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Quantum Mechanics - Addition of Angular Momentum

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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, [tex]j_1+j_2\geq J\geq|j_1-j_2|,~j_1+j_2\geq M\geq-(j_1+j_2)[/tex] 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 [tex]|1~ 2\rangle ,~|1 ~1\rangle ,~|1 ~0\rangle ,~|1 ~-1\rangle ,~and |1 ~-2\rangle ,~[/tex]?

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

[tex]|2~2\rangle=|1\rangle|1\rangle[/tex]

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

Orthog. combination gives us a J=1 state: [tex]|1 ~1\rangle=\frac{1}{\sqrt{2}}(|0\rangle|1\rangle-|1\rangle|0\rangle)[/tex]. But if we try and apply [tex]J_+[/tex] to this in order to get [tex]|1 ~2\rangle[/tex] we just get zero, why? How do we get [tex]|1 ~2\rangle[/tex]?
 
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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, [tex]j_1+j_2\geq J\geq|j_1-j_2|,~j_1+j_2\geq M\geq-(j_1+j_2)[/tex] 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 [tex]|1~ 2\rangle ,~|1 ~1\rangle ,~|1 ~0\rangle ,~|1 ~-1\rangle ,~and |1 ~-2\rangle ,~[/tex]?

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

[tex]|2~2\rangle=|1\rangle|1\rangle[/tex]

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

Orthog. combination gives us a J=1 state: [tex]|1 ~1\rangle=\frac{1}{\sqrt{2}}(|0\rangle|1\rangle-|1\rangle|0\rangle)[/tex]. But if we try and apply [tex]J_+[/tex] to this in order to get [tex]|1 ~2\rangle[/tex] we just get zero, why? How do we get [tex]|1 ~2\rangle[/tex]?
 
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  • #3
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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?
 
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vela
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Right. It has to do with the Pauli exclusion principle and the symmetry of the wave function.
 
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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?
 
  • #6
vela
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Yes, that's right.
 
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Yes, that's right.
So then how do we construct the states, we're given [tex]\psi_i(x)[/tex] and told that we have to construct states of lowest energy (i.e. [tex]E_1[/tex]) so do we only care about [tex]\psi_1^A(x)[/tex] and [tex]\psi_1^B(x)[/tex]?
 

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