According to Sakurai the solutions of the two-electron system are of the form $\psi=\phi({\bf x_1},{\bf x_2})\chi(m_{s1},m_{s2})$

Since it's a fermionic system, $\psi$ must be a linear combination of antisymmetric states. If $\phi$ is symmetric and $\chi$ is antisymmetric (or the other way around), then $\phi\chi$ is antisymmetric, and so is a linear combination.

With no spin dependence, the Hamiltonian is $\mathcal{H}=({\bf p_1}^2 + {\bf p_2}^2)/2m$, and the spatial solutions are of the form $\omega_A({\bf x_1})\omega_B({\bf x_2})$, so $\phi$ can be written as a symmetrical and antisymmetrical combination

\begin{equation} \phi_{\pm}({\bf x_1},{\bf x_2}) = \frac{1}{\sqrt{2}} \left[ \omega_A({\bf x_1})\omega_B({\bf x_2}) \pm \omega_A({\bf x_2})\omega_B({\bf x_1}) \right] \end{equation}

In the same way, $\chi$ can be a triplet or a singlet state.

But, is every possible solution a linear combination of antisymmetric terms $\phi\chi$? I don't think so, because I found the following state

\begin{equation} \psi = \omega_A({\bf x_1})\omega_B({\bf x_2})\chi_{+-} - \omega_A({\bf x_2})\omega_B({\bf x_1})\chi_{-+} \end{equation}

And I couldn't write it as a linear combination of the following:

\begin{equation} \left\lbrace \begin{array}[l] &\phi_+({\bf x_1},{\bf x_2})\frac{1}{\sqrt{2}}\left( \chi_{+-}-\chi_{-+} \right)\\ \phi_-({\bf x_1},{\bf x_2}) \left\lbrace \begin{array}[l] &\chi_{++}\\ \frac{1}{\sqrt{2}}\left( \chi_{+-}+\chi_{-+} \right)\\ \chi_{--} \end{array} \right. \end{array} \right. \end{equation}

The state $\psi$ is antisymmetric, and it is a valid state for the 2-electron system. But it isn't a combination of antisymmetric states of the form $\phi({\bf x_1},{\bf x_2})\chi(m_{s1},m_{s2})$, so these states do not form a complete basis of solutions. I would like to know a complete basis for the system.

  • $\begingroup$ Does Sakurai claim that states can be of that form (or linear combinations thereof), or that they must be of that form? If the latter, the claim is incorrect, but you should provide the specific wording. $\endgroup$ Commented Jul 2, 2018 at 6:32
  • $\begingroup$ @EmilioPisanty On page 453 of the 2nd edition he says: This together with (7.3.6) implies that if the space part of the wave function is symmetrical (antisymmetrical), the spin part must be antisymmetrical (symmetrical). That's why I think he claims the states must be of that form, or a linear combination $\endgroup$
    – slaaidenn
    Commented Jul 4, 2018 at 20:38
  • $\begingroup$ The claim you quote doesn't come anywhere near a claim that all states 'must' be of that form, or even that they must be linear combinations of states of that form. $\endgroup$ Commented Jul 4, 2018 at 21:22

1 Answer 1



$$ S \equiv \frac 1 {\sqrt 2}[\chi_{+-}-\chi_{-+}]$$


$$ T \equiv \frac 1 {\sqrt 2}[\chi_{+-}+\chi_{-+}]$$

and subbing in:

$$ \omega_A(x_1)\omega_B(x_2) = \frac 1 {\sqrt 2}[\phi^++\phi^-] $$

and likewise for the other $\omega$:

$$ \psi = \frac 1 2 [(\phi^+ + \phi^-)(T+S) - (\phi^+-\phi^-)(T-S)] $$

$$ \psi = \frac 1 2 [(\phi^+T + \phi^+S + \phi^-T +\phi^-S) - (\phi^+T - \phi^+S - \phi^-T +\phi^-S) ]$$

$$ \psi = \frac 1 2 [\phi^+T + \phi^+S + \phi^-T +\phi^-S - \phi^+T + \phi^+S + \phi^-T -\phi^-S ]$$

$$ \psi = \phi^+S + \phi^-T$$

which is the sum of both antisymmetric terms.


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