Assume a beam of electrons is emerging  from a spin filter with its "up" axis pointing into the z-direction, as shown below.

The emerging electrons all are in the |+> state, and they are traveling along the y-axis.  Assume that the electrons then enter another spin filter whose "up" axis can point along one of three different directions.  What fraction of the electrons will pass through the second filter?

Quantum mechanics tells us that if the angle between the "up" axes of two consecutive spin filters is q, then the probability that an electron that passed through the first filter will also pass through the second filter is cos²(q/2).

If the "up" axis of the second analyzer points along direction 1, 100% of the electrons will pass through the analyzer.
If the "up" axis of the second analyzer points along direction 2, cos²(q/2) = cos²(60^o) = 1/4 (25%) of the electrons will pass through the analyzer.
If the "up" axis of the second analyzer points along direction 3, cos²(q/2) = cos²(60^o) = 1/4 (25%) of the electrons will pass through the analyzer.
 

Correlations

Assume that a radioactive substance emits a pair of electrons in each decay and that the two electrons are emitted in opposite directions.  Assume that angular momentum conservation dictates that the total spin of the two electrons is zero.  The magnitude of the total spin is zero and the component of the total spin measured along any axis is zero.  The two electrons are entangled.  Entanglement means as far as spin is concerned neither particle has properties of its own, they only have common properties.  Such a state is often called a "Schroedinger Cat" state.

• With a combination of spin filters as shown below, we will try to measure the projections of the spins of both particles along various axes and look for correlations.

  

  Without having detailed information about the source, we do not know if the electrons moving to the right or left are partially polarized.  So we cannot predict what fraction of the electrons will pass either filter.  But we know that if we look at correlations, we will find that if an electron does pass through the right filter, then its companion electron does not pass through the left filters.  Similarly, if the electron does not pass through the right filter, then its companion electron does pass through the left filters.  Each radioactive decay has a total spin of zero.  If one electron is spin up its companion electron is spin down.  If we change the orientation of one of the filters by 180^o with respect to the other filter, we will observe that if a particular electron passes through its filter, then its companion electron will also pass through its filter.  Similarly, if a particular electron does not pass through its filter, neither does its companion electron.

These outcomes are independent of the absolute orientation of the filters with respect to the source or the separation of the filters.

How do these results for entangled electrons differ from results for electrons that are not entangled?

• Assume we have two independently emitted electrons.  Assume we have passed them through spin analyzers and we know that one of them has S_z = h/2, and the other has S_z = -h/2.  We do not know which electron is the spin up electron.  Since the electrons were independently produced, they are not entangled.

  What correlations will we measure when we use the apparatus described above and send one electron to the right and the other one to the left?

  

  In the orientation shown in the figure above, the result will be the same as for the entangled electrons.  The correlation will be zero.   But if we change the orientation of the filter apparatus as a whole with respect to the source, the correlation will change  If we, for example, rotate the apparatus by 90^o, the correlation will be 50%.
  For an entangled state the correlation will always be zero, no matter how far away from each other the two filters are located.  If one observer measures the spin to be "up" along a arbitrary direction, then the other observer will measure it down along the same direction.  Some properties of the two entangled electrons are inextricably linked to each other, even if the electrons are located on opposite sides of the galaxy.  In this sense, quantum mechanics is said to be "nonlocal".

Now let us consider the following experimental setup.

• A source of entangled electrons is placed between two spin filters, for which the "up" axis can be switched to point along directions 1, 2, or 3 as shown in the figure below.  The filters are mounted in boxes.  The box which contains the right filter is rotated by 180^o with respect to the box which contains the left filter.  The source and the two filters are not connected in any way and can be separated by very large distances.

  A		B

  Each filter is rigged in such a way, that if an electron passes the filter, a light on the box flashes green, and if an electron hits the beam stop and does not pass, a light on the box flashes red.  There are two observers, Alice (A) and Bob (B).  Alice can chose the orientation of filter A and can observe the flashing light on box A.  Bob can choose the orientation of filter B and can observe the flashing light on box B.  During an experiment, there is no way for Alice and Bob to communicate.

