I would like to know how to derive the wave function of two non-intering electrons in one-dimensional space by accounting their spin and Pauli's exclusion principle.

$\Psi(x_1, x_2,\sigma_1, \sigma_2)$ and $V(x_1, x_2) = V(x_1) + V(x_2)$

I'm just wondering that if I can use the form of separable wavefunction

$\Psi(x_1, x_2,\sigma_1, \sigma_2) = \psi(x_1, x_2) \cdot \chi(\sigma_1, \sigma_2) = \psi_1(x_1)\psi_2(x_2) \cdot \chi_1(\sigma_1)\chi_2(\sigma_2)$

Subtituting it in Schrodinger equation would yeild the same result as the case without spin.


1 Answer 1


You did not yet account for Pauli's exclusion principle. For this the wave function needs to be antisymmetric on exchanging the particles $1$ and $2$, i.e. $$\Psi(x_1,x_2,\sigma_1,\sigma_2)=-\Psi(x_2,x_1,\sigma_2,\sigma_1)$$

You can achieve this with the following wavefunction $$\begin{align} \Psi(x_1,x_2,\sigma_1,\sigma_2) &= \psi_1(x_1)\psi_2(x_2) \cdot \chi_1(\sigma_1)\chi_2(\sigma_2) \\ &- \psi_1(x_2)\psi_2(x_1) \cdot \chi_1(\sigma_2)\chi_2(\sigma_1) \end{align}$$ which satisfies both Schrödinger's equation and Pauli's exclusion principle.

  • $\begingroup$ Thank you for reply. So, in this case the spin state of each can be different, can’t it? $\endgroup$
    – Tommy Wang
    Jun 24 at 13:23
  • $\begingroup$ @TommyWang Yes, $\chi_1$ and $\chi_2$ may be different, $\endgroup$ Jun 24 at 13:34

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