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.  By convention we choose the
z-axis.

L+ = Lx + iLy
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?   Lets form a new state using using
L+ and measure it along the z axis.

Lz(L+|m>) = LzL+|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
raises m by h.  Similarly, we can show that L-|m> = (m-1)hL-|m>
lowers m by h.  Therefore 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.

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

Magnitude of Angular Momentum
-----------------------------

The magnitude of the angular momentum, 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>

= 0 + mmax2|mmax> + hmmax|mmax>  ... L+|mmax is not allowed.

= (mmax2 + mmax)|mmax>

Classically, we would expect that Lz2|mmax>  = mmax2|mmax.  This
corresponds to the situation where the angular moment 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 simultaneously.```