# Why is there no state of total spin 0 for spin-1 and spin-2?

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?

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