Bell’s Inequality: Part 1

by landonlehman

In the Quantum Mechanics II class I am taking, we recently had a lecture on Bell’s inequality.  I didn’t understand everything in the lecture, so I am going to write a post or two about it in an attempt at clarification.

The general physical situation to keep in mind is two spin-1/2 particles that are in the singlet state.  The two particles are distinguished using 1 and 2.  By some experimental method, the particles are caused to travel away from each other, and each passes through an apparatus that can measure its spin (an “analyzer”).  Each apparatus can be moved in order to measure spin at different angles, with the angle of apparatus 1 called a and the angle of apparatus 2 called b.

The conditional probabilities that will be used in the derivation are represented as follows: P(\sigma | a) is the probability that the operator \hat{\sigma} acting on particle 1 will be found in the eigenstate with eigenvalue \sigma , given that the analyzer for particle 1 is at the angle a.  Let’s just restrict things to spin measurements of the spin-1/2 particles, so that the set of eigenvalues will be \sigma = \{+1, -1\} , where the units are such that \hbar /2 =1 .  The operator \hat{\sigma} is just the spin operator along some specified direction.

Now, where the orthodox interpretation of quantum mechanics would write P(\sigma | a ), a hidden variable theory will write P(\sigma | a, \lambda ), where \lambda is the hidden variable.  In other words, in a hidden variable theory the probability of getting the eigenvalue \sigma depends not only on the analyzer angle a, but also on something else, namely \lambda .

Let A(a, \lambda ) be the expectation value for a spin measurement on particle 1 given analyzer angle a, and an unknown value of the hidden variable \lambda.  The equivalent definition holds for B(b, \lambda) .  Since we are working with the set of eigenvalues \sigma , it is easy to verify that |A(a, \lambda)| \leq 1 and |B(b, \lambda)| \leq 1 .  Also, the hidden variable \lambda has some unknown probability distribution f(\lambda ) .  Since \lambda can be continuous, we can impose normalization as \int d\lambda f(\lambda ) =1 .

As a final definition, let E(a, b) denote the expectation value of a spin measurement of both particles, with the first analyzer at angle a and the second analyzer at angle b.  For example, in orthodox quantum mechanics (QM):

E^{QM}(a, b) = \sum_{\sigma_1, \sigma_2} \sigma_1 \sigma_2 P(\sigma_1, \sigma_2 | a, b) .

For a hidden variable theory (HV):

E^{HV}(a,b) = \int d\lambda f(\lambda ) A(a, \lambda) B(b, \lambda ) .

Does the above expression make sense?  Well, we are integrating over the continuous variable \lambda, which can be thought of as summing up all the infinitesimal d\lambda‘s.  So each “term” in the “sum” has the expectation value of the first spin (for particle 1, given \lambda and a), the expectation value of the second spin (for particle 2, given \lambda and b), and is weighted by the probability of getting that value of \lambda, which is f(\lambda).  The thing to note is that there are two levels of averaging here.  First, the expectation values for the spin measurements at specific values of a and \lambda .  Second, the averaging over all possible values of \lambda , weighted by f(\lambda).  So it doesn’t look exactly like an ordinary expectation value, because of the introduction of the hidden variable.

I think this is enough for one post.  One thing that confused me during the lecture was that the professor used the same symbol to represent an actual value given by a single measurement and the expectation value of a single measurement.  I tried to indicate clearly which was meant in the definitions I wrote.  If I was not clear about something, please let me know.

This post was mostly just definitions.  In the next post, I should be able to use these definitions to derive one form of Bell’s inequality.  Then I can discuss the conceptual/philosophical stuff.  But as everyone knows, before you can discuss philosophy of quantum mechanics, you first have to “shut up and calculate” :).