# Average squared absolute value of inner product between some unit vectors in a Hilbert space

Let $$S$$ be a finite set of unit vectors in $$d$$-dimensional complex Hilbert space $$l_2^d$$.

I believe (from numerical experiment) that $$\frac{1}{\lvert S \rvert^2} \sum_{x,y \in S} \lvert \langle x,y \rangle \rvert^2 \geq \frac1d$$. In other words, if I choose two vectors from $$S$$ at random and take the squared absolute value of their inner product, the expectation is at least $$\frac1d$$, for any initial choice of $$S$$.

This feels like it should have a simple proof but it has eluded me.

Edited to add: a more general, but actually equivalent, statement is the following. Given an arbitrary probability distribution on the unit sphere of $$l_2^d(\mathbb{C})$$, if $$x$$ and $$y$$ are chosen independently from this distribution, then $$\mathbf{E}(\lvert \langle x, y \rangle \rvert^2) \geq \frac1d$$.

• Well it is clearly true when $|S|\leq d$. Things get tricky when $|S|>d$. Let me meditate on this a bit more.
– Mick
Mar 8, 2020 at 10:23
• @Mick Thanks. Indeed, $\lvert S \rvert > d$ is the interesting case. It's also easy to see that the bound is attainable when $\lvert S \rvert = kd$ (by taking $k$ orthonormal bases). Mar 8, 2020 at 12:18

Given a probability distribution $$P(x)$$ on the unit sphere we can consider the matrix
$$A = \int dx\, P(x)xx^\dagger.$$ This is readily seen to satisfy $$A^\dagger=A,\qquad \operatorname{tr}A=1,\qquad y^\dagger A y\ge 0\quad \forall y\in \mathbb{C}^d,$$ i.e. it can be diagonalized with postive eigenvalues $$\lambda_i$$ that sum to 1. We can then compute the trace of the square $$\operatorname{tr} A^2 = \int dx\, dy\,P(x)P(y)\operatorname{tr}(xx^\dagger y y^\dagger) = \int dx\, dy\,P(x)P(y)\, |x^\dagger y|^2=\mathbf{E}(\lvert \langle x, y \rangle \rvert^2),$$ which is the expectation value of interest. As this equals the trace of $$A^2$$ we can equally consider the sum of the squared eigenvalues, which can be bound from below by the Cauchy–Schwarz inequality $$\mathbf{E}(\lvert \langle x, y \rangle \rvert^2)=\sum_{i=1}^d\lambda_i^2\ge \frac{1}{d}\left(\sum_{i=1}^d \lambda_i\right)^2=\frac{1}{d}.$$