Multiple Electrons in Rectangular Traps — Question 9

Section 40-7 · Physics pages 6708–6709

Question 9 Partial shell filling creates many degenerate ground determinants In a square box, the lowest spatial orbital has energy 2ϵ, followed by two degenerate spatial orbitals at 5ϵ. Four noninteracting electrons are present, with no magnetic field and no interaction splitting. (1, 2) (2, 1) Schematic; labels, not drawing lengths, specify the data. (a) Find the ground energy and number of degenerate ground determinants. (b) Group the open-shell determinants by total spin projection MS in units of hbar.
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Question 9 - Solution Partial shell filling creates many degenerate ground determinants (1, 2) (2, 1) (a) Method and calculation Core: two electrons at 2ϵ; open shell: two at 5ϵ. Eg = 14ϵ, d = ( 4 2 ) = 6 . (b) Interpretation and further calculation MS = +1 : ( 2 2 ) = 1; MS = −1 : ( 2 2 ) = 1. MS = 0 : 2 × 2 = 4 ⇒d+1, d0, d−1 = 1, 4, 1 . Physical check These are determinants classified by spin projection. Some linear combinations have definite total spin; interactions can lift the degeneracy, so Hund ordering cannot be inferred from this noninteracting model alone.

Question 9

Partial shell filling creates many degenerate ground determinants

In a square box, the lowest spatial orbital has energy 2ϵ2\epsilon, followed by two degenerate spatial orbitals at 5ϵ5\epsilon. Four noninteracting electrons are present, with no magnetic field and no interaction splitting.

Diagram shown in the original SVG worksheet.

  1. Find the ground energy and number of degenerate ground determinants.

  2. Group the open-shell determinants by total spin projection MSM_S in units of hbar.

Question 9 - Solution

Partial shell filling creates many degenerate ground determinants

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Core: two electrons at 2ϵ;open shell: two at 5ϵ.\text{Core: two electrons at }2\epsilon;\quad\text{open shell: two at }5\epsilon.Eg=14ϵ,d=(42)=6.\boxed{E_g=14\epsilon,\quad d=\binom42=6}.

(b) Interpretation and further calculation

MS=+1:(22)=1;MS=−1:(22)=1.M_S=+1:\binom22=1;\qquad M_S=-1:\binom22=1.MS=0:2×2=4⇒d+1,d0,d−1=1,4,1.M_S=0:\quad2\times2=4\Rightarrow\boxed{d_{+1},d_0,d_{-1}=1,4,1}.

Physical check

These are determinants classified by spin projection. Some linear combinations have definite total spin; interactions can lift the degeneracy, so Hund ordering cannot be inferred from this noninteracting model alone.