4
$\begingroup$

An explicit formula for the entanglement of formation $E(\rho)$ for an arbitrary two-qubit state $\rho$ was given by Wooters in Entanglement of Formation of an Arbitrary State of Two Qubits. The entanglement of formation is defined as the minimal average entanglement entropy over pure state decompositions of $\rho$: $$E(\rho )\equiv \min_{\{(p_i,|\psi_i\rangle)\}_i}\sum_i p_i S(\operatorname{tr}_2(|\psi_i\rangle\!\langle\psi_i|)), \qquad \sum_i p_i |\psi_i\rangle\!\langle\psi_i|=\rho.$$ The above paper showed that the result of this minimisation can be written explicitly as $$E(\rho) = \mathcal E(C(\rho)),$$ where $\mathcal E(c)\equiv h\left(\frac{1+\sqrt{1-c^2}}{2}\right)$, $h(x)\equiv -x\log x-(1-x)\log(1-x)$, $$C(\rho) \equiv \max(0, \lambda_1 - \lambda_2 - \lambda_3 - \lambda_4),$$ with $\lambda_i$ eigenvalues in decreasing order of $R\equiv \sqrt{\sqrt\rho\tilde\rho\sqrt\rho}$, and $\tilde\rho\equiv (\sigma_y\otimes \sigma_y)\rho^*(\sigma_y\otimes \sigma_y)$.

I'm looking to get a better understanding of how this result is obtained, but the terseness of the PRL format makes it not particularly easy here.

The starting point of the argument seems to be the observation that arbitrary decompositions of a state $\rho$ can be obtained from its eigendecomposition through an isometry. More precisely, if the eigendecomposition reads $\rho=\sum_k p_k \mathbb{P}(v_k)$, $\mathbb{P}(v)\equiv|v\rangle\!\langle v|$, then for any isometry $V$, $$\rho=\sum_j \mathbb{P}\left(\sum_k V_{jk}\sqrt{p_k}|v_k\rangle\right),$$ which I think is what equation (11) is saying in the paper.

Unfortunately, things get a bit less clear immediately after that. They start to consider decompositions in terms of states $|x_i\rangle$ satisfying $\langle x_i|\tilde x_j\rangle=\delta_{ij}\lambda_i$ with $\lambda_i$ eigenvalues of $\rho$, and this "tilde-orthogonality" being defined as $\langle x_i|\tilde x_j\rangle=(U\tau U^T)_{ij}$ with $\tau_{ij}\equiv \langle v_i|\tilde v_j\rangle$. I'm not really sure what's the idea behind this.

Is there a good way to understand either the proof itself, or at least, to any degree, the idea behind it? Any alternative reference discussing the proof would also be welcome.

$\endgroup$
1

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.

Browse other questions tagged or ask your own question.