I need to prove the fact that a quantum channel (a superoperator) cannot increase the Holevo information of an ensemble $\epsilon = \{\rho_x, p_x\}$. Mathematically expressed I need to prove

$$\begin{align} \chi(\$(\epsilon)) \leq \chi(\epsilon) \end{align} \tag{1}\label{1}$$

where $\$$ represents a quantum channel (a positive trace preserving map) which works on the ensemble as $\$ \epsilon = \{ \$\rho_x, p_x \}$. This needs to be done with the property that a superoperator $\$$ cannot increase relative entropy (some may remember my previous question about this):

$$\begin{align} S(\$\rho || \$\sigma) \leq S(\rho || \sigma) \end{align}\tag{2}\label{2}$$

with relative entropy defined as $$S(\rho || \sigma) = \mathrm{tr}(\rho \log \rho - \rho \log \sigma).$$

The Holevo information is defined as (if someone would not know)

$$\begin{align} \chi(\epsilon) = S(\sum_x p_x \rho_x ) - \sum_x p_x S(\rho_x) \end{align}.$$

Does anyone know how we get from equation $\eqref{2}$ to equation $\eqref{1}$? Maybe what density operators to fill in $\eqref{2}$? Or how do we start such a proof?


1 Answer 1


Suppose that $\mathsf{X}$ is a register that can store each possible choice for $x$, as a classical state, while $\mathsf{Y}$ is a register that can store each possible state $\rho_x$. It is then natural to associate the classical-quantum state $$ \rho = \sum_x p_x |x\rangle \langle x| \otimes \rho_x $$ with the ensemble $\{(p_x,\rho_x)\}$.

Now try taking the following two steps:

  1. Verify that the Holevo information of the ensemble $\{(p_x,\rho_x)\}$ is just another name for the mutual information $\text{I}(\mathsf{X}:\mathsf{Y})_\rho$ between $\mathsf{X}$ and $\mathsf{Y}$.

  2. Verify, for an arbitrary state $\rho$ of the pair $(\mathsf{X},\mathsf{Y})$, that the mutual information $\text{I}(\mathsf{X}:\mathsf{Y})_\rho$ can be expressed as $$ \mathrm{I}(\mathsf{X}:\mathsf{Y})_\rho = \mathrm{S}\bigl(\rho\,\big\|\, \rho^{\mathsf{X}}\otimes \rho^{\mathsf{Y}}\bigr). $$

Once you have done this, it should become clear how to get (1) from (2), but do not assume that it is exactly the same channel appearing in the two inequalities.


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