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\begin{document}

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\today\\
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\title{Schr\"odinger equation in coordinates $(r_1,r_2,\theta)$ for two-electron atoms}

\maketitle

%\tableofcontents

\large

\section{Laplacian in coordinates $(r_1,r_2,\theta)$}

Let us prove that if a function depends only on three variables, \[\Psi(\vec{r}_1, \vec{r}_2)= \psi(r_1,r_2,\theta),\] then
\begin{multline}
\label{lapl}
(\vec{\nabla}_1^2+ \vec{\nabla}_2^2) \Psi(\vec{r}_1, \vec{r}_2)= 
\bigg[\frac1{r_1^{D-1}} \frac{\partial}{\partial r_1} r_1^{D-1} \frac{\partial}{\partial r_1}+ \frac1{r_2^{D-1}} \frac{\partial}{\partial r_2} r_2^{D-1} \frac{\partial}{\partial r_2}+ \\
\left(\frac1{r_1^2}+\frac1{r_2^2}\right) \frac1{\sin^{D-2}\theta} \frac{\partial}{\partial \theta} \sin^{D-2}\theta \frac{\partial}{\partial \theta} \bigg] \psi(r_1,r_2,\theta).
\end{multline}

Firstly, we calculate $\vec{\nabla}_1^2 \Psi(\vec{r}_1, \vec{r}_2)$, where $\vec{\nabla}_1^2= \sum_{i=1}^{D}\frac{\partial^2}{\partial r_{1i}^2}$. First derivatives are
\begin{equation}
\label{der1}
\frac{\partial}{\partial r_{1i}}\psi(r_1,r_2,\theta)= \left(\frac{\partial r_1}{\partial r_{1i}} \frac{\partial}{\partial r_1}+ \frac{\partial \theta}{\partial r_{1i}} \frac{\partial}{\partial \theta}\right) \psi(r_1,r_2,\theta).
\end{equation}
Second derivatives are
\begin{multline}
\label{der2}
\frac{\partial}{\partial r_{1i}^2}\psi(r_1,r_2,\theta)= \bigg[\left(\frac{\partial r_1}{\partial r_{1i}}\right)^2 \frac{\partial^2}{\partial r_1^2}+ 2 \frac{\partial r_1}{\partial r_{1i}} \frac{\partial \theta}{\partial r_{1i}} \frac{\partial^2}{\partial r_1 \partial \theta}+
\left(\frac{\partial \theta}{\partial r_{1i}}\right)^2 \frac{\partial^2}{\partial \theta^2} +\\
\frac{\partial^2 r_1}{\partial r_{1i}^2} \frac{\partial}{\partial r_1}+ \frac{\partial^2 \theta}{\partial r_{1i}^2} \frac{\partial}{\partial \theta}
\bigg] \psi(r_1,r_2,\theta).
\end{multline}
Derivatives of $r_1$ are
\begin{equation}
\label{derr1}
\begin{split}
\frac{\partial r_1}{\partial r_{1i}}&= \frac{r_{1i}}{r_1},\\
\frac{\partial^2 r_1}{\partial r_{1i}^2}&= \frac{1}{r_1}- \frac{r_{1i}^2}{r_1^3}.
\end{split}
\end{equation}
After summation over $i$, we obtain
\begin{align}
\label{derr1s1}
\sum_{i=1}^{D}\left(\frac{\partial r_1}{\partial r_{1i}}\right)^2&= \sum_{i=1}^{D}\frac{r_{1i}^2}{r_1^2}=1,\\
\label{derr1s2}
\sum_{i=1}^{D}\frac{\partial^2 r_1}{\partial r_{1i}^2}&= \sum_{i=1}^{D}\left(\frac{1}{r_1}- \frac{r_{1i}^2}{r_1^3}\right)= \frac{D-1}{r_1}.
\end{align}
Combining \eeref{derr1s1}{derr1s2}, we obtain
\begin{equation}
\label{derr1c}
\sum_{i=1}^{D}\left[\left(\frac{\partial r_1}{\partial r_{1i}}\right)^2 \frac{\partial^2}{\partial r_1^2}+ \frac{\partial^2 r_1}{\partial r_{1i}^2} \frac{\partial}{\partial r_1}\right]= \frac1{r_1^{D-1}} \frac{\partial}{\partial r_1} r_1^{D-1} \frac{\partial}{\partial r_1},
\end{equation}
which gives the first term in square brackets in \eref{lapl}.

