Possible Mistake in Bell's Book Speakable and Unspeakable in Quantum

Question

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1. What the question is asking – in plain language

In Chapter 3 of J. S. Bell’s Speakable and Unspeakable in Quantum Mechanics Bell derives his famous inequality for a local hidden‑variable (LHV) theory.
At one point he writes an expression that looks like

[ \Bigl|\,\int d\lambda\;\rho(\lambda)\, \bigl[ A({\bf a},\lambda)B({\bf b},\lambda)-A({\bf a},\lambda)B({\bf b’} ,\lambda)\bigr] \Bigr| \;\le\; 1+ \int d\lambda\;\rho(\lambda)\, B({\bf b},\lambda)B({\bf b’},\lambda) . \tag{★} ]

The student wonders why the absolute‑value sign is placed outside the integral (i.e. outside the sum over the hidden variable λ) and not inside, as they would expect a term like

[ \int d\lambda\;\rho(\lambda)\, \bigl|A({\bf a},\lambda)B({\bf b},\lambda)-A({\bf a},\lambda)B({\bf b’} ,\lambda)\bigr| ]

instead. In short: Is (★) a typo, or is the placement of the absolute value correct?

The answer is that the expression is correct. The absolute value belongs outside the integral because the inequality is obtained by applying the triangle inequality to the integrand after the factorisation that uses the fact that the hidden‑variable outcomes (A,B) are each ±1. Below we give a step‑by‑step derivation that makes the logic crystal clear.

2. Detailed derivation (every step shown)

2.1. Setting the stage

Hidden variable: a parameter λ that completely specifies the state of the pair of particles.
Distribution of hidden variables: a normalized probability density (\rho(\lambda)) with (\rho(\lambda)\ge 0) and (\int d\lambda \,\rho(\lambda)=1).
Measurement results: for any detector orientation a, the outcome of Alice’s measurement is a deterministic function
[ A({\bf a},\lambda)=\pm 1, ] and similarly for Bob, [ B({\bf b},\lambda)=\pm 1 . ]

The correlation predicted by any LHV theory for settings a, b is

[ E({\bf a},{\bf b})\equiv \int d\lambda\;\rho(\lambda)\;A({\bf a},\lambda)B({\bf b},\lambda) . \tag{1} ]

2.2. The quantity we want to bound

Bell considers the difference of two correlations that share the same Alice setting a but have different Bob settings b and b′:

[ \Delta\equiv E({\bf a},{\bf b})-E({\bf a},{\bf b’}). ]

Using (1),

[ \Delta =\int d\lambda\;\rho(\lambda) \Bigl[ A({\bf a},\lambda)B({\bf b},\lambda) -A({\bf a},\lambda)B({\bf b’},\lambda)\Bigr]. \tag{2} ]

Factor out the common factor (A({\bf a},\lambda)):

[ \Delta =\int d\lambda\;\rho(\lambda)\; A({\bf a},\lambda)\,\bigl[ B({\bf b},\lambda)-B({\bf b’},\lambda)\bigr]. \tag{3} ]

2.3. Using the fact that (A,B=\pm 1)

Because (A({\bf a},\lambda)^2 = 1), we can multiply the integrand by the factor

[ 1 = \frac{1}{2}\Bigl[1+ B({\bf b},\lambda)B({\bf b’},\lambda)\Bigr] +\frac{1}{2}\Bigl[1- B({\bf b},\lambda)B({\bf b’},\lambda)\Bigr], ]

but a simpler route—used by Bell—is to rewrite the bracket:

[ B({\bf b},\lambda)-B({\bf b’},\lambda) = B({\bf b},\lambda)\bigl[1- B({\bf b},\lambda)B({\bf b’},\lambda)\bigr]. \tag{4} ]

(Indeed, because (B({\bf b},\lambda)^2=1), multiplying the right‑hand side out gives the left‑hand side.)

