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For $j_{1} = 1 \; \text{and} \; j_{2}=1$ I know that $j = 2,1,0$ and $m = -2, -1, 0, 1, 2$.

All possible states in the grid are represented by dots:

enter image description here

Why for $m = 2 \; \text{and} \; j = 2$ there is a dot? Why not for $m = 2 \; \text{and} \; j = 1$?

How to calculate where dots should be placed?

Qmechanic
  • 201,751
Paweł Chmolewski
  • 103

1 Answers1

0

The tensor product (or the "spin sum") of two irreducible representations of spin $j_1$ and $j_2$ decomposes as the direct sum of irreps of spin $j=|j_1-j_2|$ to $j=j_1+j_2$. In other words, when you sum two spins $j_1$ and $j_2$, the total spin can take values $j=|j_1-j_2|, |j_1 - j_2| + 1, ..., j_1+j_2$. Each of these irreps of total spin $j$ host a multiplet with $J_3$ eigenvalues $m=-j, ...,j$. You can think of each of these $j$ as representing different orthogonal subspaces, in which we diagonalize $J_3$ with eigenvalues $m$ going from $-j$ to $j$

For example, here you want to take the tensor product of two spin 1 irreps $j_1=1$ and $j_2 =1$, so $j=0,1,2$ runs from 1-1=0 to 1+1=2, and for each $j$, $m$ runs from $-j$ to $j$. This is depicted in your image. There are no dots placed on $m=2$ $j=1$ because $|m|\leq j$ in our construction

Let's take another example to confirm our understanding, with $j_1 = 2$, $j_2 = 1$, then $j=2-1,2,2+1$ which is to say $j=1,2,3$ and

For $j=1$, $m=-j,...,j$ which means $m=-1,0,1$

Similarily for $j=2$, $m=-2,-1,0,1,2$

For $j=3$, $m=-3,...,3$

Adrien Martina
  • 1,358
  • I was calculating using formula $j = j_{1} + j_{2}, j_{1} + j_{2} - 1, ... , |j_{1} - j_{2}| $, $m = j, j - 1, ... , -j$ If I understand correctly, with $j_{1} = 2$, $j_{2} = 1$ for $j = 2$ it will be three dots, because m = -2, 1, 2. Am I correct? – Paweł Chmolewski Feb 19 '23 at 02:05
  • @PawełChmolewski No, that’s not correct. See my comment on your question. – Ghoster Feb 19 '23 at 05:52