Let’s take the a simple $H\_2^+$ molecule, where there is only electron which is $r\_a$ away from the first proton and $r\_b$ away from the other one.

Let’s call the separation between the two protons $R$.

As $R\rightarrow \infty$, the electron will stick to one of the two protons, so the wavefunction will be: $$ \phi = N\_{\pm}(1s\_a \pm 1s\_b).$$

I recognise the two solutions as the gerade and ungerade orbitals. The $1s$ means ground state around each proton.

I can work out the normalisation constant to be $$ N\_{\pm} = \sqrt{\frac{1}{2(1\pm S)}}, \, S = \int 1s\_a 1s\_b \mathrm{d}^3 r $$

**Now** in the limit of $R \rightarrow 0$, the gerade solution becomes just $1s$ which makes sense, but the ungerade is not defined - what happens to it?