Problem with CHSH's version of Bell's inequalities

[facenian]

Though I don't have access to the original 1969 paper, I think the way the inequality is ussually derived is not correct since the argument goes as follows:
$$-2≤ A(a,\lambda)[B(b,\lambda)-B(b',\lambda)]+A(a',\lambda)[B(b,\lambda)+B(b',\lambda)]≤2$$
the first mistake in this inequality is the use of same value of [itex]\lambda[/itex] in all the symbols, there are other objections too I would like to comment if somebody want to discuss this.I do not say that the final results aren't right I'm only objecting to the way they are derived

 

[stevendaryl]

On the contrary, I think that [itex]A(a,\lambda)[/itex] is a self-evidently correct argument.

 

[facenian]

On the contrary, I think that [itex]A(a,\lambda)[/itex] is a self-evidently correct argument.

I believe that you cannot use the four numbers in the equation with the same [itex]\lambda[/itex] because you need at least two runs of the experiment to get them

 

[rubi]

Well, this is the "realism" assumption of the theorem. That's why people say that Bell/CHSH rules out the class of "local realistic" theories. It's not problematic to assume this. It just means that the class of theories that are ruled out by the inequality violation obeys some restrictions. The name "realism" is a bit of a misnomer, since the assumption is just a restriction on the admissible mathematical models. It has nothing to do with philosophical realism. In modern literature, you will rather find names like "non-contextuality" or "counterfactual definiteness" or others.

 

[facenian]

Well, this is the "realism" assumption of the theorem. That's why people say that Bell/CHSH rules out the class of "local realistic" theories. It's not problematic to assume this. It just means that the class of theories that are ruled out by the inequality violation obeys some restrictions. The name "realism" is a bit of a misnomer, since the assumption is just a restriction on the admissible mathematical models. It has nothing to do with philosophical realism. In modern literature, you will rather find names like "non-contextuality" or "counterfactual definiteness" or others.

My objection has nothing to do with the implications of the theorem I only object the technical derivation of the CHSH version of the inequality

 

[rubi]

My objection has nothing to do with the implications of the theorem I only object the technical derivation of the CHSH version of the inequality

Well, it is technically unproblematic to assume that there exist functions ##A(\vec a, \lambda)##.

 

[stevendaryl]

Though I don't have access to the original 1969 paper, I think the way the inequality is ussually derived is not correct since the argument goes as follows:
$$-2≤ A(a,\lambda)[B(b,\lambda)-B(b',\lambda)]+A(a',\lambda)[B(b,\lambda)+B(b',\lambda)]≤2$$
the first mistake in this inequality is the use of same value of [itex]\lambda[/itex] in all the symbols, there are other objections too I would like to comment if somebody want to discuss this.I do not say that the final results aren't right I'm only objecting to the way they are derived

No, that is not a mistake.

Bell starts with the assumption of a general model for how measurement results in an EPR experiment might come about in a local way. He assumes that:

1. When a twin pair is produced, there is an accompanying state variable, [itex]\lambda[/itex], associated with the pair.
2. He assumes that there is a probability distribution, [itex]P(\lambda)[/itex] on the various possible values of this state variable.
3. He assumes that when Alice performs a measurement of spin on her particle, she deterministically gets the result [itex]A(a, \lambda)[/itex], which depends on the setting [itex]a[/itex] for her measurement device and the value [itex]\lambda[/itex] for that particle. [itex]A(a, \lambda)[/itex] can be either [itex]+1[/itex] or [itex]-1[/itex].
   
4. Similarly, when Bob performs a measurement, he gets the result [itex]B(b, \lambda)[/itex] where [itex]b[/itex] is the setting of his device. He also always get [itex]\pm 1[/itex].

That's basically the model, with several unknowns:

• Over what values does [itex]\lambda[/itex] range?
• What are the functions [itex]A(a, \lambda)[/itex] and [itex]B(b, \lambda)[/itex]
• What is the function [itex]P(\lambda)[/itex]

So with so many unknowns, what can you prove about the model? Well, he shows that you can prove something, and this something is violated by the actual EPR experiments. So he introduces a very general model only to show that this model cannot be correct, no matter how you fill in the details.

