Non-relativistic quantum description of particles with spin

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 ψ(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 = Er ⊗ Es.  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 = Er ⊗ Es.  So do the observables {Px,Py,Pz,S2,Sz}.
Since {|r>} forms a basis for Er and {|ε>,|ε> = |+>,|->} forms a basis for Es, {|r> ⊗ |ε> = |r,ε>} forms a basis for E for a spin ½ particle.
Each vector |r,ε> is a common eigenvector of X, Y, Z, S2, and Sz.  We have
X|r,ε> = x|r,ε>,  Y|r,ε> = y|r,ε>,  Z|r,ε> = z|r,ε>,  S2|r,ε> = (3/4)ħ2|r,ε>,  Sz|r,ε> = εħ|r,ε>.
<r',ε'|r,ε> = δεε'δ(r - r'), (normalization)  Σε∫d3r|r,ε><r,ε| = I,  (completeness).
For any vector |ψ> we have |ψ> = Σε∫d3r|r,ε><r,ε|ψ> = Σε∫d3r ψ(r)|r,ε>, with ψ(r) = <r,ε|ψ>.

Notation:
Let ψ+(r) = <r,+|ψ>,  ψ-(r) = <r,-|ψ>,  and let

be a two component spinor, with its adjoint [ψ] = (ψ+*(r), ψ-*(r)).
For a spin ½ particle |ψ> is completely specified by the two component spinor, just as for a spinless particle |ψ> is specified by the wave function ψ(r).  The inner product is defined through
<ψ|Φ> = Σε∫d3r|<ψ|r,ε><r,ε|Φ> = ∫d3r(ψ+*(r+(r) + ψ-*(r-(r)) = ∫d3r[ψ][Φ].
Note: Matrix multiplication of the spinors precedes the spatial integration.
The state is normalized in the following way:
<ψ|ψ> = ∫d3r(ψ+*(r+(r) + ψ-*(r-(r)) = 1.

The spinor associated with a tensor product vector |ψ> = |Φ> x |Χ>
with |Φ> = ∫d3r Φ(r)|r> in Er and |Χ> = c+|+> + c-|-> in Es is

Note: Not all vectors in E are tensor product vectors.

Let |ψ> be the state of a spin ½ particle and let |ψ’> be obtained from |ψ> through the action of a linear operator A,  A|ψ> = |ψ’>.
We can associate with each linear operator A a 2 x 2 matrix [A] such that [A][ψ](r) = [ψ’](r).

#### Examples:

• 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

.

Rotation operators for a spin ½ particle

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 φ.  If the state of the system is |α> before the rotation, it is |α>R after the rotation, where |α>R = U(R)|α>,   U(R) = exp(-iSzφ/ħ) = exp(-iσzφ/2).
Before the rotation the expectation value of the operator Sx is <Sx> = <α|Sx|α>, after the rotation it is <Sx>R = R<α|Sx|α>R.  We may expand

since σz2 = I.  The expression for U(R)SxU(R) therefore may be written

In matrix notation we have

.

Therefore  R<α|Sx|α>R = <α|U(R)SxU(R)|α> = <α|Sxcosφ - Sysinφ|α>
= <Sx>cosφ - <Sy>sinφ.
Similarly R<α|Sy|α>R = <Sy>cosφ + <Sx>sinφ.
The average value of Sz does not change, since it commutes with U(R).  The expectation value of the spin operator S behaves as though it were a classical vector under rotation.
We have <Si>R = ∑iRij<Sj>,
where R is the 3 x 3 rotation matrix for the rotation being considered.  For an arbitrary system we can generalize to
<Ji>R = ∑iRij<Jj>.
[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 |α> = c1|+> + c2|->.
|α>R = U(R)|α> = exp(-iSzφ/ħ)|α> = exp(-iφ/2)c1|+> + exp(iφ/2)c2|->.
For φ = 2π we have  |α>R = -c1|+> - c2|-> = -|α>.
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 {|ψ(1)>,|ψ(2)>} be the eigenstates of the observable that determines the path of the particle.  Assume that at point A the state vector is
|α(0)> = 2(-1/2)(|ψ(1)> + |ψ(2)>).
At point B the state vector is
|α(t)> = 2(-1/2)U(t,t0)(|ψ(1)> + |ψ(2)>).
Assume the evolution operator rotates |ψ(2)> but not |ψ(1)>.  If |ψ(2)> is rotated through 360o then the state vector at point B is
|α(t)> = 2(-1/2)(|ψ(1)> - |ψ(2)>).
The probability of finding the particle at point B is
|<rB|α(t)>|= ½|<rB|ψ(1)> - <rB|ψ(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 B = Bk is H = ωSz where ω = -γB is proportional to the magnetic field strength.
The evolution operator therefore is U(t,0) = exp(-iωSzt/ħ).
The evolution operator equals the rotation operator with φ = ωt.  This explains spin precession.  In a magnetic field B = Bk we have
<Sx>t = <Sx>t=0cos(ωt) - <Sy>t=0sin(ωt),  <Sy>t = <Sy>t=0cos(ωt) + <Sx>t=0sin(ωt),
<Sz>t = <Sz>t=0.

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   B = Bk.
Then |α(t)> = 2(-1/2)(|ψ(1)> + exp(-i(ωT/2)|ψ(2)>).  The intensity at point B is therefore is proportional to C + Dcos(ωT/2).  We can vary the angle φ = ωT by varying the magnetic field strength.  The intensity at point B is predicted to vary sinusoidally with a period 4π/ω.  These predictions have been experimentally verified.

#### Rotation of two component spinors 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 φ.  Let |ψ> be the state vector of a spin ½ particle in E = Er ⊗ Es before the system is rotated.  |ψ> can be represented by the two component spinor

.

After the rotation the state vector is |ψ>R = U(R)|ψ>, where U(R) = rU(R) ⊗ sU(R).  U(R) is a mixed operator.
The matrix for rU(R) is and the matrix for sU(R) is .
We therefore have

.

#### Group properties of rotations

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 x 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 x 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 2π rotation and a 4π 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 x 3 rotation matrix in the SO(3) language.  For a given R the corresponding U(R) is double valued.