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## Homework Statement

Consider eigenstates of the total angular momentum $$\vec{J}=\vec{J}_1+\vec{J}_2+\vec{J}_3$$ where all of the [itex] \vec{J}_i [/itex] are for [itex] j=1 [/itex] , and let [itex]J(J+1)[/itex] be the eigenvalue of [itex]\vec{J}^2[/itex].

a) What are the possible values of [itex]J[/itex]? How many linearly independent states are there for each of these values?

b) Construct the [itex]J=0[/itex] state explicitly. If [itex]\vec{a},\vec{b},\vec{c}[/itex] are ordinary 3-vectors, the only scalar linear in all of them that they can form is [itex](\vec{a} \times \vec{b}) \cdot \vec{c}[/itex]. Establish the connection between this fact and your result for the [itex]J=0[/itex] state.

## Homework Equations

[itex]J^2 \vert j,m \rangle=J(J+1) \vert j,m \rangle [/itex]

3. The Attempt at a Solution

[/B]

I am really struggling with this addition of angular momentum stuff, so please forgive me. Because we are more or less hinted to use [itex]J^2[/itex] instead of [itex]J_z[/itex], there is no dependence on [itex]m[/itex], so I'm thinking that for each value of J, since there are 3 values for [itex]m[/itex] because [itex]m=0,\pm{1}[/itex] there will be [itex]3^3=27[/itex] linearly independent states at least for each value of [itex]J[/itex].

I'm not quite sure how to relate [itex]J_{total}[/itex] to [itex]j[/itex]. Would I do something like

$$\langle j_1, j_2, j_3 ; m_1, m_2, m_3 \vert (J_1^2 +J_2^2+J_3^2) \vert j_1, j_2, j_3 ; m_1, m_2, m_3 \rangle = J(J+1) $$

where this then becomes

$$ j_1(j_1+1)+j_2(j_2+1)+j_3(j_3+1) = J(J+1) $$ ?

But now I'm confused, because I think that [itex]j_1=j_2=j_3=1[/itex].

I'm really just not sure how to proceed.