Exactly why he made the assumptions 1-4 is possibly worth thinking about. For example, why did he assume that the result was deterministic, rather than probabilistic? That's not a necessary assumption. You can redo Bell's analysis, where instead of assuming that the results are deterministic, they are probabilistic--that is, there are functions [itex]P_A(a, \lambda, A)[/itex] and [itex]P_B(b, \lambda, B)[/itex] giving probability distributions on possible outcomes [itex]A[/itex] and [itex]B[/itex]. That doesn't make any difference to Bell's conclusions--it's impossible to reproduce the perfect correlations (or anti-correlations) of EPR without deterministic functions. Also, you could generalize Bell's model by including other effects, such as the details of the measuring devices, or the environment. But again, those details can't possibly reproduce the perfect correlations.

Anyway, if the outcomes are produced according to Bell's model, then we can reason as follows: The expected value of the product of [itex]A(a, \lambda) B(b, \lambda)[/itex] will be given by:

[itex]E(a,b) = \int d\lambda P(\lambda) A(a, \lambda) B(b, \lambda)[/itex]

The prediction of quantum mechanics (which is borne out by experiment) for the spin-1/2 anticorrelated EPR experiment is:

[itex]E(a,b) = -cos(\theta)[/itex]

where [itex]\theta[/itex] is the angle between the detector orientations used by Alice and Bob (parameters [itex]a[/itex] and [itex]b[/itex]). Bell proves that no combination of functions [itex]P(\lambda)[/itex], [itex]A(a, \lambda)[/itex] and [itex]B(b, \lambda)[/itex] can produce this result for [itex]E(a,b)[/itex].

So your complaint about the lambdas being the same in the inequality is out of place. The expression for [itex]E(a,b)[/itex] involves a weighted average over cases where [itex]A[/itex] and [itex]B[/itex] have the same lambda.

 

[stevendaryl]

The argument for Bell's inequality seems pretty straight-forward to me. Well, it's straight-forward in hindsight, anyway.

Following the Wikipedia article https://en.wikipedia.org/wiki/Bell's_theorem,

Let
[itex]\mathcal{A} = A(a, \lambda)[/itex],
[itex]\mathcal{A}' = A(a', \lambda)[/itex],
[itex]\mathcal{B} = B(b, \lambda)[/itex],
[itex]\mathcal{B}' = B(b', \lambda)[/itex].
Now, define the quantity:

[itex]\mathcal{C} = \mathcal{A} \mathcal{B} +\mathcal{A} \mathcal{B}' +\mathcal{A}' \mathcal{B} -\mathcal{A}' \mathcal{B}[/itex]

We can rearrange this to:

[itex]\mathcal{C} = \mathcal{A}(\mathcal{B} + \mathcal{B}') + \mathcal{A'}(\mathcal{B} - \mathcal{B}')[/itex]

Since the function [itex]B[/itex] returns [itex]\pm 1[/itex], then either [itex]\mathcal{B}' = \mathcal{B}[/itex] or [itex]\mathcal{B}' = - \mathcal{B}[/itex].

If [itex]\mathcal{B}' = \mathcal{B}[/itex], then [itex]\mathcal{B} + \mathcal{B}' = \pm 2[/itex] and [itex]\mathcal{B} - \mathcal{B}' = 0[/itex]. So [itex]\mathcal{C} = \mathcal{A}(\pm 2) = \pm 2[/itex]. (Because [itex]\mathcal{A} = \pm 1[/itex].)

If [itex]\mathcal{B}' = -\mathcal{B}[/itex],then [itex]\mathcal{B} + \mathcal{B}' = 0[/itex] and [itex]\mathcal{B} - \mathcal{B}' = \pm 2[/itex]. So [itex]\mathcal{C} = \mathcal{A}'(\pm 2) = \pm 2[/itex]. (Because [itex]\mathcal{A}' = \pm 1[/itex].)

So for every possible value of [itex]a, b, a', b', \lambda[/itex], it's the case that [itex]\mathcal{C} = \pm 2[/itex]. Then when you average [itex]\mathcal{C}[/itex] over all possible values of [itex]\lambda[/itex], you can't possibly get a result that is greater than 2 or less than -2. So we have:

[itex]-2 \leq \int P(\lambda) \mathcal{C}(a,b,a',b',\lambda) \leq 2[/itex]

Expanding the definition of [itex]\mathcal{C}[/itex], we get:

[itex]-2 \leq \int P(\lambda) [ A(a,\lambda) B(b, \lambda) + A(a,\lambda) B(b', \lambda) + A(a', \lambda) B(b, \lambda) - A(a',\lambda) B(b',\lambda) ] \leq 2[/itex]

In terms of [itex]E(a,b)[/itex], this means:

[itex]-2 \leq (E(a,b) + E(a, b') + E(a', b) - E(a',b')) \leq 2[/itex]

 

[Nugatory]

Though I don't have access to the original 1969 paper...

This one? http://users.unimi.it/aqm/wp-content/uploads/CHSH.pdf

 

[DrChinese]

The first mistake in this inequality is the use of same value of [itex]\lambda[/itex] in all the symbols, there are other objections too...

[itex]\lambda[/itex] is the hypothetical set of "hidden" variables (or functions). What is objectionable about that? There is no requirement that *all* of those variables are inputs to all of the mechanisms in place. Just that it includes enough to specify the outcomes.

Also: what other objections do you have?

 

[facenian]

I want to thank you "stevendaryl" for your willingness to discuss the subject. This is about logic and not about physics and applying the same logic I conclude that I must be wrong because is not possible that nobody have noticed this mistake since 1969. That being said I will continue with my argument.

The argument for Bell's inequality seems pretty straight-forward to me. Well, it's straight-forward in hindsight, anyway.

Let me restate that I do not object the original 1964 Bell's derivation. My problem is with the CHSH experimental version of the inequality

Following the Wikipedia article https://en.wikipedia.org/wiki/Bell's_theorem,
Let
[itex]\mathcal{A} = A(a, \lambda)[/itex],
[itex]\mathcal{A}' = A(a', \lambda)[/itex],
[itex]\mathcal{B} = B(b, \lambda)[/itex],
[itex]\mathcal{B}' = B(b', \lambda)[/itex]

Here is where the problem stems from, because you can't use the same [itex]\lambda[/itex] for all four numbers because, as I explained at the beginning of the post, you need at least two runs of the experiment to obtain these four numbers so the following step

Now, define the quantity:

[itex]\mathcal{C} = \mathcal{A} \mathcal{B} +\mathcal{A} \mathcal{B}' +\mathcal{A}' \mathcal{B} -\mathcal{A}' \mathcal{B}[/itex]

We can rearrange this to:

[itex]\mathcal{C} = \mathcal{A}(\mathcal{B} + \mathcal{B}') + \mathcal{A'}(\mathcal{B} - \mathcal{B}')[/itex]

should rather be written down

[itex]\mathcal{C} = \mathcal{A}(a,\lambda) \mathcal{B}(b,\lambda) +\mathcal{A}(a,\lambda) \mathcal{B}'(b',\lambda') +\mathcal{A}'(a',\lambda') \mathcal{B}(b',\lambda') -\mathcal{A}'(a',\lambda') \mathcal{B}(a,\lambda)[/itex]

We can rearrange this to:

[itex]\mathcal{C} = \mathcal{A}(a,\lambda)(\mathcal{B}(b,\lambda) + \mathcal{B}'(b',\lambda')) + \mathcal{A'}(a',\lambda')(\mathcal{B}(b,\lambda) - \mathcal{B}'(b',\lambda'))[/itex]
all this under the assumption that one generates the entangled pair with the angles a,b and hidden variable [itex]\lambda[\itex] and then a second entangled pair with the angles a',b' and hidden variable [itex]\lambda'[\itex]
¿do you agree with me?

 

[rubi]

As I have explained earlier, you have stumbled upon the "realism" assumption of the theorem. It is well-known that this assumption is needed for the proof of the inequality. The assumption says that even those observables that can't be measured in the first run have well-defined values. If you don't assume this, you can't prove the inequality.

 

[facenian]

This one? http://users.unimi.it/aqm/wp-content/uploads/CHSH.pdf

Yes! thank you very much

 

[stevendaryl]

Here is where the problem stems from, because you can't use the same [itex]\lambda[/itex] for all four numbers because, as I explained at the beginning of the post, you need at least two runs of the experiment to obtain these four numbers so the following step

The proof that [itex]|E(a,b) + E(a,b') + E(a',b) - E(a',b')| \leq 2[/itex] is a fact about pure mathematics, under the assumption that

• [itex]E(a,b) = \int d\lambda P(\lambda) A(a,\lambda) B(b,\lambda)[/itex]
• [itex]P(\lambda)[/itex] is a probability distribution.
• The functions [itex]A[/itex] and [itex]B[/itex] return [itex]\pm 1[/itex]

So the logic is:

1. The model assumes that the expectation value [itex]E(a,b)[/itex] can be computed via a particular integral.
2. You prove a fact about [itex]E(a,b)[/itex] under that assumption.
3. You experimentally determine [itex]E(a,b)[/itex] to see if that fact holds. If not, then your assumption is false.

