Wolfram Alpha:

```Orbital Angular Momentum
-------------------------

Orbiting electron with mass, m, and velocity, v, can be viewed as a
current I around a loop with area A.  The magnetic moment, μ is
given by,

μ = IA and is a vector perpendicular to the plane of the loop.

The angular momentum of a particle is L = mvr (a vector)

I = e/T = ev/2πr

This can be rewritten as,

I = emvr/2πmr2

μ = IA = eL/2m since L = mvr and A = πr2

Solutions to the Schrodinger Equation for the Hydrogen atom show
that |L| is quantized.

|μ| = (e/2m)√l(l + 1)h where l is the ORBITAL QUANTUM NUMBER

= √l(l + 1)μB where μB is the BOHR MAGNETON = eh/2m

Also, from the analysis of the Hydrogen atom we get the MAGNETIC
(AZIMUTHAL) QUANTUM NUMBER, ml, which corresponds to the
quantization of the z-component of angular momentum.  ml can take
the values of,

-l, -l+1,...., l-1, l

Thus, for l = 2 we get,

-2, -1, 0, 1, 2

This can be represented as:

The magnitude of z-component of the magnetic moment is given by

|μz| = mlμB

Thus, the projection of orbital angular momentun, L, onto the z axis
is quantized in units of h.

Electron Spin
-------------

Stern-Gerlach showed experimentally that electrons possessed an intrinsic
angular momentum and a magnetic moment. Classically this could occur if
the electron were a spinning ball of charge, and this property was called
orbital angular momentum, and if this electron spin behaved the same way,
an angular momentum quantum number s = 1/2 was required to give just
two states.

Electrons have intrinsic angular momentum characterized by quantum number
1/2. In the pattern of other quantized angular momenta, this gives total
angular momentum of:

S2  = S(S + 1)h2

= (1/2)(1/2 + 1)h2

= (3/4)h2

|S| = √(3)/2h

The resulting fine structure which is observed corresponds to two
possibilities for the z-component of the angular momentum.

Sz = +/-(1/2)h

Combining Orbital and Spin Angular Momentum
-------------------------------------------

This diagram can be seen as describing a single electron,  or multiple
electrons for which the spin and orbital angular momenta have been
combined to produce composite angular momenta, S and L respectively.

The combination is a special kind of vector addition as is illustrated for
the single electron case l=1 and s=1/2.  As in the case of the orbital
angular momentum alone, the projection of the total angular
momentum along a direction in space is quantized to values differeing
by one unit of angular momentum.

Angular Momentum in a Magnetic Field
------------------------------------

Once you have combined orbital and spin angular momenta according
to the vector model, the resulting total angular momentum can be
visualized as precessing about any externally applied magnetic field
as shown in the following diagram.

Larmor Frequency
-------------------

When you have a magnetic moment directed at some finite angle
with respect to the magnetic field direction, the field will
exert a torque on the magnetic moment. This causes it to precess
about the magnetic field direction. This is analogous to the
precession of a spinning top around the gravity field.
Classically, the frequency of precession is given by

ωL = qB/2m

An isolated electron has an angular momentum and a magnetic
moment resulting from its spin. While an electron's spin is
sometimes visualized as a literal rotation about an axis, it
is in fact a fundamentally different, quantum-mechanical
phenomenon with no true analogue in classical physics.
Consequently, there is no reason to expect the above classical
relation to hold. In fact it does not, giving the wrong result
by a dimensionless factor called the electron g-factor, denoted
by ge where ge = 2(1 + α/2π + ...) and α is the fine-structure
constant (= 1/137) - a coupling constant that characterizes
the strength of the electromagnetic interaction between
elementary charged particles.  Thus,

ωL = qgeB/2m = geμBB/h

μB = qh/2m ... The Bohr Magneton

The magnetic moment is usually expressed as a multiple of the
Bohr magneton.

The result represents the angular frequency associated with the
transition between the two possible spin states.  In turn, this
can be expressed as the absorption or emission a photon with
energy according to the Planck relationship.

E = hωL

= geμBB

= 2μBB

Quantum Mechanical Treatment
----------------------------

In general H = (μ/2)σ.B

For the z direction:

Hz = (μ/2)σz.B
.
σz = (ih)[H,σz] = 0

Thus, there is no change in energy in the z direction (energy
is conserved).
.
σx = (iμBh)[σz,σx] = 0 since [σz,σx] = 2iσy

= -μBσ2/2

Likewise,
.
σy = μBσx/2

These equations have the following solutions:

σy = sinμBt/h

σx = cosμBt/h

This represents precession about the axis of B.
```