The Born–Oppenheimer aproximation is said to be adiabatic.

  • What does "adiabatic" mean?
  • What would be a non-adiabatic approximation to the Schrödinger equation?
  • $\begingroup$ Note that the Born-Oppenheimer approximation is just further approximation of the adiabatic one, and as such, it is also adiabatic. $\endgroup$
    – Wildcat
    May 14, 2015 at 7:40
  • $\begingroup$ Also, if you want a less mathematical but more intuitive answer. You can have a look at here: rpi.edu/dept/phys/Dept2/Courses/phys410/lct10.pdf $\endgroup$
    – James LT
    Feb 5, 2017 at 2:51

1 Answer 1


$\newcommand{\conj}[1]{\overline{#1}{}} \newcommand{\braket}[2]{\langle{#1}\,|\,{#2}\rangle} \newcommand{\bracket}[3]{\langle{#1}\,|\,{#2}\,|\,{#3}\rangle} \newcommand{\mat}[1]{\mathbf{#1}} \newcommand{\rel}{\vec{r}_{\mathrm{e}}} \newcommand{\rnuc}{\vec{r}_{\mathrm{n}}} \newcommand{\linop}[1]{\hat{#1}} \newcommand{\Tnuc}{\linop{T}_{\mathrm{n}}} \newcommand{\Tel}{\linop{T}_{\mathrm{e}}} \newcommand{\Velel}{\linop{V}_{\mathrm{ee}}} \newcommand{\Velnuc}{\linop{V}_{\mathrm{en}}} \newcommand{\Vnucnuc}{\linop{V}_{\mathrm{nn}}} \newcommand{\Hel}{\linop{H}_{\mathrm{e}}} \newcommand{\psiel}{\psi_{\mathrm{e}}} \newcommand{\psieli}{\psi_{\mathrm{e},i}} \newcommand{\psielj}{\psi_{\mathrm{e},j}} \newcommand{\Eel}{E_{\mathrm{e}}} \newcommand{\Eeli}{E_{\mathrm{e},i}} \newcommand{\Eelj}{E_{\mathrm{e},j}} \newcommand{\psinuc}{\psi_{\mathrm{n}}} \newcommand{\psinuci}{\psi_{\mathrm{n},i}} \newcommand{\psinucj}{\psi_{\mathrm{n},j}}$


In short, our goal is to separate the coordinates of all electrons $\rel = \{ \vec{r}_{i} \}_{i=1}^{n}$ from that of the nuclei $\rnuc = \{ \vec{r}_{\alpha} \}_{\alpha=1}^{\nu}$. Thus, we write the time-independent Schrödinger equation in the following form \begin{equation} \linop{H} \psi(\rel, \rnuc) = E \psi(\rel, \rnuc) \, , \tag{1} \end{equation} where the Hamiltonian operator is given as follows \begin{equation} \linop{H} = \Tnuc + \Tel + \Velnuc + \Vnucnuc + \Velel \, , \end{equation} or, explicitly in atomic units, \begin{equation} \linop{H} = - \sum\limits_{\alpha=1}^{\nu} \frac{1}{2 m_{\alpha}} \nabla_{\alpha}^2 - \sum\limits_{i=1}^{n} \frac{1}{2} \nabla_{i}^2 - \sum\limits_{\alpha=1}^{\nu} \sum\limits_{i=1}^{n} \frac{Z_{\alpha}}{r_{\alpha i}} + \sum\limits_{\alpha=1}^{\nu} \sum\limits_{\beta > \alpha} \frac{Z_{\alpha} Z_{\beta}}{r_{\alpha \beta}} + \sum\limits_{i=1}^{n} \sum\limits_{j > i}^{n} \frac{1}{r_{ij}} \, , \end{equation}
where $m_{\alpha}$ is the rest mass of nucleus $\alpha$ and $Z_{\alpha}$ is its atomic number, $r_{\alpha i} = |\vec{r}_{\alpha} - \vec{r}_{i}|$ is the distance between nucleus $\alpha$ and electron $i$, $r_{\alpha \beta} = |\vec{r}_{\alpha} - \vec{r}_{\beta}|$ is the distance between two nuclei $\alpha$ and $\beta$, and $r_{ij} = |\vec{r}_{i} - \vec{r}_{j}|$ is the distance between two electrons $i$ and $j$. Here for simplicity we ignore magnetic interactions and use the non-relativistic treatment.

