5
$\begingroup$

The concurrence for a state $\rho$ as defined here is \begin{equation} C(\rho) = {\rm max}\{0, \lambda_1-\lambda_2-\lambda_3-\lambda_4\}. \end{equation} Where $\lambda_i$ are the eigenvalues of matrix $ \sqrt{\sqrt{\rho} \tilde{\rho} \sqrt{\rho}}$, with $\tilde{\rho} = (\sigma_y \otimes \sigma_y) \rho^* (\sigma_y \otimes \sigma_y)$, $\sigma_y$ being the Pauli matrix and $\rho^*$ is the complex conjugate of state $\rho$.

In the case of Bell diagonal states of the form $\sum_{k=1}^{4} \lambda_k |\beta_k\rangle\langle\beta_k|$, where $|\beta_k\rangle$ is the Bell state. In this case, the concurrence, as defined here (Eq. B11), is given by

\begin{equation} C = 2~ {\rm max}[0, \Lambda-\frac{1}{2}] \end{equation} where $\Lambda = {\rm max}_k\{ \lambda_k\}$.

How are these two definitions connected?

Edit: Thanks for clearing the above doubts. Just found one more definition here (below Eq. 10). Here \begin{equation} C(\rho) = {\rm max}\{0, \lambda_1-\lambda_2-\lambda_3-\lambda_4\}. \end{equation} But this time $\lambda_k$ are $\textit{square root}$ o the eigenvalues of the matrix $\rho \tilde{\rho}$, rather than eigenvalue of the $\textit{square root}$ of matrix $ \sqrt{\sqrt{\rho} \tilde{\rho} \sqrt{\rho}}$. Even if we agree that $\sqrt{\rho} \tilde{\rho} \sqrt{\rho} = \rho$, in the Bell basis, then $\lambda_k$ should be $\textit{the eigenvalues of the square root of $\rho$, rather than square root of the eigenvalues of $\rho$}$.

$\endgroup$

1 Answer 1

3
$\begingroup$

The Bell states $|\beta_k\rangle$ all satisfy $Y\otimes Y|\beta_k\rangle=\pm|\beta_k\rangle$. Hence, $\tilde\rho=\rho$. Thus, the matrix $\sqrt{\sqrt{\rho}\tilde\rho\sqrt{\rho}}=\rho$, given that all the $\lambda_k$ are real and non-negative. Hence, the set of eigenvalues $\lambda_i$ is the same as the set of the $\tilde \lambda_k$, the Bell-diagonal components (I've denoted these by $\tilde\lambda_k$ to avoid any confusion). However, the stated definition of the concurrence explicitly orders the eigenvalues, $\lambda_1\geq\lambda_2\geq\lambda_3\geq\lambda_4$. Hence, $\lambda_1=\max_k\tilde\lambda_k=\Lambda$.

Recall that $\sum_k\tilde\lambda_k=1$, so we have that $\lambda_2+\lambda_3+\lambda_4=1-\Lambda$. Using $C(\rho)=\max[0,\lambda_1-\lambda_2-\lambda_3-\lambda_4]$, we now have $$ C=\max[0,2\Lambda-1]=2\max[0,\Lambda-\frac12]. $$


Given the edit to you question, it appears the following needs to be clarified. Recall that for a normal matrix $A$ (of which Hermitian is a special case), it has a spectral decomposition $$ A=\sum_i\lambda_i|\lambda_i\rangle\langle\lambda_i|. $$ For any function $f:\mathbb{C}\mapsto\mathbb{C}$, i.e. accepts a complex number as input and gives a complex number as output, we can define the action of $f$ on the matrix $A$ to be $$ f(A)=\sum_if(\lambda_i)|\lambda_i\rangle\langle\lambda_i|. $$ This means that the square root of the eigenvalues are equal to the eigenvalues of the square root. It also means that if you're trying to calculate $$ \sqrt{\sqrt{\rho}\rho\sqrt{\rho}}, $$ then you can start by writing this as \begin{align} \sqrt{\left(\sum_i\sqrt{\lambda_i}|\lambda_i\rangle\langle\lambda_i|\right)\left(\sum_i\lambda_i|\lambda_i\rangle\langle\lambda_i|\right)\left(\sum_i\sqrt{\lambda_i}|\lambda_i\rangle\langle\lambda_i|\right)}&=\sqrt{\sum_i\lambda_i^2|\lambda_i\rangle\langle\lambda_i|} \\ &=\sum_i|\lambda_i||\lambda_i\rangle\langle\lambda_i| \end{align} Since the $\lambda_i$ are positive, this is the same as $\rho$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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