Center of Mass coordinate

Let our Hamiltonian be of the form $H(x_1,p_1,x_2,p_2)=\frac{P_1}{2m_1}+\frac{P_2}{2m_2}+V(x_1-x_2)$. Since the Potential $V$ only depends on the relative position of the particle $x_1,x_2$, we can split the system into $2$ separate coordinate systems by considering the virtual particle with equations

$$\{\begin{array}{cccc}{x}_{CM}& =& \frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}& \\ x& =& x_1-x_2& \text{x is the relative virtual particle’s position}\\ p_{CM}& =& m_1{\dot{x}}_1+m_2{\dot{x}}_2& \\ p& =& \mu \dot{x}& \text{μ is the relative virtual particle’s mass}\end{array}$$

These two virtual particle have mass $M_{CM}=m_1+m_2$ and $M_{rel}=\frac{m_1{m}_2}{m_1+m_2}$ ( the harmonic mean of $\{m_1,m_2\}$)

With this, we are able to re-write the Hamiltonian as

$$\begin{array}{rll}H(x_1,p_1,x_2,p_2)& =H(x_{CM},p_{CM},x,p)& \\ & =H_{CM}(x_{CM},p_{CM})+H_{rel}(x,p)+V(x)& \text{This is a separable Hamiltionian}\end{array}$$

By getting a Separable Hamiltonian, we get 2 simpler systems which can be solved independently. The $CM$ system can often be ignored, as it behaves as a free particle with linear motion and constant Energy.