More Successful Attempt

As $\mathcal{H}_{\rm en}(q,X)$ depends on the positions of the nuclei $X$, let us try an ansatz

$$\Psi(q,X) = \psi_e(q,X) \phi_n (X)$$ (2.2)

where now the electronic part depends on the nuclear coordinates $X$ as well. This looks unsymmetric: why shouldn't one have $\Psi(q,X)=\psi_e(q,X) \phi_n (q,X)$? First, there is an asymmetry in the problem in the form of $M\gg m$, and $\Psi(q,X)=\psi_e(q,X) \phi_n (q,X)$ is no more better than $\Psi(q,X)$ in the first place.

The idea with writing $\Psi(q,X)= \psi_e(q,X) \phi_n (X)$ is that the electronic part $\psi_e(q,X)$ already solves part of the problem, i.e.

$$\left[\mathcal{H}_{\rm e}(q,p)+\mathcal{H}_{\rm en}(q,X)\right] \psi_e(q,X)= E(X) \psi_e(q,X),$$ (2.3)

an equation in which $X$, of course, appears as an external classical parameter that commutes with all other variables. Consequently, the eigenvalue $E(X)$ has to depend on $X$. We thus obtain

$$\mathcal{H} \psi_e \phi_n \equiv \left[ \mathcal{H}_{\rm e}+ \mathcal{H}_{\rm n} + \mathcal{H}_{\rm en}\right] \psi_e\phi_n$$

$$= \left[\mathcal{H}_{\rm n} + E(X)\right] \psi_e\phi_n \quad (?)= {\mathcal E}\psi_e\phi_n$$ (2.4)

where the last questionmark indicated what we would like to have! Since $\mathcal{H}_{\rm n}$ and $E(X)$ depend on the nuclear coordinates only, one would like to use an equation like

$$\left[\mathcal{H}_{\rm n}+ E(X) \right] \phi_n(X) = {\mathcal E} \phi_n(X),$$ (2.5)

because then we would have achieved our goal. However, the operator $\mathcal{H}_{\rm n}$ contains the nuclear momenta $P$ which operate on the $X$ in $\psi_e(q,X)$, i.e.

$$\mathcal{H} \psi_e \phi_n = \psi_e\left[ \mathcal{H}_{\rm n} + E(X)\right] \phi_n + \underlin... ...ft[ \mathcal{H}_{\rm n}\psi_e \phi_n - \psi_e\mathcal{H}_{\rm n} \phi_n\right]}$$

$$= {\mathcal E} \psi_e\phi_n + \underline{\left[ \mathcal{H}_{\rm n}\psi_e \phi_n - \psi_e\mathcal{H}_{\rm n} \phi_n\right]}.$$ (2.6)

This shows that we are almost there if it wasn't for the underlined term. One now tries to find arguments why this term can be neglected. If it can be neglected, then we have achieved the full solution of the Schrödinger equation by the two separate equations

$$\left[\mathcal{H}_{\rm e}(q,p)+\mathcal{H}_{\rm en}(q,X)\right] \psi_e(q,X) = E(X) \psi_e(q,X)$$

$$\left[\mathcal{H}_{\rm n}+ E(X) \right] \phi_n(X) = {\mathcal E} \phi_n(X) .$$ (2.7)

These two equations Eq. (V.2.7) are the central equations of the Born-Oppenheimer approximation. Even without solving them, some quite interesting observations can already be made:

• The electronic part is calculated as if the nuclei were at fixed positions $X$ (`clamped nuclei').
• The eigenvalue of the energy of the electronic part serves as a potential energy for the nuclei in the nuclear part of the equations.