Multiple Electrons in Rectangular Traps — Question 7

Section 40-7 · Physics pages 6704–6705

Question 7 Dipole absorption is blocked when the destination spin orbital is filled Six independent electrons fill n = 1, 2, 3 in a one-dimensional well, two per spatial level. A weak electric-dipole perturbation preserves spin. Box matrix elements of x vanish for equal-parity indices and can be nonzero for opposite-parity indices. Energies are En = ϵn2. n = 1 n = 2 n = 3 n = 4 E Schematic; labels, not drawing lengths, specify the data. (a) Find the lowest allowed one-electron absorption from the ground configuration. (b) Explain why the one-electron 1 →2 line is absent and find the next lowest allowed absorption energy.
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Question 7 - Solution Dipole absorption is blocked when the destination spin orbital is filled n = 1 n = 2 n = 3 n = 4 E (a) Method and calculation n = 3 →4 : ∆n = 1, n = 4 empty. ∆E1 = (16 −9)ϵ = 7ϵ . (b) Interpretation and further calculation 1 →2 : both destination spin orbitals occupied ⇒Pauli blocked . 1 →4 : 15ϵ; 2 →5 : 21ϵ; 3 →6 : 27ϵ ⇒∆E2 = 15ϵ . Physical check Both an allowed matrix element and an available final spin orbital are required. This independent-electron model excludes collective and multi-electron optical transitions.

Question 7

Dipole absorption is blocked when the destination spin orbital is filled

Six independent electrons fill n=1,2,3n=1,2,3 in a one-dimensional well, two per spatial level. A weak electric-dipole perturbation preserves spin. Box matrix elements of x vanish for equal-parity indices and can be nonzero for opposite-parity indices. Energies are En=ϵn2E_n=\epsilon n^2.

Diagram shown in the original SVG worksheet.

  1. Find the lowest allowed one-electron absorption from the ground configuration.

  2. Explain why the one-electron 1→21\to2 line is absent and find the next lowest allowed absorption energy.

Question 7 - Solution

Dipole absorption is blocked when the destination spin orbital is filled

Diagram shown in the original SVG worksheet.

(a) Method and calculation

n=3→4:Δn=1,n=4 empty.n=3\to4:\quad\Delta n=1,\quad n=4\text{ empty}.ΔE1=(16−9)ϵ=7ϵ.\boxed{\Delta E_1=(16-9)\epsilon=7\epsilon}.

(b) Interpretation and further calculation

1→2:both destination spin orbitals occupied⇒Pauli blocked.1\to2:\quad\text{both destination spin orbitals occupied}\Rightarrow\boxed{\text{Pauli blocked}}.1→4:15ϵ;2→5:21ϵ;3→6:27ϵ⇒ΔE2=15ϵ.1\to4:\ 15\epsilon;\quad2\to5:\ 21\epsilon;\quad3\to6:\ 27\epsilon\Rightarrow\boxed{\Delta E_2=15\epsilon}.

Physical check

Both an allowed matrix element and an available final spin orbital are required. This independent-electron model excludes collective and multi-electron optical transitions.