Suppose that two physical systems A and B, associated with the vector spaces V_{1} and V_{2}, 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 V_{1} and V_{2}.)

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 **a**_{1}, **a**_{2}, **a**_{3},… for the vector space associated with A and a basis **b**_{1}, **b**_{2}, **b**_{3} for the vector space associated with B such that the vector associated with two entangled systems can be written as the sum

c_{1} **a**_{1},**b**_{1} + c_{2} **a**_{2},**b**_{2} + c_{3} **a**_{3},**b**_{3} + ···

where c_{1}, c_{2}, c_{3},… 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 π^{0} 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 cos^{2}(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 cos^{2}(60°) = ^{1}⁄_{4}. The probability of obtaining opposite outcomes is therefore (1/3)×1+(2/3)×(1/4) = ^{1}⁄_{2}. In other words, the apparatuses flash different colors exactly half of the time.