# Quantum Mechanics - Addition of Angular Momentum

Just when I think I've understood addition of angular momentum, I find a problem that completely questions everything I think I know. Okay so here's the problem: question 3/II/32D on page 68 http://www.maths.cam.ac.uk/undergrad/pastpapers/2005/Part_2/list_II.pdf [Broken] .

For the possible values of J and M as I understand it, $$j_1+j_2\geq J\geq|j_1-j_2|,~j_1+j_2\geq M\geq-(j_1+j_2)$$ and therefore for two j=1 systems surely J=2,1, or 0 and M=-2,-1,0,1,2?

Thus we must find expressions for the states $$|1~ 2\rangle ,~|1 ~1\rangle ,~|1 ~0\rangle ,~|1 ~-1\rangle ,~and |1 ~-2\rangle ,~$$?

So letting $$|1~m_1\rangle |1~m_2\rangle=|m_1\rangle|m_2\rangle$$ where $$m_i=\pm 1$$, we start with the top state J=2:

$$|2~2\rangle=|1\rangle|1\rangle$$

Then apply $$J_-$$ to get $$|2~ 1\rangle=\frac{1}{\sqrt{2}}(|0\rangle|1\rangle+|1\rangle|0\rangle)$$.

Orthog. combination gives us a J=1 state: $$|1 ~1\rangle=\frac{1}{\sqrt{2}}(|0\rangle|1\rangle-|1\rangle|0\rangle)$$. But if we try and apply $$J_+$$ to this in order to get $$|1 ~2\rangle$$ we just get zero, why? How do we get $$|1 ~2\rangle$$?

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vela
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Just when I think I've understood addition of angular momentum, I find a problem that completely questions everything I think I know. Okay so here's the problem: question 3/II/32D on page 68 http://www.maths.cam.ac.uk/undergrad/pastpapers/2005/Part_2/list_II.pdf [Broken] .

For the possible values of J and M as I understand it, $$j_1+j_2\geq J\geq|j_1-j_2|,~j_1+j_2\geq M\geq-(j_1+j_2)$$ and therefore for two j=1 systems surely J=2,1, or 0 and M=-2,-1,0,1,2?
That's not quite correct. First you find the allowed values for J. Then for each value of J, M runs from -J to +J. So for this problem, since you're interested in J=1, you only have the states |1 1>, |1 0>, and |1 -1>. There are no other J=1 states (so it should make sense to you why you got 0 when you applied J+ below).
Thus we must find expressions for the states $$|1~ 2\rangle ,~|1 ~1\rangle ,~|1 ~0\rangle ,~|1 ~-1\rangle ,~and |1 ~-2\rangle ,~$$?

So letting $$|1~m_1\rangle |1~m_2\rangle=|m_1\rangle|m_2\rangle$$ where $$m_i=\pm 1$$, we start with the top state J=2:

$$|2~2\rangle=|1\rangle|1\rangle$$

Then apply $$J_-$$ to get $$|2~ 1\rangle=\frac{1}{\sqrt{2}}(|0\rangle|1\rangle+|1\rangle|0\rangle)$$.

Orthog. combination gives us a J=1 state: $$|1 ~1\rangle=\frac{1}{\sqrt{2}}(|0\rangle|1\rangle-|1\rangle|0\rangle)$$. But if we try and apply $$J_+$$ to this in order to get $$|1 ~2\rangle$$ we just get zero, why? How do we get $$|1 ~2\rangle$$?

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That's not quite correct. First you find the allowed values for J. Then for each value of J, M runs from -J to +J. So for this problem, since you're interested in J=1, you only have the states |1 1>, |1 0>, and |1 -1>. There are no other J=1 states (so it should make sense to you why you got 0 when you applied J+ below).

Ah I see, yes that makes sense. Also I don't understand the second paragraph of the question, I guess I need to understand what's the difference between identical and non-identical particles?

vela
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Right. It has to do with the Pauli exclusion principle and the symmetry of the wave function.

Right. It has to do with the Pauli exclusion principle and the symmetry of the wave function.

Ok I have read this article http://en.wikipedia.org/wiki/Identical_particles and I think I'm starting to understand. So if the particles are identical the state must be symmetric since the antisymmetric state would be zero. And if the particles are non-identical the state can be either symmetric or anti-symmetric? Is that logic correct?

vela
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Yes, that's right.

Yes, that's right.

So then how do we construct the states, we're given $$\psi_i(x)$$ and told that we have to construct states of lowest energy (i.e. $$E_1$$) so do we only care about $$\psi_1^A(x)$$ and $$\psi_1^B(x)$$?