Magnetic Dipole Moment & Gyromagnetic Ratio

Leave a comment

I always confuses on the definition, and wiki did not have any summary. so,

The Original definition is the Hamiltonian of a magnetic dipole under external magnetic field \vec{B},

H = -\vec{\mu}\cdot \vec{B},

where \vec{\mu} is magnetic dipole moment (MDM). It is

\vec{\mu} = g \frac{q}{2 m} \vec{J} = g \frac{\mu}{\hbar} \vec{J} = \gamma \vec{J}.

Here, the g is the g-factor, \mu is magneton, and \vec{J} is the total spin, which has a intrinsic factor m\hbar / 2 inside. \gamma is gyromegnetic ratio.

We can see, the g-factor depends on the motion or geometry of the MDM. For a point particle, the g-factor is exactly equal to 2. For a charged particle orbiting, the g-factor is 1.

Put everything into the Hamiltonian,

H = -\gamma \vec{J}\cdot \vec{B} = -\gamma J_z B = -\gamma \hbar \frac{m}{2} B [J],

Because energy is also equal E = \hbar f , thus, we can see the \gamma has unit of frequency over Tesla.

———————————–

Take electron as an example, the MDM is Bohr magneton \mu_{e} = e\hbar/(2m_e). The MDM is,

\vec{\mu_e} = g_e \frac{e}{2 m_e} \vec{S} = g_e \frac{\mu_e}{\hbar}\vec{S} = \gamma_e \vec{S}.

The magnitude of MDM is,

|\vec{\mu_e}|= g_e \frac{e}{2 m_e} \frac{\hbar}{2} = \gamma_e \frac{\hbar}{2} [JT^{-1}],

The gyromagnetic ratio is,

\gamma_e = g_e \frac{\mu_e}{\hbar} [rad s^{-1} T^{-1}].

Since using rad s^{-1} is not convenient for experiment. The gyromagnetic ratio usually divided by 2\pi,

\gamma_e = g_e \frac{\mu_e}{2\pi\hbar} [Hz T^{-1}].

————————————-

To evaluate the magnitude of  MDM of  single particle state, which has orbital angular momentum and spin, the total spin \vec{J} = \vec{L} + \vec{S}. However, the g-factor for \vec{L} is difference from that for \vec{S}. Thus, the MDM is not parallel to total spin. We have to use Landé Formula,

\left< JM|\vec{V}|JM'\right> = \frac{1}{J(J+1)} \left< JM|(\vec{J}\cdot\vec{V})|JM\right> \left<JM|\vec{J}|JM'\right>

or see wiki, sorry for my laziness.

The result is

g=g_L\frac{J(J+1)+L(L+1)-S(S+1)}{2J(J+1)}+g_S\frac{J(J+1)-L(L+1)+S(S+1)}{2J(J+1)}

For J = L \pm 1/2,

g = J(g_L \pm \frac{g_S-g_L}{2L+1})

Hartmann-Hahn Matching

Leave a comment

in dynamic nuclear polarization by microwave induced cross-polarization ( polarization transfer from one spin to another, for example, from ^1H to ^{13}C ), the condition is called Hartmann-Hahn matching.

The Hamiltonian of the 2 kind of spin in lab frame is:

H_0 = \omega_s S_z + \omega_I I_z + (A+B) S_z I_z

where A and B are scalar coupling when the 2 spin in contact and the dipolar interaction.

A = \frac{4\pi}{3} \gamma_s \gamma_I \rho(x)

B = \frac{1}{r_{sI}^2} \gamma_s \gamma_I ( 1- 3 cos^2(\theta_{sI}))

the angle in dipolar interaction is relative to the direction of the external B-field. when the angle is at:

cos^2(\theta_{sI}) = 1/3

the dipolar interaction disappear and the spectrum line get thinner, and higher resolution. this is called the Magic Angle.

when a transverse oscillating field applied, and only affect the S spin, then the total Hamiltonian is :

H= H_0 + \omega_R U^{-1}S_x U

U= e^{-\frac{i}{\hbar} \omega t S_z}

where U is the rotation operator. a standard method is switch to the rotating frame along with the transverse field. the Hamiltonian in the rotating frame is:

H_R = \omega S_z+U H U^{-1}

H_R =(\omega + \omega_s) S_z + \omega_R S_x + \omega_I I_z + (A+B) S_z I_z

we can see that, during the Spin-Lock, i.e. \omega + \omega_s = 0 , the longitudinal component of  S spin gone. But in general, it is not the case, thus, we have the rotation axis of S spin is titled. we can simplify the Hamiltonian by transform it in a titled coordinate, by another unitary transform which rotate on Sy axis.

