Summing free unitary Brownian motions with applications to quantum information

Abstract

Motivated by quantum information theory, we introduce a dynamical random density matrix built out of the sum of $k \ge 2$ independent unitary Brownian motions. In the large size limit, its spectral distribution equals, up to a normalising factor, that of the free Jacobi process associated with a single self-adjoint projection with trace 1/k. Using free stochastic calculus, we extend this equality to the radial part of the free average of k free unitary Brownian motions and to the free Jacobi process associated with two self-adjoint projections with trace 1/k, provided the initial distributions coincide. In the single projection case, we derive a binomial-type expansion of the moments of the free Jacobi process which extends to any $k \ge 3$ the one derived in Demni et al. (Indiana Univ Math J 61:1351–1368, 2012) in the special case $k=2$. Doing so give rise to a non normal (except for $k=2$) operator arising from the splitting of a self-adjoint projection into the convex sum of k unitary operators. This binomial expansion is then used to derive a pde satisfied by the moment generating function of this non normal operator and for which we determine the corresponding characteristic curves. As an application of our results, we compute the average purity and the entanglement entropy of the large-size limiting density matrix.

Appendices

Appendix A. Moments of stationary distribution

In this appendix, we derive another expression of

$$\begin{aligned} m_n(\infty ) = \int x^n \tilde{\mu }_{\infty }^k(dx) = \frac{1}{2\pi } \int x^{n-1/2}\frac{\sqrt{4(k-1) - k^2x}}{(1 -x)}\textbf{1}_{[0,4(k-1)/k^2]}(x) dx. \end{aligned}$$

To the best of our best knowledge, formula (34) below never appeared in literature. Compared to (21), it has the merit to separate the case $k=2$ corresponding to the arcsine distribution from other values $k \ge 3$. Our main ingredients are two properties satisfied by the Gauss hypergeometric function.

To proceed, perform the variable change $x = 4(k-1)y/k^2$ to write:

$$\begin{aligned} m_n(\infty )&= \frac{4^{n+1}(k-1)^{n+1}}{2\pi k^{2n+1}}\int y^{n-1/2}\frac{\sqrt{1-y}}{1 - 4(k-1)y/k^2} \textbf{1}_{[0,1]}(y) dy \\ {}&= \frac{4^{n}(k-1)^{n+1}}{\sqrt{\pi } k^{2n+1}}\frac{\Gamma (n+1/2)}{\Gamma (n+2)} {}_2F_1\left( 1, n+\frac{1}{2}, n+2; \frac{4(k-1)}{k^2}\right) \\ {}&= \frac{4^{n}(k-1)^{n+1}}{n!k^{2n+1}}\left\{ (-1)^n\sqrt{1-z}\frac{d^n}{dz^n} \left[ (1-z)^{n-1/2}{}_2F_1\left( \frac{1}{2},1,2; z\right) \right] \right\} _{z = 4(k-1)/k^2} \\ {}&= 2\frac{4^{n}(k-1)^{n+1}}{n!k^{2n+1}}\left\{ \sqrt{1-z}(-1)^n\frac{d^n}{dz^n} \left[ \frac{(1-z)^{n-1/2}}{1+\sqrt{1-z}}\right] \right\} _{z = 4(k-1)/k^2} \\ {}&= 2\frac{4^{n}(k-1)^{n+1}}{n!k^{2n+1}}\left\{ \sqrt{z}\frac{d^n}{dz^n} \left[ \frac{z^{n-1/2}}{1+\sqrt{z}}\right] \right\} _{z = [(k-2)/k]^2} \\ {}&= 2\frac{4^{n}(k-1)^{n+1}}{n!k^{2n+1}}\left\{ \sqrt{z}\frac{d^n}{dz^n} \left[ z^{n-1/2} - \frac{z^n}{1+\sqrt{z}} \right] \right\} _{z = [(k-2)/k]^2} \\ {}&= \frac{2(k-1)^{n+1}}{k^{2n+1}}\left\{ \left( {\begin{array}{c}2n\\ n\end{array}}\right) -\frac{4^{n}}{n!}\sqrt{z}\frac{d^n}{dz^n}\left[ \frac{z^n}{1+\sqrt{z}} \right] \right\} _{z = [(k-2)/k]^2}. \end{aligned}$$