The Experiment:

• The source is adjusted so it emits pairs of entangled electrons in approximately 5 second intervals.  Alice randomly chooses one of the possible orientations (1, 2, or 3) of the left filter (A) before each electron arrives, and Bob randomly chooses on of the possible orientations of the right filter (B) before the companion electron arrives.  The arrival of an electron triggers a light to flash.  In a table Alice and Bob record the sequence of colors (green or red) of the flashes on their respective boxes.  After a large number of flashes has been observed, the experiment is stopped and Alice and Bob are now allowed to communicate and compare their results.

How often did Alice and Bob obtain the same result, i.e. how often did they record the same color?

• The orientation of each filter is chosen randomly.  For each filter, each orientation is chosen ~ 1/3 of the time.  The larger the number of trials, the less we expect the fraction N_i/N_total to deviate from 1/3.  (N_i = number of trial with orientation i = 1, or 2, or 3, N_total = total number of trials.)
  (Link: Rolling Dice  The more often we roll, the more the distribution approaches the expected distribution.)

• For the two filters there are 9 possible combinations of orientations, each combination is equally likely and is chosen ~ 1/9 of the time.

  Possible Combinations

  		Filter  A
  	Position	1	2	3
  Filter B	1	1,1	1,2	1,3
  	2	2,1	2,2	2,3
  	3	3,1	3,2	3,3

  If both filters have the same orientation, 1,1, or 2,2, or 3,3,  then either both electrons will pass their through their respective filters, or both will hit a beam stop.  The lights on the filters A and B will flash the same color.

  For the other 6 combinations, 1,2, or 1,3, or 2,1, or 2,3, or 3,1 or 3,2, the probability that both lights flash the same color is 1/4.
  Let us take, for example, the combination 1,2.  The right filter axis points down.  If the light flashes green, then the right electron's spin points down.  Entanglement guaranties that the left electron's spin points up.  But axis 2 of the left filter is oriented 120^o away from the z-axis, so cos²(q/2) = cos²(60^o) = 1/4 is the probability that this electron will pass through its filter.  1/4 of the green flashes of analyzer B will be accompanied by green flashes of analyzer A.

  If, however, the light on the right filter box flashes red, then the right electron's spin points up and it hits the beam stop.  Entanglement guaranties that the left electron's spin points down.  But axis 2 of the left filter is oriented 60^o away from the  negative z-direction, so cos²(q/2) = cos²(30^o) = 3/4 is the probability that this electron will pass through its filter.  Therefore 1/4 of the electrons will hit the beam stop, and 1/4 of the red flashes of filter B will be accompanied by red flashes of filter A.

  We can analyze the combinations 1,3, or 2,1, or 2,3, or 3,1 or 3,2 in the same way.  The probability that the lights on the filters A and B will flash the same color is always 1/4 for these combinations.

• The total probability that both filters flash the same color therefore is

  P(same color) = (1/9)*[1 + 1 + 1 + 1/4 + 1/4 + 1/4 + 1/4 +1/4 +1/4] = (1/9)*4.5 = 1/2.

  For a large number of trials, ~50% of the time the lights on both filters will flash the same color.  Experiments similar to the one described here have now been carried out many times and have always yielded this predicted result.

What is so special about that?

Quantum mechanics predicts incompatible observables for a system.  An observer cannot know the values of two incompatible observables of a system at the same time.  Does this mean that the system really does not have well defined values for these observables before a measurement, or is it possible that the system has well defined values, but these values are hidden from the observer, and the observer just cannot obtain the complete information?
In 1964 J.S. Bell showed that the assumption of hidden variables is inconsistent with the outcome of the above described experiment.  If there were hidden variables, we would have to observe the light flashing the same color more than 50% of the time.  We will examine at a simple version of Bell's thought experiment in the next section.

Link:

J. S. Bell, On the Einstein Podolsky Rosen Paradox, Physics 1, 195 (1964)

J. S. Bell, On the problem of hidden variables in quantum mechanics, Rev. Mod. Phys. 38, 447 (1966)