Question

3. Consider two identical linear oscillators with spring constant k. The Hamiltonian is ha d k (2 + x) H 1 + + 122, 2m d. 2 where x1 and 22 are oscillator variable. (a) by changing the variables 11 = x +, 19=xY find the energies of the three lowest states of this system? (b) If the particle are with spin 1/2, which of the above three states are triplet states (S = 1) and which are singlet states (S = 0)?

Quantum Mechanics II, 'Quantum Mechanics', David H. McIntyre

Answer 1

27 2 + t klart at tdx, x₂ dy poko (a) x - x + Y , ex- / d a ax + 2 a lot Similarly A 2m bout 12 +84]+[*************] + x(x - 1) A 2 + 3 ] + (x+>> X+ +K-1) A226*(***) +[ +] For 2-D harmonic oscillator, Energ-2 (Mx + 2) the to My + # ) thug wah (kth, wyst om

For lowest 3 energy, na = my 2o seal, ny = 0 20, nyal For me, my = 0,0 Eoo - twat 4hWB *همد 3 For neal, ny 20, E lo = 3hwa +4 hwo energy states For Mazo, ny 1, Eo, a hwa + 12 hwo ) (b) For and - ne zony 20 both in other in down spin, S= - 1 = 0 , so groundstate, so one is up spin singlet state, For nealings or nyol, m or both in different state So, both either up or down spin Sot + h=1 triplet state Singeet state, a (n=0 My=0) Triplet state = 0, ny = 1) and (mx = 1, my 2o)
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