Larmor Precession

The Hamiltonian and Schrodinger equation

we use the notation that

The spin function is

Thus the population never change.

 for , ,the precession is on the back-ward, or left hand.

where g is the g-factor

for electron, e is negative, thus, .

And the magnetic moment is

when  ,

the Hamiltonian is

Thus, for  parallel to B will give minimum energy.

for  is anti-parallel to B will give maximum energy.

in order to see the rotation, we have to find the expectation of J, or the magnetization. In Classical analogy, it is equivalent to the Bloch vector.

Thus, for , it rotates on right-hand direction.

Notice : some reader may confues on the rotation direction. How come the J operator rotating backward but the expectation value moving forward? we can think that, the operator are stick with the coordinate, in fact, the operator define the coordinate axis. thus, the coordinate axis moving backward, the coordinate are moving forward.

Density Matrix

there are other way to visualize the effect of the spin polarization, by the density matrix.

the density matrix is defined as

notice, the density matrix is taken not on single spin but on ensemble, thus, the average is automatically applied.

which are the population on state up and state down.

which is the magnetization along z-axis.

which is the coherent, it is the degree of transverse magnetization.

for initial state

Thus, the magnetization along z-axis is

Let’s check the consistency with , which is the magnetization for single spin.

or

Thus, the direction of the magnetization is given by the phase different of up state and down state.

which agree. for a state with

for Larmor precession.

the state is:

thus, for a mixed state, it rotate right hand for

For no magnetization,

Rabi Resonance

Since the angular momentum is rotating right hand for . thus, our perpendicular B-field should be apply on right hand.

and it make a very good sense, since  

represents Jz rotating in positive direction.

But solving this is very inconvenient, but still possible.

when we change our frame to the rotating frame that move with the rotating B-field.

Under the rotating frame. the rotating Hamiltonian is :

Thus, when , The angular momentum is rotating on x-axis with frequency .

For , the angular momentum is rotating on the shifted axis , with effective frequency

To find the probability of the angular momentum on the each state. we have to solve the state.

(be careful, cannot use the  to this case, since Jx and Jz nto commute )

First, we find of the Hamiltonian in

The initial condition is all up state at the beginning.

to solve this, there are at least 3 methods, one is diagonalize the matrix, solve the coupled equations directly, and using time dependent eigen state as basic and find out the coupled equations.

the probability of going to down state or spin flip is

The magnetization is :

which trace a circle around the shifted axis.

For ,

which is rotate in right hand around x-axis.

To find the state in Lab frame

which is simple add a phase.

The magnetization in the Lab frame :

For ,

which is the oscillation of 2 frequencies - the Lissajous curve on sphere.

Relaxation

If we only switch on the transverse magnetic field for some time . after the field is off, the system will go back to the thermal equilibrium. it is due to the system is not completely isolated.

instead of consider a single spin, we have to consider the ensemble. and an ensemble is describe by the density matrix or the magnetization. which are visible effect by the spin.

the reason for not consider a single spin state is, we don’t know what is going on for individual spin. in fact, in the previous section, the magnetization is a Marco effect. a single spin cannot have so many states, it can only have 2 states - up or down. if we insist the above calculation is on one spin, thus, it only give the chance for having that direction of polarization. which, is from many measurements.

so, for a single spin, the spin can only have 2 states. and if the , the spin has chance to go to the other state, which probability is given by above. and when it goes to relax back to the minimum energy state, it will emit a photon. but when it happen, we don’t know, it is a complete random process.

However, an ensemble, a collection of spins, we can have some statistic on it. for example, the relaxation time, T1 and T2.

any decay is represented by

where  is relaxation time.

since T1 is the longitudinal relaxation time and T2 is transverse relaxation time.

thus, if we set the lowest energy state is the rest state.

in NMR experiment,