As instructed in Levine's Quantum Chemistry 7th edition, Chapter 13.8, page 377, for a $\pi \pi$ configuration of two nonequivalent electrons, four functions $\pi_{+1}(1) \pi_{-1}^{\prime}(2)$, $\pi_{+1}(2) \pi_{-1}^{\prime}(1)$, $\pi_{-1}(1) \pi^{\prime}_{+1}(2)$, $\pi_{-1}(2) \pi^{\prime}_{+1}(1)$ are used to construct the wave functions of the singlet $\Sigma$ terms $$ \begin{array}{ll}{^{1} \Sigma^{+}:} & {\pi_{+1}(1) \pi_{-1}^{\prime}(2)+\pi_{+1}(2) \pi_{-1}^{\prime}(1)+\pi_{-1}(1) \pi_{+1}^{\prime}(2)+\pi_{-1}(2) \pi_{+1}^{\prime}(1)} \\ {^{1} \Sigma^{-}:} & {\pi_{+1}(1) \pi_{-1}^{\prime}(2)+\pi_{+1}(2) \pi_{-1}^{\prime}(1)-\pi_{-1}(1) \pi_{+1}^{\prime}(2)-\pi_{-1}(2) \pi_{+1}^{\prime}(1)}\end{array} $$

The text following these equations says "Clearly, these two spatial functions have different energies." Could you suggest how to understand this point ?

The discussion further below suggests that when four functions $\pi_{+1}(1) \pi_{+1}^{\prime}(2)$, $\pi_{+1}(2) \pi_{+1}^{\prime}(1)$, $\pi_{-1}(1) \pi^{\prime}_{-1}(2)$, $\pi_{-1}(2) \pi^{\prime}_{-1}(1)$ are used to construct the ${}^1 \Delta^{+}$ and ${}^1 \Delta^{-}$ wave functions of the singlet $\Delta$ terms, "Since they have the same energy, there is no point in using the $+$ and $-$ superscripts". Could you suggest how to understand this point and the reason of the difference from the above point ?

Many thanks for your suggestions !

  • 2
    $\begingroup$ Brief note: we prefer to not use MathJax in the title field due to issues it gives rise to; see here for details. $\endgroup$
    – andselisk
    Dec 23, 2019 at 17:30
  • $\begingroup$ I believe the sigma singlet terms describe states with electron density on different atoms in the diatomic molecule, so if they are different elements, the energies are different. In the delta singlet states, the difference between them is only the orientation relative to the x and y axes, so the energies are the same regardless of whether you have hetero or homodiatomic. I need to draw it out to be sure. $\endgroup$
    – Andrew
    Dec 24, 2019 at 3:16

1 Answer 1


The setup here is a diatomic molecule with two $\pi$ electrons in different subshells. A typical scenario would be one in a $\pi$ orbital and the other in a higher energy $\pi^*$ orbital.

In a diatomic molecule aligned with its molecular axis on the z-axis, the $\pi$ subshell consists of two orbitals, one with a node in the x-z plane and one with a node in the y-z plane. The $\pi^*$ subshell likewise consists of two orbitals, which are the antibonding counterparts to the two $\pi$ orbitals.

In the cited text, "+1" and "-1" indicate the two orientations, so it might be simpler to replace them with x and y. Likewise, the "$\pi'$" can be changed to $\pi^*$ for clarity.

Now consider the first $^1\Sigma$ term: $$^1\Sigma^+: \pi_{+1}(1)\pi'_{-1}(2) + \pi_{+1}(2)\pi'_{-1}(1)+\pi_{-1}(1)\pi'_{+1}(2)+\pi_{-1}(2)\pi'_{+1}(1)$$

We can rewrite this in the friendlier notation and use only every other term, since the others are just exchange within the same orbitals. (What follows here is conceptual, not mathematically rigorous.): $$^1\Sigma^+: \pi_x\pi^*_y + \pi_y\pi^*_x$$

The second $^1\Sigma$ term likewise simplifies to: $$^1\Sigma^-: \pi_x\pi^*_y - \pi_y\pi^*_x$$

Now we further separate the contributions from each axis: $$^1\Sigma^+: \pi_x+ \pi^*_x+\pi_y + \pi^*_y $$ $$^1\Sigma^-: \pi_x- \pi^*_x+\pi^*_y - \pi_y $$

We see that what we have done is to "undo" the formation of the $\pi$ and $\pi^*$ orbitals and essentially restored the original atomic $p$ orbitals:

$$^1\Sigma^+: \pi_x+ \pi^*_x+\pi_y + \pi^*_y=p_{Ax} + p_{Ay} $$ $$^1\Sigma^-: \pi_x- \pi^*_x+\pi^*_y - \pi_y=p_{Bx} + p_{By}$$

where A and B are the two atoms in the molecule. If A and B are different elements, these terms will have different energy.

If you do the same exercise with the $^1\Delta$ terms, you'll get something like: $$^1\Delta^+: p_{Ax} + p_{By}$$ $$^1\Delta^-: p_{Bx} + p_{Ay}$$

Since both terms have equal contributions from A and B, they will have the same energies whether or not A and B are the same element. And since the orbitals on the two axes are orthogonal to each other, the change in sign does not affect the energy.

Conceptually, the sigma terms are diradicals where either both electrons are on A or both electrons are on B in orthogonal p orbitals. The delta terms are diradicals in which one electron is on each atom.


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