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- Example sheet 4
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### Exercise 1.

**The other Fuchs and van de Graaf inequality.**

Let \(\{E_m\}_{m \in M}\) be a POVM, where \(M\) is a finite set. Given two states \(\rho_A\) and \(\sigma_A\), use the Cauchy-Schwarz inequality to show that \[ F(\rho_A,\sigma_A)\leq \sum_{m\in M}\sqrt{\operatorname{tr}[E_m \rho] \operatorname{tr}[E_m \sigma]} = F( \tilde \rho_M, \tilde \sigma_M ) \](1) where \[ \tilde \rho_M = \sum_{m\in M} \operatorname{tr}[E_m \rho] | m \rangle\langle m |, \qquad \tilde \sigma_M = \sum_{m\in M} \operatorname{tr}[E_m \sigma]| m \rangle\langle m | \](2) are diagonal states encoding probabilities of the measurement outcomes \(m\), in a Hilbert space \(\mathcal{H}_M\) of dimension \(M\) with orthonormal basis \(\{| m \rangle\}_{m\in M}\).

*Hint: use the polar decomposition \(\sqrt{\rho}\sqrt{\sigma} = |\sqrt{\rho}\sqrt{\sigma}|U\).*Let \(\rho\) and \(\sigma\) be mixed states and assume \(\rho\) is invertible.

Show that if \(\hat M := \rho^{-1/2} | \rho^{1/2}\sigma^{1/2}| \rho^{-1/2}\) has spectral decomposition \(\hat M = \sum_m \lambda_m | m \rangle\langle m |\), then \[ | m \rangle\langle m |\sqrt{\rho} = \frac{1}{\lambda_m} | m \rangle\langle m | \sigma^{1/2} U^\dagger, \] where \(U\) is unitary and defined by the polar decomposition \(\sqrt{\rho}\sqrt{\sigma} = |\sqrt{\rho}\sqrt{\sigma}|U\).

*Hint: start by showing that \(| m \rangle\langle m | \hat M = \lambda_m | m \rangle\langle m |\).*Show that projective measurement \(\{E_m\}_{m\in M}\) defined by \(E_m = | m \rangle\langle m |\), achieves equality in equation (1).

Conclude that when \(\rho\) is invertible, \[ F(\rho,\sigma) = \min_{\{E_m\}_{m\in M}} F(\tilde \rho_M, \tilde \sigma_M) \](3) where the minimum is over POVMs \(\{E_m\}_{m\in M}\) and \(\tilde \rho_M\) and \(\tilde \sigma_M\) are defined in (2).Given two quantum states \(\rho\) and \(\sigma\) such that \(\rho\) is invertible, use equation (3) to show that \[ D(\rho,\sigma) \geq 1- F(\rho,\sigma). \]

