# Is the quantum relative entropy a Bregman divergence?

This is related to a broader question about quantum information geometry - this question is more specific.

A fundamental concept in Amari's treatment of information geometry is that of a Bregman divergence. Given a convex function $\psi(\mathbf{x})$, a Bregman divergence is defined as $$D_\psi(\mathbf{x}\|\mathbf{x}_0) = \psi(\mathbf{x}) - \psi(\mathbf{x}_0) - \nabla \psi(\mathbf{x}_0)(\mathbf{x}-\mathbf{x}_0).$$ Bregman divergences are non-negative and are zero when $\mathbf{x}=\mathbf{x}_0$, and they behave in some ways like the square of a distance between $\mathbf{x}_0$ and $\mathbf{x}$. (Though taking the square root does not form a metric.)

An example is the Kullback-Leibler divergence between two probability distributions, $D_{KL}(P\|P^0)=\sum_{i=1}^N p_i\log\frac{p_i}{p^0_i}$. In the case of finite support can be seen (for example) by parameterising each distribution by its first $N-1$ probabilities (so $p_N=1-\sum_{i=1}^{N-1}p_i$) and then taking $\psi$ to be the negative entropy, $\psi=\sum_{i=1}^{N}p_i\log p_i$.

In quantum theory, the natural analog of the Kullback-Leibler divergence is the quantum relative entropy, $$D_Q(\rho\|\rho_0) = \mathrm{Tr}\,\rho(\log\rho - \log\rho_0),$$ where $\rho$ and $\rho_0$ are density matrices and $\log$ is the matrix logarithm.

My question is, in the case of a finite Hilbert space (or indeed in general), can the quantum relative entropy be expressed as a Bregman divergence? If so, what is the right way to parameterise $\rho$ so that this becomes clear?

• $\rho = \psi \circ \lambda$, where $\lambda$ is the eigenvalue map and your underlying Hilbert space is the real-symmetric matrices with the trace inner product. – max_zorn Jun 8 '18 at 3:56

Abstract: In this paper the Bregman operator divergence is introduced for density matrices by differentiation of the matrix-valued function $$x \mapsto x log x$$. This quantity is compared with the relative operator entropy of Fujii and Kamei. It turns out that the trace is the usual Umegaki's relative entropy which is the only intersection of the classes of quasi-entropies and Bregman divergences.
• Thank you, that does indeed look like an answer - it shows that the quantum relative entropy is a Bregman divergence with $\psi=\operatorname{Tr}\rho\log\rho$ (in my notation). I'll accept this after reading it more thoroughly. – Nathaniel Dec 28 '18 at 0:18