The Clebsch-Gordan coefficients can only be non-zero if the triangle inequality holds: $$\vert j_1-j_2 \vert \le j \le j_1+j_2$$ In my syllabus they give the following proof: $$-j \le m \le j$$ $$-j_1 \le m_1 \le j_1$$ and $$-j_2 \le m_2 \le j_2$$

When $m$ takes its maximal value, $m = j$, $m_1 = j_1$ and $m_2 = j_2$, and we get:

1) $-j_1 \le j-j_2 \le j_1$ which implies $j_2-j_1 \le j \le j_1+j_2$

2) $-j_2 \le j-j_1 \le j_2$ which implies $j_1-j_2 \le j \le j_1+j_2$

which should prove the triangle inequality.

This proof looks really simple, but I don't completely understand it though. It seems that I'm missing some essential reasoning, and I can't find where. Why for instance do they take for $m_1$, $m_2$ and $m$ all maximal values? Can't I also take $m$ maximal and $m_1$ minimal? This would give bad results though. So I really don't understand it, and I hope that someone can clarify it.


First of all, if $m$ is maximal, $m_1$ cannot be minimal because $m=m_1+m_2$ so $m_{max}=m_{1, max}+m_{2, max}$ by definition.

The reason they need to use maximal values for $m_i$ is because they need to find a relationship between only the $j$-values, but they only have a relationships between $m$-values and $j$-values (namely, $-j_i \le m_i \le j_i$).

When $m_i$ is maximal, $m_i=j_i$ (by virtue of the inequality relationship above). Therefore, $j=m_{max}=m_{1, max}+m_{2, max}=j_1+j_2$ or, in other words, $j_2=m_2=j-j_1$ (and vice versa for $m_1$).

By substituting into the inequality $-j_2\le m_2\le j_2$, we get $-j_2\le j-j_1\le j_2$. The rest of the proof is a trivial extension of what is done here. The arguments are exactly the same.


This is only to add on the other answer:

Note that we have $m = m_1 + m_2$ and $-j_1 \le m_1 \le j_1$, so $$-j_1 \le m - m_2 \le j_1$$ In particular $m$ can take its maximum value $j$ and $m_2$ can take its maximum value $j_2$, which gives $$-j_1 \le j - j_2 \le j_1$$ In fact one can consider the situation where $m$ takes its minimum value $-j$ and $m_2$ takes its minimum value $-j_2$, which gives $$-j_1 \le -j + j_2 \le j_1$$ But this is in fact the same inequality after rearrangement.

The other two situations when one takes its minimum and the other takes its maximum are forbidden.


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