectors yield legitimate density operators?

Let us compute the eigenvalues of \rho.
The sum of the two eigenvalues of \rho is, of course, equal to one (\operatorname{tr}\rho=1) and the product is equal to the determinant of \rho, which can be computed from the matrix form above:
\det\rho
= \frac{1}{4}(1-s^2)
= \frac12(1+s)\frac12(1-s)
where s=|\vec{s}|.
It follows that the two eigenvalues of \rho are \frac12(1\pm s).
They have to be non-negative, and so s, the length of the Bloch vector, cannot exceed one.
We can now visualise the convex set of (2\times 2) density matrices as a unit ball in three-dimensional Euclidean space: the extremal points, which represent pure states, are the points on the boundary (s=1), i.e. the surface of the ball; the maximally mixed state \mathbf{1}/2 corresponds to s=0, i.e. the centre of the ball.
In general, the length of the Bloch vector s can be thought of as a “purity” of a state.