# Born-Oppenheimer approximation

I'm studying about formulation of Born-Oppenheimer approximation from Atkins, Molecular quantum mechanics, 5th edition. In chapter 8.1 I found the following:

I keep wondering where did this W come from in equation 8.4.? It says later in text that it is responsible for non-adiabatic effects, but I don't see how it just appeared here. If someone is willing to explain, I'd be greatful.

Also, if anyone has tips on where to find more nice and clean text about more general formulation of Born-Oppenheimer approximation, since this one is confined to particular system.

Thanks!

• The question is at least related to that one and might be even a duplicate. – Wildcat Sep 11 '16 at 11:28
• JFYI: Atkins' book is just an introduction to the big field, and as such, quite frequently it contains no derivations but just merely states something. – Wildcat Sep 11 '16 at 11:43
• Note though that in this particular case the derivation is quite straightforward. The most important thing to watch for is that the parametric dependence of the electronic wave function on nuclear coordinates is assumed to be continuous and differentiable. So, when taking the second-order derivative of the $\psi \chi$ product with respect to nuclear coordinates in $T_N \psi \chi$ term, remember to use the product rule twice. As a result you'll get three terms: $\psi T_N \chi$ and the other two collected in $W$. – Wildcat Sep 11 '16 at 11:50

Atkins did a good job on explaining what parametric dependence of the electronic wave function (and electronic energy) on nuclear coordinates is, but failed to mention that in the derivation of the Born-Oppenheimer approximation this dependence is assumed to be continuous and differentiable and that both first and second derivatives of this quantities with respect to nuclear coordinates are in general non-zero. In particular, for the system described in the text, we have $$\frac{\partial \psi}{\partial Z_j} \neq 0 \, , \quad \frac{\partial^2 \psi}{\partial Z_j^2} \neq 0\, .$$
Now, when the solution of the form (8.3) is substituted into the Schrödinger equation (8.2), the term involving $T_\mathrm{e}$ is trivial: since the nuclear wave function $\chi$ is not a function of electronic coordinates it is just a constant when differentiating with respect to them, so we get $$T_\mathrm{e} (\psi \chi) = \chi T_\mathrm{e} \psi \, .$$ But the term involving $T_\mathrm{N}$ does not trivially transform in a similar way, $$T_\mathrm{N} (\psi \chi) \neq \psi T_\mathrm{N} \chi$$ since both $\psi$ and $\chi$ depend on the nuclear coordinates. Rather, applying the product rule twice, we get \begin{align} \frac{\partial^2}{\partial Z_j^2} (\psi \chi) &= \frac{\partial}{\partial Z_j} \left( \frac{\partial}{\partial Z_j} (\psi \chi) \right) \\ &= \frac{\partial}{\partial Z_j} \left( \psi \frac{\partial \chi}{\partial Z_j} + \chi \frac{\partial \psi}{\partial Z_j} \right) \\ &= \psi \frac{\partial^2 \chi}{\partial Z_j^2} + 2 \frac{\partial \psi}{\partial Z_j} \frac{\partial \chi}{\partial Z_j} + \chi \frac{\partial^2 \psi}{\partial Z_j^2} \, , \end{align} so that $$T_\mathrm{N} (\psi \chi) = - \psi \sum\limits_{j=1,2} \frac{\hbar^2}{2 m_j} \frac{\partial^2 \chi}{\partial Z_j^2} - \sum\limits_{j=1,2} \frac{\hbar^2}{2 m_j} \left( 2 \frac{\partial \psi}{\partial Z_j} \frac{\partial \chi}{\partial Z_j} + \chi \frac{\partial^2 \psi}{\partial Z_j^2} \right) \, ,$$ where the first term is nothing but $\psi T_\mathrm{N} \chi$ and the remainings are designated as $W$.