I have two questions regarding the exercise 9.2.8 of Quantum information by Wilde, which is as follows:

Let $\rho,\sigma \in \mathcal{D}(\mathcal{H}_A)$ and let $\mathcal{N: L(H}_A)\rightarrow \mathcal{L(H}_B)$ be a quantum channel. Show that the fidelity is monotone w.r.t. the channel $\mathcal{N}$: $$F(\rho,\sigma)\leq F(\mathcal{N}(\rho),\mathcal{N}(\sigma))$$

  1. The first question is how to prove this inequality. So my approach was to use the Uhlmann theorem by defining a purification of the states and using $\max_{U} <\phi^\rho|_{RA}(U_R\otimes I_A)|\phi^\sigma> $ and compare it with the same for the output of channel. But the problem is I don't know how to decompose the purification of the channel $\mathcal{N}(\rho)$. Even if using the canonical purification, it needs the square root of $\mathcal{N}(\rho)$ which I don't know how to obtain.

  2. The second question is about the meaning of this theorem. It actually seems counter-intuitive to me. Because, as I expect it, the channel is a form of noise-increasing operation on the states. So if two density operators go through a channel, their fidelity must have reduced due to the random noise that has been added to them. Not increased! Can you explain it for me?

Thanks a lot. I appreciate any help or comment.


1 Answer 1


Part 1

Monotonicity under channels is sometimes also referred to as satisfying a data-processing inequality. One way to prove this is to use a variational formula for the fidelity function, see Theorem 3.17 and the subsequent discussions in TQI - Watrous. This is slightly cheating as you first need to prove the variational formula is correct but in my experience data-processing follows quite quickly once you have a variational formula. So the fidelity can be rewritten as the semidefinite programming problem

\begin{equation} \begin{aligned} F(\rho, \sigma) = \,&\max_{X} \quad\mathrm{Tr}[X + X^*]/2 \\ &\,\,\mathrm{s.t.} \quad \begin{pmatrix} \rho & X \\ X^* & \sigma \end{pmatrix}\geq 0 \end{aligned} \end{equation} where the maximization is over all linear operators $X$ on the Hibert space which $\rho$ and $\sigma$ act on.

Now take any quantum channel $\mathcal{N}$ and take any feasible point $X$ for the SDP characterization of $F(\rho, \sigma)$. As $\mathcal{N}$ is a completely positive map we have $$ \begin{pmatrix} \rho & X \\ X^* & \sigma \end{pmatrix}\geq 0 \implies \begin{pmatrix} \mathcal{N}(\rho) & \mathcal{N}(X) \\ \mathcal{N}(X)^* & \mathcal{N}(\sigma) \end{pmatrix}\geq 0. $$ Furthermore we have $\mathrm{Tr}[X + X^*]/2 = \mathrm{Tr}[\mathcal{N}(X) + \mathcal{N}(X)^*]/2$ as $\mathcal{N}$ is trace-preserving. Thus we have shown that for each feasible point $X$ of the SDP for $F(\rho, \sigma)$ we can define a feasible point $\mathcal{N}(X)$ of the SDP for $F(\mathcal{N}(\rho), \mathcal{N}(\sigma))$ which has the same objective value. As we are taking a maximization over all feasible points it follows that we must have $F(\rho, \sigma) \leq F(\mathcal{N}(\rho), \mathcal{N}(\sigma))$.

Part 2

First note that it is the same channel that is being applied to the two states $\rho$ and $\sigma$. So if for example $\rho = \sigma$ and they have perfect fidelity then $\mathcal{N}(\rho) = \mathcal{N}(\sigma)$ and the `noisy' outputs also have perfect fidelity. On the opposite end of the spectrum if we take a channel which produces white noise i.e., $\mathcal{N}(\rho) = \mathrm{Tr}[\rho] I/d$ then $\mathcal{N}(\rho) = \mathcal{N}(\sigma)$ for any two states $\rho$ and $\sigma$. Thus even those that previously had fidelity $0$ will have, after sending them through this maximally noisy channel, perfect fidelity.

A better way to think of this result is to think of the fidelity as a measure of how well we can distinguish two quantum states (where values closer to $0$ are more distinguishable). This interpretation is justified by the Fuchs-van de Graaf inequalities that relate the fidelity and the trace distance and the trace distance's operational characterization as a distinguishability measure that comes from Holevo-Helstrom theorem. Taking a step back, if we were to have any hope that the fidelity is a good measure of distinguishability then it would have to be the case that the fidelity satisfies a data processing inequality. For if we are to think that $F(\rho, \sigma)$ really characterizes our ability to distinguish $\rho$ from $\sigma$ then it shouldn't be the case that we can send the unknown states through some quantum channel and then distinguish them better, i.e. $F(\rho, \sigma) \not\geq F(\mathcal{N}(\rho), \mathcal{N}(\sigma))$.

  • $\begingroup$ Thanks a lot, I didn't expect such a complete and precise answer. About part 2, it is interesting that by thinking of it as distinguishability, it makes sense. And the fact that it can turn both inputs into same white noise, is a good disproof... But according to what you said, I think in this case data processing inequality will be different from above formula. it might be something like $F(\rho, \sigma ) >= F(\rho, N(\sigma))$. Just applying the channel to one of the inputs and comparing its similarity to the other. $\endgroup$
    – hafezmg48
    Commented Oct 8, 2020 at 21:25
  • $\begingroup$ No problem. Applying the channel to only one of the states is not quite right -- you should be able to cook up examples where this doesn't obey the monotonicity bound. Thinking operationally: the distinguishing task is we are sent the state $\rho$ with some probability $q$ and otherwise we are sent $\sigma$. Overall from our perspective we receive the state $q\rho+(1-q)\sigma$. Now upon receiving the state we can do whatever we want (within the limitations of QM). For example we could apply the channel $M$ which gives us $M(q\rho+(1-q)\sigma) = qM(\rho) + (1-q)M(\sigma)$... $\endgroup$
    – Rammus
    Commented Oct 9, 2020 at 10:01
  • $\begingroup$ This means after applying the channel we have changed the problem of distinguishing $\rho$ and $\sigma$ to distinguishing $M(\rho)$ from $M(\sigma)$. When we talk about how well we can distinguish two states we can think of taking the set of all possible protocols to distinguish the states and then optimizing to find the one that gives us the best probability of success. As the set of protocols that include as a first step "apply channel $M$" are a subset of all these protocols it is clear that it cannot be easier to distinguish the latter from the former. Another example... $\endgroup$
    – Rammus
    Commented Oct 9, 2020 at 10:08
  • $\begingroup$ If we take the channel to be unitary then its action is reversible. You should also find that the fidelity is invariant under unitary conjugation of the inputs. $\endgroup$
    – Rammus
    Commented Oct 9, 2020 at 10:09

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