Consider a collection of water molecules.  Each proton in each molecule is a spin ½ particle with a magnetic moment m=gS, g=g_nq_e/m_p (SI units).

If we place a sample in a uniform magnetic field B=(0,0,B₀), then, neglecting the interaction between different particles, the Hamiltonian of each proton is H=-m×B=-m_zB₀.  (Here we treat the space coordinates macroscopically.)

-gB₀S_z =-w₀S_z.

The eigenstates of H are the common eigenstates of S² and S_z, {|+>, |->}.  The eigenvalues are E₊=-w₀h/2, E_-=+w₀h/2.

Assume a collection of water molecules, placed in uniform magnetic field B=(0,0,B₀), is in thermal equilibrium with its surroundings.  We then are dealing with a statistical mixture of states.  The density operator for the system is r = p₊|+><+| + p_-|-><-|.

p_±=Nexp(-E_±/(kT))=(1/Z)exp(-E_±/(kT)).

We have Tr(r)=<+|r|+> + <-|r|-> = (1/Z)(exp(w₀h/(2kT))+exp(-w₀h/(2kT)) = 1.

The partition function Z is therefore given by Z=2cosh(w₀h/(2kT)).

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

.

How does the density matrix evolve?

.

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

.

We therefore have

dr₁₁/dt=0,  dr₂₂/dt=0,  dr₁₂/dt=iw₀r₁₂,  dr₂₁/dt= -iw₀r₂₁.

This yields 

  .

We can now calculate <S_x>, <S_y>, and <S_z> for our sample.

<S_x>=Tr(rS_x).

.

Similarly, <S_y>=0.

<S_z>=Tr(rS_xz). 

.

The macroscopic magnetization M of the water sample is given by M=ngB₀<S>, where n is the number of protons per unit volume. We therefore have M=(0,0,M_z) where M_z= ngB₀<S_z>.

Assume that at t=0 a time-varying field B₁(t) is added to B₀.  Assume that for t>0 we have B=(B₁coswt, -B₁sinwt, B₀).  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 w.

Assume that B₁<<B₀.  The proton Hamiltonian is now given by

H(t)=-gS×B(t)=-gB₀S_z-gB₁(coswt S_x-sinwt S_y)=-w₀S_z-w₁(coswt S_x-sinwt S_y)

with w₁=gB₁.

Consider the state vector |y(t)> describing a single proton.

|y(t)>=U(t,0)|y(0)>.

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

U(R)=exp(-(i/h)S_zf)

We have

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

U(R) can be viewed as an operator that rotates every state vector ccw through an angle f or an operator that changes the basis vectors and therefore rotates the coordinate system cw through an angle f.

Let f=wt, then U(R) rotates the coordinate system cw with angular frequency w.  The coordinate system thus rotates with the magnetic field cw about the z-axis.

Consider an infinitesimal rotation and an infinitesimal time interval dt.

U(R)|y(dt)> = (I-(i/h)S_zwdt)|y(dt)> = (I-(i/h)S_zwdt)(I-(i/h)Hdt)|y(0)>

U(R)|y(dt)> = (I-(i/h)S_zwdt)(I-(i/h)(-w₀S_z-w₁(coswt S_x-sinwt S_y))dt)|y(0)>

U(R)|y(dt)>=(I-(i/h)((w-w₀)S_z-w₁S_xcoswt+w₁S_ysinwt)dt)|y(0)>

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

For a cw rotation R^-1(f) we have V_x’=V_xcosf-V_ysinf.

Therefore S_x’=S_xcoswt-S_ysinwt is the x-component of the observable S in a frame that rotates cw with angular frequency w about the z-axis, i.e. in a frame that rotates with the magnetic field.  We therefore have

U(R)|y(dt)>=|y'(dt)>=(I-(i/h)((w-w₀)S_z-w₁S_x’)dt)|y(0)>

|y'(dt)>=(I-(i/h)((w-w₀)S_z’-w₁S_x’)dt)|y’(0)>

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

In this frame the evolution operator is U(t,0)=exp(-(i/h)((w-w₀)S_z’-w₁S_x’)t).

Let w=w₀. 

Then |y'(dt)>=(I+(i/h)w₁S_x’dt)|y’(0)>.  The infinitesimal evolution operator is I+(i/h)w₁S_x’dt and the evolution operator is U(t,0)=exp((i/h)w₁S_x’t).

The Hamiltonian therefore is H’=-w₁S_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’=-w₁S_x’ it is convenient to change the basis.  In the {|+>_x, |->_x} eigenbasis of S_x is given by U^Tr(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 Dp= p₊-p_-.

The matrix of S_x in the {|+>_x, |->_x} basis is >

. 

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

.

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

.

We therefore have

dr₁₁/dt=0,  dr₂₂/dt=0,  dr₁₂/dt=iw₁r₁₂,  dr₂₁/dt= -iw₁r₂₁.

This yields  r₁₁(t)=1/2,  r₂₂(t)=1/2,  r₁₂(t)=exp(iw₁t) Dp/2,  r₂₁(t)= exp(-iw₁t) Dp/2.

We can now calculate <S’_x>, <S’_y>, and <S’_z> in the rotating frame.

<S’_x(t)>=Tr(r'(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 vector observable.  The expectation values of the components of a  of a vector observable behave under rotation as though the observable were a classical vector.  We have

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

The matrix of R^-1 is  

.

We therefore have (with w=w₀)

<S_x>=<S’_x>cosw₀t+<S_y’>sinw₀t=(h/2)Dpsinw₁tsinw₀t,

<S_y>=-<S’_x>sinw₀t+<S_y’>cosw₀t=(h/2)Dpsinw₁tcosw₀t,

<S_z>=<S’_z>=(h/2)Dpcosw₁t.

The macroscopic magnetization M=ngB₀<S> of the water sample changes from ngB₀(h/2)Dp to -ngB₀(h/2)Dp and back.  Since B₁<<B₀ and the eigenstate of the unperturbed Hamiltonian are not degenerate, the eigenstates of the perturbed Hamiltonian are approximately equal to |+> and |->, and the eigenvalues are approximately -(hw₀/2) and +(hw₀/2).  When the magnetization changes from ngB₀(h/2)Dp to -ngB₀(h/2)Dp, many protons make transitions from the |+> to the |-> state.  They have to absorb energy from the field.  When the magnetization changes back to ngB₀(h/2)Dp, energy is dumped back into the field.

NMR

A continuous wave experiment can be performed by sweeping the frequency of the rotating field B₁ through the resonance frequency w₀ and by monitoring the power output of the power supply that supplies the current that produces B₁.  Oscillations in the power output indicate that the frequency is sweeping through the resonance frequency.  If g is known, w₀=gB₀ yields B₀.  In this way the strength of an unknown magnetic field can be measured.  If B₀ is known, w₀=gB₀ yields g.  Instead of sweeping the frequency w of the rotating field, a continuous wave experiment can also be performed by sweeping B₀ while holding w constant.

Experiments can also be performed with pulsed magnetic fields.  Assume that at time t₁=p/(2w₁) we turn off B₁.  Then for t>t₁ H=-w₀S_z.  In the lab frame we have

<S_x>=(h/2)Dpsinw₀t₁, <S_y>=(h/2)Dpcosw₀t₁, <S_z>=0

at t=t₁.  

At t=t₁ the magnetization lies in the x-y plane.  How does it evolve?

To find the density matrix r(t₁) in the lab frame in the {|+>, |->} basis we can use

<S_x>=Tr(r,S_x), or (h/2)Dpsinw₀t₁=(h/2)(r₁₂+r₂₁)= hRe(r₁₂),  Re(r₁₂)=½Dpsinw₀t₁,

<S_y>=Tr(r,S_y), or (h/2)Dpcosw₀t₁=i(h/2)(r₁₂-r₂₁)=-hIm(r₁₂),  Im(r₁₂)=-½Dpcosw₀t₁,

<S_z>=Tr(r,S_z), or 0=(h/2)(r₁₁-r₂₂),  r₁₁=r₂₂=½, since the trace is 1.

For t>t₁ we have dr₁₁/dt=0,  dr₂₂/dt=0,  dr₁₂/dt=iw₀r₁₂,  dr₂₁/dt= -iw₀r₂₁,

which yields

.

For t>t₁ we therefore have

<S_x(t)>=Tr(r(t),S_x)=(h/2)Dpsinwt,

<S_y(t)>=Tr(r(t)S_y)=(h/2)Dpcoswt,

<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 T2 is the time with which the transverse magnetization decays.  M_xy=M_xy0e^-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 T1.  M_z=M₀(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 T1~T2~2-3s. In biological materials T2<T1.

References: