# Gaining intuition about summing over random basis vectors in random matrix theory

I'm currently reading the following reference on eigenstate thermalization and chaos in quantum mechanics: https://arxiv.org/abs/1509.06411

I'm confused by a derivation that I think is very important to understand properly; it's the subject of section 2.2.2, beginning on page 11. To summarize, the claim is the following: consider a Hermitian operator $$\hat{O} = \sum_{i}O_{i}|i\rangle\langle i |$$. Now assume we are given a random Hamiltonian $$H$$ with eigenvectors $$|m\rangle, |n\rangle$$. Use the eigenvectors of $$H$$ to look at matrix elements of $$O$$, e.g., $$O_{mn} = \langle m|\hat{O}|n\rangle$$. The text claims that (to first order in $$\mathcal{D}$$, the Hilbert space dimension)

$$\overline{\langle m|i\rangle\langle j|n\rangle} = \frac{1}{\mathcal{D}}\delta_{mn}\delta_{ij}$$

where the overline indicates "the average ... over random eigenkets $$|m\rangle$$ and $$|n\rangle$$". This result is not obvious to me at all. Perhaps my intuition is just failing because this is implicitly dealing with a large Hilbert space dimension. Clearly this depends on the observation that eigenvectors of random Hamiltonians are random orthogonal unit vectors, but I would like something closer to a proof to better understand this claim. I'm also unclear about what it means to average over random eigenkets (there is a comment in a later paragraph on how what we are really doing is summing over an ensemble of random Hamiltonians, but that confuses me even more frankly). Can anyone offer intuition or something closer to a proof for this statement?

What they're doing is this: imagine a set of unit orthogonal vectors in $$D$$-dimensional vector space where any coordinate of a vector can be a complex number. Now consider all possible "orientations in the vector space" of this set. This is what "random $$|n\rangle$$" means - for example, 1st vector of the set can be really any unit vector in the space, with uniform probability over all the "possible orientations" in the vector space.

The vectors $$|i\rangle,|j\rangle$$ are from a set of fixed orthogonal vectors that form a basis in the above vector space.

Now, expression $$\langle j | n\rangle$$ is just a notation for $$j$$-th coordinate of the vector $$|n\rangle$$, let us denote it $$c_{jn}$$. But that vector can have any orientation, so this coordinate must have angle -independent (angularly symmetrical) probability distribution in complex plane (it is limited to a finite disk of radius 1), which has zero average. And this is valid for all coordinates, not just $$j$$-th.

What is average of these quantities?

• 1) $$c_{31}^* c_{31}$$
• 2) $$c_{31}^* c_{32}$$
• 3) $$c_{31}^* c_{41}$$

In case 1), the expression give modulus squared of a complex number $$c_{31}$$ (square of length of its radius vector). We chose the point from angularly symmetrical cloud of points inside unit disk, so the average of this over all possible points is some number between 0 and 1. The same is true for $$c_{41}^* c_{41}$$, and others. Let us sum them all: $$c_{11}^* c_{11}~+~ ...~+~c_{D1}^* c_{D1}$$ - they have to add up to value one squared, because this expression is square of length of the 1st vector, which is one by assumption.

Now, taking average of all terms in the sum and using the fact that due to symmetry of all coordinates, all the averages are the same, we obtain

$$Av\big[c^*_{j1}c_{j1}\big] = \frac{1}{D}$$ and since all vectors have the same behaviour, $$Av\big[c^*_{jn}c_{jn}\big] = \frac{1}{D}$$ for any $$j,n$$.

In case 2), we are multiplying coordinates of different but orthogonal vectors. The result of this multiplication can be anything lying in the unit disk, but with probability independent of angle, so average value of this is zero. The same can be said of 3) and 4) (when all indices are different).

So the only case where the average of

$$c^*_{im}c_{jn}$$ is nonzero is if $$i=j$$ and $$n=m$$ at the same time. Collecting all these observations, we arrive at the formula (15) from the review paper above.