Consider a spin ½ particle in a uniform magnetic field BAssume B ≈ B0k.  The potential energy of a particle with intrinsic magnetic moment m = γS in this field is
U = -mB = -mzB0 = -γSzB0 = ω0Sz, where ω0  = -γB0
If we are quantizing only the internal degrees of freedom, then the operator H, (the Hamiltonian), which is the operator whose possible eigenvalues are the total energy of the particle associated with those internal degrees of freedom, is found by letting Sz become an operator.
H = ω0Sz,  [H,Sz] = 0.
H and Sz have the same eigenfunctions.
H|+> = (ħω0/2)|+>,  H|-> = -(ħω0/2)|->.

There are two energy levels, E+ = ħω0/2 and E- = -ħω0/2, separated by ħω0.
We define the Bohr frequency f+- = |E+ - E-|/h = |ω0|/(2π).
In general, the Bohr frequencies of a system whose energy eigenvalues are {En} are defined as
fnn' = |En - En'|/(2πħ).
Note: If B ≈ B0u then H = ω0Su.  For silver atoms γ is negative, ω0 is positive , E+ > E- .

Assume a system in an arbitrary state
|ψ> = cos(θ/2)exp(-iφ/2)|+> + sin(θ/2)exp(iφ/2)|->
with  0 ≤ θ ≤ π  and  0 ≤ φ ≤ 2π  is placed into a uniform magnetic field B ≈ B0k at t = 0.
The system evolves and we have
|ψ(t)> = U(t,0)|ψ(0)> = exp(-iHt/ħ)|ψ(0)> = exp(-iω0Szt/ħ)|ψ(0)>.
|ψ(t)> = cos(θ/2)exp(-i(φ + ω0t)/2)|+> + sin(θ/2)exp(i(φ + ω0t)/2)|->
= cos(θ/2)exp(-iφ'/2)|+> + sin(θ/2)exp(iφ'/2)|->,
where φ' = φ + ω0t.
|ψ(t)> is an eigenvector of Su’ , (u' is characterized by θ and φ'), while |ψ(0)> is an eigenvector of Su , (u is characterized by θ and φ ).  The vector u' defining Su'Su' revolves around the z-axis with angular frequency ω0 ∝ B0.  Its component along the z-axis remains constant, while its component perpendicular to the z-axis rotates with angular velocity ω0.  This is the quantum mechanical analog of the rotation of the classical magnetic moment m and is called Larmor precession.

<Sz(t)> = <ψ(t)|Sz|ψ(t)> = (ħ/2)cos2(θ/2) - (ħ/2)sin2(θ/2)
= (ħ/2)½(1 + cosθ) - (ħ/2)½(1 - cosθ) = (ħ/2)cosθ.
Sz is a constant of motion.

<Sx(t)> = <ψ(t)|Sx|ψ(t)> = cos2(θ/2)<+|Sx|+> + sin2(θ/2)<-|Sx|->
+ cos(θ/2)sin(θ/2)exp(iφ')<+|Sx|-> + cos(θ/2)sin(θ/2)exp(-iφ')<-|Sx|+>.

.

<Sx(t)> = (ħ/2)cos(θ/2)sin(θ/2) 2cosφ' =  (ħ/2)sinθ cos(φ + ω0t).

Similarly,  <Sy(t)> = (ħ/2)sinθ sin(φ + ω0t).
Sx and Sy are not constants of motion.  Their expectation values oscillate with the single Bohr frequency |ω0|/(2π).  The mean values of Sx, Sy, and Sz behave like the components of the classical angular momentum undergoing Larmor Precession.

Problem:

At t = 0 the x-component of the spin of a spin ½ particle is measured and found to be ħ/2.  At t = 0 the particle is therefore in the |+>x eigenstate of the Sx operator.  The particle is confined to a region with a uniform magnetic field B = B0k, its Hamiltonian is H = ω0Sz.  The eigenstates of H are |+> and |->,
H|+> = (ħω0/2)|+> and H|-> = -(ħω0/2)|->.
|+>x can be written as a linear combination of eigenstates of H.
(a)  Find the probability of measuring Sx = ħ/2 at t = T.
(b)  What is the mean value of Sx, <Sx>, at t = T?
(c)  Find the probability of measuring Sz = ħ/2 at t = T.

Solution:

• Concepts:
The two dimensional state space of a spin ½ particle, the evolution operator, the postulates of Quantum Mechanics
• Reasoning:
The state space corresponding to the observable Sz of a spin ½ particle is two-dimensional.  We denote the eigenvectors of Sz by |+> and |->.  The Hamiltonian of the system is H = ω0Sz.  The evolution operator is exp(-iHt/ħ).  We use the evolution operator to find the state of the system at time T and then answer the questions using the postulates of Quantum Mechanics.
• Details of the calculation:
(a)  The matrix of Sx in the eigenbasis of Sz is
.
The eigenvectors of Sx are |+>x and |->x.
|±>x = (1/√2)(|+> ± |->).
|ψ(t)> = U(t,0)|ψ(0)> = U(t,0)|+>x = (1/√2)exp(-iHt/ħ)(|+> + |->).
|ψ(t)> = (1/√2)(exp(-iω0t/2)|+> + exp(-iω0t/2)|->).
P(Sx=ħ/2,T) = |x<+||ψ(t)>|2 = ¼|(<+| + <-|)(exp(-iω0T/2)|+> + exp(-iω0T/2)|->)|2
= ¼|exp(-iω0T/2)|+> + exp(-iω0T/2)|2 = cos20T/2).

(b) P(Sx=-ħ/2,T) = 1 - cos20T/2) = sin20T/2).
<Sx(T)> = (ħ/2)[cos20T/2) - sin20T/2)] = (ħ/2)cos(ω0T).

(c) P(Sz=ħ/2,T) = |<+|ψ(t)>|2 = ½|<+|(exp(-iω0t/2)|+> + exp(-iω0t/2)|->)|2 = ½.
The probability of measuring Sz = ħ/2 is independent of t.