Cusp conditions

When two Coulomb particles get close, the potential has $1/r$ singularity. We want modify the wave function in such a way to cancel this singularity. Let us consider the case of an electron close to a nucleus, the Schrödinger equation reduces to:

$$\left[- \frac{1}{2L^2} \nabla_e^2 - \frac{Ze^2}{rL}\right] \psi = E \psi$$ (C.1)

where $Z$ is the nuclear charge, notice that we used rescaled distances (see Eq. 4.23). Writing the first term in spherical coordinates, we get

$$-\frac{1}{2}\frac{\psi''}{L^2} -\frac{1}{rL}\left( Ze^2 \psi + \frac{\psi'}{L} \right) = E\psi$$ (C.2)

To cancel the singularity at small $r$ the term multiplying by $1/r$ must vanish. So we have

$$\frac{1}{\psi} \psi' = -Z L e^2$$ (C.3)

If $\psi= e^{-cr}$ we must have $c=Z L e^2$ . For the case of two electrons, when they are close each other the Schrödinger equation, using relative coordinates $r_{12} = r_1 - r_2$ , reduces to

$$\left[- \frac{\nabla_{12}^2}{L^2} + \frac{e^2}{L r_{12}}\right] \psi = E \psi$$ (C.4)

Electrons with unlike spins have an extra factor of $1/2$ in the cusp condition compared with the electron-nucleus case. So we have $c= -e^2 L/2$ . In the antisymmetric case, the electrons will be in a relative $p$ state, reducing the cusp condition by $1/2$ , so $c=-e^2 L/4$ . Since the antisymmetry requirement keeps them apart anyway having the correct cusp for like spin electrons leads to a very little in the energy or the variance(see Ref. (39).