I am trying to formulate the calculation of conditional min-entropy as a semidefinite program. However, so far I have not been able to do so. Different sources formulate it differently. For example, in this highly influential paper, it has been formulated as:

$$H_{\text{min}}(A|B)_\rho = - \underset{\sigma_B} {\text{inf}} \ D_{\infty}(\rho_{AB} \| id_A \otimes \sigma_B) $$ Where $$\rho_{AB} \in \mathcal{H_A \otimes H_B}, \sigma \in \mathcal{H_B}$$ and $$D_{\infty}(\tau \| \tau') = \text{inf} \{\lambda \in \mathbb{R}: \tau \leq 2^{\lambda} \tau' \}$$

How do I formulate it into a semidefinite program? It is possible as is mentioned in this lecture.

A possible SDP program is given in Watrous's lecture:

$$\text{maximize}: <\rho, X>$$ $$\text{subject to}$$ $$Tr_X{X} == \mathcal{1}_Y$$ $$X \in \text{Pos}(X \otimes Y)$$

How do I write it in CVX or any other optimization system?

  • 2
    $\begingroup$ What do you want to know? How to write it as an SDP, or how to solve the SDP? $\endgroup$ – Norbert Schuch Nov 10 '18 at 23:28
  • $\begingroup$ How could I write it in CVX properly? $\endgroup$ – QuestionEverything Nov 11 '18 at 3:51

I think I have an answer. The following should be the CVX code for one of the formulations found in this link.

cvx_begin sdp
variable X(2, 2) hermitian

minimize(trace(id' * X)) % id is eye(2)
subject to

kron(id, X) >= rho_ab % the tensor product of two density matrices a, b
X >= 0


The optimal value found in this program is $$\text{optval} = e^{-H_{\text{min}}(A|B)}.$$ So simple calculation would solve for ${H_{\text{min}}(A|B)}$. It turns out to be pretty simple at the end, given that the theoretical foundation leading up to this solution is not quite straightforward.

  • 1
    $\begingroup$ The implementation here is not fully correct. You should replace 'variable X(2,2) semidefinite' with 'variable X(2,2) hermitian'. Otherwise you are restricting the variable X to have real-valued entries and you will only get a lower bound on Hmin (and this lower bound can be quite a bit lower than the optimum). For example, you can find CQ states for which the above program would say Hmin(X|E) < 0, which is not possible for a cq-state. $\endgroup$ – Rammus Jan 21 '20 at 18:46

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