CHSH and the triangle inequality

[gespex]

Hello everybody,

I've been trying to understand the CHSH proof as it is listed on Wikipedia:
http://en.wikipedia.org/wiki/CHSH_inequality

I got to this without any problem:
E(a, b) - E(a, b^\prime) = \int \underline {A}(a, \lambda)\underline {B}(b, \lambda)[1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b^\prime, \lambda)]\rho(\lambda)d\lambda - \int \underline {A}(a, \lambda)\underline {B}(b^\prime, \lambda)[1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b, \lambda)]\rho(\lambda)d\lambda

However, now it mentions two things to get to the next step:
- The fact that [1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b^\prime, \lambda)]\rho(\lambda) and [1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b, \lambda)]\rho(\lambda) are non-negative (easy enough to see).
- The triangle inequality "to both sides" (how?)

And the next equation is:
|E(a, b) - E(a, b^\prime)| \leq \int [1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b^\prime, \lambda)]\rho(\lambda)d\lambda + \int [1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b, \lambda)]\rho(\lambda)d\lambda

I don't understand how it gets to this last equation from the one before. Could somebody please explain?Thanks in advance,
Gespex

 

[Delta Kilo]

Hello everybody,

I've been trying to understand the CHSH proof as it is listed on Wikipedia:
http://en.wikipedia.org/wiki/CHSH_inequality

I got to this without any problem:
E(a, b) - E(a, b^\prime) = \int \underline {A}(a, \lambda)\underline {B}(b, \lambda)[1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b^\prime, \lambda)]\rho(\lambda)d\lambda - \int \underline {A}(a, \lambda)\underline {B}(b^\prime, \lambda)[1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b, \lambda)]\rho(\lambda)d\lambda

However, now it mentions two things to get to the next step:
- The fact that [1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b^\prime, \lambda)]\rho(\lambda) and [1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b, \lambda)]\rho(\lambda) are non-negative (easy enough to see).
- The triangle inequality "to both sides" (how?)

And the next equation is:
|E(a, b) - E(a, b^\prime)| \leq \int [1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b^\prime, \lambda)]\rho(\lambda)d\lambda + \int [1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b, \lambda)]\rho(\lambda)d\lambda

I don't understand how it gets to this last equation from the one before. Could somebody please explain?


Thanks in advance,
Gespex

So you have:
E(a, b) - E(a, b^\prime) = \int \underline {A}(a, \lambda)\underline {B}(b, \lambda)[1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b^\prime, \lambda)]\rho(\lambda)d\lambda - \int \underline {A}(a, \lambda)\underline {B}(b^\prime, \lambda)[1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b, \lambda)]\rho(\lambda)d\lambda

Taking absolute value of both sides
|E(a, b) - E(a, b^\prime)| = |\int \underline {A}(a, \lambda)\underline {B}(b, \lambda)[1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b^\prime, \lambda)]\rho(\lambda)d\lambda - \int \underline {A}(a, \lambda)\underline {B}(b^\prime, \lambda)[1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b, \lambda)]\rho(\lambda)d\lambda|

Using |A \pm B| \le |A| + |B|:

|E(a, b) - E(a, b^\prime)| \le |\int \underline {A}(a, \lambda)\underline {B}(b, \lambda)[1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b^\prime, \lambda)]\rho(\lambda)d\lambda| + |\int \underline {A}(a, \lambda)\underline {B}(b^\prime, \lambda)[1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b, \lambda)]\rho(\lambda)d\lambda|

Then using | \int f(x) dx | \le \int |f(x)| dx, (which, if you think of it, is just a variation of the above):

|E(a, b) - E(a, b^\prime)| \le \int |\underline {A}(a, \lambda)\underline {B}(b, \lambda)[1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b^\prime, \lambda)]\rho(\lambda) | d\lambda + \int |\underline {A}(a, \lambda)\underline {B}(b^\prime, \lambda)[1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b, \lambda)]\rho(\lambda)|d\lambda

And since |\underline {A}(a, \lambda)\underline {B}(b, \lambda)|=1, and both 1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b^\prime, \lambda) \ge 0 and \rho(\lambda) \ge 0, we get:

|E(a, b) - E(a, b^\prime)| \leq \int [1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b^\prime, \lambda)]\rho(\lambda)d\lambda + \int [1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b, \lambda)]\rho(\lambda)d\lambda

 

[gespex]

So you have:
E(a, b) - E(a, b^\prime) = \int \underline {A}(a, \lambda)\underline {B}(b, \lambda)[1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b^\prime, \lambda)]\rho(\lambda)d\lambda - \int \underline {A}(a, \lambda)\underline {B}(b^\prime, \lambda)[1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b, \lambda)]\rho(\lambda)d\lambda

Taking absolute value of both sides
|E(a, b) - E(a, b^\prime)| = |\int \underline {A}(a, \lambda)\underline {B}(b, \lambda)[1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b^\prime, \lambda)]\rho(\lambda)d\lambda - \int \underline {A}(a, \lambda)\underline {B}(b^\prime, \lambda)[1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b, \lambda)]\rho(\lambda)d\lambda|

Using |A \pm B| \le |A| + |B|:

|E(a, b) - E(a, b^\prime)| \le |\int \underline {A}(a, \lambda)\underline {B}(b, \lambda)[1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b^\prime, \lambda)]\rho(\lambda)d\lambda| + |\int \underline {A}(a, \lambda)\underline {B}(b^\prime, \lambda)[1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b, \lambda)]\rho(\lambda)d\lambda|

Then using | \int f(x) dx | \le \int |f(x)| dx, (which, if you think of it, is just a variation of the above):

|E(a, b) - E(a, b^\prime)| \le \int |\underline {A}(a, \lambda)\underline {B}(b, \lambda)[1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b^\prime, \lambda)]\rho(\lambda) | d\lambda + \int |\underline {A}(a, \lambda)\underline {B}(b^\prime, \lambda)[1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b, \lambda)]\rho(\lambda)|d\lambda

And since |\underline {A}(a, \lambda)\underline {B}(b, \lambda)|=1, and both 1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b^\prime, \lambda) \ge 0 and \rho(\lambda) \ge 0, we get:

|E(a, b) - E(a, b^\prime)| \leq \int [1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b^\prime, \lambda)]\rho(\lambda)d\lambda + \int [1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b, \lambda)]\rho(\lambda)d\lambda

Thanks a lot! The proof's easy enough to follow like that. But this statement seems to me to contradict Wikipedia (though it's probably just my ignorance):
|\underline {A}(a, \lambda)\underline {B}(b, \lambda)|=1

As Wikipedia says "Since the possible values of A and B are −1, 0 and +1". So couldn't it also be that |\underline {A}(a, \lambda)\underline {B}(b, \lambda)|=0?

It also says "where \underline {A} and \underline {B} are the average values of the outcomes", so not just -1, 0 or 1, but any value from -1 to 1, which would mean only that: 0 \le |\underline {A}(a, \lambda)\underline {B}(b, \lambda)|\le 1

Or am I missing something here?

 

[Delta Kilo]

Or am I missing something here?

Oh, I'm sorry, you are right of course, it should have been |\underline {A}(a, \lambda)\underline {B}(b, \lambda)| \le 1. It does not change anything though.

 

[gespex]

It does not change anything though.

Oops, yeah, good point. Thanks!