So you have an experimentally determined [itex]E(a,b)[/itex]. This doesn't involve lambda at all. You have a theoretically derived [itex]E(a,b)[/itex]. This doesn't involve experiment at all. You prove a fact about the theoretically derived [itex]E(a,b)[/itex]. That doesn't involve experiment; it's just mathematics. The way that theory and experiment come together is by comparing the theoretically derived fact about [itex]E(a,b)[/itex] with the experimentally derived [itex]E(a,b)[/itex]. If they contradict each other, then your theory is wrong.

 

[stevendaryl]

[itex]\mathcal{C} = \mathcal{A}(a,\lambda) \mathcal{B}(b,\lambda) +\mathcal{A}(a,\lambda) \mathcal{B}'(b',\lambda') +\mathcal{A}'(a',\lambda') \mathcal{B}(b',\lambda') -\mathcal{A}'(a',\lambda') \mathcal{B}(a,\lambda)[/itex]

Um. That's not how [itex]\mathcal{C}[/itex] is defined. It's defined with the same value of [itex]\lambda[/itex] appearing in each expression. You can't say that's wrong, because it's just a definition. It's just a mathematical expression formed from the assumed functions [itex]A(a,\lambda)[/itex] and [itex]B(b, \lambda)[/itex]. But given the assumption that [itex]E(a,b) = \int d\lambda P(\lambda) A(a,\lambda) B(b,\lambda)[/itex], it follows that [itex]E(a,b) + E(a,b') + E(a', b) - E(a', b') = \int d\lambda \mathcal{C}(a,a',b,b',\lambda)[/itex]

 

[facenian]

[itex]\lambda[/itex] is the hypothetical set of "hidden" variables (or functions). What is objectionable about that? There is no requirement that *all* of those variables are inputs to all of the mechanisms in place. Just that it includes enough to specify the outcomes.

that is not my objection, maybe you can read my last response to "stevendaryl"

Also: what other objections do you have?

it wouldn't make sense to discuss my other objections if we don't agree on first one because if the first does not hold neither does the rest

 

[facenian]

You can't say that's wrong, because it's just a definition.

The problem is that this is not yust a definition, these symbols represent outcomes of a realized experiment. I insist that I have no problem with the theoretical derivation made by Bell.

 

[stevendaryl]

The problem is that this is not yust a definition, these symbols represent outcomes of a realized experiment. I insist that I have no problem with the theoretical derivation made by Bell.

There is no lambda in a real experiment. There is only an experimentally determined [itex]E(a,b)[/itex]. Then you use that experimentally determined function to compute [itex]C(a,b,a',b') \equiv E(a,b) + E(a,b') + E(a',b) + E(a', b')[/itex]. Again, there is no lambda involved. So talking about different lambdas for different measurements seems irrelevant.

The theoretical model predicts that [itex]|C(a,b,a',b')| \leq 2[/itex], but the experimentally derived [itex]C[/itex] does not satisfy this inequality.

 

[DrChinese]

Here is where the problem stems from, because you can't use the same [itex]\lambda[/itex] for all four numbers because, as I explained at the beginning of the post, you need at least two runs of the experiment to obtain these four numbers ...

As rubi points out, this is the realism assumption at play. You don't have to accept it. But it makes little sense to reject it unless you have a better formulation. This is in fact considered an easy-to-swallow step.

 

[PeterDonis]

these symbols represent outcomes of a realized experiment

Not ##\lambda##. The symbol ##\lambda## represents a set of "hidden variables" which are never observed or measured.

 

[facenian]

Not ##\lambda##. The symbol ##\lambda## represents a set of "hidden variables" which are never observed or measured.

Yes, that's right and has nothing to do with my objection

 

[facenian]

As rubi points out, this is the realism assumption at play. You don't have to accept it. But it makes little sense to reject it unless you have a better formulation. This is in fact considered an easy-to-swallow step.

My objection is far less profound as compared to rejecting Bell's theorem. I only pointed out a logical mistake made not by Bell but by others presenting, as you put it, an easy-to-swallow step

 

[DrChinese]

My objection is far less profound as compared to rejecting Bell's theorem. I only pointed out a logical mistake made not by Bell but by others presenting, as you put it, an easy-to-swallow step

It's an assumption. There is no logic involved. You accept it, or you don't. No one can assist you if you reject it, because the entire purpose is to test the assumption.