Insert (4) into (3):

[ \Delta =\int d\lambda\;\rho(\lambda)\; A({\bf a},\lambda)\,B({\bf b},\lambda)\, \bigl[1- B({\bf b},\lambda)B({\bf b’},\lambda)\bigr]. \tag{5} ]

Now take the absolute value of (\Delta). Since the absolute value of an integral is bounded by the integral of the absolute value (the triangle inequality),

[ |\Delta| \le \int d\lambda\;\rho(\lambda)\; \bigl| A({\bf a},\lambda)B({\bf b},\lambda) \bigl[1- B({\bf b},\lambda)B({\bf b’},\lambda)\bigr]\bigr|. \tag{6} ]

But (

A({\bf a},\lambda)B({\bf b},\lambda)

= 1) (product of two ±1 numbers), so the absolute value drops out of that factor:

[ |\Delta| \le \int d\lambda\;\rho(\lambda)\; \bigl|1- B({\bf b},\lambda)B({\bf b’},\lambda)\bigr|. \tag{7} ]

Because the quantity inside the absolute value is either (0) or (2) (again a difference of two ±1 numbers), its absolute value equals the expression itself:

[ |1- B({\bf b},\lambda)B({\bf b’},\lambda)| = 1- B({\bf b},\lambda)B({\bf b’},\lambda). \tag{8} ]

Thus

[ |\Delta| \le \int d\lambda\;\rho(\lambda)\; \bigl[1- B({\bf b},\lambda)B({\bf b’},\lambda)\bigr]. \tag{9} ]

Finally split the integral into two pieces:

[ |\Delta| \le \underbrace{\int d\lambda\;\rho(\lambda)}{=1} \;-\; \underbrace{\int d\lambda\;\rho(\lambda)\; B({\bf b},\lambda)B({\bf b’},\lambda)}{=E({\bf b},{\bf b’})}. ]

Re‑arranging gives the Bell inequality in the form that appears in the book:

[ \boxed{ \; |E({\bf a},{\bf b})-E({\bf a},{\bf b’})| \;\le\; 1 + E({\bf b},{\bf b’}) \; } . \tag{10} ]

Equation (10) is exactly the highlighted expression (★).

Key point: the absolute value surrounds the whole left‑hand side, i.e. the difference of the two integrated correlation functions. It is not inside the integral. The inequality follows from the triangle inequality after the factorisation that uses (A^2 = B^2 = 1); moving the absolute value inside the integral would be a different (and generally weaker) bound and is not what Bell wrote.

3. Final answer

The expression in Bell’s book is correct; the absolute‑value bars are meant to enclose the entire difference of the two correlation integrals, not the integrand itself. The derivation uses:

Factorisation of the difference of the two terms,
The fact that each hidden‑variable outcome is ±1,
The triangle inequality applied after the factorisation.

Hence no typo is present in the book.

4. Common mistakes when reproducing this derivation

Mistake	Why it’s wrong	How to avoid it
Placing the absolute value inside the integral (i.e. writing (\int \rho	\dots	)).	The triangle inequality is applied to the integral of the product, not to each point‑wise term. Putting the absolute value inside would give a weaker (or sometimes meaningless) bound.	Keep the absolute value outside until after you have used the identity (B({\bf b})-B({\bf b’}) = B({\bf b})[1-B({\bf b})B({\bf b’})]).
Forgetting that (A) and (B) are ±1 and therefore (	A B	= 1).	Without this you cannot drop the absolute value of the prefactor and you’ll end up with an extra (unnecessary) factor.	Explicitly note (A({\bf a},\lambda)^2 = B({\bf b},\lambda)^2 = 1) before applying the triangle inequality.
**Using (	1-xy	= 1-xy) without justification**.	This holds only because (x,y = \pm1) ⇒ (xy = \pm1) ⇒ the bracket is either 0 or 2, both non‑negative.	State the possible values of (xy) and verify the sign before dropping the absolute value.
Mixing up the order of the settings (e.g., writing (E({\bf b},{\bf a})) instead of (E({\bf a},{\bf b}))).	Correlation functions are symmetric for the singlet state, but the algebraic steps rely on a consistent labelling of the two parties.	Keep a clear table of which variable belongs to Alice and which to Bob throughout the derivation.
Skipping the step where the integral of (\rho(\lambda)) is set to 1.	The normalization (\int\rho =1) is essential to obtain the constant “1” on the right‑hand side of the inequality.	After expanding the integral, explicitly write (\int d\lambda\,\rho(\lambda)=1).

By being careful about the placement of the absolute value, the ±1 property of the hidden‑variable outcomes, and the proper use of the triangle inequality, the derivation proceeds smoothly and the result matches Bell’s original formula.

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