# Is the space of total angular momentum complete?

znbhckcs
Hi
It's easy to see that for addition of 2 angular momenta l1 and l2 , the space l1 m1 , l2 m2 is equivalent to the space of l1 l2 l m (where l is the total angular momentum).
Counting the total number of states is usually a convenient way to make sure you got the addition right.

But what about the addition of 3 angular momenta? consider for example, l1,l2,l3 all equal to 1.
It's easy to count the total number of states: 3X3X3=27.

Adding the momenta we can get l=0,1,2,3 and so the total number of states is 1+3+5+7=16.

So what happened to 27-16=11 missing states? There must be some quantum number to distinguish between them, right?

Homework Helper
You're not using the formulas properly

$$1\otimes 1\otimes 1 = (2\oplus 1\oplus 0)\otimes 1 = 3\oplus 2\oplus 1 \oplus 2\oplus 1\oplus 0\oplus 1$$

znbhckcs
Obviously there are various way to add the third momentum, as is shown in your notation.
But my question was, suppose we have a space defined by the quantum numbers l1,l2,l3 , m1,m2,m3 .
We now add the momenta and get a space of l1,l2,l3,L,M, with M=-L,...L .
But there are not enough states in this space!

What is quantum number that tells 2 states with the same L and M from each other?

The quantum numbers that tell you that are m1, m2, and m3. Given M, only two of these are independent.

If you are asking what m1, m2, m3 state corresponds to eigenstate with given M, it's actually a superposition of all possible states such that m1+m2+m3=M. You'd have to use Clebsch-Gordon coefficients to find the weights of the possible combinations.

znbhckcs