# Adding angular momenta using levi-civta symbol

1. Nov 8, 2013

### geoduck

Suppose you want to add 2 spin-1 particles.

I understand you can get the j=1 triplet by $$e_{ijk}|j\rangle |k\rangle$$ where i, j, k run from -1, 0 , 1.

The idea is that levi-civita symbol is a tensor under SO(3) rotations, so the contraction with the $$|j\rangle |k\rangle$$ tensor gives a vector under rotation, which is the triplet.

However, isn't kronecker delta also a tensor under rotation SO(3)? So why can't you get the singlet from:

$$\delta_{jk} |j\rangle |k\rangle = |1\rangle |1\rangle +|0\rangle |0\rangle+|-1\rangle |-1\rangle$$
Looking at the textbooks, the middle |00> term is with a negative sign.

Also, is there a group theoretic way to get the quintuplet?

Thanks.

2. Nov 8, 2013

### Bill_K

Aren't you confusing the Cartesian basis with the spherical basis? When you say |k>, k = ±1, 0, that's the spherical basis. The relationship between the two bases is

e+1 = - (ex + i ey)/√2
e0 = ez
e-1 = (ex - i ey)/√2

Take the Kronecker delta in the form of the identity matrix,

I = exex + eyey + ezez

rewrite it in terms of e±1 and e0, and you'll see where the minus sign comes from..

3. Nov 8, 2013

### geoduck

Thanks! I'm the grader for a graduate QM class, this will help the whole class.