Is there a Bell type inequality involving only three values?

[Zafa Pi]

There are several Bell inequalities involving 4 values (e.g. CHSH where they are sometimes denoted by Q, R, S, T). The original Bell inequality involved 6. All being refuted by QM. Is it known whether there is one with only 3 values? I can prove there isn't one with 2 values.

 

[wle]

Is it known whether there is one with only 3 values? I can prove there isn't one with 2 values.

No, there isn't one. You need at least two parties (an Alice and a Bob) each doing at least two measurements each with at least two possible outcomes in order to derive a nontrivial Bell inequality. If one of the parties only does one measurement (which is necessarily the case if there are only three measurements) then you can always come up with a local hidden variable model for the statistics. E.g. if in a Bell-type experiment Alice gets result $a$ and Bob gets result $b$ conditioned on Alice doing measurement $x$ with probability $P(ab \mid x)$, then using the definition of conditional probability, the no-signalling principle, and inserting a Kronecker delta gets you $$\begin{eqnarray*}
P(ab \mid x) &=& P_{\mathrm{A}}(a \mid b, x) P_{\mathrm{B}}(b \mid x) \\
&=& P_{\mathrm{A}}(a \mid b, x) P_{\mathrm{B}}(b) \\
&=& \sum_{\lambda} P_{\mathrm{A}}(a \mid \lambda, x) P_{\mathrm{B}}(\lambda) \delta_{b, \lambda} \\
&=& \sum_{\lambda} q_{\lambda} Q_{\mathrm{A}}(a \mid x; \lambda) Q_{\mathrm{B}}(b \mid \lambda) \,,
\end{eqnarray*}$$ with $q_{\lambda} = P_{\mathrm{B}}(\lambda)$, $Q_{\mathrm{A}}(a \mid x; \lambda) = P_{\mathrm{A}}(a \mid \lambda, x)$, and $Q_{\mathrm{B}}(b \mid \lambda) = \delta_{b, \lambda}$, which is exactly the form of a local hidden variable model for the statistics.

The original Bell inequality involved 6.

Not sure where you got that idea. The original Bell inequality is a special case of CHSH where you assume perfect anticorrelations for one pair of measurements. If you write the CHSH inequality as $$\bigl \lvert \langle A_{0} B_{0} \rangle - \langle A_{0} B_{1} \rangle \bigr\rvert + \bigl\lvert \langle A_{1} B_{0} \rangle + \langle A_{1} B_{1} \rangle \bigr\rvert \leq 2 \,,$$ then setting, e.g., $\langle A_{1} B_{0} \rangle = -1$ you get $$\bigl\lvert \langle A_{0} B_{0} \rangle - \langle A_{0} B_{1} \rangle \bigr\rvert \leq 1 + \langle A_{1} B_{1} \rangle \,,$$ which is the inequality in Bell's 1964 paper.

 

[Zafa Pi]

No, there isn't one. You need at least two parties (an Alice and a Bob) each doing at least two measurements each with at least two possible outcomes in order to derive a nontrivial Bell inequality. If one of the parties only does one measurement (which is necessarily the case if there are only three measurements) then you can always come up with a local hidden variable model for the statistics. E.g. if in a Bell-type experiment Alice gets result $a$ and Bob gets result $b$ conditioned on Alice doing measurement $x$ with probability $P(ab \mid x)$, then using the definition of conditional probability, the no-signalling principle, and inserting a Kronecker delta gets you $$\begin{eqnarray*}
P(ab \mid x) &=& P_{\mathrm{A}}(a \mid b, x) P_{\mathrm{B}}(b \mid x) \\
&=& P_{\mathrm{A}}(a \mid b, x) P_{\mathrm{B}}(b) \\
&=& \sum_{\lambda} P_{\mathrm{A}}(a \mid \lambda, x) P_{\mathrm{B}}(\lambda) \delta_{b, \lambda} \\
&=& \sum_{\lambda} q_{\lambda} Q_{\mathrm{A}}(a \mid x; \lambda) Q_{\mathrm{B}}(b \mid \lambda) \,,
\end{eqnarray*}$$ with $q_{\lambda} = P_{\mathrm{B}}(\lambda)$, $Q_{\mathrm{A}}(a \mid x; \lambda) = P_{\mathrm{A}}(a \mid \lambda, x)$, and $Q_{\mathrm{B}}(b \mid \lambda) = \delta_{b, \lambda}$, which is exactly the form of a local hidden variable model for the statistics.

