# How to understand the symmetry $\Sigma^-$ in the wavefunction of a diatomic molecule?

In Wikipedia (and elsewhere), a particular symmetry of the quantum system of a diatomic molecule is mentioned: symmetry under reflection along a plane containing the internuclear axis. The wavefunctions may be either symmetric or antisymmetric under the this symmetry.

It is then asserted that, in particular, for the $\Sigma$ wavefunctions, which have z-angular momentum $0$, both kinds exist, so that there are two distinct wavefunctions labelled $\Sigma^+$ and $\Sigma^-$ for the symmetric and antisymmetric case, respectively.

My question is, how can a wavefunction with a vanishing $z$ component of the angular momentum be antisymmetric under reflection through such a plane. In particular and w.l.o.g. take the reflection to be $y \rightarrow -y$, how can the wavefunction change sign under the action of this, when if the z-angular momentum is $0$, we know that the wavefunction is uniform in any cross-section parallel to the $xy$ plane?

• hey, but isn't that opposite case? I think that wiki says Λ>0 has two distinct states, symemtric and antisymmetric state. and Λ=0 state has only two states, symmetric or antisymmetric state. I think this is the case because the reflection of plane is same as the letting \phi \rightarrow -\phi in spherical harmonics. With this, the two states must be same in Λ=0 state.. – user42298 Jan 28 '15 at 6:31