# Question 6

### Question 6.

On the 2011 exam Q3.iii, do you act on each term in the MES product individually as separable states with the channel acting on the second qubit and the identity leaving the first, and then adding the 4 terms together? Or is there a better way without writing everything out?

This question is about the qubit depolarizing channel1, which can be written as $\Phi(\rho) = \frac{q}{2}I + (1-q)\rho$(1) where $$q = 4p/3$$. Note that the roles of $$p$$ and $$1-p$$ are interchanged in this question compared to on the revision sheet (so the relationship between $$q$$ and $$p$$ is correspondingly changed); it is the same as in the notes however. We are asked for the value of $$p$$ such that $(\operatorname{id}\otimes \Phi)(| \phi^+ \rangle\langle \phi^+ |) = \frac{I \otimes I}{4}$(2) where $$| \phi^+ \rangle = (| 00 \rangle + | 11 \rangle)/\sqrt{2}$$. It might be interesting to note that since $$| \phi^+ \rangle$$ is a maximally entangled state and $$\frac{I \otimes I}{4}$$ is separable, this implies that for the value of $$p$$ we are asked to determine, $$\Phi$$ is entanglement breaking.

This question is a bit tricky. By (1), we can guess $$q=1$$ would do the job, since then $$\Phi(\rho) = \frac{I}{2}$$ for any input state $$\rho$$. This seems like a good candidate, since if $$\Phi$$ is mapping any state to the completely mixed state then inuitively it is breaking any entanglement that the input state shares (correlations must be destroyed since the output of the map doesn’t depend on the input, so it can’t be preserving correlations; entanglement is just a particular type of correlation). Moreover, getting $$I/2$$ is already half of what we want! The value $$q=1$$ corresponds to $$p=3/4$$. So how do we check that this really works?

I think that just expanding out the definitions like you said is the way to go: $(\operatorname{id}\otimes \Phi)(| \phi^+ \rangle\langle \phi^+ |) = \frac{1}{2}\sum_{ij=0}^1 | i \rangle\langle j |\otimes \Phi(| i \rangle\langle j |).$ However, there’s complication, and also a simplification. First, the complication: we can’t use the form (1) for computing two of the values, $$\Phi(| 0 \rangle\langle 1 |)$$ and $$\Phi(| 1 \rangle\langle 0 |)$$. Why? $$| 0 \rangle\langle 1 |$$ and $$| 1 \rangle\langle 0 |$$ are not states (they’re traceless, for one thing), and (1) actually only holds when $$\rho$$ is a density matrix (and to prove that formula, you need to use that $$\rho$$ is a density matrix). We can see in fact that (1) can’t hold for any arbitrary matrix $$\rho$$, since $$\Phi$$ must be trace-preserving when acting on any matrix, but if (1) were true for arbitrary $$\rho$$ (not just density matrices), then $$\operatorname{tr}(\Phi(\rho)) = q + (1-q)\operatorname{tr}(\rho)$$ which isn’t always the same as $$\operatorname{tr}(\rho)$$! (For example, $$\rho$$ could be traceless, like with the case $$| 0 \rangle\langle 1 |$$). So the complication is to determine $$\Phi(| 0 \rangle\langle 1 |)$$ and $$\Phi(| 1 \rangle\langle 0 |)$$, we have to go back to the original definition of $$\Phi$$ in terms of the Pauli matrices: $\Phi(\rho) = (1-p) \rho + \frac{p}{3}\sum_{i=1}^3 \sigma_k \rho \sigma_k.$(3) We can see (3) is actually written as a Kraus representation in terms of four Kraus operators, $$\{\sqrt{1-p}I, \sqrt{\frac{p}{3}}\sigma_1, \sqrt{\frac{p}{3}}\sigma_2, \sqrt{\frac{p}{3}}\sigma_3\}$$, which implies $$\Phi$$ is CP. We can check that these Kraus operators satisfy $$\sum_i A_i^\dagger A_i=I$$, implying it is trace-preserving as well. So (3) really gives a CPTP map which can act on arbitrary matrices (not just density matrices).

Ok, so what’s the simplification? $$\Phi(| 0 \rangle\langle 1 |)^\dagger = \Phi( (| 0 \rangle\langle 1 | )^\dagger) = \Phi(| 1 \rangle\langle 0 |)$$, since $\Phi(X)^\dagger = \Phi(X^\dagger)$ for any positive map $$\Phi$$2. So we only need to compute one of $$\Phi(| 0 \rangle\langle 1 |)$$ or $$\Phi(| 1 \rangle\langle 0 |)$$. By writing out the action of the Pauli matrices on $$| 0 \rangle\langle 1 |$$ and adding everything up, we find $\Phi(| 0 \rangle\langle 1 |) = (1- \frac{4p}{3})| 0 \rangle\langle 1 |$ so for $$p= \frac{3}{4}$$ we have $$\Phi(| 0 \rangle\langle 1 |)=0$$, implying $$\Phi(| 1 \rangle\langle 0 |)=0$$ as well. For $$| 0 \rangle\langle 0 |$$ and $$| 1 \rangle\langle 1 |$$, since these are density matrices (positive semi-definite and trace-1), we can use the form (1) and find $$\Phi(| 0 \rangle\langle 0 |) = \Phi(| 1 \rangle\langle 1 |) = I/2$$. So indeed, for $$p=3/4$$, (2) holds.

1. Note the depolarizing channel was discussed in Exercise 4 of example sheet 4 and Exercise 2 of the revision sheet.

2. I’m not sure if this was covered in class. Don’t feel bad if you didn’t know this fact! I’m short on time here so I won’t include a proof here until someone emails me though.