3
$\begingroup$

Consider $N$ spin-1/2, for which we can define the collective spin operator $\vec{S}=\sum_i \frac{\vec{\sigma}^{(i)}}{2}$. My question is, what is the upper bound $U$ on $$ f(\rho) = \text{Var}[ S_z ] - \langle S_x\rangle $$ for separable states $\rho$? Meaning $f(\rho)\leq U$ for all $\rho$ separable, and with the bound saturable by some $\rho$. (In the above, $\text{Var}[ S_z ]=\langle S_z^2\rangle-\langle S_z\rangle^2$ is the variance).

A naive conclusion would be to bound each quantity independently, e.g. based on the spectrum of the operators, $\Delta^2 S_z\leq \frac{N^2}{4}$ and $S_x\leq \frac{N}{2}$, but I think $U^\prime=\frac{N^2}{4}+\frac{N}{2}$ cannot be attained by separable states. (One has also to fulfil the uncertainty relation).

Therefore, is there a convenient approach to tackle this kind of question?

$\endgroup$
5
  • $\begingroup$ There was a typo in the sign, yes. I meant to put just upper bounds based on the operator's spectrum. $\endgroup$
    – m137
    Commented Mar 11, 2021 at 20:59
  • $\begingroup$ Is there a motivation behind this question? (I'm asking since often there might be a better answer for the original problem.) $\endgroup$ Commented Mar 11, 2021 at 21:20
  • 2
    $\begingroup$ This doesn’t make sense unitwise as you are comparing something with units of $\hbar^2$ with something with units of $\hbar$/ $\endgroup$ Commented Mar 12, 2021 at 0:58
  • $\begingroup$ @ZeroTheHero Good point, which links back to the question about the motivation behind wanting to show such a bound (there might be scenarios if these are not actual spins - though one would indeed expect that for any physical scenario, sigma would be a unit-ful operator.) $\endgroup$ Commented Mar 12, 2021 at 14:59
  • $\begingroup$ I would check out spin-squeezing criteria for separability/entanglement, there must be something in this direction. $\endgroup$ Commented Jun 23, 2021 at 0:32

0

Your Answer

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