1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Step 4 looks a lot better if you use the binary entropy function $H_b(x) = x \lg x$. $\endgroup$ – Craig Gidney Jul 20 '16 at 7:23
  • $\begingroup$ 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. $\endgroup$ – Jo Be Jul 20 '16 at 7:34
  • $\begingroup$ The thing with the binary entropy is wrong, btw. I'll post an answer soon. $\endgroup$ – Jo Be Jul 20 '16 at 9:24
0
$\begingroup$

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 $S (T (ρ))$.

$\endgroup$
0
$\begingroup$

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: Capacity of quantum depolarizing channel in d dimensions in dependence of the noise parameter p, plotted for different values of d

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.