Course:Physics of Atoms and Molecules/Molecules/Born-Oppenheimer
approximation
The aim of this section is to study the quantum description of the
molecular bond. We are going to limit the analysis to diatomic molecules.
First of all we consider a molucule composed by two atomic nuclei A and B,
whose positions are defined by vectors \vec{R} _{A} and \vec{R} _{B}. The
electrons positions are instead identified by \vec{x} _{i}. Both nuclei and
electrons positions are refered to the center of mass of the two nuclei.
The Hamiltonian we can write for a system like this is:
\hat{H}=\hat{T_{N}}+\hat{T_{e}}+U_{eN}+U_{ee}+U_{NN}
where \hat{T_N} and \hat{Tₑ} are the kinetic energies associated
respectively to the nuclei and the electrons and U_{eN}, U_{ee} and U_{NN}
represent the Coulomb interactions. Now it's possible to isolate from this
Hamiltonian an electron Hamiltonian:
\hat{H_e}=\hat{T_e}+U_{eN}+U_{ee}
This \hat{Hₑ} is a function of the nuclei positions and we can obtain an
eigenvalues equation for the electrons:
\hat{H_e}\left(\left\lbrace
\vec{R}_{i}\right\rbrace\right)\psi^{el}_{n}\left(\left\lbrace
\vec{R}_{i}\right\rbrace;\left\lbrace\vec{x}_{i}\right\rbrace\right)=
E^{el}_{n}\psi^{el}_{n}\left(\left\lbrace
\vec{R}_{i}\right\rbrace;\left\lbrace\vec{x}_{i}\right\rbrace\right)
It's possible to demonstrate that solving this equation for the electrons
Hamiltonian leads to a solution ψ for the original complete eigenvalue
equation which has the form:
\psi=\sum_{n}\chi_{n}\left(\left\lbrace\vec{R}_{i}\right\rbrace\right)\psi_{n}\left(\left\lbrace
\vec{R}_{i}\right\rbrace;\left\lbrace\vec{x}_{i}\right\rbrace\right)
where χ_{n} is a wave function depending only on the nuclei positions. Such
an equation is difficult to solve, but it can be semplified through the
Born-Oppenheimer adiabatic approximation. This model is based on the fact
that the mass of the nucleus is much greater than the electron one
(m_{p}/m_{e}≈1800). Being so, we can state that electrons move way faster
than nuclei. Hence we can assume that electrons follow the motion of the
nuclei without changing the energy level they belong to. In other words,
there is no energy exchange between nuclei and electrons. In this adiabatic
approximation the solution ψ becomes:
\psi=\chi_{n}\left(\left\lbrace\vec{R}_{i}\right\rbrace\right)\psi_{n}\left(\left\lbrace
\vec{R}_{i}\right\rbrace;\left\lbrace\vec{x}_{i}\right\rbrace\right)
The sum over the quantum number n disappears, because of the impossibility
for an electron to jump to an excited state. Now we can use theese results
to write an equation for the total Hamiltonian:
\left(\hat{T_N}+U_{NN}+\hat{H_e}\right)\chi_{n}\left(\left\lbrace\vec{R}_{i}\right\rbrace\right)\psi^{el}_{n}\left(\left\lbrace\vec{R}_{i}\right\rbrace;\left\lbrace\vec{x}_{i}\right\rbrace\right)=E_{tot}\chi_{n}\left(\left\lbrace\vec{R}_{i}\right\rbrace\right)\psi^{el}_{n}\left(\left\lbrace\vec{R}_{i}\right\rbrace;\left\lbrace\vec{x}_{i}\right\rbrace\right)
At this point we can observe that \hat{Hₑ} only acts on ψ^{el}, which is
actually an eigenstate of \hat{Hₑ} , so we can substitute \hat{Hₑ} in the
equation above with the corresponding eigenvalue E^{el}\left (\vec{R}
_{i}\right ). Moreover, again thanks to the greatness of nuclei masses in
respect to electrons ones, we can assume that the application of the
kinetic term \hat{T_N} to the ψ^{el} component of the total wave function
gives a negligible contribution, then we can consider that \hat{T_N} acts
only on χ. Relying on this last approximation, it's possibile to semplify
the ψ^{el} term in order to obtain an eigenvalues equation for the sole χ
function:
\left[\hat{T_N}+U_{NN}+E^{el}_n\left(\left\lbrace\vec{R}_{i}\right\rbrace\right)\right]\chi\left(\left\lbrace\vec{R}_{i}\right\rbrace\right)=E_{tot}\chi\left(\left\lbrace\vec{R}_{i}\right\rbrace\right)
The last equation describes the nuclei dynamics and this means that the
equations for ψ^{el} and χ are now uncoupled. E_{tot} depends only on the
distance between the nuclei:
E_{tot}=E_{tot}\left(\left|\vec{R}_{A}-\vec{R}_{B}\right|\right)
Now we can redefine the quantity E^{el}\left (\left \lbrace \vec{R}
_i\right \rbrace \right ) considering that U_{NN} is just a moltiplicative
operator:
E^{el}\left(\left\lbrace\vec{R}_i\right\rbrace\right)=E^{el}_n\left(\left\lbrace\vec{R}_i\right\rbrace\right)+U_{NN}
In other words, E^{el}\left (\left \lbrace \vec{R} _i\right \rbrace \right
) is the eigenvalue for the operator \hat{H} _e\left (\left \lbrace \vec{R}
_i\right \rbrace \right )+U_{NN} corresponding to the eigenfunction ψ^{el}.
It's possible to plot E^{el}\left (\left \lbrace \vec{R} _i\right \rbrace
\right ) in respect to the nuclear distance R=\left |\vec{R} _A-\vec{R}
_B\right |:
In the graph, the value of the energy in correspondence of the x axis is
E_A+E_B, where E_A and E_B are the energy associated to the idrogenoid
Hamiltonian for nucleus A and B. It can be seen that there is an R_{min} at
which the energy is minimal. This is called dissociation energy and it's
the energy that has to be given to the molecule in order to break the bond.