# Calculating the product state capacity of the quantum depolarizing channel

This is homework, so just let me know if I'm on the right track or where I went wrong, please.

So, we are asked to compute the product state capacity $$C_1(T) = \max_{\{p_j,\,\vert \psi_j\rangle\}}\left[ S\left(\sum_j p_j T(\vert \psi_j\rangle\langle \psi_j\vert)\right) - \sum_j p_jS(T(\vert \psi_j\rangle\langle \psi_j\vert))\right]$$ of the depolarizing channel $T(\rho)\equiv(1-p)\rho + p\frac{1}{d}\textbf{I}$ in $d$ dimensions.

Here is how I proceded:

1. See that for each pure state $\vert \psi\rangle$, we could write $$T(\vert \psi\rangle\langle \psi\vert) \doteq \text{diag}\left(1-p + \frac{p}{d},\frac{p}{d},\ldots,\frac{p}{d}\right),$$

2. which would then lead to $$S(T(\vert \psi\rangle\langle \psi\vert))=-\left[\left(1-p+\frac{p}{d}\right)\log\left(1-p+\frac{p}{d}\right)+\frac{d-1}{d}p\log\left(\frac{p}{d}\right) \right]$$

3. Further, by the same token as in 1., we get $$\sum_x q_x T(\vert \psi\rangle\langle \psi\vert)\doteq (1-p)\,\text{diag}\left(\sum_x q_x \lvert\langle\psi_x\vert 1\rangle \rvert^2,\ldots,\sum_x q_x \lvert\langle\psi_x\vert d\rangle \rvert^2 \right)+p\frac{1}{d}\textbf{I}.$$

4. Thus, we need to maximize the expression \begin{align} &-\sum_{i=1}^d\left[(1-p)\sum_x q_x \lvert\langle\psi_x\vert i\rangle \rvert^2 + \frac{p}{d}\right]\cdot\log\left((1-p)\sum_x q_x \lvert\langle\psi_x\vert i\rangle \rvert^2 + \frac{p}{d}\right)\\ &+\left(1-p+\frac{p}{d}\right)\log\left(1-p+\frac{p}{d}\right) + \frac{d-1}{d}p\log\left(\frac{p}{d}\right) \end{align}

This looks kinda bogus and I don't even know what to do now.

• Step 4 looks a lot better if you use the binary entropy function $H_b(x) = x \lg x$. – Craig Gidney Jul 20 '16 at 7:23
• You are right, thanks! My solution now is $C_1(T) = H_b\left(\frac{d-1}{d}\cdot p\right)$, I'll update my post once I know if it's correct. – Jo Be Jul 20 '16 at 7:34
• The thing with the binary entropy is wrong, btw. I'll post an answer soon. – Jo Be Jul 20 '16 at 9:24

As far as I can tell, you didn't do anything wrong, but your straightforward attempt to optimisation is not very enlightning. Let's try to simplify this.

## Hint 1:

As often in Physics, symmetries simplify the problem.

It does not help you ? Then let us move to hint 2

## Hint 2:

At step 2., you can see that $$S(T (|ψ\rangle))$$ doesn't depend on $$|ψ\rangle$$. That allows you to simplify the quantity to optimise.

Still not enough ? Then move on to

## Hint 3:

You now have to find $$ρ=\sum_x p_x |ψ_x\rangle\langle ψ_x|$$ maximising the output entropy $$S (T (ρ))$$.

Ok, so. Nothing wrong with what I did, and it thus follows (with the binary entropy $\mathcal{h}_2$)

\begin{alignat}{2} C_1(T)&=\max_{\{q_x,\,\vert \psi_x\rangle\}} \left\{ S\left(\sum_xq_x T\left(\vert \psi_x\rangle\langle \psi_x\vert\right)\right) +\left(1-\frac{d-1}{d}\cdot p\right)\log\left(1-\frac{d-1}{d}\cdot p\right)\right. \\[1em] &\qquad\qquad\qquad \left. + \frac{d-1}{d}\cdot p\log\frac{p}{d}\right\}\\[2em] &=\left(1-\frac{d-1}{d}\cdot p\right)\log\left(1-\frac{d-1}{d}\cdot p\right) + \frac{d-1}{d}\cdot p\log\frac{p}{d}\\[1em] &\quad+\max_{\{q_x,\,\vert \psi_x\rangle\}} \left\{S\left(\sum_xq_x T\left(\vert \psi_x\rangle\langle \psi_x\vert\right)\right)\right\} \end{alignat} And since $0\leq S \leq \log d$, it is easily seen that the maximum of the last term is obtained if the state $$\sum_x q_x T(\vert \psi_x\rangle\langle \psi_x\vert)$$ is the maximally entangled state $\frac{1}{d}\textbf{I}$, which can be achieved by pure states $\vert \psi_x\rangle$.

All in all, \begin{align}C_1(T)&=\log d + \left(1-\frac{d-1}{d}\cdot p\right)\log\left(1-\frac{d-1}{d}\cdot p\right) + \frac{d-1}{d}\cdot p\log\frac{p}{d}\\[1em] &= \log d - h_2\left(\frac{d-1}{d}\cdot p\right) + \frac{d-1}{d}\cdot p\left(\log\frac{p}{d} -\log\left(\frac{d-1}{d}\cdot p\right)\right)\\[1em] &= \log d - h_2\left(\frac{d-1}{d}\cdot p\right)-\frac{d-1}{d}\cdot p \log (d-1) \end{align}

And finally, a quick plot: