Physics

V. F. Brattsev

On the Energy of the Ground State of a Molecule in the Adiabatic Approximation

(Presented by Academician V. A. Fock on 27 VII 1964)

The paper proves the inequality \(E_A \leq E\), which relates the exact eigenvalue of the molecular Hamiltonian \(E\) and its adiabatic approximation \(E_A\). The variational principle gives, as is known, for \(E\) the approximate value \(E_V \geq E\). Thus, for \(E\) we have both an upper and a lower bound.

The Hamiltonian operator determining the internal state of a molecule may be represented in the form

\[ H = T_{\mathbf r} + T_{\mathbf R} + V(\mathbf r,\mathbf R), \tag{1} \]

where

\[ T_{\mathbf r} = -{}^{1}/_{2}\sum_i \Delta_{r_i} \]

is the kinetic-energy operator of the electrons;

\[ T_{\mathbf R} = -{}^{1}/_{2}\sum_j \frac{1}{m_j}\Delta_{R_j} \]

is the kinetic-energy operator of the nuclei;

\[ V(\mathbf r,\mathbf R) = \sum_{i<j}\frac{1}{|r_i-r_j|} - \sum_{i,j}\frac{Z_j}{|R_j-r_i|} + \sum_{i<j}\frac{Z_iZ_j}{|R_i-R_j|} \]

is the operator of the potential energy of interaction of all particles. Here \(m_j\) denotes the masses of the nuclei, and \(Z_j\) their charges.

In the adiabatic approximation, the solution of the Schrödinger equation

\[ \{T_{\mathbf R}+T_{\mathbf r}+V(\mathbf r,\mathbf R)\}\Psi(\mathbf r,\mathbf R) = E\Psi(\mathbf r,\mathbf R) \]

is reduced, as is known, to the successive solution of two equations:

\[ \{T_{\mathbf r}+V(\mathbf r,\mathbf R)\}\Phi_A(\mathbf r,\mathbf R) = \mathscr E(\mathbf R)\Phi_A(\mathbf r,\mathbf R), \]

\[ \{T_{\mathbf R}+\mathscr E(\mathbf R)\}\varphi_A(\mathbf R) = E_A\varphi_A(\mathbf R), \]

and the quantity \(E_A\) is taken as the approximate value of the energy, while the product \(\Phi_A(\mathbf r,\mathbf R)\varphi_A(\mathbf R)\) is taken as the approximate corresponding function.

The normalization of \(\Phi_A\) and \(\varphi_A\) corresponds to the normalization of \(\Psi\). If \(\Psi\) is normalized to unity,

\[ \iint |\Psi(\mathbf r,\mathbf R)|^2\,d\mathbf r\,d\mathbf R = 1, \]

then one may set

\[ \int |\Phi_A(\mathbf r;\mathbf R)|^2\,d\mathbf r = 1, \qquad \int |\varphi_A(\mathbf R)|^2\,d\mathbf R = 1. \]

Let \(E\) be the least eigenvalue, and let \(\Psi\) be the corresponding eigenfunction of the operator (1), normalized to unity. In this case \(\Phi_A\) and \(\varphi_A\) also correspond to the least eigenvalues \(\mathscr E(\mathbf R)\) and \(E_A\). The functions \(\Psi\), \(\Phi_A\), and \(\varphi_A\) may obviously be considered real.

We shall prove that

\[ E_A \leqslant E . \tag{2} \]

First note that the function \(\Psi(\mathbf r,\mathbf R)\) can be represented (uniquely up to sign) in the form of the product

\[ \Psi(\mathbf r,\mathbf R)=\Phi(\mathbf r,\mathbf R)\varphi(\mathbf R), \]

in which

\[ \int \Phi^2(\mathbf r,\mathbf R)\,d\mathbf r=1,\qquad \int \varphi^2(\mathbf R)\,d\mathbf R=1. \tag{3} \]

For this it is sufficient (and necessary) to put

\[ \varphi^2(\mathbf R)=\int \Psi^2(\mathbf r,\mathbf R)\,d\mathbf r,\qquad \Phi(\mathbf r,\mathbf R)=\frac{\Psi(\mathbf r,\mathbf R)}{\varphi(\mathbf R)}. \]

Differentiating (3) with respect to \(R_j\), we obtain

\[ \int \Phi(\mathbf r,\mathbf R)\frac{\partial\Phi(\mathbf r,\mathbf R)}{\partial R_j}\,d\mathbf r=0, \tag{4} \]

\[ -\int \Phi(\mathbf r,\mathbf R)\frac{\partial^2\Phi(\mathbf r,\mathbf R)}{\partial R_j^2}\,d\mathbf r\,d\mathbf R = \int \left[\frac{\partial\Phi(\mathbf r,\mathbf R)}{\partial R_j}\right]^2\,d\mathbf r . \tag{5} \]

Let us now express \(E\) through \(\Psi(\mathbf r,\mathbf R)=\Phi(\mathbf r,\mathbf R)\varphi(\mathbf R)\):

\[ \begin{aligned} E&=\iint \Psi(\mathbf r,\mathbf R)\{T_{\mathbf r}+T_{\mathbf R}+V(\mathbf r,\mathbf R)\}\Psi(\mathbf r,\mathbf R)\,d\mathbf r\,d\mathbf R\\ &=\iint \Phi(\mathbf r,\mathbf R)\varphi(\mathbf R)\{T_{\mathbf r}+V(\mathbf r,\mathbf R)\}\Phi(\mathbf r,\mathbf R)\varphi(\mathbf R)\,d\mathbf r\,d\mathbf R\\ &\quad+\iint \Phi^2(\mathbf r,\mathbf R)\varphi(\mathbf R)T_{\mathbf R}\varphi(\mathbf R)\,d\mathbf r\,d\mathbf R +\iint \varphi^2(\mathbf R)\Phi(\mathbf r,\mathbf R)T_{\mathbf R}\Phi(\mathbf r,\mathbf R)\,d\mathbf r\,d\mathbf R\\ &\quad-\iint \Phi(\mathbf r,\mathbf R)\varphi(\mathbf R)\sum_j\frac{1}{m_j} \frac{\partial\varphi(\mathbf R)}{\partial R_j} \frac{\partial\Phi(\mathbf r,\mathbf R)}{\partial R_j}\,d\mathbf r\,d\mathbf R . \end{aligned} \]

