# Bell inequality with triplet state

Is it possible to prove Bell inequality starting from a state formed from triplet states, i.e. $\frac{1}{\sqrt{2}}(|\uparrow>_A|\uparrow>_B+|\downarrow>_A|\downarrow>_B)$?

If not, why?

I do not see why not, but somewhere it is mentioned something about non-rotational invariance. Morevoer I have always seen singlet state as starting point. Thanks.

• @MonkeysUncle triplet does not mean 3 particles, but aligned spins! – Arnaldo Maccarone Apr 13 '15 at 16:19

It's "no, you cannot violate Bell's inequality with this state", if you refer to what according to wikipedia is "the" Bell inequality: $$\rho(a, c) -\rho(b, a) - \rho(b, c) \le 1$$ where $a,b,c$ are three measurement settings. This inequality (as stated) seems to be only violated by states that are totally anti-correlated with parallel measurements. Your state, however, is totally correlated.