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 $0\le z\le 1/3$.

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?

  • $\begingroup$ "A nice way" -- For separable states in general? For the Werner state? For qubit states satisfying the PPT criterion? $\endgroup$ Apr 13 '20 at 9:51
  • $\begingroup$ I searched on google for "werner state separable decomposition". The first hit was arxiv.org/abs/quant-ph/0703240: "A Decomposition of Separable Werner States". $\endgroup$ Apr 13 '20 at 9:52
  • $\begingroup$ (As a quick observation: For the Werner state, which is $U\otimes U$ invariant, a "nice" way might be to take some separable state and twirl over $U\otimes U$. Of course, whether this gives the full range depends on the separable state you pick, so the argument might be a bit circular.) $\endgroup$ Apr 13 '20 at 9:54
  • $\begingroup$ @NorbertSchuch for Werner states, but obviously the more general the method the better (though I vaguely remember the general case to be NP-hard or something). I saw that paper but I was hoping for a more elementary derivation in at least the two-qubit case. I found explicit decompositions discussed in a couple of other papers (see answer). They work, but to be honest I still don't fully understand the method used to derive them. I'll try and go through the papers more thoroughly when I get the time. Mostly, here I was interested in what kinds of tricks would work to solve such problems $\endgroup$
    – glS
    Apr 13 '20 at 10:59
  • $\begingroup$ I guess the general case is hard since if you can find a decomposition, you can use the same method to decide if there exists one (i.e. if the state is separable), which is NP-hard. $\endgroup$ Apr 13 '20 at 11:35

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 $$ \frac{1-3z}{4}I+\frac{z}{2}(|00\rangle\langle00|+|11\rangle\langle 11|)+\frac{z}{2}(|+-\rangle\langle +-|+|-+\rangle\langle -+|)+\frac{z}{2}(|y_+y_-\rangle\langle y_+y_-|+|y_-y_+\rangle\langle y_-y_+|). $$


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.