This is a sort of appendix to a blog post I’m writing, and will make a lot more sense with that context. I want to find stationary points of the sum of marginal probabilities $$\sum p_i = p_z + p_x + p_y$$, where the $$p_i$$s are defined as in my qubit phase space cheat sheet:

$$\sum p_i = \frac{3}{2} - \frac{1}{2}(\cos\theta + \sin\theta\cos\phi + \sin\theta\sin\phi)$$,

where $$\theta \in [0, \pi ]$$ and $$\phi \in [0, 2\pi )$$.

The rest of this page is just the details of finding and classifying the stationary points, which occur when:

1. $$\frac{\partial \sum p_i}{\partial \phi} = 0$$ and
2. $$\frac{\partial \sum p_i}{\partial \theta} = 0$$.

The details are in the next section, then there’s a summary and graphs, then I find the corresponding qubit states.

Details

Equation 1. gives

$$\sin \theta \sin \phi = \sin \theta \cos \phi$$.

So either:

• $$\sin\theta = 0$$, so that $$\theta = 0, \pi$$
• $$\sin\phi = \cos\phi$$, so that $$\phi = \frac{\pi}{4}, \frac{5\pi}{4}$$.

Deal with the $$\sin\theta = 0$$ case first. In that case $$\sum p_i$$ just becomes a constant. At $$\theta = 0$$, $$\sum p_i = \frac{3}{2} - \frac{1}{2} = 1$$, and at $$\theta = \pi$$, $$\sum p_i = \frac{3}{2} + \frac{1}{2} = 2$$. So there are lines of constant $$\sum p_i$$ at these values of $$\theta$$.

Then take the remaining $$\sin\phi = \cos\phi$$ case, and apply equation 2 as well:

$$\cos\theta\cos\phi + \cos\theta\sin\phi = \sin\theta$$,

or (as $$\cos\theta \neq 0$$ at any of the stationary points)

$$\cos\phi + \sin\phi = \tan\theta$$.

At $$\phi = \frac{\pi}{4}$$, $$\tan\theta = \sqrt{2}$$, so

$$\sin\theta = \frac{\sqrt{2}}{\sqrt{3}}$$ and $$\cos\theta = \frac{1}{\sqrt{3}}$$

This happens when $$\sum p_i = \frac{1}{2}(3 - \sqrt{3}) \sim 0.634$$.

At $$\phi = \frac{5\pi}{4}$$, $$\tan\theta = -\sqrt{2}$$, so

$$\sin\theta = \frac{\sqrt{2}}{\sqrt{3}}$$ and $$\cos\theta = -\frac{1}{\sqrt{3}}$$

This happens when $$\sum p_i = \frac{1}{2}(3 + \sqrt{3}) \sim 2.366$$.

Summary

• Maximum line at $$\theta = \pi$$, value $$\sum p_i = 2$$
• Minimum line at $$\theta = 0$$, value $$\sum p_i = 1$$
• Maximum point at $$\phi = \frac{5\pi}{4}$$, $$\sin\theta = \frac{\sqrt{2}}{\sqrt{3}}$$, $$\cos\theta = -\frac{1}{\sqrt{3}}$$, value of $$\sum p_i = \frac{1}{2}(3 + \sqrt{3})$$
• Minimum point at $$\phi = \frac{\pi}{4}$$, $$\sin\theta = \frac{\sqrt{2}}{\sqrt{3}}$$, $$\cos\theta = \frac{1}{\sqrt{3}}$$, value of $$\sum p_i = \frac{1}{2}(3 - \sqrt{3})$$.

The last one is the case we’re really interested in, as we want to maximise $$\sum p_i$$. Will find the corresponding qubit state below.

Overall the function $$\sum p_i$$ looks like this (thanks WolframAlpha):

The qubit phase space state for the maximum point is the following:

$\begin{equation} \left[ \begin{array}{c|c} \frac{1}{4}\left(1+ \frac{1}{\sqrt{3}}\right) & \frac{1}{4}\left(1+ \frac{1}{\sqrt{3}}\right) \\ \hline \frac{1}{4}\left(1 - \sqrt{3}\right) & \frac{1}{4}\left(1+ \frac{1}{\sqrt{3}}\right) \\ \end{array} \right] \end{equation}$