Consider a two-qubit space, and a Hermitian operator $R\in\mathrm{Herm}(\mathbb{C}^2\otimes\mathbb{C}^2)$ in this space.

The operator is positive semidefinite iff $\langle u,Ru\rangle\ge0$ for all $u\in\mathbb{C}^2\otimes\mathbb{C}^2$. We can equivalently write this condition as $\langle \mathbb{P}_u,R\rangle\ge0$, where $\mathbb{P}_u\equiv uu^\dagger\equiv |u\rangle\!\langle u|$ and $\langle \cdot,\cdot\rangle$ denotes here the trace inner product between operators (which equals the standard complex inner product between their vectorisations).

On the other hand, $R$ is an entanglement witness if $R$ is not semidefinite positive, but satisfies $\langle \mathbb{P}_u\otimes\mathbb{P}_v,R\rangle\ge0$ for all $u,v\in\mathbb{C}^2$.

Some easy consequences of this requirement are that $R$ must have some negative eigenvalue corresponding to some non-separable vector. A standard example of such an object is the Swap operator: if $$W \equiv \begin{pmatrix}1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{pmatrix}\equiv \mathbb{P}_{00}+\mathbb{P}_{11} + E_{01}\otimes E_{10} + E_{10}\otimes E_{01} = I\otimes I - 2\mathbb{P}_{\Psi^-},$$ where $|\Psi^-\rangle\equiv\frac{1}{\sqrt2}(|01\rangle-|10\rangle)$, then $W$ is not positive semidefinite, as $\langle \mathbb{P}_{\Psi^-},W\rangle=-1$, but we nonetheless have $$\langle \mathbb{P}_u\otimes \mathbb{P}_v,W\rangle = |u_0 v_0|^2 + |u_1 v_1|^2 + 2\operatorname{Re}(\bar u_1 u_2 v_1 \bar v_2) = a^2 + b^2 + 2 ab \cos\alpha, $$ where $a\equiv |u_0 v_0|, b\equiv |u_1 v_1|$, and $\alpha$ is some angle depending on the relative phases of $|u\rangle$ and $|v\rangle$. We conclude that $\langle \mathbb{P}_u\otimes \mathbb{P}_v,W\rangle\ge (a-b)^2\ge0$, hence $W$ is a witness. We can equivalently reach the same conclusion observing that $W$ is the Choi operator of the transpose map, which is positive.

My question is: while I know that characterising witness operators (equivalently, characterising positive non-CP maps) is nontrivial, is there a "good" set of conditions that we can use to characterise two-qubit witness operators? In particular, we know how to characterise the positive semidefiniteness of Hermitian matrices, and for 2x2 matrices we know that $A\ge0$ iff diagonal elements and determinant are non-negative. Is there any similar kind of condition that can be used to characterise witnesses? Given that characterising the positive semidefiniteness of $R$ itself is relatively easy via Sylvester's criteria, I'd settle with conditions on $R$ characterising $\langle \mathbb{P}_u\otimes \mathbb{P}_v,R\rangle\ge0$.

An easy necessary condition for $R$ to be a witness is that its top-left and bottom-right 2x2 submatrices must be positive semidefinite. These come from requiring $\langle \mathbb{P}_0\otimes \mathbb{P}_v,R\rangle,\langle \mathbb{P}_1\otimes \mathbb{P}_v,R\rangle\ge0$ for all $v\in\mathbb{C}^2$. More generally, for any fixed $u\in\mathbb{C}^2$, requiring $\langle \mathbb{P}_u\otimes \mathbb{P}_v,R\rangle\ge0$ for all $v$ amounts to the condition $$|u_0|^2 R_{00} + |u_1| R_{11} + (\bar u_0 u_1 R_{01} + u_0 \bar u_1 R_{01}^\dagger)\ge0,$$ where I'm denoting with $R_{ij}$ the $(i,j)$-th 2x2 submatrix of $R$. This seems to give a N&S set of conditions for $R$ being a witness (technically, for it having non-negative expectation value on separable states), but if we have to check the condition for all possible $u\in\mathbb{C}^2$ it's not tremendously useful.

Of course, any other kind of characterisation that is not in the form of a similar set of inequalities would also be great.

