# Prove that an entanglement witness satisfies $\operatorname{tr}(W)>0$ and $\operatorname{tr}(W)^2\ge \operatorname{tr}(W^2)$

If $$W$$ is an entanglement witness ($$W \neq 0$$), prove that

(a) $$tr(W) >0$$

(b) $$tr(W)^2 > tr(W^2)$$

For (a), by definition, since $$|ab\rangle$$ is separable, thus $$tr(\rho W)=\langle ab | W| ab \rangle \ge 0$$. Then $$tr(W) = \sum_{i,j} \langle a_i b_j | W | a_i b_j \rangle \ge 0$$ But I don't know how to prove it cannot be 0.

For (b), I have a hint: any state $$\rho$$ s.t. $$tr(\rho^2)\le 1/(d-1)$$, where $$d$$ is the dimension of $$\rho$$, is separable. But I have no idea how to use this property, and I also can't prove it.

• for (a), you already wrote that the trace is a sum of expectation values over product states. If $W$ is a witness those are all non-negative, so the only way for the sum to be zero is that every single term is zero, which would imply $W=0$.
– glS
Commented Dec 15, 2023 at 13:55
• I don't think it is trivial that the diagonal terms are zero implying the whole $W$ is zero, I tried to prove this but failed. Commented Dec 16, 2023 at 16:29
• is this an exercise from some textbook? also, I just realised in the title I put the inequality as $\operatorname{tr}(W)^2> \operatorname{tr}(W^2)$ rather than a strict inequality. However, I think that's how the statement should look like, considering as a counterexample the two-qubit swap operator, which is a witness, and has $\operatorname{tr}(W^2)=4=\operatorname{tr}(W)^2$.
– glS
Commented Dec 17, 2023 at 17:54

For (a), you've started well - taking a separable orthonormal basis and showing that the sum over all those expectations is non-negative. And, yes, you're right to worry that all those diagonal elements could be 0 given your starting point. So, let's assume they are all 0. But we're told that $$W\neq 0$$. So that would imply there's some off-diagonal element that's non-zero.

There are a couple of cases to deal with. Firstly, assume there's a non-zero element $$\langle a_1b|W|a_2b\rangle=re^{i\theta}.$$ In this case, choose a new orthonormal basis that includes $$|v\rangle=(|a_1\rangle+e^{-i\theta}|a_2\rangle)|b\rangle/\sqrt{2}$$. Now we have $$\langle v|W|v\rangle=r$$ (don't forget that $$W$$ is Hermitian). So the diagonal elements cannot all be 0 unless $$W=0$$.

The other case is that there are no such choices of $$a_1,a_2,b$$. This must bean that there's a choice such that $$\langle a_1b_1|W|a_2b_2\rangle=re^{i\theta}.$$ Now you can pick a new basis including the element $$|v\rangle=(|a_1\rangle+e^{-i\theta}|a_2\rangle)(|b_1\rangle+|b_2\rangle)/2.$$ Again, evaluate $$\langle v|W|v\rangle=r$$. There are lots of cross terms which are 0 by assumption. The same conclusion holds.

I'm still thinking about the second part...

• Thanks for your answer, I just cannot construct such $|v \rangle$ to find contradictions. Commented Dec 16, 2023 at 16:33

(a) holds since $$\mathrm{tr}(W)$$ is the value the witness takes on the maximally mixed state. Since there is an open ball of separable states around the maximally mixed state, this implies that we can add $$\varepsilon$$ of any hermitian operator $$H$$ the maximally mixed state and $$\mathrm{tr}((1\!\!1+\varepsilon H)W)$$ is still non-negative. This is only possible if $$\mathrm{tr}(W)$$ is strictly positive. (To this this, you can e.g. choose $$H=-W$$.)

• I think you are wrong, it is just like $tr(\rho^2)\le 1$, for Hermitian operators this still holds. Commented Dec 16, 2023 at 16:52
• @Fireond My bad. I have instead now provided a simple answer to (a) ;) Commented Dec 16, 2023 at 17:11
• It could be possible to prove (b) using a quantitative version of the above statement: There is a specific $\varepsilon$ s.th. $\mathrm{tr}(\mathrm{tr}W1\!\!1/d -\varepsilon W)>0$; then, $(\mathrm{tr}(W)^2) > d\varepsilon \mathrm{tr}(W^2)$. If this still works for $d\varepsilon=1$, you are good. Commented Dec 16, 2023 at 17:55
• to clarify: you're saying that $\operatorname{tr}((I+\epsilon H)W)=\operatorname{tr}(W)+\epsilon \operatorname{tr}(HW)$, and if $W\neq0$ one can choose $H$ to be a projection along a negative eigenvalue of $W$, thus $\operatorname{tr}((I+\epsilon H)W)<\operatorname{tr}(W)$, and thus you must have $\operatorname{tr}(W)=0$, lest there being entangled states arbitrarily close to the maximally mixed one?
– glS
Commented Dec 17, 2023 at 17:39
• @glS If you meant to say "must have $\mathrm{tr}(W)>0$", then yes. But $H=-W$ should also work. -- This can definitely be made quantitative as well and will give sth. along the lines of (b). Whether it actually gives (b) one would have to work out. Commented Dec 17, 2023 at 17:48