# Example Sheet 4

## May 14, 2018*

Example sheet 4
Example sheet 4 solutions
Example sheet 4 solutions (printing)

### Exercise 1.

The other Fuchs and van de Graaf inequality.

1. 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 $$\sqrt{\rho}\sqrt{\sigma} = |\sqrt{\rho}\sqrt{\sigma}|U$$.

2. 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 $$| m \rangle\langle m | \hat M = \lambda_m | m \rangle\langle m |$$.

3. 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).
4. Given two quantum states $$\rho$$ and $$\sigma$$ such that $$\rho$$ is invertible, use equation (3) to show that $D(\rho,\sigma) \geq 1- F(\rho,\sigma).$

1. 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 $$U^* U = I$$.
2. 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.
3. 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.
4. 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 proof1.
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:

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

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

1. 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.
2. 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}
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 concave1, 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.