## Wigner-Eckart theorem

The simplest way to say is:

a operator can be projected on another one, for example, The orbital angular momentum cab be projected on the total angular momentum.

$L = L\cdot J \frac {J}{j(j+1)}$

a simple application is on the Zeeman effect on spin-orbital coupling. the Hamiltonian is:

$H_B = - \mu \cdot B = - ( \mu_l L + \mu_s S ) \cdot B$

by the Wigner-Eckart theorem:

$L = L\cdot J \frac {J}{j(j+1)}$

$S = S\cdot J \frac {J}{j(j+1)}$

then the Hamiltonian becomes:

$H_B = - \frac{1}{j(j+1)} ( \mu_l (L \cdot J) + \mu_s (S \cdot J) ) J\cdot B$

and introduce the Bohr Magneton and g-factor:

$H_B = - g \mu_B J \cdot B$

$g = - \frac{1}{j(j+1)} ( g_l (L \cdot J) + g_s (S \cdot J) )$

# Motivation

In order to convert the NMR single to the absolute magnetic field strength of the polarization. the thermal polarization ( polarization at thermal equilibrium ) should be measure and used to calibrate the NMR signal.

the thermal polarization is given by the Boltzmann statistic. The excited and the ground state population is:

$\frac{ N_{\uparrow}} { N_{\downarrow} } = Exp \left( - 2 \frac { \mu_p B } {k_B T } \right)$

the thermal polarization is the ratio of the different of spin-up and spin-down to the total spin.

$P_{thermal} (B , T ) = tanh \left( \frac{ \mu_p B }{ k_B T } \right)$

where $\mu=p$ is the proton magnetic moment.

$\mu_p = g_p \mu_N = g_p \frac{e \hbar }{2 m_p} = 1.410606662 \times 10^{-26} J T^{-1}$

$k_B$ is the Boltzmann constant.

$k_B = 1.3806504 \times 10^{-23} J K^{-1}$

The proton magnetic moment is small, the thermal polarization can be approximated as a linear function:

$P_{thermal} ( B, T) = \frac{\mu_p B}{ k_B T}$

since our polarization is on $T = -5 ^oC$ and $B = 0.05 T$, thus, the thermal polarization is:

$P_{thermal} (0.05, 268.15) = 1.90509 \times 10^{-7}$

which is very small to be detected, or to say, the signal is smaller then the noise level.

the small system has a better sensitivity, down to $10^{-7}$.

# TO-do

• Connect the Static field power and water cooling system
• connect the Controler Unit
• connect the magnetic field sweeping
• Test Run

## 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.