D.23 Molecular ion wave function symmetries

Let $z$ be the horizontal coordinate measured from the symmetry plane towards the right nucleus. Let $M$ be the “mirror operator” that changes the sign of $z$, in other words,

$$M\Psi(x,y,z)=\Psi(x,y,-z)$$

This operator commutes with the Hamiltonian $H$ since the energy evaluates the same way at positive and negative $z$. This means that operators $H$ and $M$ must have a complete set of common eigenfunctions. That set must include the ground state of lowest energy: so the ground state must be an eigenfunction of $M$ too. Now the eigenvalues of $M$, which can be seen to be a Hermitian operator, are either $+$1 or $-$1: if $M$ is applied twice, it gives back the same wave function, i.e. $1\Psi$, so the square of the eigenvalue is 1, so that the eigenvalue itself can only be 1 and $-$1. Eigenfunctions with eigenvalue 1 are called “symmetric”, eigenfunctions with eigenvalue $-$1 are called “antisymmetric”. Since the previous note found that the ground state must be everywhere positive, it can only be a symmetric eigenfunction of $M$.

Similarly, let $R$ be the operator that rotates $\Psi$ over a small angle $\phi$ around the axis of symmetry. The magnitude of the eigenvalues of $R$ must be 1, since $\Psi$ must stay normalized to 1 after the rotation. Complex numbers of magnitude 1 can be written as $e^{{\rm i}a}$ where $a$ is a real number. Number $a$ must be proportional to $\phi$, since rotating $\Psi$ twice is equivalent to rotating it once over twice the angle, so the eigenvalues are $e^{{\rm i}{m}\phi}$, where $m$ is a constant independent of $\phi$. (In addition, $m$ must be integer since rotating over 360 degrees must give back the original wave function.) In any case, the only way that $\Psi$ can be real and positive at all angular positions is if $m=0$, and then the eigenvalue of $R$ is 1, implying that the ground state $\Psi$ does not change when rotated; it must be the same at all angles. That means that the wave function is axially symmetric.

For future reference, one other symmetry must be mentioned, for the ground state of the neutral hydrogen molecule that will be covered in the next chapter. The neutral molecule has two electrons, instead of one, with positions ${\skew0\vec r}_1$ and ${\skew0\vec r}_2$. The Hamiltonian will commute with the operation of “exchanging the electrons,” i.e. swapping the values of ${\skew0\vec r}_1$ and ${\skew0\vec r}_2$, because all electrons are identical. So, for the same reasons as for the mirror operator above, the spatial wave function will be symmetric, unchanged, under particle exchange.

(Regrettably this argument stops working for more than two electrons due to the antisymmetrization requirement of chapter 5.6. It does keep working for bosons, like helium atoms, [17, p. 321])