The two-body problem · Quantum Mechanics

Intuition

A hydrogen atom is two particles, not one electron round a fixed nucleus. When two particles interact through a force depending only on their separation, the problem splits: the centre of mass moves freely, as a particle of the total mass, and the separation moves as a single particle of the reduced mass in the potential. Since the proton is 1836 times heavier than the electron, the reduced mass is almost the electron’s, and the correction it makes is small but measured — it distinguishes hydrogen from deuterium in a spectrum.

Two dancers holding hands spin about a point between them, closer to the heavier one, while the pair as a whole drifts across the floor. The drift and the spin are separate motions, and each can be described on its own.

Two particles with $m_{1}=3m_{2}$ . The centre of mass $\mathbf{R}$ lies on the line between them, three times closer to the heavier one; the relative position $\mathbf{r}=\mathbf{r}_{1}-\mathbf{r}_{2}$ is the arrow from one to the other. The Hamiltonian separates into the free motion of $\mathbf{R}$ and the motion of $\mathbf{r}$ in the potential.

Centre of mass and relative motion

For $\hat{H}=\frac{\hat{p}_{1}^{2}}{2m_{1}}+\frac{\hat{p}_{2}^{2}}{2m_{2}}+V(|\mathbf{r}_{1}-\mathbf{r}_{2}|)$ , change to the centre of mass and the relative coordinate. The two motions decouple.

\begin{aligned}\mathbf{R}&=\frac{m_{1}\mathbf{r}_{1}+m_{2}\mathbf{r}_{2}}{M},\qquad \mathbf{r}=\mathbf{r}_{1}-\mathbf{r}_{2}\\ \hat{H}&=\frac{\hat{\mathbf{P}}^{2}}{2M}+\frac{\hat{\mathbf{p}}^{2}}{2\mu}+V(r),\qquad \mu=\frac{m_{1}m_{2}}{m_{1}+m_{2}}\end{aligned}

Properties

$M=m_{1}+m_{2}$ is the total mass and $\mu$ the reduced mass, always smaller than either mass.

The kinetic energy separates

\frac{\hat{p}_{1}^{2}}{2m_{1}}+\frac{\hat{p}_{2}^{2}}{2m_{2}}=\frac{\hat{\mathbf{P}}^{2}}{2M}+\frac{\hat{\mathbf{p}}^{2}}{2\mu}

Express the two momenta through the total momentum and the relative momentum, square, and add with the masses as weights. The cross terms cancel, and what remains is the kinetic energy of the total mass plus that of the reduced mass.

Proof steps

\hat{\mathbf{P}}=\hat{\mathbf{p}}_{1}+\hat{\mathbf{p}}_{2},\qquad \hat{\mathbf{p}}=\frac{m_{2}\hat{\mathbf{p}}_{1}-m_{1}\hat{\mathbf{p}}_{2}}{M}

The momenta conjugate to $\mathbf{R}$ and $\mathbf{r}$ .

\hat{\mathbf{p}}_{1}=\frac{m_{1}}{M}\hat{\mathbf{P}}+\hat{\mathbf{p}},\qquad \hat{\mathbf{p}}_{2}=\frac{m_{2}}{M}\hat{\mathbf{P}}-\hat{\mathbf{p}}

Solve for the particle momenta.

\begin{aligned}&\frac{\hat{p}_{1}^{2}}{2m_{1}}+\frac{\hat{p}_{2}^{2}}{2m_{2}}\\ &=\frac{m_{1}+m_{2}}{2M^{2}}\hat{P}^{2}+\Bigl(\frac{1}{2m_{1}}+\frac{1}{2m_{2}}\Bigr)\hat{p}^{2}\end{aligned}

The cross terms $\pm\frac{1}{M}\hat{\mathbf{P}}\cdot\hat{\mathbf{p}}$ cancel.

=\frac{\hat{P}^{2}}{2M}+\frac{\hat{p}^{2}}{2\mu},\qquad \frac{1}{\mu}=\frac{1}{m_{1}}+\frac{1}{m_{2}}

Simplify the two coefficients.

Applications

Practice

The Reduced Mass

The relative motion of two bodies behaves like one particle with the reduced mass, which is the product of the masses over their sum.

\mu=\frac{m_{1}m_{2}}{m_{1}+m_{2}}

Try it

Two particles have masses 2 and 6. What is their reduced mass?