How Can Bell's Inequality Be Transformed into the CHSH Inequality?

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Homework Help Overview

The discussion revolves around the transformation of Bell's inequality into the CHSH inequality, a topic within quantum mechanics and the study of correlations in entangled states.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the mathematical steps needed to derive the CHSH inequality from Bell's inequality, with one participant expressing confusion about the process and another suggesting a potential method involving algebraic manipulation and ensemble averaging.

Discussion Status

The conversation is ongoing, with participants sharing their attempts and questioning the assumptions underlying the inequalities. One participant has provided a detailed algebraic approach, while another has posed a follow-up question regarding the limitations of the EPR state in satisfying the Bell inequality.

Contextual Notes

Participants acknowledge the constraints of homework rules, which may limit the depth of exploration into the topic.

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Homework Statement



I need to get the CHSH inequality from Bell's inequality

Homework Equations



|C(a, b) - C(a, b')| + |C(a', b) - C(a', b')| <= 2

to

-2 <= C(a, b) - C(a, b') + C(a', b) + C(a', b') <= 2


The Attempt at a Solution



I know the CHSH allows for no correlation of 0 but I can't get anywhere. I think my math with inequalities just sucks.
 
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cscott said:

Homework Statement



I need to get the CHSH inequality from Bell's inequality

Homework Equations



|C(a, b) - C(a, b')| + |C(a', b) - C(a', b')| <= 2

to

-2 <= C(a, b) - C(a, b') + C(a', b) + C(a', b') <= 2


The Attempt at a Solution



I know the CHSH allows for no correlation of 0 but I can't get anywhere. I think my math with inequalities just sucks.

The above looks a bit confused. How about this trick?

(1) (a + a')b + (a - a')b' = +/-2;

since (a + a') = 0 AND (a - a') = +/-2, XOR (a + a') = +/-2 AND (a - a') = 0.

So, expanding (1):

(2) ab + a'b + ab' - a'b' = +/-2.

Then, ensemble-averaging:

(3) <ab> + <a'b> + <ab'> - <a'b'> </= +/-2.

Or:

(4) |<ab> + <a'b> + <ab'> - <a'b'>| </= 2;

where <ab> = C(ab), etc. (4) is known as the CHSH Inequality; and the real trick is to see why EPR-Bell tests cannot (in general) satisfy (4).
 
JenniT said:
The above looks a bit confused. How about this trick?

(1) (a + a')b + (a - a')b' = +/-2;

since (a + a') = 0 AND (a - a') = +/-2, XOR (a + a') = +/-2 AND (a - a') = 0.

So, expanding (1):

(2) ab + a'b + ab' - a'b' = +/-2.

Then, ensemble-averaging:

(3) <ab> + <a'b> + <ab'> - <a'b'> </= +/-2.

Or:

(4) |<ab> + <a'b> + <ab'> - <a'b'>| </= 2;

where <ab> = C(ab), etc. (4) is known as the CHSH Inequality; and the real trick is to see why EPR-Bell tests cannot (in general) satisfy (4).

Very nice answer, sir. Now I'd like to ask what you lead to: why cannot the EPR state satisfy the Bell inequality generally?
 
mmmrrrrrrr said:
Very nice answer, sir. Now I'd like to ask what you lead to: why cannot the EPR state satisfy the Bell inequality generally?

Good question, but it's a homework question at our school. So please follow the PF homework rules.
 

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