- Since \(\sum_m E_m = I\), we have that \[\begin{aligned} F(\rho,\sigma) &= \operatorname{tr}[ | \sqrt\rho \sqrt\sigma|] = \operatorname{tr}[ \sqrt\rho \sqrt\sigma U^*] \\ &= \sum_m \operatorname{tr}[ \sqrt\rho E_m \sqrt\sigma U^*] = \sum_m \operatorname{tr}[ \sqrt\rho \sqrt E_m \sqrt E_m \sqrt\sigma U^*]\\ &= \sum_m \langle \sqrt E_m \sqrt \rho, \sqrt E_m \sqrt\sigma U^* \rangle_\text{HS}\\ &\leq \sum_m \sqrt{ \langle \sqrt E_m \sqrt \rho,\sqrt E_m \sqrt \rho \rangle_\text{HS}} \sqrt{\langle \sqrt E_m \sqrt\sigma U^* , \sqrt E_m \sqrt\sigma U^* \rangle_\text{HS}} \end{aligned}\] by the Cauchy-Schwarz inequality, where \(\langle A,B \rangle_\text{HS} = \operatorname{tr}[A^* B]\) is the Hilbert-Schmidt inner product. Thus, \[\begin{aligned} F(\rho,\sigma) &\leq \sum_m \sqrt{ \operatorname{tr}[\sqrt \rho E_m \sqrt \rho]} \sqrt{\operatorname{tr}[ U \sqrt\sigma E_m \sqrt\sigma U^* ]}\\ &= \sum_m \sqrt{ \operatorname{tr}[ E_m \rho]} \sqrt{\operatorname{tr}[ E_m \sigma ]} \end{aligned}\] by the cyclicity of the trace and that \(U^* U = I\).
- We have that \[\begin{aligned} | m \rangle\langle m | \hat M &= | m \rangle\langle m | \sum_{m'}\lambda_{m'} | m' \rangle\langle m' | \\ &= \sum_{m'} \lambda_{m'} \delta_{m,m'} | m \rangle\langle m' | \\ &= \lambda_m | m \rangle\langle m |. \end{aligned}\](4) Subsituting the definition of \(\hat M\), \[\begin{aligned} | m \rangle\langle m | \hat M &= | m \rangle\langle m | \rho^{-1/2} | \rho^{1/2}\sigma^{1/2}| \rho^{-1/2} \\ &=| m \rangle\langle m | \rho^{-1/2} \sqrt{\rho}\sqrt{\sigma} U^\dagger \rho^{-1/2} \\ &=| m \rangle\langle m | \sqrt{\sigma} U^\dagger \rho^{-1/2}. \end{aligned}\] Thus, by (4), \(| m \rangle\langle m | \sqrt{\sigma} U^\dagger \rho^{-1/2} = \lambda_m | m \rangle\langle m |\). Thus, \[ | m \rangle\langle m | \sqrt{\rho} = \lambda_m^{-1}| m \rangle\langle m | \sqrt{\sigma} U^\dagger \] as desired.
- We have equality is Cauchy-Schwarz if and only if the two vectors are linearly dependent. Since that is the only inequality we used, we have that \(F(\rho,\sigma) = F(\tilde \rho, \tilde \sigma)\) if and only if there exists constants \(\alpha_m \in \mathbb{C}\) such that \[ \sqrt E_m \sqrt \rho = \alpha_m \sqrt E_m \sqrt\sigma U^* \] for each \(m\). Choosing \(E_m = | m \rangle\langle m |\), this condition becomes \[ | m \rangle\langle m | \sqrt \rho = \alpha_m | m \rangle\langle m | \sqrt\sigma U^*. \] By choosing \(\alpha_m = \lambda_m^{-1}\) we see that indeed, \(F(\rho,\sigma) = F(\tilde \rho, \tilde \sigma)\). Since for any measurement \(\{E_m\}\) we have \(F(\rho,\sigma) \leq F(\tilde \rho, \tilde \sigma)\), and that equality is achieved for a specific measurement, (3) follows.
- Let \(\{E_m\}\) be a measurement which achieves \(F(\rho,\sigma) = F(\tilde \rho, \tilde \sigma)\). Moreover, \(D(\tilde \rho ,\tilde \sigma) \leq D(\rho,\sigma)\) by the data-processing inequality, since the ‘’measure-and-prepare’’ map \(\Lambda: \rho \mapsto \sum_m \operatorname{tr}[E_m \rho] | m \rangle\langle m |\) is CPTP. Thus, it remains to prove that \[
1- F(\tilde \rho, \tilde \sigma) \leq D(\tilde \rho ,\tilde \sigma).
\](5) These states are both diagonal in the basis \(\{| m \rangle\}\) and thus we’ve reduced to the classical case. We define \(p_m = \operatorname{tr}[\rho E_m]\) and \(q_m = \operatorname{tr}[\sigma E_m]\). Then \(F(\tilde \rho ,\tilde \sigma) = \sum_m \sqrt{p_m q_m}\), and \(\{p_m\}_m\) and \(\{q_m\}_m\) are each probability distributions. Then \[\begin{aligned}
\sum_m (\sqrt{p_m} -\sqrt{q_m})^2 &= \sum_m p_m + \sum_m q_m - 2\sum_m \sqrt{p_m q_m}\\
&= 2(1 - F(\tilde \rho, \tilde \sigma)).
\end{aligned}\] On the other hand, we can use the inequality \(|\sqrt{p_m} - \sqrt{q_m}| \leq |\sqrt{p_m} + \sqrt q_m|\) to see \[\begin{aligned}
\sum_m (\sqrt{p_m} -\sqrt{q_m})^2 &=\sum_m |\sqrt{p_m} -\sqrt{q_m}|^2 \leq \sum_m |\sqrt{p_m} -\sqrt{q_m}||\sqrt{p_m} + \sqrt q_m|\\
&= \sum_m |p_m - q_m| = 2 D(\tilde \rho, \tilde \sigma)
\end{aligned}\] which completes the proof
^{1}.

