# Prove that $-\log{d} \leq H(A|B) \leq \log{d}$ for von Neumann entropy

I'm trying to prove that $-\log{d} \leq H(A|B) \leq \log{d}$ for von Neumann entropy.

Now, for this to make sense I should give some definitions. System $A$ lives in Hilbert space $\mathcal{H}_A$, system $B$ in $\mathcal{H}_B$, and these are finite dimensional systems. Moreover, system A has dimension $d$. $H(A|B)$ is the conditional entropy, which I here take to be $H(A|B) = H(AB) - H(B)$, where $H(A) = -Tr(\rho_A \log{\rho_A})$, $\rho_A$ being the density operator of system $A$, so that $H(AB)$ is the density operator of the joint system.

Now, I can prove the upperbound quite easily:

$H(A|B) = H(AB) - H(B) \leq H(A) + H(B) - H(B) \leq \log{d}$, where both inequalities follow from the positivity of the relative entropy $H(A||B) = Tr(\rho_A\log{\rho_A}-\rho_A\log{\rho_B})$ quite easily. If needed, I can write those two line proofs in an update.

However, for the lowerbound I'm quite lost. I'm pretty sure I need to make some clever choice of $A$ and $B$ for the relative entropy again and abuse its positivity, but I can't figure out what to do. Could anyone give me a hint?

• What is $d$ here?
– Danu
Jan 10, 2015 at 15:13
• The dimension of Hilbert space $A$, so the number of elements in its basis. Jan 10, 2015 at 15:28
• Oh, don't know why I missed that initially. Thanks.
– Danu
Jan 10, 2015 at 16:13
• One can show this using strong subadditivity of the von Neumann entropy using a purification, but this seems like a bit of overkill. Jan 10, 2015 at 16:14
• As in, $H(AB|C) \leq H(A|C) + H(B|C)$, I suppose? I can look at that for a bit. What purification would you use? Jan 10, 2015 at 16:35

The lower bound $-\log\,d\le H(A|B)$ follows from strong subadditivity, $$H(ABC)+H(B)\le H(AB)+H(BC)\ .$$ To this end, choose a purification $\lvert\psi_{ABC}\rangle$ of $\rho_{AB}$. Then, $H(ABC)=0$, and $H(BC)=H(A)$, and thus, we have $H(B)\le H(AB) + H(A)$ (this is also known as the Araki-Lieb inequality), which implies $$-\log\,d\le -H(A)\le H(AB)-H(B)=H(A|B)\ .$$
• Thanks for the answer. I see indeed that it follows from strong subadditivity, however I don't see exactly how this is the Araki Lieb inequality. I tried doing it from that myself, and failed. I thought the Araki Lieb was the fact that $|H(A) - H(B)| \leq H(AB) \leq H(A) + H(B)$. How does one rewrite that to $H(B) \leq H(AB) + H(A)$? Maybe I'm just making a simple mistake, but the way I see it it states that $H(AB) - H(A) \leq H(B)$, which is not easily rewritten to the form you state as far as I can see. In fact, taking H(AB) to the other side and multiplying by -1 seems to give the opposite. Jan 11, 2015 at 13:09
• $|H(A)-H(B)|\le H(AB)$ implies $H(B)-H(A)\le H(AB)$, which is equivalent to $H(B)\le H(AB)+H(A)$. Jan 11, 2015 at 13:26
• The triangle equality follows from subadditivity by making use of duality as follows: Let R be a third purifying reference system, so that $|\psi\rangle_{RAB}$ is a purification of $\rho_{AB}$. Then $H(B) = H(RA) \leq H(A) + H(R) = H(A) + H(AB)$. This implies $H(AB) \geq H(B) - H(A)$, and you get the absolute values by doing the proof with A and B reversed. Jan 12, 2015 at 19:39