The operator associated with the spin of a particle is a vector observable. Its components satisfy the commutation relations that define an angular momentum operator. The quantum state of a particle with spin is not completely specified by specifying its wave function y(r) in orbital space Er. We must also specify its spin variables in spin space Es. A given particle is characterizes by a unique value of s, which is a positive integer or half integer. (2s+1) basis vectors span the spin space Es of this particle. We can choose these basis vectors to be {|s,ms>,ms= -s,-s+1,...,s}, the eigenvectors of S2 and Sz. The state space of the particle is the tensor product space E=ErÄEs. The spin observables commute with all the orbital observables.
For the electron Es is two-dimensional. The observables {X,Y,Z,S2 Sz} form a C.S.C.O. in E=ErÄEs. So do the observables {Px,Py,Pz,S2,Sz}. Since {|r>} forms a basis for Er and {|e>,|e>=|+>,|->} forms a basis for Es, {|r>Ä|e>=|r,e>} forms a basis for E for a spin ½ particle. Each vector |r,e> is a common eigenvector of X, Y, Z, S2, and Sz. We have
X|r,e>=x|r,e>, Y|r,e>=y|r,e>, Z|r,e>=z|r,e>,
For any vector |y> we have
,
with
.
Notation:
Let y+(r)=<r,+|y>, y-(r)=<r,-|y>, and let
be a two component spinor, with its adjoint
.
For a spin ½ particle |y> is completely specified by the two component spinor, just as for a spinless particle |y> is specified by the wave function y(r). The inner product is defined through
Note: Matrix multiplication of the spinors precedes the spatial integration.
The state is normalized in the following way:
The spinor associated with a tensor product vector |y>=|f>´|c>
with 
in Er
and 
in Es
is
Note: Not all vectors in E are tensor product vectors.
Let |y> be the state of a spin ½ particle and let |y’> be obtained from |y> through the action of a linear operator A, A|y>=|y’>. We can associate with each linear operator A a 2´2 matrix [A] such that [A][y](r)=[y’](r).
 | Let A be a spin operator, say S+. We have
The matrix of a spin operator is its matrix in the { |+>, |-> } basis in Es. Similarly, if A = Sz then
|
| Let A be an orbital operator, say X. We have
The matrix of an orbital operator is diagonal. |
| Let A be a mixed operator, say LzSz.
Similarly, if A=XS+ , then
|
Consider a system consisting of a single spin ½ particle. Neglect the spatial degrees of freedom. Rotate the system ccw about the z-axis through an angle f. If the state of the system is |a> before the rotation, it is |a>R after the rotation, where |a>R=U(R)|a>,
Before the rotation the expectation value of the operator Sx is <Sx>=<a|Sx|a>, after the rotation it is <Sx>R=R<a|Sx|a>R. We may expand
since 
The expression for UT(R)SxU(R)
therefore may be written
In matrix notation we have
.
Therefore
Similarly
.
The average value of Sz does not change, since it commutes wit U(R). The expectation value of the spin operator S behaves as though it were a classical vector under rotation. We have
where R is the 3´3 rotation matrix for the rotation being considered. For an arbitrary system we can generalize to
.
[Remember S is a vector operator. While the expectation values of the components of S behave as the components of a classical vector under rotation, the components of S itself behave like a classical vector rotated backward, S'=R-1S.]
How does the state vector of the spin ½ particle behave under rotation?
Let |a>=c1|+>+c2|->.
For f=2p we have
|a>R=-c1|+>-c2|->=-|a>.
A 360o rotation does not give us the original ket back. We need to rotate through 720o to get back our original state vector. This is a classically totally unexpected result.
Can it be observed?
The minus sign is a change in the phase of the state vector. If all state vectors are multiplied by a minus sign there will be no observable effect. We can only observe changes in relative phase. We must compare a rotated ket to an unrotated ket and look for interference effects. Assume a spin ½ particle can move from point A to point B along two different path. Assume that when we observe the particle at point B we do not know which path it actually took to get there. Assume that path 1 does not rotate the state vector and path 2 does rotate the state vector.
Let {|y(1)>,|y(2)>} be the eigenstates of the observable that determines the path of the particle.
Assume that at point A the state vector is
|a(0)>=2(-1/2)(|y(1)>+|y(2)>).
At point B the state vector is
|a(t)>=2(-1/2)U(t,t0)(|y(1)>+|y(2)>).
Assume the evolution operator rotates |y(2)> but not |y(1)>. If |y(2)> is rotated through 360o then the state vector at point B is
|a(t)>=2(-1/2)(|y(1)>-|y(2)>).
The probability of finding the particle at point B is
|<rB|a(t)>|2=(1/2)|<rB|y(1)>-<rB|y(2)>|2.
If without rotation the probability of finding the particle at point B is 1, then with rotation we have destructive interference.
How can we rotate the state vector?
The Hamiltonian of a spin ½ particle in a magnetic field 
is H=wSz where w=-gB is proportional to the magnetic field strength.
The evolution
operator therefore is
The evolution operator equals the rotation operator with f=wt.
This explains spin precession. In a magnetic field we have
.
Assume that in a neutron interferometry experiment path 1 goes through a field free
region and path 2 goes through a region with a static magnetic field .
Then |a(t)>=2(-1/2)(|y(1)>+exp(-i(wT/2)|y(2)>).
The intensity at
point B is therefore is proportional to 
.
We can vary the angle f=wT
by varying the magnetic field strength. The intensity at point B is predicted to vary
sinusoidally with a period 
These
predictions have been experimentally verified.
We now take into account both the internal and the external degrees of freedom of a spin ½ particle. Let us rotate the system ccw about the z-axis through an angle f. Let |y> be the state vector of a spin ½ particle in E=ErÄEs before the system is rotated. |y> can be represented by the two component spinor
.
After the rotation the state vector is |y>R=U(R)|y>, where U(R)=rU(R)Ä sU(R). U(R) is a mixed operator.
The matrix for rU(R) is 
and the matrix for sU(R) is 
.
We therefore have
.
How can we characterize a rotation? We need three real numbers to characterize a general rotation. We can, for example, specify the direction of the rotation axis ( two angles ) and the rotation angle. We can also specify the 3´3 rotation matrix R. R has 9 elements. R is an orthogonal matrix. RRt=RtR=I. The orthogonality condition results in a set of 6 independent equations for 9 unknowns. Therefore R contains 3 independent numbers.
The set of all multiplication operation with orthogonal matrices forms a group. This group has the name SO(3). (S stands for special (determinant=1), O stands for orthogonal, 3 stands for three dimensions.)
We can also characterize a rotation by specifying the 2´2 unitary matrix U(R) which is the matrix representations of the rotation operator in Es. The most general form of such a rotation matrix is
with |a|2+|b|2=1 and a and b complex numbers. Again we have 3 independent elements. The matrices U(R) also form a group. This group has the name SU(2). (S stands for special, U stands for unitary, 2 stands for dimensionality 2.) The groups SO(3) and SU(2) are not isomorphic. A 2p rotation and a 4p rotation are represented by the same R matrix but by different U(R) matrices. U(a,b) and U(-a,-b) correspond to the same 3´3 rotation matrix in the SO(3) language. For a given R the corresponding U(R) is double valued.
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