5
$\begingroup$

I was reading Wilde's 'Quantum Information Theory' and saw the following theorem at chapter 11 $(11.7.2)$:

$$ I(A; B | C) \ge 0, $$ where, $$ I(A;B|C) := H(A|C) + H(B | C) - H(AB|C). $$

I know that the mutual information is non-negative, i.e. $$ I(A;B) \ge 0, $$ where, $$ I(A;B) := H(A) + H(B) - H(AB). $$ Now if we have access to an additional subsystem $C$, this can't decrease the mutual information of $A$ and $B$. But I was looking for sort of formal proof of this. I was trying to apply the non-negativity of mutual info. into this, but not sure how to proceed. Thanks in advance!

$\endgroup$
1
3
$\begingroup$

Here's a relatively simple proof just based on the data processing inequality (DPI) for the relative entropy $D(\rho\|\sigma) = \mathrm{tr}[\rho (\log \rho - \log \sigma)]$ -- if you're willing to accept the DPI as a basis for a formal proof. Recall that the DPI says that for any channel $\Phi$ we have $$ D(\rho \|\sigma) \geq D(\Phi(\rho)\|\Phi(\sigma)). $$

Now $$ \begin{aligned} I(A:B|C) &= H(A|C) + H(B|C) - H(AB|C) \\ &= H(AC) - H(C) + H(BC) - H(C) - H(ABC) + H(C) \\ &= H(AC) + H(BC) - H(ABC) - H(C) \\ &= H(A|C) - H(A|BC) \\ &= -D(\rho_{AC} \| I_A \otimes \rho_C) + D(\rho_{ABC} \| I_A \otimes \rho_{BC}). \end{aligned} $$ Thus $I(A:B|C) \geq 0$ is equivalent to $$ D(\rho_{ABC} \|I_A \otimes \rho_{BC}) \geq D(\rho_{AC} \|I_A \otimes \rho_C), $$ but this follows immediately from the DPI by taking the channel $\Phi$ to be the partial trace over the $B$ system.

Note also that the fourth line of the derivation shows this result is equivalent to strong subadditivity of the von Neumann entropy, as mentioned in the comments by @Purva Thakre.

$\endgroup$
1
$\begingroup$

To expand on @Purva Tharke's comment, the strong subadditivity inequality states: $$H(ABC)+H(C) \le H(AC) + H(BC)$$ $$=H(ABC)+H(C) +H(C) -H(C) \le H(AC) + H(BC)$$ $$=H(AB|C) \le H(A|C) + H(B|C)$$ $$=0\le H(A|C) + H(B|C) - H(AB|C)=H(A;B|C)$$

Edit: A good proof of SS for entropies can be found in Nielsen and Chuang, in case you wanted to take a look.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.