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}

I do not understand the last step $- \sum_ip_i\log p_i = H\left(p_i\right)$. I thought the definition of binary entropy is $H(p_{i}) = -p_{i}\log p_{i} - (1-p_{i})\log (1-p_{i})$. This does not give the desired equality.

Moreover, it is rather confusing that $i$ is used as a running index, and suddently it becomes a fixed index in the last equality.

Thanks for any helps.


It is not just the binary entropy that is denoted $H(p_i)$. The quantity that is relevant here is the Shannon entropy of the distribution $\{p_i\}$ which is defined as $$ H(p_i) = - \sum_i p_i \log p_i. $$ Note that when the distribution has only two elements we recover the binary entropy.

  • $\begingroup$ Thank you. This makes sense now. I got confused with the notation. $\endgroup$
    – frank
    Jun 23 at 5:43

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.