 

[stevendaryl]

Yes, that's rihgt and has nothing to do with my objection

Well, your objection was:

you cannot use the four numbers in the equation with the same [itex]\lambda[/itex] because you need at least two runs of the experiment to get them.

So you are associating different values of lambda with different runs of an experiment. The runs don't involve lambda at all.

I really don't understand your objection, at all. Do you agree that if

[itex]E(a,b) = \int d\lambda P(\lambda) A(a,\lambda) B(b, \lambda)[/itex]

then

[itex]E(a,b) + E(a, b') + E(a', b) - E(a',b') = \int d\lambda P(\lambda) (A(a,\lambda) B(b, \lambda) + A(a,\lambda) B(b', \lambda) + A(a',\lambda) B(b, \lambda) - A(a',\lambda) B(b', \lambda))[/itex]

That's just a fact.

 

[PeterDonis]

that's right and has nothing to do with my objection

Sure it does. Your objection was:

the use of same value of ##\lambda## in all the symbols

And that objection is invalid because ##\lambda## doesn't represent a value, it represents the set of all possible values for the hidden variables, which are then integrated over to get the probabilities for the possible measurement results. ##\lambda## never takes on any particular value because the hidden variables are never measured.

 

[zonde]

all this under the assumption that one generates the entangled pair with the angles a,b and hidden variable [itex]\lambda[/itex] and then a second entangled pair with the angles a',b' and hidden variable [itex]\lambda'[/itex]

This isn't right. One does not generate entangled pair with the angles a,b. Angles "a" and "b" are supposed to be independent measurement settings that have nothing to do with ##\lambda##. Function ##A(\vec a,\lambda)## is defined for any possible combination of values of ##\vec a## and ##\lambda##. So you can take fixed value of ##\lambda## (say ##\lambda_1##) and calculate function ##A(\vec a,\lambda_1)## for different values of ##\vec a##, say ##A(\vec a_1,\lambda_1)## and ##A(\vec a_2,\lambda_1)##

 

[jk22]

If this point using the same parameter disturbs you you can make the independence assumption namely that AB and AB' are independent events which correspond in your question to choose two different lambda for each pair.

Then you can easily compute $$p (Chsh=-4)=p (AB=-1)p (AB'=1)p (A'B=-1)p (A'B'=-1)=(1-\theta_{ab})(\theta_{ab'})(1-\theta_{a'b})(1-\theta_{a'b'})$$

Maximizing the probability of -4 gives you the Bell angles configuration.

You can then compute a bit more complicatedly the other values of Chsh namely p (Chsh=-2) p (Chsh=0) and so on.

The average of course gives <Chsh>=-2 but all possible values are here computed not making the "same lambda" assumption.

In this way of computing you obtain Bell's theorem as $$ |<Chsh>|=2 <=<|Chsh|>=\frac {71}{32}\approx 2.22 <<Chsh>_{qm}=2\sqrt {2}\approx 2.82$$

 

[facenian]

If this point using the same parameter disturbs you you can make the independence assumption namely that AB and AB' are independent events which correspond in your question to choose two different lambda for each pair.

I don't understand the details yet but seems very interesting, thank you

 

[DrChinese]

I don't understand the details yet but seems very interesting, thank you

Looked at another way: the hidden variables can be considered as containing the information necessary to dictate the results at ALL angle settings. Perhaps certain angle settings only require a subset of the total set of HVs.

But one issue you run into quickly: you must also be able to reproduce the so-called "perfect correlations" which occur. So E(a, a)=E(a',a')=... This is a powerful constraint.

 

[facenian]

But one issue you run into quickly: you must also be able to reproduce the so-called "perfect correlations" which occur. So E(a, a)=E(a',a')=... This is a powerful constraint.

Yes, in fact this constraint was explicitly used by Bell in his original paper(of course I know you know that)

 

[Nicky665]

So what are now supposed to be the limits of bell inequation +-2 or +-2.82

 

[facenian]

So what are now supposed to be the limits of bell inequation +-2 or +-2.82

2.82 is limit for the quantum mechanical calculation and this value was not at discussion

 

[stevendaryl]

Here's another response to the original poster:

If, as he suggests, we consider different values of [itex]\lambda[/itex] for the different terms, we would have the expression:

[itex]\mathcal{C}(a,b,a',b',\lambda, \lambda', \lambda'', \lambda''') = A(a,\lambda) B(b,\lambda) + A(a', \lambda') B(b, \lambda') + A(a, \lambda'') B(b', \lambda'') - A(a', \lambda''') B(b', \lambda''')[/itex]

