REE from EOF

Abstract

It is well known that entanglement of formation (EOF) and relative entropy of entanglement (REE) are exactly identical for all two-qubit pure states even though their definitions are completely different. We think this fact implies that there is a veiled connection between EOF and REE. In this context, we suggest a procedure, which enables us to compute REE from EOF without relying on the converse procedure. It is shown that the procedure yields correct REE for many symmetric mixed states such as Bell-diagonal, generalized Vedral–Plenino, and generalized Horodecki states. It also gives a correct REE for less symmetric Vedral–Plenio-type state. However, it is shown that the procedure does not provide correct REE for arbitrary mixed states.

Appendix

In this section, we will show that REE and EOF are identical for two-qubit pure states. This fact was already proven in Theorem $3$ of Ref. [3]. We will prove this again more directly, because explicit Schmidt bases are used in the main body of the paper.

Let us consider a general two-qubit pure state $|\psi _2 \rangle _{AB} = \alpha _1 |00 \rangle + \alpha _2 |01 \rangle + \alpha _3 |10 \rangle + \alpha _4 |11 \rangle$ with $|\alpha _1|^2 + |\alpha _2|^2 + |\alpha _3|^2 + |\alpha _4|^2 = 1$. Then, its concurrence is ${\mathcal {C}} = 2 |\alpha _1 \alpha _4 - \alpha _2 \alpha _3|$. Now, we define

$$\begin{aligned} x_{\pm } = \frac{\alpha _1^* \alpha _2 + \alpha _3^* \alpha _4}{{\mathcal {N}}_{\pm }} \quad y_{\pm } = \frac{\lambda _{\pm } - (|\alpha _1|^2 + |\alpha _3|^2)}{{\mathcal {N}}_{\pm }} \end{aligned}$$

(7.1)

where

$$\begin{aligned} \lambda _{\pm } = \frac{1}{2} \left[ 1 \pm \sqrt{1 - {\mathcal {C}}^2} \right] \quad {\mathcal {N}}_{\pm }^2 = |\alpha _1^* \alpha _2 + \alpha _3^* \alpha _4|^2 + |\lambda _{\pm } - (|\alpha _1|^2 + |\alpha _3|^2)|^2. \end{aligned}$$

(7.2)

Now, we consider $2 \times 2$ matrix $u$, whose components $u_{ij}$ are

$$\begin{aligned} u_{11}&= \alpha _1 \left( \frac{|x_+|^2}{\sqrt{\lambda _+}} + \frac{|x_-|^2}{\sqrt{\lambda _-}} \right) + \alpha _2 \left( \frac{x_+^* y_+}{\sqrt{\lambda _+}} + \frac{x_-^* y_-}{\sqrt{\lambda _-}} \right) \nonumber \\ u_{12}&= \alpha _1 \left( \frac{x_+ y_+^*}{\sqrt{\lambda _+}} + \frac{x_- y_-^*}{\sqrt{\lambda _-}} \right) + \alpha _2 \left( \frac{|y_+|^2}{\sqrt{\lambda _+}} + \frac{|y_-|^2}{\sqrt{\lambda _-}} \right) \\ \nonumber u_{21}&= \alpha _3 \left( \frac{|x_+|^2}{\sqrt{\lambda _+}} + \frac{|x_-|^2}{\sqrt{\lambda _-}} \right) + \alpha _4 \left( \frac{x_+^* y_+}{\sqrt{\lambda _+}} + \frac{x_-^* y_-}{\sqrt{\lambda _-}} \right) \nonumber \\ u_{22}&= \alpha _3 \left( \frac{x_+ y_+^*}{\sqrt{\lambda _+}} + \frac{x_- y_-^*}{\sqrt{\lambda _-}} \right) + \alpha _4 \left( \frac{|y_+|^2}{\sqrt{\lambda _+}} + \frac{|y_-|^2}{\sqrt{\lambda _-}} \right) . \nonumber \end{aligned}$$

(7.3)

Then, Schmidt bases for each party are defined as

$$\begin{aligned} |i_A \rangle = \sum _{j=0}^1 v_{ji} |j \rangle \quad |i_B \rangle = \sum _{k=0}^1 w_{ik} |k \rangle \quad (i = 0, 1) \end{aligned}$$

(7.4)

where

$$\begin{aligned} v = \left( \begin{array}{c@{\quad }c} u_{11} &{} u_{12} \\ u_{21} &{} u_{22} \end{array} \right) \left( \begin{array}{c@{\quad }c} x_+ &{} x_- \\ y_+ &{} y_- \end{array} \right) \quad w = \left( \begin{array}{c@{\quad }c} x_+^* &{} y_+^* \\ x_-^* &{} y_-^* \end{array} \right) . \end{aligned}$$

(7.5)

Using Eq. (7.4), one can show straightforwardly that $|\psi _2 \rangle _{AB}$ reduces to $|\psi _2 \rangle _{AB} = \sqrt{\lambda _+} |0_A 0_B \rangle + \sqrt{\lambda _-} |1_A 1_B \rangle$. Thus, its CSS $\sigma _*$ are simply expressed in terms of the Schmidt bases as

$$\begin{aligned} \sigma _{*} = \lambda _+ |0_A 0_B \rangle \langle 0_A 0_B | + \lambda _- |1_A 1_B \rangle \langle 1_A 1_B |. \end{aligned}$$

(7.6)

Applying Eq. (1.1), one can show easily ${\mathcal {E}}_R (|\psi _2 \rangle ) = - \lambda _+ \ln \lambda _{+} - \lambda _- \ln \lambda _{-}$, which is exactly the same with EOF.