Two Particles in Three Dimensions

The generalization of the Hamiltonian to three dimensions is simple.

\begin{displaymath}\bgroup\color{black}H={\vec{p}_1^2\over 2m}+{\vec{p}_2^2\over 2m}+V(\vec{r}_1-\vec{r}_2)\egroup\end{displaymath}

We define the vector difference between the coordinates of the particles.
\bgroup\color{black}$\displaystyle \vec{r}\equiv\vec{r}_1-\vec{r}_2 $\egroup
We also define the vector position of the center of mass.
\bgroup\color{black}$\displaystyle \vec{R}\equiv{m_1\vec{r}_1+m_2\vec{r}_2\over m_1+m_2} $\egroup

We will use the chain rule to transform our Hamiltonian. As a simple example, if we were working in one dimension we might use the chain rule like this.

\begin{displaymath}\bgroup\color{black}{d\over dr_1}={\partial r\over\partial r_...
...+{\partial R\over\partial r_1}{\partial\over\partial R} \egroup\end{displaymath}

In three dimensions we would have.

\begin{displaymath}\bgroup\color{black}\vec{\nabla}_1=\vec{\nabla}_r+{m_1\over m_1+m_2}\vec{\nabla}_R \egroup\end{displaymath}


\begin{displaymath}\bgroup\color{black}\vec{\nabla}_2=-\vec{\nabla}_r+{m_2\over m_1+m_2}\vec{\nabla}_R \egroup\end{displaymath}

Putting this into the Hamiltonian we get

\begin{displaymath}\bgroup\color{black}H={-\hbar^2\over 2m_1}\left[\vec{\nabla}_...
...1\over m_1+m_2}\vec{\nabla}_r\cdot\vec{\nabla}_R\right] \egroup\end{displaymath}


\begin{displaymath}\bgroup\color{black}\qquad +{-\hbar^2\over 2m_2}\left[\vec{\n...
...m_2}\vec{\nabla}_r\cdot\vec{\nabla}_R\right]+V(\vec{r}) \egroup\end{displaymath}


\begin{displaymath}\bgroup\color{black}H=-\hbar^2\left[ \left({1\over 2m_1}+{1\o...
...2\over 2(m_1+m_2)^2}\vec{\nabla}_R^2\right]+V(\vec{r}) .\egroup\end{displaymath}

Defining the reduced mass \bgroup\color{black}$\mu$\egroup

\bgroup\color{black}$\displaystyle {1\over\mu}={1\over m_1}+{1\over m_2} $\egroup
and the total mass
\bgroup\color{black}$\displaystyle M=m_1+m_2 $\egroup
we get.
\bgroup\color{black}$\displaystyle H=-{\hbar^2\over 2\mu}\vec{\nabla}_r^2-{\hbar^2\over 2M}\vec{\nabla}_R^2+V(\vec{r}) $\egroup

The Hamiltonian actually separates into two problems: the motion of the center of mass as a free particle

\begin{displaymath}\bgroup\color{black}H={-\hbar^2\over 2M}\vec{\nabla}_R^2\egroup\end{displaymath}

and the interaction between the two particles.
\bgroup\color{black}$\displaystyle H=-{\hbar^2\over 2\mu}\vec{\nabla}_r^2+V(\vec{r}) $\egroup

This is exactly the same separation that we would make in classical physics.

Jim Branson 2013-04-22