What is a separable decomposition for the Werner state?

Question

Consider the two-qubit Werner state, defined as $$\rho_z = z |\Psi_-\rangle\!\langle \Psi_-| + \frac{1-z}{4}I, \quad |\Psi_-\rangle\equiv\frac{1}{\sqrt2}(|00\rangle-|11\rangle),$$ for $z\ge0$. Using the PPT criterion, one can see that this state is separable iff $0\le z\le 1/3$. Equivalently, the same state can be written as a mixture over symmetric and antisymmetric (normalised) projections: $$\rho_p \equiv p \frac{\Pi_{\rm sym}}{\binom{d+1}{2}} + (1-p) \frac{\Pi_{\rm asym}}{\binom{d}{2}},$$ with $d=2$ for the two-qubit case, and $2\Pi_{\rm sym}\equiv I+\operatorname{SWAP}$, $2\Pi_{\rm asym}=I-\operatorname{SWAP}$. The two definitions are connected setting $p=\frac34 (1-z)$, and thus the separability threshold in this parameterisation is at $p=\frac12$.

I couldn't, however, find a source discussing explicit separable decompositions (in the $z\le1/3$ regime, of course). Is there a "nice" way to find such decompositions?

Answer 1

The way that I think about this is to take a decomposition in Paulis, $$ \rho_z=(I+zZ\otimes Z-zX\otimes X-zY\otimes Y)/4. $$ I can group these as $$ ((1-3z)I+z(I+Z\otimes Z)+z(I-X\otimes X)+z(I-Y\otimes Y))/4. $$ Each of the 4 terms is diagonal in a separable basis, and positive semi-definite (provided $z\leq 1/3$). This directly implies a separable decomposition $$ \frac{1-3z}{4}I+\frac{z}{2}(|00\rangle\langle00|+|11\rangle\langle 11|)+\frac{z}{2}(|+-\rangle\langle +-|+|-+\rangle\langle -+|)+\frac{z}{2}(|y_+y_-\rangle\langle y_+y_-|+|y_-y_+\rangle\langle y_-y_+|). $$

Answer 2

Separable decompositions for high-dimensional Werner states are discussed in Unanyan et al. (2007). They find decompositions in terms of infinite terms, but from their Eqs. (10) and (11) it seems that these are reducible to finite decompositions (which we know must always exist for finite-dimensional states).

In the same paper, they also mention that decompositions for the two-qubit case are given in Wootters (1998) and Azuma and Ban (2006). In the latter in particular they give an explicit decomposition in the appendix. I'll report this decomposition here, only changing slightly the way the decomposition is written to make the expressions more compact.

We have:

$$\rho_q = \sum_{i=1}^4 |z_i\rangle\!\langle z_i|, \,\,\text{ where }\,\, |z_i\rangle = \sum_{k=1}^4 (H^{\otimes 2})_{ik}e^{i\theta_k}|x_k\rangle.$$ The (unnormalised) states $|x_k\rangle$ are defined as: $$\begin{gathered}|x_1\rangle = -i\sqrt{\lambda_+}|\Psi^-\rangle, \\ |x_2\rangle = \sqrt{\lambda_-}|\Psi^+\rangle, \qquad |x_3\rangle = \sqrt{\lambda_-}|\Phi^-\rangle, \\ |x_4\rangle = -i\sqrt{\lambda_-}|\Phi^+\rangle, \end{gathered}$$ where $\lambda_\pm$ are the eigenvalues of $\rho_q$: $\lambda_+=(1+3q)/4$ and $\lambda_-=(1-q)/4$, $H$ is the Hadamard matrix, and $\theta_k$ are phases satisfying $$e^{-2i\theta_1} \lambda_+ + (e^{-2i\theta_2}+e^{-2i\theta_3}+e^{-2i\theta_4})\lambda_-=0$$

This decomposition follows the method outlined in Wootters 1998 (mostly the second page of the PRL version), but I can't say I fully understand it.