The Hardy State is another state that shows predetermined variables can’t reproduce the predictions of quantum theory. This works in a different way than GHZ. It’s not that you can’t find a predetermined setting which isn’t ruled out, but rather that you can get a result that no “legal” predetermined setting could provide.
Details
- hardy$_{zz} = \frac{1}{\sqrt{3}}|+z,+z\rangle
+ \frac{1}{\sqrt{3}}|+z,-z\rangle
+ \frac{1}{\sqrt{3}}|-z,+z\rangle$ - hardy$_{zx} = \frac{\sqrt{2}}{\sqrt{3}}|+z,+x\rangle
+ \frac{1}{\sqrt{3}}|-z,+x\rangle
+ \frac{1}{\sqrt{3}}|-z,-x\rangle$ - hardy$_{xz} = \frac{\sqrt{2}}{\sqrt{3}}|+x,+z\rangle
+ \frac{1}{\sqrt{3}}|+x,-z\rangle
+ \frac{1}{\sqrt{3}}|-x,-z\rangle$ - hardy$_{xx} = \frac{\sqrt{3}}{2} |+x,+x\rangle
+ \frac{\sqrt{3}}{6} |+x,-x\rangle
+ \frac{\sqrt{3}}{6} |-x,+x\rangle
– \frac{\sqrt{3}}{6} |-x,-x\rangle$
- Unlike the GHZ experiment, it’s easy for Alice and Bob to choose predetermined values which will work for a single experiment. For example, they can just set all their variables to plus.
- It’s simple to show, however, that no predetermined values can reproduce the results of repeated experiments. Just consider the $|xx\rangle$ state. If both Alice and Bob do repeated X spin measurements, they will both see spin down (minus) results one out of twelve times.
- But, when you consider the other three states, there is no way that Alice and Bob can both set their X variables to minus:
az ax bz bx
-- -- -- --
+ - + - +z,-xruled out by $|zx\rangle$
+ - - - +z,-xruled out by $|zx\rangle$
- - + - -x,+zruled out by $|xz\rangle$
- - - - -z,-zruled out by $|zz\rangle$
Notebooks
-
hardy-state.ipynb
References
- Wikipedia – Search for: Possibilistic nonlocality
This is the version I’m using. - Mermin QC book, Appendix D (p.175 in my copy)
- Nonlocality for Two Particles without Inequalities for Almost All
Entangled States
Lucien Hardy
PHYSICAL REVIEW LETTERS | VOLUME 71, NUMBER 11 | 13 SEPTEMBER 199
(I find the original paper hard to understand.)