Lecture 21.
We used step operators 
to find the eigenstates 
.
We first showed that the step operators 
are
eigenoperators under commutation with 
, calculating the
commutation relations 
. In analysis
that should be becoming familiar, we considered an initial
eigenstate 
(with eigenvalue 
). We
showed that the state 
is also an eigenstate, with
eigenvalue 
. Thus 
increases a state's
value by 
, while 
decreases it by .
We next showed that 
commutes with (since it is just a
linear combination of 
and 
). Thus when we consider a
eigenstate (whose eigenvalue 
is unknown), acting
on that eigenstate with a step operator gives back an
eigenstate of , with the same eigenvalue f(j).
We thus know how the step operators act on the fully specified
eigenstates : they leave j invariant, while raising or
lowering m. This gives us a ladder of eigenstates for fixed
j. As in the simple harmonic oscillator case, the ladder does not
continue forever. Again the constraint that raised and lowered states,
, have nonnegative norm restricts the allowed states on the
ladder. We found that the constraint resulted in a bound 
, placing both upper and lower bounds on m. This means that the
ladder has both an upper and a lower rung; for consistency, must
annihilate the highest state, while annihilates the lowest. We
showed by explicitly calculating the norm of 
that
this occurs only if 
, with 
giving
the eigenvalue of . 
imposes the same
eigenvalue on , with 
.
Finally, the ladder goes from 
to 
in integral
steps. This means that there must be 2j+1 steps that go from the
highest to lowest state, which can occur only if 2j+1 is an integer
(otherwise, we do not hit the lowest state as we descend the ladder).
We have thus found all the angular momentum eigenstates . For
each half-integer j, we have the truncated ladder of states from m=
j to m=-j, with eigenvalue 
and
eigenvalues 
.
-- KB