Spin ${\bf S}=0$

This is the simplest case. The total angular momentum of the nuclei is then

$${\bf J} = {\bf K} - {\bf L}.$$ (1.16)

Since we have neglected geometric phase terms, we can replace $\Delta_{\bf r}$ by its expectation value in the electronic state $\alpha$ under consideration,

$$\Delta_{\bf r} = \langle \psi_\alpha\vert \Delta_{\bf r}\vert \psi_\alpha\rangle$$ (1.17)

$$\leadsto {\bf J}^2 = \langle \psi_\alpha\vert {\bf J}^2 \vert \psi_\alpha\rangle = \langle \psi_\alpha\vert ({\bf K} - {\bf L})^2 \vert \psi_\alpha\rangle.$$ (1.18)

This allows one to express everything in terms of total angular quantum numbers $K$ as follows: We first write

$${\bf J}^2 = \langle \psi_\alpha\vert ({\bf K} - {\bf L})^2 \vert \psi_\alpha\rangle$$ (1.19)

$$= \langle \psi_\alpha\vert{\bf K}^2 \vert\psi_\alpha\rangle - 2 \la... ...psi_\alpha\rangle + \langle \psi_\alpha \vert{\bf L}^2 \vert\psi_\alpha\rangle.$$

First, ${\bf K}^2$ is conserved and can be replaced by its eigenvalue $K(K+1)$ whence

$$\langle \psi_\alpha\vert{\bf K}^2 \vert\psi_\alpha\rangle = K(K+1).$$ (1.20)

Second, $\langle \psi_\alpha\vert {\bf L}^2 \vert\psi_\alpha\rangle$ only depends on the electronic degrees of freedom and can therefore be simply added to the potential $E_\alpha(r)$.

Finally, we assume that the electronic state $\alpha$ is an eigenstate of the $z$ component $L_z$ with eigenvalue $\Lambda$ of the electronic angular momentum. Then,

$$\langle \psi_\alpha\vert{\bf K}{\bf L} \vert\psi_\alpha\rangle = {\bf K} \langle \psi_\alpha\vert {\bf L}\vert \psi_\alpha\rangle = {\bf K}{\bf e}_z \Lambda.$$ (1.21)

On the other hand, we have ${\bf J}{\bf e}_z=0$ since the angular momentum of the two nuclei is perpendicular to the molecule axis ${\bf e}_z\propto {\bf r}$, thus

$$(K_z-L_z) =0 \leadsto K_z = L_z$$ (1.22)

and

$$\langle \psi_\alpha\vert{\bf K}{\bf L} \vert\psi_\alpha\rangle = {\bf K}{\bf e}_z \Lambda = L_z \Lambda$$

$$= \langle \psi_\alpha\vert L_z \vert \psi_\alpha\rangle \Lambda = \Lambda^2.$$ (1.23)

Summarizing, we now have for the radial part

$$- \frac{\hbar^2}{2\mu} \left(\frac{\partial^2}{\partial r^2}+\fra... ...gle \psi_\alpha\vert{\bf L}^2\vert \psi_\alpha\rangle}{2\mu r^2} + E_\alpha(r)}$$

$$\equiv - \frac{\hbar^2}{2\mu} \left(\frac{\partial^2}{\partial r^2}+\frac{2}{r}\frac{\partial}{\partial r}\right) + \frac{K(K+1)}{2\mu r^2} + U_\alpha(r).$$

Thus we have finally arrived at the form for the effective potential energy,

$$\frac{K(K+1)}{2\mu r^2} + U_\alpha(r).$$ (1.24)

The first term $\frac{K(K+1)}{2\mu r^2}$ is the centrifugal energy as in the hydrogen problem. Since $K_z = L_z$ with fixed eigenvalue $\Lambda$ for the given state $\alpha$, the eigenvalues of the total angular momentum must fulfill

$$K \ge \Lambda.$$ (1.25)