Wigner-Eckart theorem

Leave a comment

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) )

Larmor Precession (quick)

1 Comment

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.