  • $\begingroup$ I gotta say, your questions are always well written and informative - although I have almost no chance of providing any helpful observation, I feel like I learned something just by reading and understanding! Apart from SWAP $X\otimes X$ is another permutation matrix but yet is not an entanglement witness, as by definition it factors, correct? $\endgroup$ Aug 12, 2022 at 23:35
  • $\begingroup$ @MarkS It's not entanglement witness, since by choosing $\mathbb{P} _u=|+\rangle \langle +|$ and $\mathbb{P} _v=|-\rangle \langle -|$, we have $\langle \mathbb{P} _u\otimes \mathbb{P} _v,X\otimes X\rangle =-1< 0$. $\endgroup$
    – narip
    Aug 13, 2022 at 9:28
  • $\begingroup$ @MarkS And mind that, entanglement witness does not say witness itself is entangled. You may notice that the condition required in the problem is a separable pure state, in fact, all the separable state is a convex combination of separable pure states, so the condition is equivalently saying that, for all separable state the inner product will be positive, so once you found a negative inner product, the state must be entangled state. $\endgroup$
    – narip
    Aug 13, 2022 at 9:38
  • $\begingroup$ @MarkS glad to hear it =). As already pointed out, mind that the standard definition of witness op is slightly different, if equivalently, to the one I wrote here (you generally ask for positive expvals on all separable states, and negative expval for some non-separable state). A product operator will never be a witness, because it will admit product eigenstates, and it'd have to be in the form $A\otimes B$ with $A,B\ge0$ to get non-neg expval on product states. I think $W$ being a permutation matrix is mostly by accident: most other such matrices won't be witnesses (nor Hermitian, really) $\endgroup$
    – glS
    Aug 13, 2022 at 10:18
  • 1
    $\begingroup$ @Sherlock it's just a way to specify the partition we're referring to. While $\mathbb{C}^4$ and $\mathbb{C}^2\otimes\mathbb{C}^2$ are clearly the same space, when discussing things like entanglement you're always referring to some choice of bipartition for the space. So writing $\mathbb{C}^2\otimes\mathbb{C}^2$ I'm saying that we're talking about entanglement between two two-dimensional spaces (in this case the bipartition is trivial, as there's nothing to say about $\mathbb{C}\otimes\mathbb{C}^3$, but in more general cases it's not) $\endgroup$
    – glS
    Oct 12, 2022 at 13:50

1 Answer 1


You can try to use the Størmer-Woronowicz theorem for that (it's used to prove the sufficiency of the Peres–Horodecki criterion in $2 \times 2$ and $2 \times 3$ cases).

The theorem states that any positive map $\Lambda: \mathbb{C}^{2 \times 2} \rightarrow \mathbb{C}^{2 \times 2}$ can be written as a combination $$ \Lambda = \Lambda_1 + \Lambda_2 \circ T, $$ where $\Lambda_i$ are completely positive and $T$ is the transposition.

From the Choi isomorphism it follows that any Hermitian $R \in \mathbb{C}^{4\times 4}$, with non-negative expectation value on separable states, can be written as $$ R = R_1 + R_2^{p_T}, $$ where $R_i$ are positive semidefinite, and $p_T$ is the partial transpose. For $R$ being a witness you just need to check that it is not positive semidefinite.

  • $\begingroup$ good point. It's also nicely consistent with the example of the swap, as $W^{T_B}=\mathbb{P}_{\Phi^+}$ with $|\Phi^+\rangle\equiv\frac{1}{\sqrt2}(|01\rangle+|10\rangle)$. Still, this characterisation is great to generate examples, but is there a good way to check for this condition given an operator $R$? $\endgroup$
    – glS
    Aug 13, 2022 at 10:15
  • $\begingroup$ I'm not aware of such a criterion. AFAIK, checking if a map is positive is NP-hard in general situation. $\endgroup$
    – Danylo Y
    Aug 13, 2022 at 11:13
  • $\begingroup$ it's NP-hard, sure, but that doesn't necessarily mean it's not doable in small dimensions, no? The hardness of checking it might scale badly with number of qubits or space dimensions, but it might still be doable in small instances. $\endgroup$
    – glS
    Aug 13, 2022 at 11:16
  • $\begingroup$ Well, there must be a kind of brute force search via semidefinite programming. Similarly to problem 1 in arxiv.org/abs/1011.2751, we can try to solve weak membership problem (for witnesses) for smaller and smaller values of $\epsilon$. $\endgroup$
    – Danylo Y
    Aug 13, 2022 at 13:00

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