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.

  • $\begingroup$ @MonkeysUncle triplet does not mean 3 particles, but aligned spins! $\endgroup$ Commented Apr 13, 2015 at 16:19

1 Answer 1


The answer is yes and no, but first, let me point out that you cannot "prove Bell's inequality", the whole point is that you violate the inequality in quantum mechanics.

Now, let me come to the yes/no part:

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.

It's "yes, of course you can violate Bell's inequality with this state", if you refer to what is nowadays understood as Bell's inequality. Not one, but literally an infinite amount of inequalities that can be violated by states that do not admit a joint probability distribution for all setups. This is very loosely speaking, to get a clearer picture let me refer to Asher Peres and the Braunschweig/Hanover question site with progress on the matter of Bell inequalities.


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