## The Bloch ball

We have already talked in some depth about the Bloch sphere, 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.

An arbitrary (2\times 2) Hermitian matrix has four real parameters and can be expanded in the basis \{\mathbf{1}, \sigma_x, \sigma_y, \sigma_z\} consisting of the identity and the three Pauli matrices.
Since the Pauli matrices are traceless, the coefficient of \mathbf{1} in the expansion of a density matrix \rho must be \frac{1}{2}, in order to have \operatorname{tr}\rho=1.
Thus \rho may be expressed as
\begin{aligned}
\rho
&= \frac{1}{2}\left( \mathbf{1}+\vec{s}\cdot\vec{\sigma} \right)
\\&= \frac{1}{2}
\begin{bmatrix}
1+s_z & s_x-is_y
\\s_x+is_y & 1-s_z
\end{bmatrix}.
\end{aligned}
where \vec{s}=(s_x,s_y,s_z) and \vec{\sigma}=(\sigma_x,\sigma_y,\sigma_z).
The vector \vec{s} is called the **Bloch vector** for the density operator \rho.
Any real Bloch vector \vec{s} defines a Hermitian operator \rho with \operatorname{tr}\rho=1, but in order for \rho to be a density operator it must *also* be non-negative.
Which Bloch vectors yield legitimate density operators?
That is, what does the non-negative condition on \rho translate to in terms of the Bloch vector \vec{s}?

To answer this, let us compute the eigenvalues of \rho.
The trace of a matrix is equal to the *sum* of its eigenvalues, and the determinant is equal to the *product* of its eigenvalues.
We know that \operatorname{tr}\rho=1, and we can calculate \det\rho from the matrix form above:
\begin{aligned}
\det\rho
&= \frac{1}{4}(1-s^2)
\\&= \frac{1}{2}(1+s)\frac{1}{2}(1-s)
\end{aligned}
where s=|\vec{s}|=\sqrt{|s_x|^2+|s_y|^2+|s_z|^2}.
It follows that the two eigenvalues of \rho are \frac{1}{2}(1\pm s).
For \rho to be non-negative, its eigenvalues have to be non-negative, and so s (the length of the Bloch vector) cannot exceed 1.

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 (\vec{s} such that s=1), i.e. the surface of the ball (the Bloch sphere, which we have already seen!); 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 the “purity” of a state.

One might hope that there is an equally simple visualisation of the density operators in higher dimensions.
Unfortunately, there is *not*: things become *much* more complicated, very quickly.

Qubits are 2-dimensional and give rise to the Bloch ball, which is a 3-dimensional object.
In general, n-dimensional quantum systems give rise to (n^2-1)-dimensional state spaces, often denoted \mathcal{Q}_n; for n=3, where we study **qutrits**, we would need to study an 8-dimensional object \mathcal{Q}_3.

It turns out, quite surprisingly, that there exists a 3-dimensional object that has many (but not all) of the properties that we would want from \mathcal{Q}_3.
For example, the rank-1 pure states form a connected set on the surface, which lies a maximum distance of \sqrt{2} from the maximally mixed state \frac13\mathbf{1}; the other points on the surface correspond to rank-1 and rank-2 operators; the points strictly inside correspond to rank-3 (i.e. full rank) operators.
However, since it is only 3-dimensional, it can never satisfy *all* the properties that we would like, since \mathcal{Q}_3 *has to be* 8-dimensional.
Nevertheless, the construction is both interesting and useful (and very recent!) — see C Eltschka, M Huber, S Morelli, and J Siewert, “The shape of higher-dimensional state space: Bloch-ball analog for a qutrit”, *Quantum* **5** (2021), DOI: 10.22331/q-2021-06-29-485.

One has to be careful when trying to use the Bloch ball to talk about multiple qubits, precisely for the reason that “most” states are not separable states, but instead have some amount of entanglement.
If we have n qubits, then we can describe the corresponding product state in terms of n vectors in the Bloch ball, but this method only lets us describe *product* states of the n qubits — we saw in Section 5.5 that, as n grows larger, “most” states are *not* separable!

For example, say that we have a system with two qubits, and we wish to understand how they move around the Bloch sphere under some unitary evolution.
If our qubits are initially in state |a\rangle|b\rangle, then evolve to the state U|a\rangle|b\rangle.
Simple!
But now say that, before applying our unitary U, we first *rotated* the Bloch ball so that our qubits were in some other state |a'\rangle|b'\rangle, and *then* applied our unitary U to this rotated state.
A natural question to ask is if there exists some rotation that takes the first result U|a\rangle|b\rangle to the second result U|a'\rangle|b'\rangle.
In other words, if we denote our rotation by R, then does there exist a rotation S such that U\circ R = S\circ U?

The answer is most definitely *no*, as shown by a reasonably simple example: consider the controlled-\texttt{NOT} gate acting on two qubits initially in some state |0\rangle|\psi\rangle, and where the rotation R takes |0\rangle|\psi\rangle to |\psi'\rangle|0\rangle.
Then (U\circ R)|a\rangle|b\rangle=|\psi'\rangle|\psi'\rangle, and U|a\rangle|b\rangle=|0\rangle|\psi\rangle.
But we cannot transform the latter into the former by a simple rotation of the sphere, since the latter has two distinct Bloch vectors, whereas the former has a single repeated one, and rotations never “collapse” two distinct vectors into one.
The key point here is that the angles between the Bloch vectors can *change* upon applying unitary operations, and the amount by which they change can depend on the Bloch vectors themselves, whereas rotations keep these relative angles *constant*.