When you have the eigenvalues of angular momentum states in quantum mechanics, you can solve the Hamiltonian and get the allowed energy levels of an object with angular momentum. The eigenvalues of the angular momentum are the possible values the angular momentum can take.

Here’s how to derive eigenstate equations with

Note that L^{2} – L_{z}^{2} = L_{x}^{2} + L_{y}^{2}, which is a positive number, so

That means that

And substituting in

and using the fact that the eigenstates are normalized, gives you this:

So there’s a maximum possible value of

which you can call

You can be clever now, because there has to be a state

such that you can’t raise

any more. Thus, if you apply the raising operator, you get zero:

Applying the lowering operator to this also gives you zero:

And because

that means the following is true:

Putting in

gives you this:

At this point, it’s usual to rename

You can say even more. In addition to a

there must also be a

such that when you apply the lowering operator, L_{–}, you get zero, because you can’t go any lower than

And you can apply L_{+} on this as well:

From

you know that

which gives you the following:

And comparing this equation to

gives you

Note that because you reach

by *n* successive applications of

you get the following:

Coupling these two equations gives you

Therefore,

can be either an integer or half an integer (depending on whether *n* is even or odd).

Because

and *n* is a positive number, you can find that

So now you have it:

The eigenstates are |

*l*,*m*>*.*The quantum number of the total angular momentum is

*l*.The quantum number of the angular momentum along the

*z*axis is*m*.

For each *l*, there are 2*l* + 1 values of *m*. For example, if *l* = 2, then *m* can equal –2, –1, 0, 1, or 2.

You can see a representative L and L* _{z}* in the figure.

*.*

_{z}L is the total angular momentum and L* _{z}* is the projection of that total angular momentum on the

*z*axis.