Two Particles in 3 Dimensions

So far we have been working with states of just one particle in one dimension. The extension to two different particles and to three dimensions is straightforward. The coordinates and momenta of different particles and of the additional dimensions commute with each other as we might expect from classical physics. The only things that don't commute are a coordinate with its momentum, for example,

$$\bgroup\color{black}[p_{(2)z},z_{(2)}]={\hbar\over i}\egroup$$

while

$$\bgroup\color{black}[p_{(1)x},x_{(2)}]=[p_{(2)z},y_{(2)}]=0.\egroup$$

We may write states for two particles which are uncorrelated, like $u_0(\vec{x}_{(1)})u_3(\vec{x}_{(2)})$, or we may write states in which the particles are correlated. The Hamiltonian for two particles in 3 dimensions simply becomes

$$\bgroup\color{black}H={-\hbar^2\over 2m_{(1)}}\left( {\parti... ...artial z_{(2)}^2}\right) +V(\vec{x}_{(1)},\vec{x}_{(2)})\egroup$$

$$\bgroup\color{black}H={-\hbar^2\over 2m_{(1)}}\nabla^2_{(1)}+... ... 2m_{(2)}}\nabla^2_{(1)} +V(\vec{x}_{(1)},\vec{x}_{(2)})\egroup$$

If two particles interact with each other, with no external potential,

$$\bgroup\color{black}H={-\hbar^2\over 2m_{(1)}}\nabla^2_{(1)}+... ... 2m_{(2)}}\nabla^2_{(1)} +V(\vec{x}_{(1)}-\vec{x}_{(2)})\egroup$$

the Hamiltonian has a translational symmetry, and remains invariant under the translation $\vec{x}\rightarrow \vec{x}+\vec{a}$. We can show that this translational symmetry implies conservation of total momentum. Similarly, we will show that rotational symmetry implies conservation of angular momentum, and that time symmetry implies conservation of energy.

For two particles interacting through a potential that depends only on difference on the coordinates,

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

we can make the usual transformation to the center of mass made in classical mechanics

$$\bgroup\color{black}\vec{r}\equiv\vec{r}_1-\vec{r}_2\egroup$$

$$\bgroup\color{black}\vec{R}\equiv{m_1\vec{r}_1+m_2\vec{r}_2\over m_1+m_2}\egroup$$

and reduce the problem to the CM moving like a free particle

$$\bgroup\color{black}M=m_1+m_2 \egroup$$

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

plus one potential problem in 3 dimensions with the usual reduced mass.

$$\bgroup\color{black}{1\over\mu}={1\over m_1}+{1\over m_2}\egroup$$

$$\bgroup\color{black}H=-{\hbar^2\over 2\mu}\vec{\nabla}_r^2+V(\vec{r})\egroup$$

So we are now left with a 3D problem to solve (3 variables instead of 6).