Wolfram Alpha:

```Ladder Operators
----------------

The following calculations equally apply for
angular momentum, L; spin angular momentum, S;
and the total angular momentum, J where J = L + S
(vector addition).  Therefore, we can substitute L
with either S or J in the following equations.
Consider L.

Classically:

L = R x P  - L points along spin axis.

In cartesian coords

Lx = ypz - zpy

Ly = zpx - xpz

Lz = xpy - ypx

QM Commutators:

[xi,xj] = 0

[pi,pj] = 0

[x,px] = ih

or in general

[xi,pj] = ihδij

[Lx,Ly] = [ypz - zpy, zpx - xpz]

= -ihypx + ihxpy

= ih(xpy - ypx)

= ihLz

[Ly,Lz] = ihLx

[Lz,Lx] = ihLy

What this means is that you cannot simultaneously
measure angular momentum in more than one direction.

Introduce raising and lowering (ladder) operators
similar to annihilation and creation operators.

L+ = Lx + iLy and L- = Lx - iLy

[L+,Lz] = [Lx + iLy,Lz]

= [Lx,Ly] + i[Ly,Lz]

= ihLy + iihLx

= ihLy - hLx

= -h(Lx + iLy)

= -hL+

Similarly, we can show:

[L-,Lz] = +hL-

What do these operators do?   Consider a state,
m, such that Lz|m> = m|m>.  The eigenvalue, m,
is the value of the angular momentum along the
z axis (i.e. the magnetic quantum number, ml).
Lets now form a new state by applying L+ to |m>
and again measuring the value of the angular
momentum along the z axis.  Therefore, we can
write:

Lz(L+|m>) = ?

Now consider the commutator [L+,Lz].  From
before,

[L+,Lz] = L+Lz - LzL+ = -hL+

∴ LzL+ = L+Lz + hL+

Therefore,

Lz(L+|m>) = L+Lz|m> + hL+|m>

Lz(L+|m>) = L+mh|m> + hL+|m> since Lz|m> = mh|m>

= (m+1)hL+|m>

Therefore, L+|m> = (m+1)hL+|m> which means that
the operator L+ raises m by h.  Similarly, we
can show that L-|m> = (m-1)hL-|m> lowers m by h.
Therefore he spectrum of Lz is gapped by integers.
Obviously, m cannot go on forever so we need to
define what is the maximum and min value for a
specific particle.

z                   z
+                   +  <-- mmax
|                   |
+3/2                + 2
|                   |
+1/2                + 1
|                   |
+0                  + 0
|                   |
+-1/2               +-1
|                   |
+-3/2               +-2
|                   |
+                   +  <-- mmin
Fermion              Boson

The L2 Operator
---------------

The square of the magnitude of the angular momentum
operator, L, is given by:

L2 = Lx2 + Ly2 + Lz2

Now L-L+ = (Lx - iLy)(Lx + iLy)

= Lx2 + Ly2 + i(LxLy - LyLx)

= Lx2 + Ly2 + iihLz

= Lx2 + Ly2 - hLz

Therefore,

L2 = L-L+ + Lz2 + hLz

Now operate on mmax

L2|mmax> = L-L+|mmax> + Lz2|mmax> + hLz|mmax>

L+|mmax is not allowed, therefore,

L2|mmax> = 0 + mmax2|mmax> + hmmax|mmax>

= (mmax2 + mmax)|mmax>

Classically, we would expect that Lz2|mmax> = mmax2|mmax>.
This corresponds to the situation where the angular
momentum is aligned along the positive z axis.
Likewise, the minimum occurs when the momentum is
aligned along the negative z axis. The total number
of states is therefore 2mmax + 1 where mmax
defines the type of particle.

The fact that we have an extra mmax term in the
above reflects the commutation relationships that
state we cannot measure more than one component
of momentum at a time.  In essence, the term
reflects the uncertainty associated with the
angular momentum in the x and y directions that
cannot be measured.

Can you measure the z component of angular
momentum and the magnitude of the angular
momentum simultaneously? Consider,

[L2,Lz]|mmax>

This is equivalent to,

[L-L+ + Lz2 + Lz,Lz]|mmax>

[(0 + mmax2 + mmax),mmax]

Since all real numbers commute, the commutator
is equal to 0.  Therefore, it is possible to
measure L2 and Lz at the same time.

The L Matrices
--------------

From the note on Clebsch-Gordan coefficients, for
l = 1, the operator matrices are:

-     -
| 0 1 0 |
Lx = h/√2| 1 0 1 |
| 0 1 0 |
-     -

-      -
|  0 1 0 |
Ly = h/√2| -1 0 1 |
|  0 1 0 |
-      -

-      -
| 1 0  0 |
Lz = h| 0 0  0 |
| 0 0 -1 |
-      -

Using the notation |l,m> we can define the 3
eigenstates as follows:

- -            - -             - -
| 1 |          | 0 |           | 0 |
|1,1> = | 0 |  |1,0> = | 1 |  |1,-1> = | 0 |
| 0 |          | 0 |           | 1 |
- -            - -             - -

For Lz, m = -1, 0, 1 we get:

-      -  - -       - -
| 1 0  0 || 1 |     | 1 |
h| 0 0  0 || 0 | = +1| 0 |h
| 0 0 -1 || 0 |     | 0 |
-      -  - -       - -

-      -  - -      - -
| 1 0  0 || 0 |    | 0 |
h| 0 0  0 || 1 | = 0| 1 |h
| 0 0 -1 || 0 |    | 0 |
-      -  - -      - -

-      -  - -       - -
| 1 0  0 || 0 |     | 0 |
h| 0 0  0 || 0 | = -1| 0 |h
| 0 0 -1 || 1 |     | 1 |
-      -  - -       - -

L2 = Lx2 + Ly2 + Lz2 = 2h2I

In the process of solving the Schrodinger equation
for the hydrogen atom, it is found that the orbital
angular momentum is quantized according to the
relationship:

L2 = l(l + 1)h2 = 2h2I

Where L now represents the orbital angular momentum
in units of h.

The operators for spin 1/2 are the Pauli matrices.
Similar calculations to the above yield:

S2 = Sx2 + Sy2 + Sz2 = (3h2/4)I

Where Si = (h/2)σi

also,

S2 = s(s + 1)h2 = 3h2/4```