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I am trying to understand Griffiths explanation of Bell's theorem in chapter 12 of 2nd edition Intro to QM (or the afterword section A.2 in the 1st edition).

He starts with given two detectors measuring the spin component along axes denoted by the vector $\vec{a}$ and right detector along $\vec{b}$, define a function that is the average of the product of the spin of both particles called $P(\vec{a}, \vec{b})$

He then defines two more functions $A(\vec{a},\lambda)=\pm 1$ and $B(\vec{b},\lambda)=\pm 1$ where these two functions express the component of a particle's angular momentum as a function of a hidden variable $\lambda$ and along the above mentioned axes.

For entangled particles, if the detectors are aligned then the spins are anti-correlated such that

$$ A(\vec{a},\lambda)=-B(\vec{a},\lambda) $$

Next he defines the average of the product of the spins using the hidden variable with a probability distribution $\rho(\lambda)$ as follows

$$ \begin{align*} P(\vec{a}, \vec{b}) &= \int \rho(\lambda)\, A(\vec{a},\lambda)\, B(\vec{b},\lambda) \, d\lambda\\ &= -\int \rho(\lambda) \, A(\vec{a},\lambda) \, A(\vec{b},\lambda) \, d\lambda \end{align*} $$

Now given any other unit vector $\vec{c}$, then

$$ \begin{align*} P(\vec{a}, \vec{b}) - P(\vec{a}, \vec{c}) &= -\int \rho(\lambda) \Big[ \, A(\vec{a},\lambda) \, A(\vec{b},\lambda) - A(\vec{a},\lambda) \, A(\vec{c},\lambda) \Big] \, d\lambda\\ &= -\int \rho(\lambda) \Big[ 1 - A(\vec{b},\lambda) \, A(\vec{c},\lambda) \Big] \, A(\vec{b},\lambda) \, A(\vec{c},\lambda) \, d\lambda \hspace{5mm}\textbf{ since } [A(b,\lambda)]^2=1 \end{align*} $$

The next part is where he loses me. He then states that since $-1 \leq A(\vec{b},\lambda) \, A(\vec{c},\lambda)\leq +1$ and that since $\rho(\lambda) \Big[ 1 - A(\vec{b},\lambda) \, A(\vec{c},\lambda) \Big] \geq 0$ (which I get both of these points) he then jumps to this (which I don't get):

$$ |P(\vec{a}, \vec{b}) - P(\vec{a}, \vec{c})| \leq \int \rho(\lambda) \Big[ 1 - A(\vec{b},\lambda) \, A(\vec{c},\lambda) \Big] d\lambda $$

So what steps is he skipping here? The only thing I can think of is that he is effectively taking the absolute value the right side? And he is doing it inside the integral by effectively discarding the $A(\vec{b},\lambda) \, A(\vec{c},\lambda)$ term (since is either just $\pm 1$$ and discarding the negative sign in front of the integral.

Is that it or am I off base?

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closed as off-topic by Martin, CuriousOne, ACuriousMind, Gert, Ryan Unger Mar 18 '16 at 0:53

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    $\begingroup$ It's just the triangle inequality: You take the absolute value on the LEFT side, which is equal to the absolute value on the right side and then $|\int f |\leq \int |f|$. Since the term $1-A(b,\lambda)A(c,\lambda)$ is positive in all cases as is $\rho$, you are done. $\endgroup$ – Martin Mar 15 '16 at 11:19
  • $\begingroup$ that said, it is a really simple math-question and not really a conceptual question, so I vote to close. $\endgroup$ – Martin Mar 15 '16 at 11:19
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First off, you do have a typo.

$$ \begin{align*} P(\vec{a}, \vec{b}) &= -\int \rho(\lambda) \, A(\vec{a},\lambda) \, A(\vec{b},\lambda) \, d\lambda \end{align*} $$

Therefore

$$ \begin{align*} P(\vec{a}, \vec{b}) - P(\vec{a}, \vec{c}) &= -\int \rho(\lambda) \Big[ \, A(\vec{a},\lambda) \, A(\vec{b},\lambda) - A(\vec{a},\lambda) \, A(\vec{c},\lambda) \Big] \, d\lambda\\ &= -\int \rho(\lambda) \Big[ 1 - A(\vec{b},\lambda) \, A(\vec{c},\lambda) \Big] \, A(\vec{a},\lambda) \, A(\vec{b},\lambda) \, d\lambda \end{align*} $$ Instead of $$ -\int \rho(\lambda) \Big[ 1 - A(\vec{b},\lambda) \, A(\vec{c},\lambda) \Big] \, A(\vec{b},\lambda) \, A(\vec{c},\lambda) \, d\lambda, $$

which equals zero since since $[A(\vec b,\lambda)]^2=1=[A(\vec c,\lambda)]^2$.

As for your question recall the triangle inequality:

$$|a+b|\leq |a|+|b|$$

Therefore

$$\left|\sum a_n\right|\leq\sum\left|a_n\right|$$ And finally $$\left|\int a_n\right|\leq\int\left|a_n\right|.$$ So if you take the absolute value of both sides of the correct equality (the one in Bell's paper, not the one you have with no $\vec a$) and then push that absolute value inside you get an inequality and then use $|ab|=|a|\,|b|$ and that $|\pm 1|=1$ and you are good.

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