But we don't actually observe [itex]\lambda[/itex], (by definition, it is a "hidden variable"). So what we can observe is the average of [itex]\mathcal{C}[/itex] over all values of [itex]\lambda, \lambda', \lambda'', \lambda'''[/itex]:

[itex]\mathcal{C}_{av}(a,b,a',b') = \int d\lambda d\lambda' d\lambda'' d\lambda''' P(\lambda) P(\lambda') P(\lambda'') P(\lambda''') \mathcal{C}(a,b,a',b',\lambda, \lambda', \lambda'', \lambda''')[/itex]

It's elementary to prove that this is equal to:

[itex]\mathcal{C}_{av}(a,b,a',b') = \int d\lambda P(\lambda) \mathcal{C}(a,b,a',b',\lambda, \lambda, \lambda, \lambda)[/itex]

with all four lambdas the same.

 

[facenian]

Here's another response to the original poster:It's elementary to prove that this is equal to:

[itex]\mathcal{C}_{av}(a,b,a',b') = \int d\lambda P(\lambda) \mathcal{C}(a,b,a',b',\lambda, \lambda, \lambda, \lambda)[/itex]

with all four lambdas the same.

Sorry, but I don't know that elemtary proof

 

[stevendaryl]

Sorry, but I don't know that elemtary proof

Well, we have the definition:

[itex]\mathcal{C}(a,b,a',b',\lambda, \lambda', \lambda'', \lambda''')[/itex]
[itex]= A(a,\lambda) B(b,\lambda) + A(a, \lambda') B(b', \lambda') + A(a', \lambda'') B(b, \lambda'') - A(a', \lambda''') B(b', \lambda''')[/itex]

Now, integrate over [itex]\lambda'''[/itex]:

[itex]\int d\lambda''' P(\lambda''') \mathcal{C}(a,b,a',b',\lambda,\lambda', \lambda'', \lambda''')[/itex]
[itex]= \int d\lambda''' P(\lambda''') A(a,\lambda) B(b,\lambda) + \int d\lambda''' P(\lambda''') A(a, \lambda') B(b', \lambda') + \int d\lambda''' P(\lambda''') A(a', \lambda'') B(b, \lambda'') - \int d\lambda''' P(\lambda''') A(a', \lambda''') B(b', \lambda''')[/itex]
[itex]= A(a,\lambda) B(b,\lambda) + A(a, \lambda') B(b', \lambda') + A(a', \lambda'') B(b, \lambda'') - \int d\lambda''' P(\lambda''') A(a', \lambda''') B(b', \lambda''')[/itex]

This just uses the fact that since [itex]A(a, \lambda) B(b, \lambda)[/itex] doesn't depend on [itex]\lambda'''[/itex], it can be pulled outside the integral: [itex]\int d\lambda''' P(\lambda''') A(a,\lambda) B(b,\lambda) = A(a,\lambda) B(b,\lambda) \int d\lambda''' P(\lambda''')[/itex]. And [itex]\int d\lambda''' P(\lambda''') = 1[/itex]. Similarly for all the other terms that do not involve [itex]\lambda'''[/itex].

Now, integrate over [itex]\lambda'', \lambda'[/itex] and [itex]\lambda[/itex]. This gives:
[itex]\int d\lambda P(\lambda) A(a,\lambda) B(b,\lambda) + \int d\lambda' P(\lambda') A(a, \lambda') B(b', \lambda') + \int d\lambda'' P(\lambda'') A(a', \lambda'') B(b, \lambda'') - \int d\lambda''' P(\lambda''') A(a', \lambda''') B(b', \lambda''')[/itex]

But [itex]\lambda, \lambda', \lambda'', \lambda'''[/itex] are just dummy integration variables now. You can rename them all to [itex]\lambda[/itex], to get:
[itex]\int d\lambda P(\lambda) A(a,\lambda) B(b,\lambda) + \int d\lambda A(a, \lambda) B(b', \lambda) + \int d\lambda P(\lambda) A(a', \lambda) B(b, \lambda) - \int d\lambda P(\lambda) A(a', \lambda) B(b', \lambda)[/itex]

By linearity of integration, this is the same as:
[itex]\int d\lambda P(\lambda) [A(a,\lambda) B(b,\lambda) + A(a, \lambda) B(b', \lambda) + A(a', \lambda) B(b, \lambda) - A(a', \lambda) B(b', \lambda)][/itex]