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Given a nonempty finite subset $S$ of the unit sphere of $d$-dimensional complex Hilbert space, let $\lambda(S) = \frac{1}{\lvert S \rvert^2} \sum_{x,y \in S} \lvert \langle x,y \rangle \rvert^2$ be the mean squared absolute value of the inner product of two vectors chosen from $S$.

A basic observation from experiment is that $\lambda(S) \geq \frac1d$. This looks simple enough, but I don't know how to prove it, and as I've had no answers on math.stackexchange I am asking the question here.

A second experimental observation is that if $S_1$ and $S_2$ are disjoint then $\lambda(S_1 \cup S_2) \leq \max(\lambda(S_1), \lambda(S_2))$.

If these conjectures are true, it follows that $\mathcal{E}_d = \{S: \lambda(S) = \frac1d\}$ is closed under disjoint unions, and it would be interesting to characterize its "basic" sets, i.e. those sets in $\mathcal{E}_d$ that are not a disjoint union of other sets in $\mathcal{E}_d$ (or alternatively, those with no proper subset in $\mathcal{E}_d$). Clearly these include the orthonormal bases, but there are plenty of others, e.g. $S = \{(1, 0), (\frac12, \frac{\sqrt3}{2}), (\frac12, -\frac{\sqrt3}{2})\}$ for $d = 2$.

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The Welch bound gives

$$\lambda(S) = \frac{1}{\lvert S \rvert^2} \sum_{x,y \in S} \lvert \langle x,y \rangle \rvert^2 \geq \frac{(\sum_{x \in S} \lvert \langle x,x \rangle \rvert)^2}{d \lvert S \rvert^2}=\frac{1}{d}$$ which is what you want.

There are Welch bound equality sets (do a google search) but achieving equality for general set size $|S|$ is difficult.

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  • $\begingroup$ Thank you, that was the reference I needed! $\endgroup$ – Alec Edgington Mar 14 at 15:47

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