\tilde{U} = e^{\frac{i}{\hbar} \phi S_y }

tan(\phi) = \frac{\omega_R}{\omega+\omega_s}

in this tilted axis, the rotating axis is on the \tilde{S_z} axis with magnitude:

\omega_{eff} = \sqrt{ (\omega + \omega_s)^2 + \omega_R^2 }

the tilted Hamiltonian is :

\tilde{H} = \omega_{eff} \tilde{S_z} + \omega_I I_z + (A+\tilde{B}) I_z ( \tilde{S_z} cos(\phi) - \tilde{S_x} sin(\phi) )

For the interaction terms are small, the energy level is just like ordinary 2 spin system. but when

 \omega_{eff} = \omega_I

which is the Hartmann-Hahn matching, the flip-flop exchange of the spin no need any energy and then the spin transfer. on the other hand, if:

 \omega_{eff} = - \omega_I

the flip-flip forbidden transition happen.

In the case of electron spin to proton spin, if we apply a ESR freqeuncy, which is GHz order, so it is microwave, the power of the microwave have to be matched to the proton Larmor frequency.

(\omega_{\mu w} - \gamma_s H )^2 + k P_{\mu w} = \gamma_I^2 H^2

here i used the microwave wave B-field strength is proportional to the  voltage applied, and power is proportional to the square of voltage.

On NMR signal

Leave a comment

The NMR signal is obtained by the coil, which also generate the Rabi field or a radio frequency to flip the spin.

the origin of the NMR signal is the transverse magnetization. for spin-½ system. the transverse component of the magnetization is:

M_T = ( M_x, M_y ) = A ( cos(\omega_0 t), sin(\omega_0 t))

where A is the amplitude and \omega_0 is the Larmor frquency. for consistency and cross reference in this blog, i keep the 0 with the \omega .

the magnetization is proportional a changing magnetic field. a changing magnetic field will induce an e.m.f on a coil. if the coil is perpendicular to an oscillating magnetic field a maximum e.m.f will be obtained. however, since the magnetization is rotating, the coil can be point at any direction to give the same e.m.f. . without lost of generality, the coil will define the x-axis of the system.

B = B_{NMR} ( cos (\omega_0 t ), sin ( \omega_0 t) )

and the Maxwell’s equation:

\nabla \times E = \frac { d}{dt} B

\nabla \times E = B_{NMR} \omega_0 ( - sin (\omega_0 t), cos(\omega_0 t))

we can see that the amplitude of the E field in the coil, which is the NMR signal strength, is depending on the Larmor frequency \omega_0 . That explained why NMR always looking for strong magnetic field, now can go to 22 Tesla ( earth magnetic field is just 5 \times 10^{-5} Tesla ), a higher magnetic field strength, the higher Larmor frequency, and a stronger signal.

Moreover, the magnetic field produced by the sample is proportional to number of NMR center, the polarization and a factor on how the spin ensemble to combine to be a giant single field. and also, the change of the flux of the NMR coil is depends on how the area was integrated. These all factor are not just related to the NMR coil but also on the particular sample.

detail treatment on Larmor Precession and Rabi Resonance

Leave a comment

a treatment on Larmor Precession and Rabi resonance

the pdf is a work on this topic. it goes through Larmor Precession and give example on spin-½ and spin-1 system.

then it introduce Density matrix and gives some example.

The Rabi resonance was treated by rotating frame method and using density matrix on discussion.

the last topic is on the relaxation.

the purpose of study it extensively, is the understanding on NMR.

the NMR signal is the transverse component of the magnetization.

on angular momentum adding & rotation operator

Leave a comment

the angular momentum has 2 kinds – orbital angular momentum L , which is caused by a charged particle executing orbital motion, since there are 3 dimension space. and spin S , which is an internal degree of freedom to let particle “orbiting” at there.

thus, a general quantum state for a particle should not just for the spatial part and the time part. but also the spin, since a complete state should contains all degree of freedom.

\left| \Psi \right> = \left| x,t \right> \bigotimes \left| s \right>

when we “add” the orbital angular momentum and the spin together, actually, we are doing:

J = L \bigotimes 1 + 1 \bigotimes S

where the 1 with L is the identity of the spin-space and the 1 with S is the identity of the 3-D space.

the above was discussed on J.J. Sakurai’s book.

the mathematics of L and S are completely the same at rotation operator.

