# Exercise 8

## from Example Sheet 4

### Exercise 8.

Quantum Fano inequality. Prove the quantum Fano inequality: $S(\rho,\Lambda)\le h(F_e(\rho,\Lambda))+(1-F_e(\rho,\Lambda))\,\log(d^2-1)$ where

• $$\Lambda$$ denotes a quantum operation with Kraus representation $\Lambda(\rho)=\sum_{i=0}^{d^2} V_i \rho V_i^\dagger,$ with $$\rho$$ being the state of a quantum system $$Q$$ with Hilbert space of dimension $$d$$.
• $$h(\cdotp)$$ denotes the binary Shannon entropy, i.e., for any $$0<p<1$$, $h(p) = - p \log p - (1-p) \log (1-p).$
• $$S(\rho, \Lambda)=-\operatorname{tr}(W \log W)$$, with $$W$$ being a matrix with elements $$W_{ij}=\operatorname{tr}(V_i\rho V_j)$$. The quantity $$S(\rho, \Lambda)$$ is called entropy exchange.
• $F_e(\rho, \Lambda):=\langle \psi_{RQ}^\rho|({\rm{id}}\otimes \Lambda)\psi^\rho_{RQ}| \psi_{RQ}^\rho\rangle,$ is the entanglement fidelity. Here $$|\psi_{RQ}^\rho\rangle$$ is a purification of the state $$\rho$$, with $$R$$ denoting the reference system used for the purification.

What is the implication of the quantum Fano inequality as regards entanglement?

First, we have the useful result that the entropy is increasing under unital maps. That is, if $$\Lambda$$ is a unital CPTP map, then $$S(\Lambda(\rho)) \geq S(\rho)$$ for any state $$\rho$$. We can see this simply by using the data-processing inequality: $S(\rho) = - D(\rho \| I ) \leq - D( \Lambda(\rho) \| \Lambda(I)) = - D(\Lambda(\rho)\| I) = S(\Lambda(\rho)).$ Now, from the previous exercise, we know that $$S(\rho,\Lambda) = S(\rho'_{RQ})$$ for $$\rho'_{RQ} = (\operatorname{id}_R\otimes \Lambda)(\psi^\rho_{RQ})$$. Let us consider an orthonormal basis $$\{| f_i \rangle\}_{i=1}^{d^2}$$ for the joint space $$\mathcal{H}_R\otimes \mathcal{H}_Q$$ such that $$| f_1 \rangle = | \psi_{RQ}^\rho \rangle$$. That is, we take the first vector to be $$| \psi_{RQ}^\rho \rangle$$ and then use Grahm-Schmidt to complete to an orthonormal basis. Then we define the unital CPTP map $\Gamma( \omega_{RQ}) = \sum_i \langle f_i | \omega_{RQ} | f_i \rangle\, | f_i \rangle\langle f_i |.$ This is sometimes called a pinching map for the basis $$\{| f_i \rangle\}$$. We can check it is CPTP as it has Kraus operators $$\{| f_i \rangle\langle f_i |\}_{i=1}^{d^2}$$, and unital by substituting $$\omega_{RQ} = I_{RQ}$$. By our “useful result”, we have $S(\rho,\Lambda) = S(\rho'_{RQ}) \leq S( \Gamma(\rho'_{RQ})) = H( \{\langle f_i | \rho'_{RQ} | f_i \rangle \}_{i=1}^{d^2})$ which is the Shannon entropy of the probability distribution $$p := \{\langle f_i | \rho'_{RQ} | f_i \rangle \}_{i=1}^{d^2}$$ (as these are the eigenvalues of $$\Gamma(\rho'_{RQ})$$ which can see by the definition of $$\Gamma$$). Moreover, we know the first entry is $$\langle f_i | \rho'_{RQ} | f_i \rangle = F_e(\rho,\Lambda)$$ by construction. Thus, if we define $$\varepsilon:= 1 - F_e(\rho,\Lambda)$$, the distribution $$p$$ has probability $$1-\varepsilon$$ for the first entry. Therefore, by classical Fano’s inequality1, $H(p) \leq h(\varepsilon) + \varepsilon\log (d^2-1 ) = h(1-\varepsilon) + \varepsilon\log (d^2-1)$ where to obtain the equality we use that the binary entropy is symmetric (in the sense that $$h(x) = h(1-x)$$ for $$x\in [0,1]$$). Substituting $$\varepsilon$$ yields thus result.

By the previous exercise, we can see $$S(\rho,\Lambda)$$ as the entropy of $$\operatorname{tr}_E [ \rho'_{RQE} ]$$ (and also of $$\operatorname{tr}_{RQ}[\rho'_{RQE}]$$). Since $$\rho'_{RQE}$$ is a pure state, this quantity is the entropy of entanglement of the state $$\rho'_{RQE}$$ across the partition $$RQ | E$$. This is an entanglement monotone (i.e. non-increasing under local operations and classical communication, so-called LOCC operations), and you can check that it is zero for product states and non-zero for entangled states. Since we can see $$\rho'_{RQE}$$ as induced by the action of $$\Lambda$$ (in Stinespring form) on a product state $$\psi^\rho_{RQ}\otimes | 0_E \rangle\langle 0_E |$$, the quantity $$S(\rho,\Lambda)$$ is a measurement of the entanglement generated by $$\Lambda$$ by its action on this state. On the other hand, $$F_e(\rho,\Lambda)$$ is the fidelity between $$\psi^\rho_{RQ}$$ and $$\Lambda(\psi^\rho_{RQ})$$, which is close to $$1$$ when $$\psi^\rho_{RQ}$$ is close to $$\Lambda(\psi^\rho_{RQ})$$ (so $$\varepsilon$$ is close to zero in this case). So we can upper bound the entanglement generated by $$\Lambda$$ in terms of how much it changes the state in fidelity (and in particular, if $$\Lambda$$ does not change the state at all, then it can’t generate any entanglement).

1. Exercise 8 of Example sheet 1