# What is the conditional min-entropy of a pure bipartite state?

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.

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.

Remember that the max relative entropy satisfies, in general, $$D_{\rm max}(\rho\|\sigma) = \log \inf\{\eta\ge0:\,\, \rho\le \eta \sigma\}.$$

In particular, $$D_{\rm max}(\rho\|I\otimes \sigma)=\log \min\{\eta\ge0: \,\, \rho\le \eta( I\otimes \sigma)\}, %= \log \min\{ \operatorname{tr}(Y): \,\, Y\ge0, \,\, \rho\le I\otimes Y \} \\[20px] H_{\rm min}(A|B)_\rho \equiv - \min_\sigma D_{\rm max}(\rho\|I\otimes \sigma) = -\log \min\{\operatorname{tr}(Y): \,\, Y\ge0,\, \rho\le I\otimes Y\}.$$ These identities are all discussed in details for example in https://cs.uwaterloo.ca/~watrous/QIT-notes/.

The above tells us that the task is essentially to find $$Y\ge0$$ such that $$\rho\le I\otimes Y$$ that minimises $$\operatorname{tr}(Y)$$, for some pure bipartite state $$|\Psi\rangle$$, which we can always write in its Schmidt decomposition as $$|\Psi\rangle=\sum_k \sqrt{s_k} |u_k,v_k\rangle$$. Let then $$\rho=|\Psi\rangle\!\langle\Psi|$$. I'm pretty sure there is a decent general proof of this, but it's eluding me right now. The following is a pretty ugly way to prove this that works for 2-qubit systems, assuming the solution for $$Y$$ is diagonal in the basis of the Schmidt decomposition of $$|\Psi\rangle$$.

We can assume without loss of generality that $$|\Psi\rangle=\sqrt{s_0}|00\rangle+\sqrt{s_1}|11\rangle$$, as this only amounts to a local change of basis. Let's also assume the solution corresponds to a diagonal $$Y$$, meaning $$Y=\sum_i Y_{ii} \mathbb{P}_i$$, where $$\mathbb{P}_i\equiv|i\rangle\!\langle i|$$. Clearly, we'd also need to show that this is the case, so this proof is incomplete in this regard.

Using this assumption, our task is to minimise $$\operatorname{tr}(Y)=Y_{00}+Y_{11}$$ under the constraint $$I\otimes Y-\rho\ge0$$. Explicitly, this amounts to the condition $$\begin{pmatrix} Y_{00} -s_0 & 0 & 0 & -\sqrt{s_0 s_1} \\ 0 & Y_{11} & 0 & 0 \\ 0 & 0 & Y_{00} & 0 \\ -\sqrt{s_0 s_1} & 0 & 0 & Y_{11} - s_1 \end{pmatrix} \ge 0.$$ Using Sylvester's criterion, this is equivalent to the set of conditions $$Y_{00} \ge s_0, \qquad Y_{11} \ge s_1, \qquad Y_{00} Y_{11} \ge s_0 Y_{11} + s_1 Y_{00} .$$ Given that we need to minimise $$Y_{00}+Y_{11}$$ under these constraints, we can focus on the last constraint and use Lagrange's multipliers, obtaining the condition $$Y_{11} -s_1 = Y_{00}-s_0.$$ Putting this condition back into the constraint, we get the following equation for $$Y_{00}$$: $$Y_{00}(Y_{00}+s_1-s_0) \ge s_0 (Y_{00}+s_1-s_0) + s_1 Y_{00} \\ \iff Y_{00}^2 - 2s_0 Y_{00} + s_0 (s_0-s_1) \ge 0 \\ \implies Y_{00} \ge s_0 + \sqrt{s_0 s_1} \,\,\vee\,\, Y_{00} \le s_0 - \sqrt{s_0 s_1}.$$ Remembering we must also have $$Y_{00}\ge s_0$$, we see that only the first condition is in our feasible set. Thus using $$Y_{00}=s_0+\sqrt{s_0 s_1}$$, the relation found above between $$Y_{00}$$ and $$Y_{11}$$, and $$s_0+s_1=1$$, we conclude that the trace is given by $$\operatorname{tr}(Y) = Y_{00} + Y_{11} = 2Y_{00} + s_1 - s_0 = 1 + 2\sqrt{s_0 s_1}.$$

This corresponds to the result stated in terms of $$[\operatorname{tr}(\sqrt\rho_A)]^2$$, because $$\sqrt{\rho_A}=\sqrt{s_0} \mathbb{P}_0+\sqrt{s_1} \mathbb{P}_1$$, hence $$[\operatorname{tr}(\sqrt\rho_A)]^2 = (\sqrt{s_0}+\sqrt{s_1})^2 = 1 + 2 \sqrt{s_0 s_1}.$$

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.