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

Question

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.

Answer 1

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...

Answer 2

(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 $H=-W$.)