R_J (\theta) = Exp( - \frac {i}{\hbar} \theta J)

where J can be either L or S.

the L can only have effect on spatial state while S can only have effect on the spin-state. i.e:

R_L(\theta) \left| s \right> = \left| s\right>

R_S(\theta) \left| x \right> = \left| x\right>

the L_z can only have integral value but S_z can be both half-integral and integral. the half-integral value of Sz makes the spin-state have to rotate 2 cycles in order to be the same again.

thus, if the different of L and S is just man-made. The degree of freedom in the spin-space is actually by some real geometry on higher dimension. and actually, the orbital angular momentum can change the spin state:

L \left| s \right> = \left | s' \right > = c \left| s \right>

but the effect is so small and

R_L (\theta) \left| s\right > = Exp( - \frac {i}{\hbar} \theta c )\left| s \right>

but the c is very small, but if we can rotate the state for a very large angle, the effect of it can be seen by compare to the rotation by spin.

\left < R_L(\omega t) + R_S(\omega t) \right> = 2 ( 1+ cos ( \omega ( c -1 ) t)

the experiment can be done as follow. we apply a rotating magnetic field at the same frequency as the Larmor frequency. at a very low temperature, the spin was isolated and T_1 and T_2 is equal to \infty . the different in the c will come up at very long time measurement and it exhibit a interference pattern.

if c is a complex number, it will cause a decay, and it will be reflected in the interference pattern.

if we find out this c, then we can reveal the other spacial dimension!

___________________________________

the problem is. How can we act the orbital angular momentum on the spin with out the effect of spin angular momentum? since L and S always coupled.

one possibility is make the S zero. in the system of electron and positron. the total spin is zero.

another possibility is act the S on the spatial part. and this will change the energy level.

__________________________________

an more fundamental problem is, why L and S commute? the possible of writing this

\left| \Psi \right> = \left| x,t \right> \bigotimes \left| s \right>

is due to the operators are commute to each other. by why?

if we break down the L in to position operator x and momentum operator p, the question becomes, why x and S commute or p and S commute?

[x,S]=0 ?

[p,S]=0 ?

[p_x, S_y] \ne 0 ?

i will prove it later.

___________________________________

another problem is, how to evaluate the Poisson bracket? since L and S is not same dimension. may be we can write the eigenket in vector form:

\begin {pmatrix} \left|x, t \right> \\ \left|s\right> \end {pmatrix}

i am not sure.

 

___________________________________

For any vector operator, it must satisfy following equation, due to rotation symmetry.

[V_i, J_j] = i \hbar V_k   run in cyclic

Thus,

where J is rotation operator. but i am not sure is it restricted to real space rotation. any way, spin is a vector operator, thus

$latex [S_x, L_y] = i \hbar S_z = – [S_y, L_x] $

so, L, S is not commute.

on Relaxation in NMR

Leave a comment

If we only switch on the transverse magnetic field for some time \tau . 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.

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 transverse B field frequency is not equal to the Larmor frequency , and the pule is not a π-pulse, the spin has chance to go to the other state, which probability is given by a formula. 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.

NMR (nuclear magnetic resonance)

Leave a comment

NMR is a technique to detect the state of nuclear spin. a similar technique on electron spin is call ESR ( electron spin resonance)

The principle of NMR is simple.

  1. apply a B-field, and the spin will align with it due to interaction with surrounding and precessing along the B-field with Larmor frequency, and go to Boltzmann equilibrium. the time for the spin align with the field is call T1, longitudinal relaxation time.
  2. Then, we send a pule perpendicular to the B-field, it usually a radio frequency pulse. the frequency is determined by the resonance frequency, which is same as the Larmor frequency. the function of this pulse is from the B-field of it and this perpendicular B-field with perturb the spin and flip it 90 degrees.
  3. when the spin are rotate at 90 degrees with the static B-field, it will generate a strong enough signal around the coil. ( which is the same coil to generate the pule ) and this signal is called NMR signal.
  4. since the spins will be affected by its environment, and experience a slightly different precession frequency. when the time goes, they will not aligned well, some precess faster, some slower. thus, the transverse magnetization will lost and look as if it decay. the time for this is called T2, transverse relaxation time.

by analyzing the T1 and T2 and also Larmor frequency, we can known the spin, the magnetization, the structure of the sample, the chemical element, the chemical formula, and alot many others thing by different kinds of techniques.

For nuclear physics, the use of NMR is for understand the nuclear spin. for example, the polarization of the spin.

 

 

Older Entries