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For $j=\frac{1}{2}, j'=\frac{1}{2}$ we have $$|11\rangle=|\frac{1}{2}\frac{1}{2}\rangle$$ $$|10\rangle=\frac{1}{\sqrt{2}}(|-\frac{1}{2}\frac{1}{2}\rangle+ |\frac{1}{2}-\frac{1}{2}\rangle)$$ $$|10\rangle=\frac{1}{\sqrt{2}}(|-\frac{1}{2}\frac{1}{2}\rangle- |\frac{1}{2}-\frac{1}{2}\rangle)$$ $$|1-1\rangle=|-\frac{1}{2}-\frac{1}{2}\rangle$$

My book says that this proves that

$$\frac{1}{2}\otimes \frac{1}{2} = 1\oplus0$$

Now the way that I'd read this is that when we combine two spin $\frac{1}{2}$ particles, we can get total spin of either 1 or 0. I'm confused how the above shows this. I see that $m_{total}=1$ in the first equation, $m_{total}=0$ in the two middle equations, and $m_{total}=-1$ in the last equation. Which would make me think the direct sum should really be

$$\frac{1}{2}\otimes \frac{1}{2} = 1\oplus0\oplus -1$$

What's wrong here? Any help would be greatly appreciated!

Qmechanic
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DS08
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2 Answers2

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The addition rule has an absolute value sign. That is

$$ s_{tot}= |s_1 + s_2|,|s_1 + s_2 - 1|, \ldots, |s_1 - s_2| $$

And so the smallest value you can have is $1/2 - 1/2 = 0$. $s_{tot} $ is the thing we use to label the irreducible representations, not $m$.

InertialObserver
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The values of $0$ and $1$ in $\frac{1}{2}\otimes \frac{1}{2}=0\oplus 1$ are values of total angular momentum $J$, not projections $M$.

Thus, you have three states with $J=1$. You can verify that the $M=1$ state is killed by $J_+$, and this defines the state with $J=1, M=1$. Lowering from this will take you to one of the $M=0$ states, and lowering from this will give the $J=1,M=-1$ state. You can also verify these are eigenstates of $J_x^2+J_y^2+J_z^2$ with eigenvalue $\sqrt{2}$.

There remains one $M=0$ state: you can verify that it is killed by $J_+$ so this defines the state with $J=0,M=0$. This is an eigenstate of $J_x^2+J_y^2+J_z^2$ with eigenvalue $0$.

ZeroTheHero
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