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In this paper, it is stated that the conditional min-entropy $H(A|B)_{\rho_{AB}}$ of $A$ conditioned on $B$ for any $\textbf{pure}$ quantum system $\rho_{AB}=|\psi_{AB} \rangle \langle \psi_{AB} |$ is $$ H_{\textrm{min}}(A|B)_{|\psi_{AB} \rangle \langle \psi_{AB} |} = - \textrm{log}_{2} \Big(\textrm{tr}\sqrt{\rho_{A}} \Big)^{2}\,, $$ where $\rho_{A}:= \textrm{tr}_{A}\rho_{AB}$ and $H_{\textrm{min}}(A|B)_{\rho_{AB}}$ is defined by $$ H_{\textrm{min}}(A|B)_{\rho_{AB}} := - \inf_{\sigma_B}\, {D}_{\infty}(\rho_{A B} \| \mathbb{I}_A \otimes \sigma_B)\,, $$ where $$D_{\infty}(\tau\|\tau') := \inf\{\lambda \in \mathbb{R}: \, \tau \leq 2^\lambda \tau' \} \ . $$ The result for pure states is stated without any proof or reference, so I assume it comes straight from the definition, but I haven't been able to derive it. I found on Wikipedia that $H_{\textrm{min}}(A|B)_{\rho_{AB}}$ can be defined via an SDP as $$ H_{\textrm{min}}(A|B)_{\rho_{AB}} = - \textrm{log}_{2}\, \min_{\sigma_{B}}\, \textrm{tr}\sigma_{B}\,\,\,\,\,\,\,\, \textrm{subject to}\,\,\,\,\,\,\,\, \mathbb{I}_A \otimes \sigma_B - \rho_{AB} \geq 0\,, $$ so the optimal $\sigma_{B}$ must satisfy $\textrm{tr}\,\sigma_{B}= \big(\textrm{tr}\sqrt{\rho_A} \big)^{2}$, but I don't see how to prove this. Any help appreciated.

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2 Answers 2

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I'll use an equivalent definition of the min-entropy $$ \begin{aligned} H_{\min}(A|B) = - \log_2 \min& \quad \lambda \\ \mathrm{s.t.}& \quad \rho_{AB} \leq \lambda I_A \otimes \sigma_B \\ & \quad \mathrm{tr}[\sigma_B] = 1 \\ & \quad \sigma_B \geq 0 \end{aligned} $$ The interesting condition is $\rho_{AB} \leq \lambda I_A \otimes \sigma_B$. Let's assume $\sigma_B >0$ is positive definite, I'll leave it as an exercise to show that this also works when things are only positive semidefinite. Then this condition is equivalent to $\sigma_B^{-1/2} \rho_{AB} \sigma_{B}^{-1/2} \leq \lambda I_{AB}$ which is in turn equivalent to $\|\sigma_B^{-1/2} \rho_{AB} \sigma_{B}^{-1/2}\|_{\infty} \leq \lambda$. Thus the above program could be equivalently written as $$ \begin{aligned} H_{\min}(A|B) = - \log_2 \min& \quad \|\sigma_{B}^{-1/2} \rho_{AB} \sigma_{B}^{-1/2}\|_{\infty} \\ \mathrm{s.t.}& \quad \mathrm{tr}[\sigma_B] = 1 \\ & \quad \sigma_B \geq 0 \end{aligned} $$

Now from the question we know that $\rho_{AB}$ is pure and hence $$ \|\sigma_{B}^{-1/2} \rho_{AB} \sigma_{B}^{-1/2}\|_{\infty} = \|\sigma_{B}^{-1/2} |\psi\rangle\langle \psi|_{AB} \sigma_{B}^{-1/2}\|_{\infty} = \langle \psi|\sigma_B^{-1}|\psi\rangle = \mathrm{tr}[\rho_B\sigma_B^{-1}] $$ Taking $\sigma_B = \frac{\sqrt{\rho_B}}{\mathrm{tr}[\sqrt{\rho_B}]}$ we see this feasible point of the SDP gives the objective value you expected, namely $\mathrm{tr}[\sqrt{\rho_B}]^2$.

