Quantum Mechanics for Engineers 

© Leon van Dommelen 

12.3 Ladders
This section starts the quest to figure out everything that the
fundamental commutation relations mean for angular momentum. It will
first be verified that any angular momentum can always be described
using eigenstates with definite values of square angular
momentum and angular momentum . Then it will be
found that these angular momentum states occur in groups called
ladders
.
To start with the first one, the mathematical condition for a complete
set of eigenstates to exist is that the angular momentum
operators and commute. They do; using the commutator
manipulations of chapter 4.5.4), it is easily found
that:
So mathematics says that eigenstates of and
exist satisfying
and that are complete in the sense that any state can be described in
terms of these .
Unfortunately the eigenstates , except for
states, do not satisfy relations like (12.2) for or
. The problem is that and do not commute
with . But and do commute with
, and you might wonder if that is still worth something.
To find out, multiply, say, the zero commutator by
:
Now take the second term to the right hand side of the equation,
noting that with just a number that can
be moved upfront, to get:
Looking a bit closer at this equation, it shows that the combination
satisfies the same eigenvalue problem for as
itself. In other words, the multiplication by does
not affect the square angular momentum at all.
To be picky, that is not quite true if would be zero,
because zero is not an eigenstate of anything. However, such a thing
only happens if there is no angular momentum; (it would make
an eigenstate of with eigenvalue zero in addition to an
eigenstate of {D.63}). Except for that trivial case,
does not affect square angular momentum. And neither does
or any combination of the two.
Angular momentum in the direction is affected by and
by , since they do not commute with like they do
with . Nor is it possible to find any linear combination
of and that does commute with . What is the
next best thing? Well, it is possible to find two
combinations, to wit

(12.4) 
that satisfy the “commutator eigenvalue problems”:
These two turn out to be quite remarkable operators.
Like and , their combinations and
leave alone. To examine what the operator does with the
linear momentum in the direction, multiply its commutator relation
above by an eigenstate :
Or, taking the second term to the right hand side of the equation and
noting that by definition ,
That is a stunning result, as it shows that is an
eigenstate with angular momentum instead
of . In other words, adds exactly one unit
to the angular momentum, turning an state into a
one!
Figure 12.1:
Example bosonic ladders.

Figure 12.2:
Example fermionic ladders.

If you apply another time, you get a state of still higher
angular momentum , and so on, like the
rungs on a ladder. This is graphically illustrated for some examples
in figures 12.1 and 12.2. The process
eventually comes to an halt at some top rung
where 0. It has to, because the
angular momentum in the direction cannot just keep growing
forever: the square angular momentum in the direction only must
stay less than the total square angular momentum in all three
directions {N.27}.
The second ladder operator
works in much the
same way, but it goes down the ladder; its deducts one unit
from the angular momentum in the direction at each application.
provides the second stile to the ladders, and must terminate at
some bottom rung .