Connection between the definitions of concurrence for a two-qubit states

Question

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 $\rho$}$.

Answer 1

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 $$ C=\max[0,2\Lambda-1]=2\max[0,\Lambda-\frac12]. $$

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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 $\rho$.