## 21 Entanglement, Bell revisited

Sup­pose that two phys­ical sys­tems A and B, asso­ci­ated with the vector spaces V1 and V2, have been sub­jected to mea­sure­ments, and that the out­comes are (rep­re­sented by) the respec­tive unit vec­tors a and b (in lieu of the 1-​​dimensional sub­spaces con­taining a and b). What is the prob­a­bility with which two sub­se­quent mea­sure­ments yield the respec­tive out­comes c and d? If the ini­tial mea­sure­ments are repeat­able, then by Born’s rule this prob­a­bility is the product |<c|a>|2 |<d|b>|2.

If we rep­re­sent the ini­tial state of the com­posite system by the pair a,b and define the scalar product <c,d|a,b> in the vector space con­taining such pairs of vec­tors as equal to the product of the scalar prod­ucts <c|a> and <d|b>, then this prob­a­bility can also be written as |<c,d|a,b>|2. (The vector space con­taining such pairs of vec­tors is tech­ni­cally known as the direct product of V1 and V2.)

If the out­comes of mea­sure­ments per­formed on the two sys­tems are cor­re­lated, the sys­tems are said to be “entan­gled”. (This is all there is to the meaning of entan­gle­ment.) In this case the state of the com­posite system made up of A and B cannot be rep­re­sented by a single pair of vec­tors. It is, how­ever, always pos­sible to find a basis a1, a2, a3,… for the vector space asso­ci­ated with A and a basis b1, b2, b3 for the vector space asso­ci­ated with B such that the vector asso­ci­ated with two entan­gled sys­tems can be written as the sum

c1 a1,b1 + c2 a2,b2 + c3 a3,b3 + ···

where c1, c2, c3,… are com­plex num­bers whose squared mag­ni­tudes add up to 1. The most famous example of such a state is the so-​​called sin­glet state of two spin-​​1/​2 systems:

S = (z↑,z↓ − z↓,z↑)/√2.

A sin­glet state can come about in a variety of ways: by inducing a spin­less hydrogen mol­e­cule to dis­so­ciate into a pair of hydrogen atoms, by let­ting two pro­tons scatter each other at low ener­gies, or by let­ting a spin­less par­ticle decay into two spin-​​1/​2 par­ti­cles — for example, a π0 meson into an elec­tron and a positron. One readily checks that <z↑,z↑|S> = <z↓,z↓|S> = 0, while <z↑,z↓|S> = 1/√2 and <z↓,z↑|S> = −1/√2. Thus the prob­a­bility of finding both spins “up” or “down” with respect to the z-​​axis is 0, while the prob­a­bility of oppo­site out­comes is (1/√2)2 + (−1/√2)2 = 1. The same prob­a­bil­i­ties are obtained if the spin com­po­nents of the two sys­tems are mea­sured with respect to any other axis.

The sin­glet state S can be used to repro­duce the sta­tis­tical prop­er­ties of the exper­i­ment described on this page. The devise at the center of Fig. 1.1.1 launches pairs of spin-​​1/​2 par­ti­cles “in” the sin­glet state. The three quan­ti­ties that the appa­ra­tuses are designed to mea­sure are spin com­po­nents with respect to three coplanar axes — they lie in the same plane, and the angle between any two of them is 120°.

It is imme­di­ately clear that when­ever the spins are mea­sured with respect to the same axis (iden­tical appa­ratus set­tings), equal out­comes (two reds or two greens) are never observed. What remains to be shown is that the out­comes are com­pletely random if the appa­ratus set­tings are not taken into account.

Pre­vi­ously we noted that if the spin of a spin-​​1/​2 par­ticle is found “up” with respect to a given axis, then the prob­a­bility of sub­se­quently finding it “up” with respect to an axis that is rotated by an angle α is cos2(α/​2). If each mea­sure­ments is per­formed on one of two par­ti­cles “in” the sin­glet state, this is also the prob­a­bility with which the two mea­sure­ments yields oppo­site out­comes. Because the appa­ratus set­tings are ran­domly selected, the prob­a­bility with which the same spin com­po­nent is mea­sured is 1/​3, and in these cases the prob­a­bility of obtaining oppo­site out­comes is 1. The prob­a­bility with which the par­ticle spins are mea­sured with respect to dif­ferent axes is 2/​3, and in these cases the prob­a­bility of obtaining oppo­site out­comes is cos2(60°) = 14. The prob­a­bility of obtaining oppo­site out­comes is there­fore (1/3)×1+(2/3)×(1/4) = 12. In other words, the appa­ra­tuses flash dif­ferent colors exactly half of the time.

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