Question

Discuss (qualitatively) the energy level scheme for helium if (a) electrons were identical bosons, and (b) if electrons were distinguishable particles (but with the same mass and charge). Pretend these “electrons” still have spin 1/2, so the spin configurations are the singlet and the triplet.

Short answer

(a)The ground state (Eq. 5.30) is spatially symmetric, so it goes with the symmetric (triplet) spin configuration.

(b) The ground state (Eq. 5.30) and all excited states (Eq. 5.32) come in both ortho and para form.

Step-by-step solution

(a) If electrons were identical bosons

The ground state (Equation 5.30) fits into a symmetric (triplet) spin configuration because it is spatially symmetric.

Thus, orthohelium, a degenerate triple, is the ground state. Ortho (triplet) and para excited states (Equation 5.32) are two types of excited states (singlet). The energy level of the orthohelium state is higher than that of the comparable (non-degenerate) para-state since the former is linked to the symmetric space wavefunction.

${\psi }_0\left(r_1,r_2\right)={\psi }_{100}\left(r_1\right){\psi }_{100}\left(r_2\right)=\frac{8}{\pi a^3}{e}^{-2\left(r_1+r_2\right)/a}$ …(5.30)

${\psi }_{n\ell m}{\psi }_{100}$ ...(5.32).

(b) Spinning the configurations are the singlet and the triplet

Both orthogonal and paragon variants of the ground state (Eq. 5.30) and all stimulated states (Eq. Everything is degenerate four times, or alternatively, we don't know what happens in a symmetric spatial composition, so we can't determine which is more energetic—at least ortho or para.

${\psi }_0\left(r_1,r_2\right)={\psi }_{100}\left(r_1\right){\psi }_{100}\left(r_2\right)=\frac{8}{\pi a^3}{e}^{-2\left(r_1+r_2\right)/a}$ …(5.30).

${\psi }_{n\ell m}{\psi }_{100}$ ...(5.32).