I am reading about entanglement witnesses from here. In section 2.5.2, it is written that

Furthermore, a witness $\mathcal{W}$ is called optimal, if there is no other witness, which is finer than $\mathcal{W}$. This implies that for any positive operator $P$ the observable $X = \mathcal{W} − P$ is not a witness anymore. From this it can be easily seen that $\mathcal{W}$ is optimal, if and only if the product vectors $| \phi_i \rangle \ = | a_ib_i \rangle \text{ with } \langle \phi_i | \mathcal{W}| \phi_i \rangle \ = 0$ span the whole space.

I did not understand how do we arrive at the bold part, from the definition of optimality. Can anyone please explain?


1 Answer 1


Definition of optimal witness — Start recalling that an entanglement witness $W^{\rm opt}$ is said to be optimal if there is no other witness $W'$ with detects a strict superset of entangled states — that is, a $W'$ such that $\langle\rho,W^{\rm opt}\rangle<0\implies \langle\rho,W'\rangle<0$, and there is some $\sigma$ such that $\langle\sigma,W'\rangle<0< \langle\sigma,W^{\rm opt}\rangle$. Note that I'll use the shorthand notation $\langle A,B\rangle\equiv \operatorname{tr}(A^\dagger B)$ throughout this answer, as I think it better highlights some of the geometric ideas involved in these calculations.

Prove that if $W-P$ is a witness then $W$ isn't optimal

Subtracting a PSD operator can't do worse — Let $W$ be an entanglement witness, and let $P\ge0$ be a PSD operator such that $W-P$ is still an entanglement witness. It's easy to see that $W-P$ is at least as strong a witness as $W$, because for any $\rho$, $\langle W,\rho\rangle<0$ implies $\langle W-P,\rho\rangle<0$. We want to prove that $W-P$ is strictly stronger than $W$, that is, that there is some entangled state $\eta^e$ such that $\langle W,\eta^e\rangle\ge0$ but $\langle W-P,\eta^e\rangle<0$.

Proof idea — One way to prove this is to show that there is always some state $\eta$ such that $\langle W,\eta\rangle=0$ and $\langle P,\eta\rangle>0$. If you can find this state, then it immediately follows that $W-P$ detects it while $W$ doesn't, hence $W$ isn't optimal.

Prove existence of an entangled state needing $P$ to be witnessed — Note that unless $P=0$, there must be some state $\eta$ such that $\langle P,\eta\rangle>0$. Let's then consider the expectation value of $W$ on this state, and distinguish between the following three possibilities:

  1. If $\langle W,\eta\rangle=0$, then $\langle W-P,\eta\rangle<0$, and thus $\eta$ is entangled, and detected as such by $W-P$ but not by $W$. Thus $W-P$ is stronger than $W$, and $W$ isn't optimal.
  2. If $\langle W,\eta\rangle<0$, let $\eta^+$ be some state such that $\langle W,\eta^+\rangle>0$ (which, again, must exist if $W\neq0$). But then, there must be some $p\in(0,1)$ such that $$\bar\eta = p \eta+(1-p)\eta^+$$ satisfies $\langle W,\bar\eta\rangle=0$. Furthermore, we still have $\langle P,\bar\eta\rangle>0$, hence again $W$ isn't optimal.
  3. If $\langle W,\eta\rangle>0$, let $\eta^-$ be some (necessarily entangled) state such that $\langle W,\eta^-\rangle<0$. Then as above, there must be some $p\in(0,1)$ such that $$\bar\eta = p\eta + (1-p)\eta^-$$ gives $\langle W,\bar\eta\rangle=0$, and still $\langle P,\bar\eta\rangle>0$, hence again the conclusion.

For more related results you might want to have a look at (Lewenstein et al. 2000, quant-ph/0005014). For example, they show in Lemma 2 that an entanglement witness $W_2$ is finer than $W_1$ iff there's $P\ge0$ with $\operatorname{tr}(P)=1$ and $0\le \epsilon<1$ such that $W_1=(1-\epsilon)W_2+\epsilon P$. And then soon after that a witness $W$ is optimal iff for all $P$ and $\epsilon>0$ the operator $(1+\epsilon)W-\epsilon P$ isn't a witness.

$W$ is optimal if no $P\ge0$ can be subtracted from it

This is an intermediate step that will help proving the next result. It's essentially a rewording of what is proved in Lemma 2 and Theorem 1 of https://arxiv.org/abs/quant-ph/0005014.

They show in Lemma 2 that $W_2$ is finer than $W_1$ iff there's $P\ge0$ with $\operatorname{tr}(P)=1$ and $\epsilon\in[0,1)$ such that $W_1=(1-\epsilon)W_2+\epsilon P$. But observing that any $W$ is a witness iff $\alpha W$ is a witness for any $\alpha>0$, this is the same as saying that $W_2$ is finer than $W_1$ iff it's a positive multiple of $W_1-P$ for some $P\ge0$ with $\operatorname{tr}(P)\in[0,1)$. So in summary, they're saying that all witnesses finer than $W_1$ must have (modulo positive multiples) the form $W_1-P$ for some $P\ge0$. Which in turn means that if $W-P$ isn't a witness for all $P\ge0$, then $W$ must be optimal.

Let's try to more directly show that if $W'$ is a finer entanglement witness than $W$, then $\alpha W'=W-P$ for some $\alpha>0$ and some $P\ge0$. This amounts to proving that any such $W'$ is such that $W-\alpha W'\ge0$ for some $\alpha>0$, that is, that there is $\alpha>0$ such that $\langle W,\rho\rangle\ge \alpha \langle W',\rho\rangle$. for all states $\rho$. To show this, let's consider a few different cases:

  1. If $\langle W,\rho\rangle=0$, then $\langle W',\rho\rangle\le0$, and thus the inequality is satisfied. This follows from Lemma 1(i) in the paper.
  2. If $\langle W,\rho\rangle<0$, then $\langle W',\rho\rangle\le\langle W,\rho\rangle$, and thus the inequality is again satisfied. This follows from Lemma 1(2) in the paper.
  3. If $\langle W,\rho\rangle>0$ then $\langle W',\rho\rangle\le\lambda\langle W,\rho\rangle$ for some fixed $\lambda\ge1$ that only depends on $W$ and $W'$. This is Lemma 1(3) in the paper.

We thus have the result with $\alpha=1/\lambda$.

Optimality vs product states such that $\langle e, f|W|e,f\rangle=0$

The other point is again discussed in https://arxiv.org/abs/quant-ph/0005014. Define $$P_W\equiv \{|e,f\rangle : \,\, \langle W,\mathbb{P}_e\otimes\mathbb{P}_f\rangle=0\}.$$ They then show in the paper that:

  1. (Lemma 3) If $P\ge0$ is such that $PP_W\neq0$, then $W-P$ isn't a witness. In other words, if there's $|e,f\rangle\in P_W$ such that $P|e,f\rangle\neq0$, then $W-P$ isn't a witness. This is immediate remembering that this would mean $\langle W,\mathbb{P}_e\otimes \mathbb{P}_f\rangle=0$ and $\langle P,\mathbb{P}_e\otimes \mathbb{P}_f\rangle>0$, and thus $\langle W-P,\mathbb{P}_e\otimes\mathbb{P}_f\rangle<0$.

  2. (Corollary 2) If $P_W$ spans the whole space, then $W$ is optimal. Using the above results this is now also easy to see: if $P_W$ spans the space, then any operator $P\neq0$ is such that $W-P$ isn't a witness anymore. Thus $W$ must be optimal.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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