$N(\frac{1}{2},2)=3$ for vectors in a Hilbert Space

Question

Came across This question regarding the maximum number of almost orthogonal vectors one can embed in a Hilbert space. They state that $N(\frac{1}{2},2)=3$, and that explicit construction of the vectors using the Bloch sphere shows this. However, I cannot seem to grasp what they mean by this. Their further example of $N(\frac{1}{\sqrt{2}},2)=6$ does make sense to me, as these are simply the eigenvectors of the pauli operators. But how does one show that the number of vectors which meet the following criteria is only 3?

$$\langle V_i|V_i\rangle = 1$$

$$|\langle V_i|V_j\rangle| \leq \epsilon, i \neq j$$

Answer 1

Here's a very visual way to think about this (I make no claim about it being a rigorous proof). Let $$ |V_1\rangle=|0\rangle,|V_2\rangle=\frac12|0\rangle+\frac{\sqrt{3}}{2}|1\rangle,|V_3\rangle=\frac12|0\rangle-\frac{\sqrt{3}}{2}|1\rangle. $$ These each have overlaps of 1/2. Now draw these on the Bloch sphere. They are three equally spaced vectors around a great circle. You cannot push one closer to another because that would increase their overlap.

Now, can I add a fourth vector? Whatever vector I add into the sphere, it must make an angle of $\pi/2$ or less with one of the existing vectors, and hence would have overlap $1/\sqrt{2}$ or greater. So, at least for this choice of three vectors, I cannot add a fourth and maintain the value of $\epsilon$.

With this picture in mind, you can probably also convince yourself that these vectors have to be selected this way. $|V_1\rangle$ is arbitrary, I can just orient the view so that it's at the top of the sphere. For $|V_2\rangle$ I've got an arbitrary freedom of rotation about the $V_1\rangle$ axis, so I just picked the orthogonal component to be real and positive. At that point, my choice of $|V_3\rangle$ was fixed - there was only one possible choice that could have the correct overlap.

If the visual version doesn't do it for you, I'm sure someone will formalise this mathematically...