Not sure where you got that idea. The original Bell inequality is a special case of CHSH where you assume perfect anticorrelations for one pair of measurements. If you write the CHSH inequality as $$\bigl \lvert \langle A_{0} B_{0} \rangle - \langle A_{0} B_{1} \rangle \bigr\rvert + \bigl\lvert \langle A_{1} B_{0} \rangle + \langle A_{1} B_{1} \rangle \bigr\rvert \leq 2 \,,$$ then setting, e.g., $\langle A_{1} B_{0} \rangle = -1$ you get $$\bigl\lvert \langle A_{0} B_{0} \rangle - \langle A_{0} B_{1} \rangle \bigr\rvert \leq 1 + \langle A_{1} B_{1} \rangle \,,$$ which is the inequality in Bell's 1964 paper.

Thanks. I'll study this.

 

[billschnieder]

P(AB|xy) ≠ P(A|x)P(B|y)

Three values.

 

[DrChinese]

Three. Where A, B and C are any 3 different* angles (for Type I photon entanglement):

X = correlations between measurements at A and B
Y = correlations between measurements at A and C
Z = correlations between measurements at B and C

Then there isthe following inequality (although depending on the selected angles, A/B/C may be substituted differently than above):

(X + ~Y - Z) >= 0

But QM predicts the value for (X + ~Y - Z) is less than zero. See the following link for more detail:

http://www.drchinese.com/David/Bell_Theorem_Negative_Probabilities.htm*Except where any difference is a multiple of 90 degrees.

 

[wle]

Thanks. I'll study this.

If you want to learn more about the more practical aspects of investigating Bell's theorem, I'd recommend a review article that was published last year: arXiv:1303.2849 [quant-ph].

 

[wle]

P(AB|xy) ≠ P(A|x)P(B|y)

No, but $P(AB \mid xy) = P(A \mid B, xy) P(B \mid xy)$, and, in the case I was considering, Bob only does one measurement so the index $y$ is redundant.

 

[billschnieder]

No, but $P(AB \mid xy) = P(A \mid B, xy) P(B \mid xy)$, and, in the case I was considering, Bob only does one measurement so the index $y$ is redundant.

P(AB|xy) ≠ P(A|x)P(B|y)

I stand by my expression, not yours. Three experimentally measurable values. The expression embodies Bell's theorem.

 

[wle]

P(AB|xy) ≠ P(A|x)P(B|y)

I stand by my expression, not yours. Three experimentally measurable values. The expression embodies Bell's theorem.

You are honestly not making any sense. You post an inequation for apparently no reason -- nobody here claimed that $P(AB \mid xy) = P(A \mid x) P(B \mid y)$, so I don't see what your point is supposed to be. As for the "number of quantities" in the 1964 Bell inequality, if you want to count them differently than I do then good for you, but it doesn't change the points I was making: 1) Bell's 1964 inequality can be derived as a special case of CHSH, and 2) the situation that these inequalities apply to is the simplest for which you can derive a nontrivial Bell inequality.

 

[DrChinese]

P(AB|xy) ≠ P(A|x)P(B|y)

Three experimentally measurable values. The expression embodies Bell's theorem.

Hmmm, how would you experimentally measure P(A|x)? What would the QM prediction be for that?

 

[Mentz114]

Three. Where A, B and C are any 3 different* angles (for Type I photon entanglement):
..
http://www.drchinese.com/David/Bell_Theorem_Negative_Probabilities.htm

*Except where any difference is a multiple of 90 degrees.

It sems to me that the only way that could happen is something like this;

Suppose we have two machines that each periodically produces a lemon or an orange with equal probability. These machines record how many of each they produce. The outputs are channeled to a counting machine which counts lemon/lemon coincidences. The number of lemon/lemon coincidences cannot be greater than the minimum of the two lemon counts from the outputs. Assuming the machines never lie, the only way for that to happen is if one or more oranges has been changed to a lemon between output and counting. Probably a carefully selected orange(s) so a high-level conspiracy may be required.

Is this the kind of thing that has to happen for a violation of a CHSH type limit ?

 

[Zafa Pi]

The reason that DrChinese and wle disagree on their answers is that with the 3 devices A, B, and C DrChinese allows interactions AB, AC, and BC, where as wle allows 2, e,g. AB and AC while B and C don't interact (don't get simultaneously measured). Alice has just A, while Bob has B and C.

The proof given in the link provided by DrChinese is elementary, i.e. fit for a college freshman. However that is not the case for wle's proof. Is there an elementary proof?
Thanks to both of you.