Here, the second equality follows from the Euler integral representation of the Gauss hypergeometric function, the third and fourth ones follow from the variational formula (25), p.102 and formula (6), p.101 in [13]. Using direct computations, we readily see that

$$\begin{aligned} \frac{d^n}{dz^n}\left[ \frac{z^n}{1+\sqrt{z}} \right] = \frac{\mathscr {P}_n(\sqrt{z})}{2^n(1+\sqrt{z})^{n+1}} \end{aligned}$$

for some polynomial of degree n. For instance

$$\begin{aligned} \mathscr {P}_0(x)= & {} 1, \mathscr {P}_1(x) = x+2, \mathscr {P}_2(x) = 3x^2+9x+8, \\ \mathscr {P}_3(x)= & {} 15x^3+60x^2+87x+48. \end{aligned}$$

Consequently,

$$\begin{aligned} m_n(\infty ) = \frac{2(k-1)^{n+1}}{k^{2n+1}}\left\{ \left( {\begin{array}{c}2n\\ n\end{array}}\right) -\frac{(k-2)}{2n!(k-1)^{n+1}}k^n\mathscr {P}_n\left( \frac{k-2}{k}\right) \right\} , \end{aligned}$$

(34)

which reduces for $k=2$ to the known formula:

$$\begin{aligned} m_n(\infty ) = \frac{(2n)!}{2^{2n}(n!)^2}. \end{aligned}$$

Appendix B. Free cumulants of an orthogonal projection

The first part of the proof of (29) is a routine computation in free probability theory and we refer the reader to [18] for further details on this machinery. Start with the Cauchy transform of P:

$$\begin{aligned} \tau [(z-P)^{-1}] = \frac{\alpha }{z-1} + \frac{1-\alpha }{z} = \frac{z + \alpha -1}{z(z-1)}, \quad z \notin \{0,1\}. \end{aligned}$$

Next, consider the equation

$$\begin{aligned} yz^2 -z(y+1) + 1-\alpha = 0, \end{aligned}$$

for y lying in a neighborhood of zero. Then the K-transform of P reads:

$$\begin{aligned} K(y) = \frac{y+1 + \sqrt{y^2+1 -2y(1-2\alpha )}}{2y}, \end{aligned}$$

and in turn, its R-transform is given by

$$\begin{aligned} R(y) = K(y) -\frac{1}{y} = \frac{1}{2}\left[ 1 + \frac{\sqrt{y^2+1 -2y(1-2\alpha )}-1}{y}\right] . \end{aligned}$$

It remains to write down the Taylor expansion of the function:

$$\begin{aligned} f_{\alpha }: y \mapsto \frac{\sqrt{y^2+1 -2y(1-2\alpha )}-1}{y}. \end{aligned}$$

To this end, we appeal to the generating series of Legendre polynomials:

$$\begin{aligned} \sum _{n=0}^{\infty } P_n(x)y^n = \frac{1}{\sqrt{1+y^2-2xy}}, \quad |y| < 1. \end{aligned}$$

Indeed, setting $\beta := 1-2\alpha \in [-1,1]$, one has

$$\begin{aligned}{}[yf_{\alpha }(y)]' = (y - \beta ) \sum _{n\ge 0} P_n(\beta ) y^n \end{aligned}$$

so that

$$\begin{aligned} f_{\alpha }(y)&= \sum _{n \ge 1}\frac{y^{n}}{n+1} [P_{n-1}(\beta ) - \beta P_n(\beta )] - \beta \\ {}&= \sum _{n \ge 1}\frac{y^{n}}{2n+1} \left[ P_{n-1}(\beta ) - P_{n+1}(\beta )\right] - \beta \end{aligned}$$

where the last equality follows from the recurrence relation:

$$\begin{aligned} (2n+1)xP_n(x) = (n+1)P_{n+1}(x) + nP_{n-1}(x). \end{aligned}$$

Extracting the Taylor coefficients of $f_{\alpha }$ and recalling the definition

$$\begin{aligned} R(y) =\sum _{n \ge 0}\kappa _{n+1}(P)y^n \end{aligned}$$

we get (29).

Note that since Legendre polynomials are orthogonal with respect to the uniform distribution in $[-1,1]$, they are parity preserving. In particular,

$$\begin{aligned} P_{2n+1}(0) = 0, \quad P_{2n}(0) = (-1)^n\frac{(1/2)_n}{n!}, \end{aligned}$$

so that one recovers (28) after some computations.