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Tangent87
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Hi, I am doing question 32D on page 18 here:
http://www.maths.cam.ac.uk/undergrad/pastpapers/2005/Part_2/PaperII_3.pdf
and I am stuck on the second paragraph where we have to explain how to construct the two-particle states of lowest energy for (i). identical spin-1 particles with combined total spin J=1, and (ii). non-identical spin-1 particles with combined total spin J=1.
I found in the first part of the question that all three of the J=1 spin states are antisymmetric, and thus for (i) since we have identical spin 1 particles the total state must be symmetric overall therefore the only possible state is [tex]\psi_1(A)\psi_1(B)[/tex]. But I'm not sure about this because I don't think it takes into account the fact that we're in a total spin J=1 state.
For (ii), I said that since the particles are non-identical there is no exchange symmetry and so we can have any of the three states: [tex]\psi_1(A)\psi_1(B)|1 M>[/tex] for M=-1,0 or 1. Is that correct?
http://www.maths.cam.ac.uk/undergrad/pastpapers/2005/Part_2/PaperII_3.pdf
and I am stuck on the second paragraph where we have to explain how to construct the two-particle states of lowest energy for (i). identical spin-1 particles with combined total spin J=1, and (ii). non-identical spin-1 particles with combined total spin J=1.
I found in the first part of the question that all three of the J=1 spin states are antisymmetric, and thus for (i) since we have identical spin 1 particles the total state must be symmetric overall therefore the only possible state is [tex]\psi_1(A)\psi_1(B)[/tex]. But I'm not sure about this because I don't think it takes into account the fact that we're in a total spin J=1 state.
For (ii), I said that since the particles are non-identical there is no exchange symmetry and so we can have any of the three states: [tex]\psi_1(A)\psi_1(B)|1 M>[/tex] for M=-1,0 or 1. Is that correct?
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