The trick of considering \(\sum_m (\sqrt{p_m} -\sqrt{q_m})^2\) and the subsequent bounds are from Nielsen and Chuang, p. 415.↩

### Exercise 2.

Let \(|\psi\rangle_{ABE}\) be a pure state of a tripartite system \(ABE\). define the *coherent information* from \(A\) to \(B\) of \(\psi\) to be \[
I_c^{A>B}(\psi) = - S(A|B)_\psi.
\] Here \(S(A|B)_\psi\) denotes the conditional entropy of the subsystem \(A\) with respect to subsystem \(B\), given that the composite system \(ABE\) is in the pure state \(|\psi\rangle_{ABE}\). Henceforth we shall omit \(\psi\).

Prove the following identities:

\(\frac{1}{2} I(A:B) + \frac{1}{2} I(A:E)= S(A)\)

\(\frac{1}{2} I(A:B) - \frac{1}{2} I(A:E)= I_c^{A>B}\)

- By definition of the quantum mutual information, \[\begin{aligned} \frac{1}{2} I(A:B) + \frac{1}{2} I(A:E) &= \frac{1}{2} [ S(A) + S(B) - S(AB) + S(A) + S(E) - S(AE)] \\ &= S(A) + \frac{1}{2} [ S(B) - S(AB) + S(E) - S(AE)]. \end{aligned}\] Since \(| \psi \rangle_{ABE}\) is pure, \(S(B) = S(AE)\) and \(S(E) = S(AB)\) (since for any bipartition of the systems ABE, the entropies of both reduced density matrices are equal). Thus, the term in brackets vanishes and we obtain \(S(A)\) as desired.
- Likewise, \[\begin{aligned} \frac{1}{2} I(A:B) - \frac{1}{2} I(A:E) &= \frac{1}{2} [ S(A) + S(B) - S(AB) - S(A) - S(E) + S(AE)] \\ &= \frac{1}{2} [ S(B) - S(AB) - S(E) + S(AE)] \\ &= \frac{1}{2} [ S(B) - S(AB) - S(AB) + S(B)] \\ &= S(B) - S(AB) = - S(A|B) = I_c^{A>B}. \end{aligned}\]

### Exercise 3.

A bipartite quantum state \(\rho_{AB}\) is said to be *separable* if it can be written as a convex combination of product states, i.e., if there exists an ensemble \(\{p_i, \sigma_A^{(i)} \otimes \tau_B^{(i)}\}\), with \(\sigma_A^{(i)} \in \mathcal{B}(\mathcal{H}_A)\) and \(\tau_B^{(i)}\in \mathcal{B}(\mathcal{H}_B)\), such that \[
\rho_{AB} = \sum_i p_i \sigma_A^{(i)} \otimes \tau_B^{(i)}.
\] This allows us to extend the definition of entanglement to mixed states: *a mixed state is entangled if it is not separable.*

Show that if \(\rho_{AB}\) is separable then \(I_c^{A>B} \le 0\).

What implication does this have on the conditional entropy \(S(A|B)\)?

Since any separable state can be written as a convex combination of product pure states, any separable state \(\rho_{AB}\) may be written as \(\rho_{AB} = \sum_i p_i \psi_A^{(i)} \otimes \phi_B^{(i)}\) for some pure states \(\psi_A^{(i)}\) and \(\phi_B^{(i)}\). Next, because the quantum conditional entropy is concave^{1}, the coherent information is convex. Thus \[
I_c^{A>B}(\rho) \leq \sum_i p_i I_c^{A>B} ( \psi_A^{(i)} \otimes \phi_B^{(i)}) = \sum_i p_i [ S(\phi_B^{(i)}) - S(\psi_A^{(i)} \otimes \phi_B^{(i)})].
\] But each entropy is the entropy of a pure state and thus vanishes. Thus, \(I_c^{A>B}(\rho) \leq 0\). Thus conditional entropy is therefore non-negative on separable states.

Exercise 9 of Example Sheet 3↩