Rolling up the Bloch Ball

In an earlier post, we discussed how to do quantum mechanics for the simplest possible quantum system, a single qubit, using expectation values. What if we want to apply quantum theory to a bigger system, like multiple qubits put together? This is where the standard mathematical language of the subject starts to pay off. It is possible to keep working with expectation values the way we were, and in some applications it is even beneficial. However, expressing what the valid set of preparations looks like is difficult to do without bringing in more of the linear algebra.

I’ve taught this to college students, after first reviewing how complex numbers work and some basics about how to manipulate matrices — adding them, multiplying them, taking the trace and the determinant, what eigenvalues and eigenvectors are.

For our own purposes, our next step will be to develop the framework in which we can consider multiple qubits together. It might not seem obvious now, but a good way to make progress is to combine our three expected values $(x,y,z)$ into a matrix, like so:
$$ \rho = \frac{1}{2} \begin{pmatrix} 1 + z & x – iy \\ x + iy & 1 – z \end{pmatrix} \, . $$
This matrix has some nice properties of the sort that we can generalize to bigger matrices. For example, its trace is 1, which feels kind of like how a list of probabilities sums up to 1. Meanwhile, the determinant is the pleasingly Pythagorean quantity
$$ \det\rho = \frac{1}{4}(1 – x^2 – y^2 – z^2) \, . $$
This will be nonnegative for all the valid preparation points. So, the product of the two eigenvalues of $\rho$ will be positive for every point in the interior; we can only get a zero eigenvalue by picking a point on the surface. Using the trace and the determinant, we can find the eigenvalues thanks to a nifty application of the quadratic formula:
$$ \lambda_\pm = \frac{\mathrm{tr}\rho \pm \sqrt{(\mathrm{tr}\rho)^2 – 4\det\rho}}{2} = \frac{1}{2}(1 \pm \sqrt{x^2 + y^2 + z^2}) \, . $$
And indeed, this will always give us positive real numbers, except on the surface of the Bloch ball where the $\lambda_+$ solution is 1 while the $\lambda_-$ solution is 0. Requiring that a matrix’s eigenvalues be nonnegative is another property we can generalize.

Another interesting thing happens if we take the square of $\rho$:
$$ \rho^2 = \frac{1}{4}
\begin{pmatrix} 1 + x^2 + y^2 + z^2 + 2z
& 2x – 2iy \\
2x + 2iy
& 1 + x^2 + y^2 + z^2 – 2z
\end{pmatrix} \, . $$
If the point $(x,y,z)$ is on the surface of the sphere, then $\rho^2 = \rho$. This will turn out to be a way to characterize the extreme elements in our set of valid preparations, no matter how big we make our matrices.

This gets us almost to the point of being able to do the quantum math for the parable of the muffins.