## Magnetic Dipole Moment & Gyromagnetic Ratio

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.

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

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

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

## First experiment of 6He with a polarized proton target

this paper reported a first spin polarized proton solid target under low magnetic field ( 0.08 T ) and hight temperature ( 100K )

the introduction overview the motivation of a solid target.

• a polarized gas target is ready on many nuclear experiment.
• on the radioactive beam ( IR beam ), the flux of a typical IR beam is small, since it is produced by 2nd scattering.
• a solid target has highest density of solid.
• most solid target can only be polarized on low temperature ( to avoid environmental interaction to reduced the polarization )
• increase the experimental difficult, since a low temperature should be applied by a cold buffer gas.
• high field ( the low gyromagnetic  ratio ).
• high magnetic field make low energy scattered proton cannot get out from the magnetic field and not able to detect.
• a solid target can be polarized at high temperature and low magnetic field is very useful

the material on use is a crystal of naphthalene doped with pentacene.

the procedure of polarizing the proton is :

1. use optical pumping the polarize the electron of pentacene
• the population of the energy states are independent of temperature and magnetic field.
2. by Dynamic Nuclear Polarization (DNP) method  to transfer  the polarization of the electron to the proton.
• if the polarization transfer is 100% and the relaxation time is very long. the expected polarization of proton will be 72.8%

The DNP method is archived under a constant microwave frequency with a sweeping magnetic field. when the magnetic field and  microwave frequency is coupled. the polarization transfer will take place.

the next paragraph talks about the apparatus’s size and dimension, in order to fit the scattering experiment requirements.

the polarization measurement is on a scattering experiment with 6He at 71 MeV per nucleons. By measuring the polarization asymmetry $\epsilon$, which is related to the yield. and it also equal to the polarization of the target $P_t$  times the analyzing power $A_y$. $\epsilon = P_t \times A_y$

with a reasonable guess of the target polarization. the analyzing power of  6He was found.

the reason why the polarization-asymmetry is not equal to the analyzing power is that, the target is not 100% polarized, where the analyzing power is defined. when the polarization of the target is 100%, both are the same.

in the analysis part. it used optical model and Wood-Saxon central potential to simulate the result. And compare the result from 6He to 6Li at same energy. the root mean square of 6Li is larger then 6He. it suggest the d-α core of 6Li may responsible for that.

they cannot go further discussion due to the uncertainly on the polarization of the target.

## Larmor Precession (quick)

Magnetic moment ( $\mu$) :

this is a magnet by angular momentum of charge or spin. its value is: $\mu = \gamma J$

where $J$ is angular momentum, and $\gamma$ is the gyromagnetic rato $\gamma = g \mu_B$

Notice that we are using natural unit.

the g is the g-factor is a dimensionless number, which reflect the environment of the spin, for orbital angular momentum, g = 1. $\mu_B$ is Bohr magneton, which is equal to $\mu_B = \frac {e} {2 m}$ for positron

since different particle has different mass, their Bohr magneton value are different. electron is the lightest particle, so, it has largest value on Bohr magneton.

Larmor frequency:

When applied a magnetic field on a magnetic moment, the field will cause the moment precess around the axis of the field. the precession frequency is called Larmor frequency.

the precession can be understood in classical way or QM way.

Classical way:

the change of angular momentum is equal to the applied torque. and the torque is equal to the magnetic moment  cross product with the magnetic field. when in classical frame, the angular momentum, magnetic moment, and magnetic field are ordinary vector. $\vec {\Gamma}= \frac { d \vec{J}}{dt} = \vec{\mu} \times \vec{B} = \gamma \vec {J} \times \vec{B}$

solving gives the procession frequency is : $\omega = - \gamma B$

the minus sign is very important, it indicated that the J is precessing by right hand rule when $\omega >0$.

QM way:

The Tim dependent Schrödinger equation (TDSE) is : $i \frac {d}{d t} \left| \Psi\right> = H \left|\Psi\right>$

H is the Hamiltonian, for the magnetic field is pointing along the z-axis. $H = -\mu \cdot B = - \gamma J\cdot B = -gamma B J_z = \omega J_z$

the solution is $\left|\Psi(t) \right> = Exp( - i \omega t J_z) \left| \Psi(0) \right>$

Thus, in QM point of view, the state does not “rotate” but only a phase change.

However, the rotation operator on z-axis is $R_z ( \theta ) = Exp( - i \frac {\theta}{\hbar} J_z )$

Thus, the solution can be rewritten as: $\left|\Psi (t)\right> = R_z( \omega t) \left|\Psi(0)\right>$

That makes great analogy on rotation on a real vector.