# Is the space of total angular momentum complete?

## Main Question or Discussion Point

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?

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dextercioby
Homework Helper
You're not using the formulas properly

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

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?

K^2