In Steven Weinberg's Lecture on Quantum Mechanics (p. 342), he writes:

The correlation between the spins of the two particles can be expressed as the average value of the product of the $\hat{a}$ component of the spin of particle 1 and the $\hat{b}$ component of the spin of particle 2:

$\langle (\textbf{s1} \cdot \hat{a}) (\textbf{s2} \cdot \hat{b}) \rangle = -\frac{\hbar^2}{4} \int{d\lambda\ \rho(\lambda) S(\hat{a}, \lambda) S(\hat{b}, \lambda)}$

where $\hat{a}$ and $\hat{b}$ are any two unit vectors.

In quantum mechanics, we have:

$\langle (\textbf{s1} \cdot \hat{a}) (\textbf{s2} \cdot \hat{b}) \rangle = -\frac{\hbar^2}{4} \hat{a} \cdot \hat{b}$

There is no obstacle to constructing a function $S$ and a probability density $\rho$ for which are equal for any single pair of directions $\hat{a}$ and $\hat{b}$.

Can someone explain to me precisely how this is done and what $S(\hat{a}, \lambda)$ and $\rho(\lambda)$, along with the hidden variables $\lambda$ would be?

My guess is that since we want something that is rotationally invariant since it depends only on the dot product, we want some $S$ and hidden variables $\lambda$ that act nicely under rotations.

On the other hand, since the first equation is a local hidden variable theory, I thought this was not possible by Bell's inequalities...

  • $\begingroup$ I edited it to try and make it clear which parts are quoted from Weinberg and which parts are your own text - but I'm not 100% sure I got it right - can you check? $\endgroup$
    – N. Virgo
    Apr 22, 2013 at 9:23

1 Answer 1


You should read Bell's original paper, where he constructs such a model explicitly. The key point is that for any single pair of directions one can construct such a model. Bell's theorem states that there is no single local hidden variable model that can reproduce the quantum mechanical correlation for every pair of directions.

(If you are interested in reading more about ontological models, see also Harrigan and Spekkens and Lewis et al..)


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