The Relative Distribution of Two Unitary Operators

A weight measure, $w_{A,B}$ may be defined over the relative angles between the spectra of unitary operators as a function of respective projectors using the Hilbert-Schmidt inner product. In this article, it is proved to be a probability measure. Importantly, this allows $w_{A,B}$ to be treated as a legitimate statistical object where expectation is defined and the integration of functions relative to it are valid.

Proposition Let $A,B \in U_d(\mathbb{C})$ and let the spectral decompositions be $A = \sum_s e^{i \pi \phi_s} \Pi_{A,s}$ and $B = \sum_t e^{i \pi \psi_t} \Pi_{B,t}$, respectively. A weight measure $w_{A,B}:\mathcal{B}((0, 2\pi]) \rightarrow \mathbb{R}_{\ge 0}$, for all $\psi_t – \phi_s \mod 2 \pi \in [0,2\pi)$, may be defined as:

$w_{A,B}(E) = \sum_{s,t : \psi_t – \phi_s \in E} \langle \Pi_{A,s}, \Pi_{B,t} \rangle$

Note that $\psi_t – \phi_s \mod 2 \pi \in [0,2\pi)$.

This weight measure $w_{A,B}(E)$ is a probability distribution.

Proof

In order for this to be a probability distribution, the following must be shown:

$w_{A,B}([0, 2\pi)) = 1$

Begin with the definition of this weight measure.

$w_{A,B}([0,2\pi)) = \sum_{s,t : \psi_t – \phi_s \in [0, 2\pi)} \langle \Pi_{A,s}, \Pi_{B,t} \rangle$

Then, using the Hilbert-Schmidt inner product:

$w_{A,B}([0,2\pi)) = \sum_{s,t : \psi_t – \phi_s \in [0, 2\pi)} \frac{1}{m} \text{tr} \left( \Pi_{A,s} \Pi_{B,t} \right)$

In the above line, $m$ is dimension of the Hilbert space the projection operators are acting upon.

$\Pi \in M_m(\mathbb{C}) \Leftrightarrow \text{dim}(H) = m$

Dividing by $m$ normalizes the trace and turns the measure from that of total mass to that of a probability distribution.

Next, using linearity of the trace operator:

$w_{A,B}([0,2\pi)) = \frac{1}{m} \text{tr} \left( \left( \sum_s \Pi_{A,s} \right) \left( \sum_t \Pi_{B,t} \right) \right)$

Using the completeness relation of projectors:

$w_{A,B}([0,2\pi)) = \frac{1}{m} \text{tr} \left( \left( I \right) \left( I \right) \right) = \frac{1}{m} \text{tr} \left( I \right)$

Finally:

$w_{A,B}([0,2\pi)) = \frac{1}{m} m = 1$

Thus, $w_{A,B}$ taken over the unit circle has been proved to be a valid probability distribution. $\square$