Exact separation

We know that nuclei are much heavier as compared to electrons, $m_{\mathrm{n}} \gg m_{\mathrm{e}}$, and assuming that at average the kinetic energy of electrons and nuclei is of the same order of magnitude (see the equipartition theorem) \begin{equation} \frac{m_{\mathrm{n}} v_{\mathrm{n}}^2}{2} \sim \frac{m_{\mathrm{e}} v_{\mathrm{e}}^2}{2} \, , \end{equation} we conclude that even for case of the lightest nucleus (a single proton) \begin{equation} \frac{v_{\mathrm{e}}}{v_{\mathrm{n}}} \sim \sqrt{\frac{m_{\mathrm{n}}}{m_{\mathrm{e}}}} \sim \sqrt{1836} \sim 43 \, , \end{equation} i.e., nuclei at average are moving considerably slower than electrons (talking in terms of classical mechanics). For a coupled motion of two subsystems with significantly different speeds it is often assumed that one could treat the motion of a relatively fast moving subsystem (electrons) as if it depends only on the static configuration of a relatively slow moving one (nuclei). This assumption results in a procedure with two major steps, described below.

First step

First, we solve an equation describing the motion of electrons for a number of static configurations of the nuclei. The equation is known as the electronic Schrödinger equation and it can be written as follows \begin{equation} \Hel \psieli(\rel, [\rnuc]) = \Eeli([\rnuc]) \psieli(\rel, [\rnuc]) \quad \text{for} \quad i = 1, 2, \dotsc \, . \end{equation} where $\Hel = \Tel + \Velnuc + \Velel$ is the electronic Hamiltonian, $\psieli$ is the electronic wave function describing $i$-th electronic state, and $\Eeli$ is the electronic energy of the $i$-th electronic state --- the total energy of a molecule with nuclei being fixed at particular coordinates less the nuclear repulsion potential energy $\Eel = U - V_{\mathrm{nn}}$.

Note that the electronic wave function is a function of electronic coordinates only, though parametrically it still depends on nuclear coordinates. The parametric dependence of $\psieli$ and $\Eeli$ on $\rnuc$ means that for different sets of fixed nuclear coordinates $\rnuc$ we have different electronic Schrödinger equations with different solutions $\psieli$ and $\Eeli$, or, in other words, that the electronic Schrödinger equation written above is a family of equations parametrized by $\rnuc$, rather then one equation.

Second step

Once solutions of the electronic Schrödinger equation $\{ \psieli(\rel) \}_{i=1}^{\infty}$ and $\{ \Eeli \}_{i=1}^{\infty}$ are found for a number of different fixed nuclear configurations $\rnuc$, we took the second step: the task is to solve an equation of motion for nuclei which is obtained as follows. Note first that since $\Hel$ is self-adjoint, its eigenfunctions $\{ \psieli(\rel, [\rnuc]) \}_{i=1}^{\infty}$ form a complete orthonormal set. Now we use the assertion that a function of two independent variables $\psi(\rel, \rnuc)$ can be expanded over a complete set of functions of one variable $\{ \psieli(\rel, [\rnuc]) \}_{i=1}^{\infty}$ in the following way \begin{equation} \psi(\rel, \rnuc) = \sum\limits_{i=1}^{\infty} \psieli(\rel, [\rnuc]) \psinuci(\rnuc) \, . \tag{2} \end{equation} Substitution of (2) in the Schrödinger equation (1), multiplication of both sides by $\conj{\psi}_{\mathrm{e},j}$, and integration over electronic coordinates $\rel$ only, leads to \begin{equation} \Big( \Tnuc + \Eelj + \Vnucnuc - E \Big) \psinucj - \sum\limits_{i=1}^{\infty} \linop{\Lambda}_{ji} \psinuci = 0 \quad \text{for } j = 1, 2, \dotsc \, , \tag{3} \end{equation} where operator $\linop{\Lambda}_{ji}$ is defined as follows \begin{equation} \linop{\Lambda}_{ji} = \sum\limits_{\alpha=1}^{\nu} \frac{1}{2 m_{\alpha}} \big( 2 \bracket{ \psielj }{ \nabla_{\alpha} }{ \psieli } \nabla_{\alpha} + \bracket{ \psielj }{ \nabla_{\alpha}^2 }{ \psieli } \big) \, . \end{equation} and arguments $\rel$ and $\rnuc$ were dropped for brevity. The last equation describes a system of differential equations (one equation for each value of $j$), and once the system is solved and a set of functions $\psinuci$ is obtained, one could write down the exact solution of the Schrödinger equation (1) in the form we introduced above (2).


