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

The concurrence for a state $$\rho$$ as defined here is $$$$C(\rho) = {\rm max}\{0, \lambda_1-\lambda_2-\lambda_3-\lambda_4\}.$$$$ Where $$\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

$$$$C = 2~ {\rm max}[0, \Lambda-\frac{1}{2}]$$$$ where $$\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 $$$$C(\rho) = {\rm max}\{0, \lambda_1-\lambda_2-\lambda_3-\lambda_4\}.$$$$ But this time $$\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}$$.

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 $$\rho$$.