How many states for two spin 1 particles?

When dealing with angular momentum of a combined system you have two possible basis. Let $mathscr{H}_1$ be the Hilbert space of particle 1 and $mathscr{H}_2$ be the Hilbert space of particle 2.

On $mathscr{H}_1$ you have the basis $|j_1 m_1rangle$ where $j_1 = 1$ and $m_1 = -1,0,1$. On $mathscr{H}_2$ you have the basis $|j_2 m_2rangle$ where $j_2 = 1$ and $m_2 = -1,0,1$. In that sense both $mathscr{H}_1$ and $mathscr{H}_2$ are three-dimensional Hilbert spaces.

On the other hand, the combined system is $mathscr{H}_1otimes mathscr{H}_2$. One obvious basis for this space is that associated to the complete set of commuting observables $J_1^2,J_2^2,J_{1z}J_{2z}$. This is the basis

$$|j_1,j_2;m_1,m_2rangle=|j_1,m_1rangleotimes |j_2,m_2rangle.$$

It is clear that $j_1,j_2=1$ and $m_1,m_2=-1,0,1$. You thus have $9$ basis states. But all basis of a finite dimensional Hilbert space have the same number of element which is its dimension, whatever basis set you use, it will have $9$ states.

For completeness, another natural basis is that of the total angular momentum. You define

$$mathbf{J}=mathbf{J}_1otimes mathbf{1}+mathbf{1}otimes mathbf{J}_2$$

to get operators $J_z$ and $J^2$. They commute with $J_1^2$ and $J_2^2$ so that you get a basis

$$|j_1,j_2;j,mrangle$$

This is the basis of total angular momentum. It is a result then, that you can consult for example in Cohen's book Volume 2 on the "Addition of Angular Momentum" chapter, that the possible values for $j$ the eigenvalues of $J^2$ are

$$j=|j_1+j_2|,|j_1+j_2-1|,dots, |j_1-j_2|$$

Here $j_1=j_2=1$ hence the possible values for $j$ are $$j=2,1,0.$$

Now for $j = 0$ you have just $m = 0$ (one state), for $j = 1$ you have $m = -1,0,1$ (three states) and for $j = 2$ you have $ m = -2,-1,0,1,2$ (five states). This gives a total of $9$ states on the basis of total angular momentum as anticipated.

A completely abstract view of this would be to use Young diagrams (which I can't draw). You define your basis vector state as a box:

$$[ ]$$

It is 3 dimensional.

Then take the tensor product (which means sticking 2 boxes together in all forms that make a Young diagram):

$$[ ]otimes[ ] = [ ][ ]oplus[ ]$$

$$ [ ]$$

where the horizontal (vertical) boxes are the (anti)symmetric combinations. To find out the dimensions of the representations, one applies the Hook Length Formula:

$$ dim,W(n;r) = prod_{i,jin Y(n)}frac{r+j-i}{hook(i,j)}$$

with $n=2$ being the number of boxes in the diagram and $r=3$ is the dimension of the basis vector space. $(i,j)$ label the row number and the column number of the box in the diagram (counting from top left) and $hook(i,j)$ is the number of cells in the Hook for that cell--which is the number of cells equal or right of and equal or below the cell.

Applying that to the diagrams yields $6$ and $3$ for the symmetric and antisymmetric representations--so the total number of dimensions is 9.

Note that this agrees with the symmetries of a rank-2 cartesian tensor:

$$ T^S = frac 1 2 (T_{ij} + T_{ji})$$

has six independent components, and

$$ T^A = frac 1 2 (T_{ij} - T_{ji})$$

has three. (Hence spin-1 are called vector Bosons).

Also: if you define your fundamental object as a spinor (dimension $r=2$) and use the same hook length formula for the same diagrams (now imbued with different meaning), you will find:

$$ 2 otimes 2 = 3_S oplus 1_A $$

That is: two spin 1/2 particle can combine into a symmetric triplet or an antisymmetric singlet.

Likewise, apply it 4-vectors ($r=4$) you will discover:

$$ 4 otimes 4 = 10_S oplus 6_A $$

so that symmetric tensors like the stress energy tensor have 10 components and antisymmetric tensors like the electromagnetic field strength have 4.

This works for any number of combinations, so for instance if you want to combine 3 vectors (or spin ones or quark flavors), you get:

$$ 3 oplus 3 oplus 3 = 10_S oplus 8_M oplus 8_M oplus 1_A $$

were for instance the one dimensional antisymmetric part is the famed Levi-Civta tensor. (You may also notice some similarities to the spectrum of baryons there, too).

Only a observation on comment that starts as "When dealing with angular momentum of a combined system...". The relations,

j=|j1+j2|,|j1+j2−1|,…,|j1−j2|

are wrong, although they seem to be right. As a counterexample, for j1=1, j2=2 one has,

j=|3|, |3-1|=2, |3-2|=1, |3-3|=0, |3-4|=1 which is kind of "smelly" because the result 1 appears 2 times. Moreover, for j1=2, j2=1 (changing the values), you get

j=|3|, |3-1|=2, |3-2|=1 and you never get the result 0. The correct inequalities are

|j1-j2| <= j <= |j1+j2|, and that is j satisfying

j=|j1+j2|,|j1+j2|−1,|j1+j2|−2…,|j1−j2|, with the -1, -2, -3 and so on OUTSIDE and not inside the absolute value. Otherwise the result of coupling a j1 with j2 would be different of the coupling of j2 with j1, and it has to be symmetric.