To show this is optimal we take a look at the dual problem. We know by strong duality that the they will be equal (see the paper) and so if we can find a feasible point of the dual giving the same objective value then we are done. Note the dual problem can be written as $$ \begin{aligned} H_{\min}(A|B) = - \log_2 \max & \quad \mathrm{tr} [\rho_{AB} E_{AB}] \\ & \quad \mathrm{tr}_A [E_{AB}] = I_B \\ & \quad E_{AB} \geq 0 \end{aligned} $$ But now consider the feasible point $E_{AB} =\rho_{B}^{-1/2}\rho_{AB} \rho_{B}^{-1/2} $. Using the assumption in the question that $\rho_{AB} $ is pure we see that $$ \mathrm{tr} [\rho_{AB} E_{AB}] = \langle \psi|\rho_B^{-1/2}|\psi\rangle \langle \psi|\rho_B^{-1/2}|\psi\rangle = \mathrm{tr} [\rho_B^{1/2}]^2 $$ Which is exactly what we wanted.

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Reading again the paper you linked, I think the way the authors were thinking about the result was of showing this via the relations between conditional min- and max-entropies, see discussion at the end of section A, page 2 in the arXiv version. This is clearly a much more elegant and general way to show the result, and completely different from the other approach, so I'm posting it as a different answer.

  1. Given any bipartite $\rho\equiv\rho_{AB}$, consider a purification $\rho_{ABC}$ on some auxiliary purification space $C$, and define the conditional max-entropy as $$H_{\rm max}(A|B)_\rho \equiv - H_{\rm min}(A|C)_\rho.$$

  2. Observe that for product states, $\rho = \rho_A\otimes\rho_B$, we have $$H_{\rm min}(A|B)_\rho = -\log\min\{\operatorname{tr}(Y): \,\, Y\ge0,\,\rho_A\otimes\rho_B\le I\otimes Y\} \\= -\log\min\{\alpha: \,\,\alpha\ge0,\,\, \rho_A\le \alpha\} = -\log\|\rho_A\|_\infty \equiv H_{\rm min}(A)_{\rho_A},$$ where in the last 2 steps we observed that the $Y$ that saturates the inequality will have the form $Y=\alpha\rho_B$ for some $\alpha=\operatorname{tr}(Y)$.

    To compute $H_{\rm max}(A|B)_\rho$, we can use Lemmas 5,6, and Theorem 3 of the paper you linked, where the authors prove that $$H_{\rm max}(A|B)_\rho = \log d_A + \log \max_{\sigma_B} F(\rho,\tau_A\otimes\sigma_B)^2,$$ where $\tau_A\equiv I_A/d_A$ is the maximally mixed state, and $F(A,B)\equiv\|\sqrt A\sqrt B\|_1$ is the fidelity. In the case of $\rho=\rho_A\otimes\rho_B$, the fidelity is clearly maximal when $\sigma_B=\rho_B$, and thus we get $$H_{\rm max}(A|B)_\rho = \log d_A + \log F(\rho_A,I_A/d_A)^2, \\ F(\rho_A,I_A/d_A) = \frac{1}{\sqrt{d_A}}\|\sqrt{\rho_A}\|_1 = \operatorname{tr}(\sqrt{\rho_A})/\sqrt{d_A}.$$ Putting these together you get $$H_{\rm max}(A|B)_\rho = H_{\rm max}(A)_{\rho_A} = \log (\operatorname{tr}(\sqrt{\rho_A}))^2 = 2\log \operatorname{tr}(\sqrt{\rho_A}).$$

  3. We conclude using the complementary relations that if $\rho=\rho_{AB}$ is pure, $$H_{\rm min}(A|B)_\rho = -H_{\rm max}(A|C)_\rho,$$ where $C$ is a trivial (one-dimensional) space, and therefore $$H_{\rm max}(A|C)_\rho = H_{\rm max}(A)_{\rho_A} = 2\log\operatorname{tr}(\sqrt{\rho_A}),$$ which is the result we wanted.

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