Consider a spin ½ particle in a uniform magnetic field

U = -

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 S

H = ω

H and S

H|+> = (ħω

There are two energy levels, E_{+} = ħω_{0}/2 and E_{-}
= -ħω_{0}/2, separated by ħω_{0}.

We define the
Bohr frequency f_{+-} = |E_{+} - E_{-}|/h = |ω_{0}|/(2π).

In general, the Bohr frequencies of a system whose energy eigenvalues are {E_{n}}
are defined as

f_{nn'} = |E_{n} - E_{n'}|/(2πħ).

Note: If **B** ≈ B_{0}**u** then H = ω_{0}S_{u}. For silver atoms γ
is negative, ω_{0} 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** ≈ B_{0}**k**
at t = 0.

The system evolves and we have

|ψ(t)> = U(t,0)|ψ(0)> = exp(-iHt/ħ)|ψ(0)> =
exp(-iω_{0}S_{z}t/ħ)|ψ(0)>.

|ψ(t)> = cos(θ/2)exp(-i(φ + ω_{0}t)/2)|+> + sin(θ/2)exp(i(φ
+ ω_{0}t)/2)|->

= cos(θ/2)exp(-iφ'/2)|+> + sin(θ/2)exp(iφ'/2)|->,

where φ' = φ + ω_{0}t.

|ψ(t)> is an eigenvector of S_{u’}
, (**u**' is characterized by θ and φ'), while |ψ(0)>
is an eigenvector of S_{u} , (**u** is characterized by θ and φ ).
The vector **u**' defining S_{u'} =
**S**∙**u**' revolves around the z-axis with
angular frequency ω_{0 }∝ B_{0}.
Its component
along the z-axis remains constant, while its component perpendicular to the z-axis
rotates with angular velocity ω_{0}. This is the
quantum mechanical analog of the rotation of the classical magnetic moment **m**
and is called
Larmor precession.

<S_{z}(t)> = <ψ(t)|S_{z}|ψ(t)> = (ħ/2)cos^{2}(θ/2)
- (ħ/2)sin^{2}(θ/2)

= (ħ/2)½(1 + cosθ) - (ħ/2)½(1 - cosθ) = (ħ/2)cosθ.

S_{z} is a constant of motion.

<S_{x}(t)> = <ψ(t)|S_{x}|ψ(t)> = cos^{2}(θ/2)<+|S_{x}|+>
+ sin^{2}(θ/2)<-|S_{x}|->

+ cos(θ/2)sin(θ/2)exp(iφ')<+|S_{x}|-> + cos(θ/2)sin(θ/2)exp(-iφ')<-|S_{x}|+>.

.

<S_{x}(t)> = (ħ/2)cos(θ/2)sin(θ/2) 2cosφ' = (ħ/2)sinθ
cos(φ + ω_{0}t).

Similarly, <S_{y}(t)> = (ħ/2)sinθ sin(φ + ω_{0}t).

S_{x} and S_{y} are not constants of motion.
Their
expectation values oscillate with the single Bohr frequency |ω_{0}|/(2π).
The mean values of S_{x}, S_{y},
and S_{z} 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 S_{x} operator. The particle is confined to a region with
a uniform magnetic field **B** = B_{0}**k**,
its Hamiltonian is H = ω_{0}S_{z}. The
eigenstates of H are |+> and |->,

H|+> = (ħω_{0}/2)|+>
and H|-> = -(ħω_{0}/2)|->.

|+>_{x}
can be written as a linear combination of eigenstates of H.

(a) Find the probability of measuring S_{x} = ħ/2 at t = T.

(b) What is the mean value of S_{x}, <S_{x}>, at
t = T?

(c) Find the probability of measuring S_{z} = 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 S_{z}of a spin ½ particle is two-dimensional. We denote the eigenvectors of S_{z}by |+> and |->. The Hamiltonian of the system is H = ω_{0}S_{z}. 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 S_{x}in the eigenbasis of S_{z}is

.

The eigenvectors of S_{x}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(-iω_{0}t/2)|+> + exp(-iω_{0}t/2)|->).

P(S_{x}=ħ/2,T) = |_{x}<+||ψ(t)>|^{2}= ¼|(<+| + <-|)(exp(-iω_{0}T/2)|+> + exp(-iω_{0}T/2)|->)|^{2 }= ¼|exp(-iω_{0}T/2)|+> + exp(-iω_{0}T/2)|^{2}= cos^{2}(ω_{0}T/2).(b) P(S

_{x}=-ħ/2,T) = 1 - cos^{2}(ω_{0}T/2) = sin^{2}(ω_{0}T/2).

<S_{x}(T)> = (ħ/2)[cos^{2}(ω_{0}T/2) - sin^{2}(ω_{0}T/2)] = (ħ/2)cos(ω_{0}T).(c) P(S

_{z}=ħ/2,T) = |<+|ψ(t)>|^{2}= ½|<+|(exp(-iω_{0}t/2)|+> + exp(-iω_{0}t/2)|->)|^{2}= ½.

The probability of measuring S_{z}= ħ/2 is independent of t.