Bell’s inequality

Bell’s inequality, developed by John Bell in 1964, is a non-equal relation (between two expressions) that is based on the ERP paradox. It provides a way to determine whether quantum theory or the EPR paradox is correct. Bell suggested a unique way of measuring the spins of two spin-\frac{1}{2} particles, which are generated from the decay of a spin-0 particle at rest: the two particles are to be passed through two Stern-Gerlach devices, each oriented along one of three non-orthogonal coplanar axes, which are specified by the unit vectors \boldsymbol{a}, \boldsymbol{b} and \boldsymbol{c}.

The average values of the product of the spins in units of \frac{\hbar}{2} (denoted by P(\boldsymbol{a},\boldsymbol{b}), P(\boldsymbol{b},\boldsymbol{c}), and P(\boldsymbol{a},\boldsymbol{c})) are then calculated. As derived in an earlier article (see eq227), quantum mechanics expresses the average values as:

P(\boldsymbol{a},\boldsymbol{b})=P(\boldsymbol{b},\boldsymbol{c})=P(\boldsymbol{a},\boldsymbol{c})=-cos\theta\; \; \; \; \; \; \; \; 228

For the ERP paradox, let’s equate a spin-up measurement to +1 and a spin-down measurement to -1. The average value of the product of the measured spins, for example in the \boldsymbol{a} and \boldsymbol{b} directions, is:

P(\boldsymbol{a},\boldsymbol{b})=\int_{\lambda_i}^{\lambda_f}\rho(\lambda)A(\boldsymbol{a},\lambda)B(\boldsymbol{b},\lambda)d\lambda\; \; \; \; \; \; \; \; 229

where

i) \lambda is a hidden variable.

ii) \rho(\lambda) is the probability density that is a function of \lambda, with \int_{\lambda_i}^{\lambda_f}\rho(\lambda)d\lambda=1 and \rho(\lambda)\geq 0.

iii) A(\boldsymbol{a},\lambda) is a function of the axis of measurement and \lambda. It is associated with the measurement made by the first Stern-Gerlach device, and has output values of \pm1.

iv) B(\boldsymbol{b},\lambda) is a function of the axis of measurement and \lambda. It is associated with the measurement made by the second Stern-Gerlach device, and has output values of \pm1.

From experiments, we know that

A(\boldsymbol{a},\lambda)=-B(\boldsymbol{b},\lambda)\; \; \; \; \; \; \; \; 230

Substitute eq230 in eq229

P(\boldsymbol{a},\boldsymbol{b})=-\int_{\lambda_i}^{\lambda_f}\rho(\lambda)A(\boldsymbol{a},\lambda)A(\boldsymbol{b},\lambda)d\lambda\; \; \; \; \; \; \; \; 231

Similarly, P(\boldsymbol{a},\boldsymbol{c})=-\int_{\lambda_i}^{\lambda_f}\rho(\lambda)A(\boldsymbol{a},\lambda)A(\boldsymbol{c},\lambda)d\lambda and P(\boldsymbol{b},\boldsymbol{c})=-\int_{\lambda_i}^{\lambda_f}\rho(\lambda)A(\boldsymbol{b},\lambda)B(\boldsymbol{c},\lambda)d\lambda. Therefore,

P(\boldsymbol{a},\boldsymbol{b})-P(\boldsymbol{a},\boldsymbol{c})=-\int_{\lambda_i}^{\lambda_f}\rho(\lambda)A(\boldsymbol{a},\lambda)A(\boldsymbol{b},\lambda)d\lambda+\int_{\lambda_i}^{\lambda_f}\rho(\lambda)A(\boldsymbol{a},\lambda)A(\boldsymbol{c},\lambda)d\lambda

Since [A(\boldsymbol{b},\lambda)]^{2}=1, we can rearrange the above equation to

P(\boldsymbol{a},\boldsymbol{b})-P(\boldsymbol{a},\boldsymbol{c})=-\int_{\lambda_i}^{\lambda_f}\rho(\lambda)[1-A(\boldsymbol{b},\lambda)A(\boldsymbol{c},\lambda)]A(\boldsymbol{a},\lambda)A(\boldsymbol{b},\lambda)d\lambda

