Consider a collection of water molecules. Each proton in each molecule is a spin ½ particle with a
magnetic moment **m **= γ**S**,
γ = g_{n}q_{e}/m_{p}
(SI units).

If we place a sample in a uniform magnetic field **B **= (0,0,B_{0}),
then, neglecting the interaction between different particles, the Hamiltonian of
each proton is H = -**m∙B **= -m_{z}B_{0}.
(Here we treat the space coordinates macroscopically.)

H = -γB_{0}S_{z}
= -ω_{0}S_{z}.

The
eigenstates of H are the common eigenstates of S^{2} and S_{z},
{|+>, |->}. The eigenvalues
are E_{ +} = -ω_{0}ħ/2, E_{-} = +ω_{0}ħ/2.

Assume
a ** collection of water molecules**, placed in uniform magnetic field **B **= (0,0,B_{0}),
is in **thermal equilibrium** with its
surroundings. We then are dealing
with a ** statistical mixture of states**. The
**density
operator** for the system is ρ =
p_{+}|+><+| + p_{-}|-><-|.

p_{± }= N exp(-E_{±}/(kT)) = (1/Z) exp(-E_{±}/(kT)).

We
have Tr(ρ) = <+|ρ|+>
+ <-|ρ|-> = (1/Z)(exp(ω_{0}ħ/(2kT)) + exp(-ω_{0}ħ/(2kT))
= 1.

The
**
partition
function Z** is therefore given by Z = 2 cosh(ω_{0}ħ/(2kT)).

The
density matrix at t = 0 in the {|+>, |->} basis is given by

.

**How
does the density matrix evolve?**

(d/dt)ρ(t) = (1/(iħ))[H,ρ(t)]

We therefore have

dρ_{11}/dt = 0,
dρ_{22}/dt = 0,
dρ_{12}/dt = iω_{0}ρ_{12},
dρ_{21}/dt = -iω_{0}ρ_{21}.

This yields

We can now calculate <S_{x}>, <S_{y}>,
and <S_{z}> for our sample.

<S_{x}> = Tr(ρS_{x}).

Similarly, <S_{y}> = 0.

<S_{z}> = Tr(ρS_{xz}).

.

The macroscopic magnetization **M**
of the water sample is given by **M **= nγ<**S**>,
where n is the number of protons per unit volume. We therefore have **M **= (0,0,M_{z})
where M_{z} = nγ<S_{z}>.

Assume that at t = 0 a **time-varying
field** **B**_{1}(t) is added to **B**_{0}.
Assume that for t > 0 we have **B **= (B_{1}cosωt,
-B_{1}sinωt,
B_{0}). The z-component of **B**
is constant, while the component of **B**
perpendicular to the z-axis rotates cw about the z-axis with frequency ω.

Assume that B_{1 }<< B_{0}.
The proton Hamiltonian is now given by

H(t) = -γ**S**∙**B**(t) = -γB_{0}S_{z }- γB_{1}(cosωt S_{x }- sinωt
S_{y}) = -ω_{0}S_{z }- ω_{1}(cosωt S_{x }- sinωt
S_{y}),

with ω_{1 }= γB_{1}.

Consider
the state vector |ψ(t)>
describing a **single proton**, |ψ(t)> = U(t,0)|ψ(0)>.

Let
R(φ)
describe a ccw rotation about the z-axis and let U(R) be the rotation operator
for the ccw rotation R(φ).

U(R) = exp(-(i/ħ)S_{z}φ).

We
have

|ψ(t)> = U(t,0)|ψ(0)>, U(R)|ψ(t)> = U(R)U(t,0)|ψ(0)>.

Consider an **infinitesimal
rotation** and an infinitesimal time interval dt.

U(R)|ψ(dt)>
= (I - (i/ħ)S_{z}ωdt)|ψ(dt))>
= (I - (i/ħ)S_{z}ωdt)(I - (i/ħ)Hdt)|ψ(0)>.

U(R)|ψ(dt)>
= (I - (i/ħ)S_{z}ωdt)(I - (i/ħ)(-ω_{0}S_{z }- ω_{1}(cosωt S_{x}
- sinωt
S_{y}))dt)|ψ(0)>.

U(R)|ψ(dt)> = (I - (i/ħ)((ω - ω_{0})S_{z }- ω_{1}S_{x}cosωt
+ ω_{1}S_{y}sinωt)dt)|ψ(0)>.

**S** is a vector observable, its components transform
under a rotation R as V’ = R^{-1}V.

For a cw rotation R^{-1}(φ)
we have V_{x}’ = V_{x}cosφ - V_{y}sinφ.

_{x}’ = S_{x}cosωt - S_{y}sinωt
is the x-component of the observable **S**
in a frame that rotates cw with angular frequency ω
about the z-axis, i.e. in a frame that rotates with the magnetic field.

We therefore have U(R)|ψ(dt)> = |ψ'(dt)> = (I - (i/ħ)((ω - ω_{0})S_{z }
- ω_{1}S_{x}’)dt)|ψ(0)>.

|ψ'(dt)> = (I -(i/ħ)((ω - ω_{0})S_{z}’ - ω_{1}S_{x}’)dt)|ψ’(0)>

|ψ'> denotes the state vector in a frame rotating cw about the z-axis with angular
frequency ω.

In this frame the evolution operator is U(t,0) = exp(-(i/ħ)((ω - ω_{0})S_{z}’
- ω_{1}S_{x}’)t).

Let ω = ω_{0}.

Then |ψ'(dt)> = (I + (i/ħ)ω_{1}S_{x}’dt)|ψ’(0)>.
The infinitesimal evolution operator is I + (i/ħ)ω_{1}S_{x}’dt
and the evolution operator is U(t,0) = exp((i/ħ)ω_{1}S_{x}’t).

