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Two identical particles are in an isotropic harmonic potential. Show that, if the particles do not interact and there are no spin-orbit forces, the degeneracies of the three lowest energy values are 1, 12, 39 if the particles have spin 1/2, and 6, 27, 99 if the particles have spin 1.


My attempt at a solution: For a 3D harmonic potential, the energy levels are: $$E = (n+3/2) ℏ ω$$ where $ n = n_x + n_y + n_z $

Our first energy level is when n = 0. Here, the spatial degeneracy is 1. Now, we can have two identical particles of spin 1/2 in that state as we have two possibilities for the spin. so the degeneracy is 2.

For the next value of n, n=1, we have three possible degeneracies, (001), (010) and (100) for nx, ny, and nz, and we have two spin states. So the spatial degeneracy is 3, spin degeneracy 2, so total states at the energy level is $3\times 2$ = 6 different states at that energy. Degeneracy is 6. And so on. Now my reasoning must obviously be wrong as I don't get the values stated in the question.

Should the energy states be given by the two particles' energies added? Two fermions cannot be in the same state, but what if (100) for one and (010) for the other? Surely their energies agree, but the quantum numbers are not the same. Can anybody enlighten me please?

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closed as off-topic by John Rennie, Emilio Pisanty, Jon Custer, Kyle Kanos, Phonon Mar 17 '18 at 21:56

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  • $\begingroup$ Hey just out of curiosity, are you sure that the exercise supposes a 3D oscillator? I am asking since the results should be different according to the dimensions. $\endgroup$ – G K Mar 14 '18 at 22:20
  • $\begingroup$ Hint: for the ground state, n=0, so ignore it. Two spin 1/2 particles in the same state must be antisymmetrized, so only the spin singlet combination survives: degeneracy 1. Two spin 1 ones must be symmetrized, so you add the singlet and the quintet: degeneracy 6.... $\endgroup$ – Cosmas Zachos Mar 15 '18 at 0:09
  • $\begingroup$ @G.K. I supposed the question meant a 3D oscillator because it said "isotropic", however I might be wrong. Does it work out better with a 2D or 1D one? $\endgroup$ – Jhonny Mar 15 '18 at 8:48
  • $\begingroup$ @CosmasZachos what is a quintet spin state? This far I have only seen single and triplet states for two fermions. Also, the answer is supposed to be 12... $\endgroup$ – Jhonny Mar 15 '18 at 10:25
  • $\begingroup$ I am referring to the ground state of two spin 1 particles. A symmetric state of two such is either a singlet, or a spin 2 state, 1+5=6. Sort out the ground state respective degeneracies before moving on to the first excited state. $\endgroup$ – Cosmas Zachos Mar 15 '18 at 13:59
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This is a standard symmetry-constrained counting problem that has frequently appeared on other web homework sites. There are elegant and compact answers, but first you must develop counting skill at the ganglionic level, so to speak, in the dark.

Spin 1/2 particles are fermions, so their over-all wavefunction is antisymmetric (A). Their spin and space wavefunctions then must be A-S or S-A. Two spin 1/2 particles combine symmetrically to a spin triplet and anti symmetrically to a spin singlet.

The ground state is thus (000)(000)[$\uparrow\downarrow-\downarrow\uparrow$], S-A. Degeneracy $D=1$.

The first excited state has one of the two fermions in {(100),(010),(001)} and the other in the ground state (000). So you may have these 3 arrangements in either the S or the A form, to suitably match the spin (anti)symmetry. So match space S with spin A (singlet) and space A with spin S (triplet), hence $D=3\times (1+3)=12$.

The second excited state either has one fermion in the ground state and the other in {(200),(020),(002)}, combinatorially as above, contributing $d=12$, again; or else, one fermion in the ground state and the other in {(110),(101),(011)}, contributing another 12. Or else, both fermions are in first excited states: either {(100)(100), (010)(010), (001)(001)}, S, so the antisymmetrization must come for the spin wf, and hence a $d=3$ contribution. Or, finally, in {(100)(010), (100)(001), (001)(010)}, 3 combinations, which can be rewritten as either S or A, to match respective spin wavefunctions so as to achieve total A, so $d=3(1+3)$. Adding these up, $D=3\times 4+3\times4+3+3\times 4=39$.

Analogously, for the two spin 1 bosons, where the combined wf has to be S, and recalling two spin 1s combine to a triplet (spin 1, A), a singlet (spin 0, S), and a quintet (spin 2, S), $\mathbf{3}\otimes\mathbf{3}=\mathbf{1}_s\oplus \mathbf{3}_a\oplus\mathbf{5}_s$, we find:

For the ground state, space S and spin S, hence $D=1+5=6$.

For the first excited space, as above, 3 space S and 3 space A, combined with spin S (1+5) and spin A (3), so that $D=3\times (1+5)+3\times 3=27$.

For the second excited state, {(200)(000), (020)(000), (002)(000)} as above, $d=3\times (1+5+3)=27$. Or {(110)(000), (101)(000), (011)(000)}, yielding another 27. Else {(100)(100) , (010)(010), (001)(001)}, S, so combined with spin S, contributing $d=3\times 6=18$. Else, finally, {(100)(010), (100)(001), (001)(010)} 3 combinations, which can be made either S or A to match the corresponding S or A spin wf. Consequently, $d=3\times 9=27$. In total, $D=27+27+18+27=99$.

Having reassured yourself of the symmetry counting principle, you may find more compact and general formulas for the sequences 1,12,39,.... and 6,27,99 ...

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  • $\begingroup$ Sorry for the late comment, but what about the (000) and (011) combination for the second excited state of fermions? Why is that not valid/does contribute to the degeneracy? $\endgroup$ – Jhonny Mar 30 '18 at 23:01
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    $\begingroup$ Right you are. (000)(011) and its 2 permutants is there, contributing 12 when multiplied by spin states. In fact, this 12 comes out of the overcounted 6, which should have been 3. I'll tweak the answer. $\endgroup$ – Cosmas Zachos Mar 31 '18 at 1:52

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