Taking the absolute value on both sides of the above equation and using the relation \left | \int f(t)dt\right |\leq \int \left | f(t) \right |dt

\vert P(\boldsymbol{a},\boldsymbol{b})-P(\boldsymbol{a},\boldsymbol{c})\vert\leq \int_{\lambda_i}^{\lambda_f}\left | \rho(\lambda)[1-A(\boldsymbol{b},\lambda)A(\boldsymbol{c},\lambda)]\right | \vert A(\boldsymbol{a},\lambda)A(\boldsymbol{b},\lambda)\vert d\lambda

 

Question

Why is \left | \int_{a}^{b}f(t)dt \right |\leq \int_{a}^{b}\vert f(t)\vert dt?

Answer

For all t\in [a,\cdots,b], we have -\vert f(t)\vert\leq f(t)\leq \vert f(t)\vert. Therefore, -\int_{a}^{b}\vert f(t)\vert dt\leq\int_{a}^{b} f(t) dt\leq \int_{a}^{b}\vert f(t)\vert dt or simply \left | \int_{a}^{b}f(t)dt \right |\leq \int_{a}^{b}\vert f(t)\vert dt, where we have use the identity of \vert x\vert\leq a\; \; \; \; \Leftrightarrow \; \; \; \; -a\leq x\leq a.

 

Since \vert A(\boldsymbol{a},\lambda)A(\boldsymbol{b},\lambda)\vert =1

\vert P(\boldsymbol{a},\boldsymbol{b})-P(\boldsymbol{a},\boldsymbol{c})\vert\leq \int_{\lambda_i}^{\lambda_f}\left | \rho(\lambda)[1-A(\boldsymbol{b},\lambda)A(\boldsymbol{c},\lambda)]\right | d\lambda

Since \rho(\lambda)\geq 0 and 1-A(\boldsymbol{b},\lambda)A(\boldsymbol{c},\lambda)\geq 0, we can ignore the absolute value sign on RHS of the above equation:

\vert P(\boldsymbol{a},\boldsymbol{b})-P(\boldsymbol{a},\boldsymbol{c})\vert\leq \int_{\lambda_i}^{\lambda_f} \rho(\lambda)d\lambda-\int_{\lambda_i}^{\lambda_f}\rho(\lambda)A(\boldsymbol{b},\lambda)A(\boldsymbol{c},\lambda) d\lambda

Substituting \int_{\lambda_i}^{\lambda_f} \rho(\lambda)d\lambda=1 and P(\boldsymbol{b},\boldsymbol{c})=-\int_{\lambda_i}^{\lambda_f}\rho(\lambda)A(\boldsymbol{b},\lambda)A(\boldsymbol{c},\lambda) d\lambda into the above equation gives

\vert P(\boldsymbol{a},\boldsymbol{b})-P(\boldsymbol{a},\boldsymbol{c})\vert\leq 1+P(\boldsymbol{b},\boldsymbol{c})\; \; \; \; \; \; \; \; 232

Eq232 is the Bell’s inequality, which is based on the ERP paradox.

To show that quantum mechanics is incompatible with Bell’s inequality, we let \angle \boldsymbol{a}\boldsymbol{c}=\angle \boldsymbol{b}\boldsymbol{c}=45^{\circ}, i.e. \boldsymbol{c} bisects \angle \boldsymbol{a}\boldsymbol{b}. From eq228,

P(\boldsymbol{a},\boldsymbol{b})=-cos90^{\circ}=0\; \; \; \; \; \; \; \; 233

P(\boldsymbol{b},\boldsymbol{c})=P(\boldsymbol{a},\boldsymbol{c})=-cos45^{\circ}=-\frac{1}{\sqrt{2}}\; \; \; \; \; \; \; \; 234

Substituting eq233 and eq234 in eq232 yields \frac{1}{\sqrt{2}}\leq1-\frac{1}{\sqrt{2}}, which is inconsistent with Bell’s inequality. Therefore, it is possible to experimentally measure the spins of the two particles at non-orthogonal angles to test the predictions of quantum theory versus the ERP paradox. In fact, the results of all experiments conducted at non-orthogonal angles were in agreement with quantum mechanics. This implies that all local hidden-variable hypotheses are invalid.

 

 

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