6.4 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 as90 \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.91 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.

  1. Physicists usually still refer to the Bloch ball as the Bloch sphere, even though it really is a ball now, not a sphere.↩︎

  2. One might hope that there is an equally nice visualisation of the density operators in higher dimensions. Unfortunately there isn’t.↩︎