Example Sheet 4

May 14, 2018*

*Last modified 25-Aug-18

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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, \[\begin{aligned} \frac{1}{2} I(A:B) - \frac{1}{2} I(A:E) &= \frac{1}{2} [ S(A) + S(B) - S(AB) - S(A) - S(E) + S(AE)] \\ &= \frac{1}{2} [ S(B) - S(AB) - S(E) + S(AE)] \\ &= \frac{1}{2} [ S(B) - S(AB) - S(AB) + S(B)] \\ &= S(B) - S(AB) = - S(A|B) = I_c^{A>B}. \end{aligned}\]

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 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.

  1. Exercise 9 of Example Sheet 3