In order to find $\partial \theta/\partial r_{1i}$, we use an identity
\begin{equation}
\label{rrcos}
(\vec{r}_1,\vec{r}_2)= r_1r_2\cos\theta.
\end{equation}
Differentiating \eref{rrcos} once, we obtain
\begin{equation}
\label{rrcos1}
r_{2i}=\frac{r_{1i}}{r_1}r_2\cos\theta -r_1r_2\sin\theta \frac{\partial\theta}{\partial r_{1i}}.
\end{equation}
Solving \eref{rrcos1} in respect to $\frac{\partial\theta}{\partial r_{1i}}$, we obtain
\begin{equation}
\label{dertheta}
\frac{\partial\theta}{\partial r_{1i}}= \frac{r_{1i} r_2\cos\theta-r_{2i}r_1}{r_1^2 r_2 \sin\theta}.
\end{equation}
Using \eeref{derr1}{dertheta}, we calculate
\begin{equation}
\label{mix}
\sum_{i=1}^{D}\frac{\partial r_1}{\partial r_{1i}} \frac{\partial\theta}{\partial r_{1i}}= \sum_{i=1}^{D}\frac{r_{1i}^2 r_2\cos\theta-r_{1i}r_{2i}r_1}{r_1^2 r_2 \sin\theta}= \frac{r_1^2 r_2\cos\theta-r_1 r_2 \cos\theta r_1}{r_1^2 r_2 \sin\theta}= 0,
\end{equation}
and
\begin{multline}
\label{dth1}
\sum_{i=1}^{D}\left(\frac{\partial\theta}{\partial r_{1i}}\right)^2= \sum_{i=1}^{D}\frac{r_{1i}^2 r_2^2\cos^2\theta-2r_{1i}r_{2i}r_1r_2\cos\theta+ r_{2i}^2r_1^2}{r_1^4 r_2^2 \sin^2\theta}=\\
\frac{r_1^2 r_2^2\cos^2\theta-2r_1^2 r_2^2\cos^2\theta+ r_1^2 r_2^2}{r_1^4 r_2^2 \sin^2\theta}= \frac{r_1^2 r_2^2\sin^2\theta}{r_1^4 r_2^2 \sin^2\theta}= \frac1{r_1^2}.
\end{multline}
Differentiating \eref{rrcos} twice, we obtain
\begin{equation}
\label{rrcos2}
0= \frac{\partial^2 r_1}{\partial r_{1i}^2}r_2\cos\theta -2 \frac{\partial r_1}{\partial r_{1i}}r_2 \sin\theta \frac{\partial \theta}{\partial r_{1i}}
-r_1r_2 \cos\theta \left(\frac{\partial \theta}{\partial r_{1i}}\right)^2- r_1r_2\sin\theta \frac{\partial^2 \theta}{\partial r_{1i}^2}.
\end{equation}
After summation over $i$ and with use of (\ref{derr1s2}), (\ref{mix}), and (\ref{dth1}) we obtain
\begin{equation}
\label{rrcos2s}
0= \frac{D-1}{r_1}r_2\cos\theta- r_1r_2\cos\theta \frac1{r_1^2}-r_1r_2\sin\theta \sum_{i=1}^D \frac{\partial^2 \theta}{\partial r_{1i}^2},
\end{equation}
from which we obtain
\begin{equation}
\label{dtheta2}
\sum_{i=1}^D \frac{\partial^2 \theta}{\partial r_{1i}^2}= \frac{D-2}{r_1^2}\,\frac{\cos\theta}{\sin\theta}.
\end{equation}
Combining \eeref{dth1}{dtheta2}, we obtain
\begin{equation}
\label{derthetac}
\sum_{i=1}^{D}\left[\left(\frac{\partial\theta}{\partial r_{1i}}\right)^2 \frac{\partial^2}{\partial \theta^2}+ \frac{\partial^2 \theta}{\partial r_{1i}^2} \frac{\partial}{\partial \theta}\right]= \frac1{r_1^2 \sin^{D-2}\theta} \frac{\partial}{\partial \theta} \sin^{D-2}\theta \frac{\partial}{\partial \theta},
\end{equation}
which gives a part of the third term in square brackets in \eref{lapl}.

Using \eref{der2} and combining (\ref{derr1c}), (\ref{mix}), (\ref{derthetac}) we find
\begin{equation}
\label{nab1}
\vec{\nabla}_1^2 \Psi(\vec{r}_1, \vec{r}_2)= \frac1{r_1^{D-1}} \frac{\partial}{\partial r_1} r_1^{D-1} \frac{\partial}{\partial r_1}+
\frac1{r_1^2 \sin^{D-2}\theta} \frac{\partial}{\partial \theta} \sin^{D-2}\theta \frac{\partial}{\partial \theta}.
\end{equation}
Adding a similar expression for $\vec{\nabla}_2^2 \Psi(\vec{r}_1, \vec{r}_2)$, we finally prove (\ref{lapl}).

