To my understanding, decomposing the tensor product of two particles with spins $s_1$ and $s_2$ works as follows:

$$\mathcal{H_{s_1}}\otimes \mathcal{H_{s_2}}=\mathcal{H_{s_1+s_2}}\oplus\mathcal{H_{s_1+s_2-1}}\oplus \cdots \oplus \mathcal{H_{|s_1-s_2|}}.$$

So, for two spin-1/2's, there is going to be two possible total spins: 1 and 0. Likewise, for spin-1/2 and spin-1 there is 1/2+1 = 3/2 and 1-1/2 = 1/2, so the two possible total spins are 1/2 and 3/2.

These two examples are confirmed by the Glebsch-Gordan tables. Yet, there are some systems in which a total spin is missing. That is to say, there are some systems where there is (at least) one possible total spin $s$, for which there is no state $|sm\rangle$ with that total spin.

The $\mathcal{H_{2}}\otimes \mathcal{H_{1}}= \mathcal{H_3}\otimes \mathcal{H_2}\otimes \mathcal{H_1}\otimes \mathcal{H_0}$ system is an example. Theory says that the decomposition is into total spins 3, 2, 1 and 0. In the tables I do find 3, 2 and 1, but no 0. Likewise, $\mathcal{H_{2}}\otimes \mathcal{H_{1/2}}$ lacks states with total spin $1/2$.

Why is there no $|0m\rangle\in\mathcal{H_1}\otimes \mathcal{H_2}$?

I suppose this also means that there should be no $\mathcal{H_0}$ in the direct sum-decomposition? Why is that so?


1 Answer 1


So, the premise of my question was an arithmetic mistake. But for the sake of clarity:

In a system with two particles of spins $s_1$ and $s_2$, the total spin of the system obeys the triangle-inequality

$$|s_1-s_2|\leq s \leq s_1 + s_2,$$

where integer jumps are made.

So, in the spin-1, spin-2 system, the lowest total spin is |1-2| = 1, not zero.


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