The Hamiltonian therefore is H’ = -ω_{1}S_{x}’.

The matrix of the Hamiltonian is

in the {|+>, |->} basis.

At t = 0 the
lab frame and the frame that rotates with **B**
coincide. The density matrix at t = 0
in the {|+>, |->} basis is given by

.

Since H’ = -ω_{1}S_{x}’
it is convenient to **change the basis**.
In the
{|+>_{x}, |->_{x}} eigenbasis of S_{x} the density
matrix is given by U^{†}ρ(0)U,
where U is the unitary transformation that transforms {|+>, |->} into
{|+>_{x}, |->_{x}}.

The
matrix of U is

.

In the {|+>_{x},
|->_{x}} basis we have

.

Here Δp =
p_{+ }- p_{-}.

The
matrix of S_{x} in the {|+>_{x}, |->_{x}} basis is

.

Let
us now calculate ρ(t)
in the reference frame rotating with **B**.

(d/dt)ρ'(t) = (1/(iħ))[H',ρ'(t)].

The
matrix of H' in the {|+>_{x}, |->_{x}} basis is

We therefore have

dρ_{11}/dt = 0, dρ_{22}/dt = 0, dρ_{12}/dt
= iω_{1}ρ_{12}, dρ_{21}/dt = -iω_{1}ρ_{21}.

This yields ρ_{11}(t) = 1/2, ρ_{22}(t) = 1/2,
ρ_{12}(t) = -exp(iω_{1}t)
Δp/2, ρ_{21}(t) = exp(-iω_{1}t)
Δp/2.

We can now calculate <S’_{x}>, <S’_{y}>,
and <S’_{z}> in the rotating frame.

<S’_{x}(t) > =Tr(ρ'(t)S’_{x}).

.

In the {|+>_{x},
|->_{x}} basis the matrix of S_{y} is
and the matrix of S_{z} is
.

In the rotating reference frame we therefore have

In the rotating frame the magnetization rotates clockwise
about the x’-axis. There are
times when the magnetization has only a z-component, and times when it entirely
lies in the plane perpendicular to the z-direction.

**What do we observe in the lab frame?
S**
is a

<S

R(φ) is a ccw rotation about the z-axis.

The matrix of R

We therefore have (with ω = ω

<S

<S

<S

The macroscopic magnetization

**NMR
**A continuous wave experiment can be performed by sweeping
the frequency of the rotating field B

Experiments can also be performed with pulsed magnetic
fields. Assume that at time t_{1 }= π/(2ω_{1})
we turn off B_{1}. Then for
t > t_{1} H = -ω_{0}S_{z}.
In the lab frame we have

<S_{x}> = (ħ/2)Δp sinω_{0}t_{1},
<S_{y}> = (ħ/2)Δp cosω_{0}t_{1},
<S_{z}> = 0 at t = t_{1}.

At t = t_{1} the magnetization lies in the x-y
plane. How does it evolve?

To find the density matrix ρ(t_{1})
in the lab frame in the {|+>, |->} basis we can use

<S_{x}> = Tr(ρ,S_{x}), or (ħ/2)Δp sinω_{0}t_{1 }= (ħ/2)(ρ_{12}
+ ρ_{21}) = ħ Re(ρ_{12}), Re(ρ_{12}) = ½Δpsinω_{0}t_{1},

<S_{y}> = Tr(ρ,S_{y}), or (ħ/2)Δp cosω_{0}t_{1 }=
i(ħ/2)(ρ_{12 }- ρ_{21}) = -ħ Im(ρ_{12}), Im(ρ_{12})
= -½Δpcosω_{0}t_{1},

<S_{z}> = Tr(ρ,S_{z}),
or 0 = (ħ/2)(ρ_{11 }- ρ_{22}), ρ_{11 }= ρ_{22 }= ½, since the trace is 1.

.

For t > t_{1} we have dρ_{11}/dt = 0, dρ_{22}/dt
= 0, dρ_{12}/dt = iω_{0}ρ_{12}, dρ_{21}/dt
= -iω_{0}ρ_{21},

which yields

For t > t_{1} we therefore have

<S_{x}(t)> = Tr(ρ(t),S_{x}) = (ħ/2)Δp sinωt,

<S_{y}(t)> = Tr(ρ(t)S_{y}) = (ħ/2)Δp cosωt,

<S_{z}(t)> = 0.

The magnetization lies in the x-y plane and rotates cw
about the z-axis. As the
magnetization rotate about the z-axis, it will induce a current in a pick-up coil
located with its axis along the x-axis. Plotting
this current versus time yields a sinusoidal wave form.
This current will decay as a function of time due to dephasing.
This is called ** free-induction decay**.
The spin-spin relaxation time T_{2} is the time with which
the transverse magnetization decays. M_{xy} = M_{xy0}e^{-t/T2}.
Loss of transverse magnetization is due to molecular interactions and
field inhomogeneity.

At equilibrium the net magnetization is **M **= (0,0,M_{z}). As
we have seen, it is possible to make M_{z} = 0.
The time constant with which M_{z} returns back to its
equilibrium value is called the spin-lattice relaxation constant T_{1}. M_{z }= M_{0}(1 - e-^{t/T1}). Spin lattice
relaxation is the result of the nuclei transferring energy to the surrounding
molecules as thermal energy.

In pure water
T_{1 }~ T_{2 }~ 2s - 3s. In biological materials T_{2} < T_{1}.

**Reference:**

**The
Basics of MRI**

Note
the chapters on spin physics and NMR spectroscopy