The task to solve the system of equations (3) is extremely difficult and the reason is that equations in the system are coupled in a sense that the solution of any $i$-th equation, $\psinuci(\rnuc)$, enters all other equations as the $\linop{\Lambda}_{ji} \psinuci$ term. One obvious way to decouple equations in the system (3) is by assuming that $\linop{\Lambda}_{ji} = 0$ either for all $i \neq j$, or for all $i$ without any exception. Operators $\linop{\Lambda}_{ji}$ may be regarded as the elements of a square matrix $\mat{\Lambda}$ and of course the fact that we equate these elements with zero is an approximation: in the first case, when we equate with zero all off-diagonal elements $\linop{\Lambda}_{ji}$, we introduce the adiabatic approximation, while in the second case, when we equate with zero all elements $\linop{\Lambda}_{ji}$, we introduce the Born–Oppenheimer approximation.

Both these approximations lead to a system of the so-called nuclear Schrödinger equations \begin{equation} \linop{H}_{\mathrm{n}} \psinucj(\rnuc) = E \psinucj(\rnuc) \quad \text{for } j = 1, 2, \dotsc \, , \end{equation} where $\psinucj(\rnuc)$ is known as the nuclear wave function and $\linop{H}_{\mathrm{n}}$ as the nuclear Hamiltonian, with the only difference being that in the first case the nuclear Hamiltonian takes the following form \begin{equation} \linop{H}_{\mathrm{n}} = \Tnuc + \Eelj([\rnuc]) + \Vnucnuc - \linop{\Lambda}_{jj} \, , \end{equation} while in the second case the last term $\linop{\Lambda}_{jj}$ is absent \begin{equation} \linop{H}_{\mathrm{n}} = \Tnuc + \Eelj([\rnuc]) + \Vnucnuc \, . \end{equation}

The first thing to notice is that in both the above mentioned approximations the only terms in the nuclear Schrödinger equation for $j$-th nuclear state $\psinucj$ that depend on electronic states, $\Eelj$ and also $\linop{\Lambda}_{jj}$, are related to the same and the only electronic state $\psielj$. This means that in both these approximations each nuclear state $\psinucj$ depends only on one electronic state $\psielj$, while in the accurate physical picture (3), electronic states are coupled through off-diagonal $\linop{\Lambda}_{ji}$ terms, known as non-adiabatic coupling terms (NACT).

Note also that the first term in the nuclear Hamiltonian in both adiabatic and Born–Oppenheimer approximations represents the kinetic energy of nuclear motion, and consequently, the remaining terms collectively should stand for the potential energy of nuclear motion. It is said thus that nuclei move on the so-called potential energy surface (PES) defined by the sum of these remaining terms: $\Eelj([\rnuc]) + \Vnucnuc - \linop{\Lambda}_{jj}$ in the adiabatic approximation and just $\Eelj([\rnuc]) + \Vnucnuc$ in the Born–Oppenheimer one.

