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The slater determinant of helium atom is :

  • A.Psi(1,2) = 1/sqrt(2!) times a determinant with rows 1S(2)alpha(2), 1S(1)alpha(1) and 1S(2)beta(2), 1S(1)beta(1)
  • B.Psi(1,2) = 1/sqrt(2!) times a determinant with rows 1S(1)alpha(1), 1S(2)alpha(2) and 1S(1)beta(1), 1S(2)beta(2)
  • C.Psi(1,2) = -1/sqrt(2!) times a determinant with rows 1S(1)alpha(1), 1S(2)alpha(2) and 1S(1)beta(1), 1S(2)beta(2)
  • D.Psi(1,2) = 1/2 times a determinant with rows 1S(1)alpha(1), 1S(2)beta(1) and 1S(1)alpha(2), 1S(2)beta(2)

Answer: B

The keyed Slater determinant is (B).

For helium's ground state, both electrons occupy 1s, one with spin α and one with β. The antisymmetric wavefunction is a 2 × 2 determinant built from the spin-orbitals 1sα and 1sβ, with electron 1 and electron 2 as the other index, and the normalising factor 1/√2!. Expanding (B):

\[ \begin{aligned} \Psi &= \tfrac{1}{\sqrt{2}}\,1s(1)1s(2) \\ &\times [\alpha(1)\beta(2) - \beta(1)\alpha(2)] \end{aligned} \]

a symmetric space part times the antisymmetric singlet spin function. Writing electrons by columns rather than rows does not change a determinant.

(A) swaps the two columns and (C) puts a minus sign in front; each equals −Ψ, which differs only by an overall sign. The key takes (B), the conventional form with the positive normalisation constant and electron 1 first. (D) is wrong in substance: its 1/2 is not the right normalisation, and its entries mix the electron labels inside the spin functions.

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