\section{Elimination of first-order derivatives}

Because the Laplacian could be expressed as (\ref{lapl}), the kinetic energy operator has the form
\begin{multline}
\label{kin}
T= -\frac12(\vec{\nabla}_1^2+ \vec{\nabla}_2^2)=
-\frac12 \bigg[\frac1{r_1^{D-1}} \frac{\partial}{\partial r_1} r_1^{D-1} \frac{\partial}{\partial r_1}+ \frac1{r_2^{D-1}} \frac{\partial}{\partial r_2} r_2^{D-1} \frac{\partial}{\partial r_2}+ \\
\left(\frac1{r_1^2}+\frac1{r_2^2}\right) \frac1{\sin^{D-2}\theta} \frac{\partial}{\partial \theta} \sin^{D-2}\theta \frac{\partial}{\partial \theta} \bigg],
\end{multline}
or alternatively
\begin{multline}
\label{kina}
T= 
-\frac12 \frac{\partial^2}{\partial r_1^2} -\frac{D-1}2 \frac{\partial}{\partial r_1}
-\frac12 \frac{\partial^2}{\partial r_2^2} -\frac{D-1}2 \frac{\partial}{\partial r_2}\\
-\frac12 \left(\frac1{r_1^2}+\frac1{r_2^2}\right) \frac{\partial^2}{\partial \theta^2} -\frac{D-2}2 \,\frac{\cos\theta}{\sin\theta}\, \frac{\partial}{\partial \theta}.
\end{multline}

Let us define 
\begin{equation}
\label{phi}
\Phi(r_1,r_2,\theta)\equiv J^{1/2}(r_1,r_2,\theta) \psi(r_1,r_2,\theta),
\end{equation}
where $J$ is some positive function. Then, Schr\"odinger equation for the function $\psi$,
\begin{equation}
\label{spsi}
\left[T+V-E\right] J^{-1/2}(r_1,r_2,\theta) \Phi(r_1,r_2,\theta)=0,
\end{equation}
after multiplying by $J^{1/2}$ could be rewritten as an equation for the function $\Phi$,
\begin{equation}
\label{sphi}
\left[\bar{T}+V-E\right]\Phi(r_1,r_2,\theta)=0,
\end{equation}
where
\begin{equation}
\label{bart}
\bar{T}= J^{1/2}\,T\,J^{-1/2}.
\end{equation}

Let us choose 
\begin{equation}
\label{j}
J(r_1,r_2,\theta)= \left(r_1 r_2\right)^{D-1} \sin^{D-2}\theta.
\end{equation}
Then, using \eref{kin}, we obtain
\begin{multline}
\label{kin1}
\bar{T}= 
-\frac12 \bigg[r_1^{-\frac{D-1}2} \frac{\partial}{\partial r_1} r_1^{D-1} \frac{\partial}{\partial r_1}r_1^{-\frac{D-1}2}
+ r_2^{-\frac{D-1}2} \frac{\partial}{\partial r_2} r_2^{D-1}\frac{\partial}{\partial r_2} r_2^{-\frac{D-1}2}+ \\
\left(\frac1{r_1^2}+\frac1{r_2^2}\right) \sin^{-\frac{D-2}2}\theta \frac{\partial}{\partial \theta} \sin^{D-2}\theta \frac{\partial}{\partial \theta}\sin^{-\frac{D-2}2}\theta \bigg].
\end{multline}
After taking derivatives of products, it is evaluated as
\begin{multline}
\label{kin1a}
\bar{T}=
-\frac12 \bigg\{ \frac{\partial^2}{\partial r_1^2}- \frac{(D-1)(D-3)}{4 r_1^2} +
\frac{\partial^2}{\partial r_2^2}- \frac{(D-1)(D-3)}{4 r_2^2} \\
+\left(\frac1{r_1^2}+\frac1{r_2^2}\right) \frac{\partial^2}{\partial \theta^2}+
\left(\frac1{r_1^2}+\frac1{r_2^2}\right) \left[-\frac{(D-2)(D-4)}{4}\,\frac{\cos^2\theta}{\sin^2\theta}+ \frac{D-2}2\right]
\bigg\}.
\end{multline}
This expression has no first-order derivatives. Let us rewrite it as 
\begin{equation}
\label{kintc}
\bar{T}=\bar{T}_0+ V_\mathrm{c},
\end{equation}
where
\begin{equation}
\label{kint}
\bar{T}_0=-\frac12 \left[ \frac{\partial^2}{\partial r_1^2}+ \frac{\partial^2}{\partial r_2^2}+\left(\frac1{r_1^2}+\frac1{r_2^2}\right) \frac{\partial^2}{\partial \theta^2}\right],
\end{equation}
and
\begin{equation}
\label{kinc}
V_\mathrm{c}=-\frac12 \left(\frac1{r_1^2}+\frac1{r_2^2}\right) \left[-\frac{(D-1)(D-3)}{4}
-\frac{(D-2)(D-4)}{4}\,\frac{\cos^2\theta}{\sin^2\theta}+ \frac{D-2}2\right].
\end{equation}
The potential $V_\mathrm{c}$ could be further simplified as 
\begin{equation}
\label{kinc1}
V_\mathrm{c}=\frac12 \left(\frac1{r_1^2}+\frac1{r_2^2}\right) \left[\frac{(D-2)(D-4)}{4\sin^2\theta}-\frac14\right].
\end{equation}
Finally, we have the same expression for kinetic energy as in equations (18a), (18b) of Herschbach's paper \cite{1986JChPh..84..838H}.

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\end{document}
