3
$\begingroup$

From section 11.3.2 of Nielsen & Chuang:

(4) let $\lambda_i^j$ and $\left|e_i^j\right>$ be the eigenvalues and corresponding eigenvectors of $\rho_i$. Observe that $p_i\lambda_i^j$ and $\left|e_i^j\right>$ are the eigenvalues and eigenvectors of $\sum_ip_i\rho_i$ and thus \begin{align}S\left(\sum_ip_i\rho_i\right) &= -\sum_{ij}p_i\lambda_i^j\log p_i\lambda_i^j \\ &= - \sum_ip_i\log p_i - \sum_ip_i\sum_j\lambda_i^j\log\lambda_i^j \\ &= H\left(p_i\right) + \sum_ip_iS\left(\rho_i\right)\end{align}

So this lemma is used to prove the joint entropy of $S(\rho_{AB})$ with $\rho_{AB}$ equal to

$\rho_{AB} = \sum_i p_i (\rho_i)_A \otimes (|e_i\rangle\langle e_i|)_B$

In the attached picture (the lemma), they do it first for only the A-system $\rho_A = \sum_i p_i (\rho_i)_A$, and then they say that this result directly leads to the same result for the above described AB-system (exactly the same?). First of all, I don't understand how they go from the first to the second line. I understand the fact that $p_i \lambda^j_i $ are the eigenvalues, but I really don't see how... Second question being, how can you use this result to infer the same result for $S(\rho_{AB})$? I mean you have your extra system in $H_B$ "hanging" on your A-system and I don't see how this just magically disappears or eventually comes down to a factor of 1 when explicitly calculating $S(\rho_AB)$. I feel pretty silly because this is already the proof aka explanation in the book Nielsen & Chuang and I don't even get it. So maybe explain it really simple and down to Earth for me please.

$\endgroup$
3
$\begingroup$

I don't understand how they go from the first to the second line

So, you're starting from $-\sum_{ij}p_i\lambda^j_i\log(p_i\lambda^j_i)$. Remember that $\log(ab)=\log(a)+\log(b)$, so this is the same as $$ -\sum_{ij}p_i\lambda^j_i\log(p_i)-\sum_{ij}p_i\lambda^j_i\log(\lambda^j_i) $$ For the first term, do the sum over $j$: $\sum_j\lambda^j_i=\text{Tr}(\rho_i)=1$. This gives you $$ -\sum_{i}p_i\log(p_i)-\sum_{i}p_i\sum_j\lambda^j_i\log(\lambda^j_i), $$ which was the second line you were after.

how can you use this result to infer the same result for $S(\rho_{AB})$

Observe that the eigenvectors of $\rho_{AB}$ are $|e^j_i\rangle|e_i\rangle$ with eigenvalue $p_i\lambda^j_i$. Hence, $$ S(\rho_{AB})=-\sum_{ij}p_i\lambda^j_i\log(p_i\lambda^j_i), $$ and you're back to that first line again...

PS for anybody else coming in and trying to make sense of the question, it is important to state that a condition of the theorem that is being proven is that the $\rho_i$ have support on orthogonal subspaces. This way, if a given $\rho_i$ has an eigenvector $|e^j_i\rangle$ with a non-zero eigenvalue $\lambda^j_i$, then $|e^j_i\rangle$ is also an eigenvector of all other $\rho_k$ for $k\neq i$, but $\lambda^j_k=0$. That is the key reason why we can write that the eigenvalues of $\sum_ip_i\rho_i$ are $p_i\lambda^j_i$.

$\endgroup$
3
$\begingroup$

That lemma has condition that all $p_i$ ($1\le i \le n$) have support on orthogonal subspaces. This means that there is a decomposition $H = \oplus_i H_i$ of the space $H$ such that $p_i(H_j) = 0$ if $i\neq j$ and $p_i(v_i)\neq 0$ if $v_i \in H_i$. Note that there can be component $H_0$ such that $p_i(H_0)=0$ for all $i \ge 1$.
Now there is a little oversimplification in their explanation. In fact we pick only nonzero eigenvalues $\lambda_i^j$, so vectors $|e_i^j\rangle$ lay in $H_i$. Now it is easy to see that $\sum_ip_i\rho_i |e_k^j\rangle = p_k\lambda_k^j|e_k^j\rangle$. Moreover, the union of all vectors $|e_i^j\rangle$ span the support of $\sum_ip_i\rho_i$. So this union is a complete set of orthogonal eigenvectors (excluding those from $H_0$).

As for your second question, we apply lemma to the states $\rho_i^\prime = (\rho_i)_A \otimes (|e_i\rangle\langle e_i|)_B$ on the system $AB$. That lemma is general, it is not "just for the A-system". You can check that all $\rho_i^\prime$ indeed have orthogonal support.

$\endgroup$

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.