Thanks to (3) and (4),

\[ \iint \Phi^2(\mathbf r,\mathbf R)\varphi(\mathbf R)T_{\mathbf R}\varphi(\mathbf R)\,d\mathbf r\,d\mathbf R = \int \varphi(\mathbf R)T_{\mathbf R}\varphi(\mathbf R)\,d\mathbf R, \]

\[ \begin{aligned} \iint \Phi(\mathbf r,\mathbf R)\varphi(\mathbf R)\sum_j\frac{1}{m_j} \frac{\partial\varphi(\mathbf R)}{\partial R_j} \frac{\partial\Phi(\mathbf r,\mathbf R)}{\partial R_j}\,d\mathbf r\,d\mathbf R &= \sum_j\frac{1}{m_j}\int \varphi(\mathbf R)\frac{\partial\varphi(\mathbf R)}{\partial R_j} \left[\int \Phi(\mathbf r,\mathbf R)\frac{\partial\Phi(\mathbf r,\mathbf R)}{\partial R_j}\,d\mathbf r\right]\,d\mathbf R\\ &=0, \end{aligned} \]

and the expression for \(E\) is somewhat simplified:

\[ \begin{aligned} E&=\int \varphi(\mathbf R)\left\{T_{\mathbf R} +\int \Phi(\mathbf r,\mathbf R)[T_{\mathbf r}+V(\mathbf r,\mathbf R)]\Phi(\mathbf r,\mathbf R)\,d\mathbf r\right\} \varphi(\mathbf R)\,d\mathbf R\\ &\quad+\iint \varphi^2(\mathbf R)\Phi(\mathbf r,\mathbf R)T_{\mathbf R}\Phi(\mathbf r,\mathbf R)\,d\mathbf r\,d\mathbf R . \end{aligned} \]

Using (5), we have

\[ \iint \varphi^2(\mathbf R)\Phi(\mathbf r,\mathbf R)T_{\mathbf R}\Phi(\mathbf r,\mathbf R)\,d\mathbf r\,d\mathbf R = \iint \varphi^2(\mathbf R)\sum_j\frac{1}{m_j} \left(\frac{\partial\Phi(\mathbf r,\mathbf R)}{\partial R_j}\right)^2 \,d\mathbf r\,d\mathbf R\geqslant 0. \]

Consequently,

\[ E\geqslant \int \varphi(\mathbf R)\left\{T_{\mathbf R} +\int \Phi(\mathbf r,\mathbf R)[T_{\mathbf r}+V(\mathbf r,\mathbf R)]\Phi(\mathbf r,\mathbf R)\,d\mathbf r\right\} \varphi(\mathbf R)\,d\mathbf R . \]

For any \(\mathbf R\), according to the definition of \(\mathcal E(\mathbf R)\) as the smallest eigenvalue of the operator \(T_{\mathbf r}+V(\mathbf r,\mathbf R)\),

\[ \mathcal E(\mathbf R)\leqslant \int \Phi(\mathbf r,\mathbf R)\{T_{\mathbf r}+V(\mathbf r,\mathbf R)\}\Phi(\mathbf r,\mathbf R)\,d\mathbf r . \]

Hence

\[ E\geqslant \int \varphi(\mathbf R)\{T_{\mathbf R}+\mathcal E(\mathbf R)\}\varphi(\mathbf R)\,d\mathbf R . \]

But \(E_A\) is the least eigenvalue of the operator \(T_R+\mathscr{E}(\mathbf{R})\) and, consequently,

\[ \int \varphi(\mathbf{R})\{T_R+\mathscr{E}(\mathbf{R})\}\varphi(\mathbf{R})\,d\mathbf{R}\geq E_A . \]

Thus, indeed, \(E\geq E_A\). On the other hand,

\[ E\leq \iint \Phi_A(\mathbf{r},\mathbf{R})\varphi_A(\mathbf{R})\{T_r+T_R+V(\mathbf{r},\mathbf{R})\}\Phi_A(\mathbf{r},\mathbf{R})\varphi_A(\mathbf{R})\,d\mathbf{r}\,d\mathbf{R}= \]

\[ =E_A+\iint \varphi_A^2(\mathbf{R})\Phi_A(\mathbf{r},\mathbf{R})T_R\Phi_A(\mathbf{r},\mathbf{R})\,d\mathbf{r}\,d\mathbf{R}. \]

Together with (2) this gives the following estimate for the difference \(E-E_A\):

\[ 0\leq E-E_A\leq \iint \varphi_A^2(\mathbf{R})\Phi_A(\mathbf{r},\mathbf{R})T_R\Phi_A(\mathbf{r},\mathbf{R})\,d\mathbf{r}\,d\mathbf{R}. \]

Hence, in particular, it follows that as \(m=\min_j\{m_j\}\to\infty\) the difference

\[ E-E_A\to 0 \]

no more slowly than \(\dfrac{1}{m}\).

Leningrad Branch
of the V. A. Steklov Mathematical Institute
Academy of Sciences of the USSR

Received
1 VII 1964