CHSH and the triangle inequality

1. May 4, 2012

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?

Gespex

Last edited: May 4, 2012
2. May 5, 2012

Delta Kilo

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:

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

3. May 5, 2012

gespex

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?

4. May 5, 2012

Delta Kilo

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.

5. May 5, 2012

gespex

Oops, yeah, good point. Thanks!