Non-adiabatic coupling terms usually (but not always) can indeed be neglected, especially around the equilibrium nuclear configuration. Besides, often the errors due to neglecting these terms are less than the errors resulting from some further approximation introduced when solving electronic and nuclear equations. However, when two potential energy surfaces come close to each other or even cross, non-adiabatic coupling terms becomes large and are no longer negligible.


What adiabatic means?

The term "adiabatic" from Ancient Greek αδιαβατος (α - "not", δια - "through", βατος - "passable") literally means the situation when something "is not passing through" something else. In thermodynamics adiabatic process is a process occurring without exchange of heat of a system with its environment, i.e. a process in which heat is not passing through system enclosure. In quantum mechanics adiabatic refers to a process in which no abrupt transition from one state to another with respect to continuous changes of some parameters happens, and thus, energy of a system changes continuously with respect to continuous changes of these parameters.

And the reason why we call the above mentioned approximation adiabatic lies in the fact that in this approximation when solving the electronic Schrödinger equation for different fixed nuclear configurations by continuously varying the coordinates of the nuclei we indeed assume that electronic state of the system continues to be the same. In other words, we assume that we are "staying" at the same PES. This could be the case, of course, only when potential energy surfaces for different electronic states are well separated, so, in the regions of nuclear coordinates where potential energy surfaces come close to or even cross each other adiabatic approximation is inadequate.

What would be the case of a non adiabatic appox. to Schrodinger eq.?

Non-adiabatic solution refers to the solution of the system of coupled equations (3).

  • $\begingroup$ This answer is clearer than textbooks I have read on the BO approximation, and on adiabatic approximations in general! $\endgroup$
    – Yoda
    Nov 11, 2016 at 15:39
  • $\begingroup$ Wildcat, thanks for your answer, it was very instructive. I just want to make sure I understood you properly. Regarding the part on nuclear Schrodinger equations, this is a system of Schrodinger equations, as each electronic state $\psi_{\mathrm{e},j}$ leads to a different Hamiltonian $\hat{H}_{\mathrm{n},j}$, and each of these Schrodinger equations generates a set of nuclear wavefunctions $\{\psi_{\mathrm{n},k}\}$ which only depend on one electronic state $\psi_{\mathrm{e},j}$ - did I get that right? $\endgroup$
    – orthocresol
    May 19, 2017 at 22:47
  • 2
    $\begingroup$ @orthocresol, yep. Each $j$-th nuclear SE has a set of solutions, so for each and every electronic state $\psi_{\mathrm{e},j}$ you indeed get a bunch of nuclear states $\{\psi_{\mathrm{n},k}^{j}\}$ where $k=1, 2, \dotsc$. These states can be further (approximately) subdivided into vibrational, translational, and rotational. $\endgroup$
    – Wildcat
    May 20, 2017 at 11:49
  • 1
    $\begingroup$ Firstly, may I ask, why is the assumption that the nuclei and electrons have approximately the same momentum is valid? Secondly, in the Gaussian DFT procedure to determine molecular vibrational frequencies, is the nuclear SE solved? I understand that DFT essentially solves the electronic SE and the ground state energy (PES) is sum of the internuclei repulsion energy and the electronic energy. And that the DFT procedure solves for the electronic energy (EE) first, then determines the PES later. I know the procedure for determining EE well, but not so sure as to how DFT gives PES. $\endgroup$ Jan 1, 2019 at 14:54
  • 1
    $\begingroup$ @Tan Yong Boon Regarding the assertion that 'the nuclei and electrons have approximately the same momentum', I don't think it's true. In classical terms, if you have a gas of two particles with different masses then the equipartition theorem will lead to each particle having approximately the same energy, not momentum. See, e.g., arxiv.org/pdf/physics/0610189.pdf . As a consequence, the ratios of the velocities are not proportional to m1/m2−−−−−−√ . So, in classical terms, the nuclei do move slower than the electrons, but by a factor rather closer to 40 than to 2000. $\endgroup$
    – Lorents
    May 19, 2021 at 17:22

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