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Adding 3 Spin 1s

  • Thread starter Dishsoap
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1. 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.

2. Homework Equations

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

3. The Attempt at a Solution

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.
 

kuruman

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You are correct in saying that there are 27 linearly independent states.

The general way to proceed is to add two of the angular momenta to get and intermediate angular momentum and then add the third. For example you can add j1 + j2 = j12 and then j12 + j3 = J. But there is another way, j2 + j3 = j23 and then j23 + j1 = J. This kind of addition is achieved using Wigner 6-j symbols. However, your case is simpler because all angular momenta are j = 1 so there is only one way of adding them. Say you add j1 and j2 to get j12. What are the possible j12 values? Then add j3 and j12. Again what are the possible values?

As for part (b), think of J = 0 as a scalar (one thing needed to describe it, J =1 as a vector (three things needed to describe it) when the time comes to answer this part.
 
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Thank you for clarifying, but I think that I am still missing something conceptually. I do not understand why there is not only one possible value for [itex]J[/itex], namely, [itex]J=3[/itex], since we are adding three states with [itex]j=1[/itex].
 

kuruman

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Thank you for clarifying, but I think that I am still missing something conceptually. I do not understand why there is not only one possible value for ##J##, namely, ##J=3##, since we are adding three states with ##j=1##.
For the same reason that you don't get just 2 when you add two angular momenta with j = 1. Please read this http://www2.ph.ed.ac.uk/~ldeldebb/docs/QM/lect15.pdf to understand how angular momenta add.
Also, if you only got J = 3, that is 2×3 + 1 = 7 states when you know you have 27. So where are the other 20? Answer: If you account correctly for the other values alluded to in part (a), you will see.
 

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