# 5.3: Two-Particle Systems

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Consider a system consisting of two particles, mass $$m_1$$ and $$m_2$$, interacting via a potential $$V(x_1-x_2)$$ that only depends on the relative positions of the particles. According to Equations ([ex3]) and ([ex10]), the Hamiltonian of the system is written $H(x_1,x_2) = -\frac{\hbar^{\,2}}{2\,m_1}\frac{\partial^{\,2}}{\partial x_1^{\,2}} - \frac{\hbar^{\,2}}{2\,m_2}\frac{\partial^{\,2}}{\partial x_2^{\,2}}+ V(x_1-x_2).$ Let $x' = x_1-x_2$ be the particles’ relative position coordinate, and $X = \frac{m_1\,x_1+m_2\,x_2}{m_1+m_2}$ the coordinate of the center of mass. It is easily demonstrated that \begin{aligned} \frac{\partial}{\partial x_1}& = \frac{m_1}{m_1+m_2}\frac{\partial}{\partial X} + \frac{\partial}{\partial x'},\\[0.5ex] \frac{\partial}{\partial x_2} &= \frac{m_2}{m_1+m_2}\frac{\partial}{\partial X} - \frac{\partial}{\partial x'}. \end{aligned} Hence, when expressed in terms of the new variables, $$x'$$ and $$X$$, the Hamiltonian becomes $\label{ex6.24} H(x',X) = -\frac{\hbar^{\,2}}{2\,M} \frac{\partial^{\,2}}{\partial X^{\,2}} -\frac{\hbar^{\,2}}{2\,\mu}\frac{\partial^{\,2}}{\partial x'^{\,2}} + V(x'),$ where $M = m_1+ m_2$ is the total mass of the system, and $\mu = \frac{m_1\,m_2}{m_1+m_2}$ the so-called reduced mass . Note that the total momentum of the system can be written $\label{exa} P= -{\rm i}\,\hbar\left(\frac{\partial}{\partial x_1} + \frac{\partial}{\partial x_2}\right) = -{\rm i}\,\hbar\,\frac{\partial}{\partial X}.$

The fact that the Hamiltonian ([ex6.24]) is separable when expressed in terms of the new coordinates [i.e., $$H(x',X) = H_{x'}(x') + H_X(X)]$$ suggests, by analogy with the analysis in the previous section, that the wavefunction can be factorized: that is, $\label{exb} \psi(x_1,x_2,t) = \psi_{x'}(x',t)\,\psi_X(X,t).$ Hence, the time-dependent Schrödinger equation ([ex7]) also factorizes to give ${\rm i}\,\hbar\,\frac{\partial\psi_{x'}}{\partial t} = -\frac{\hbar^{\,2}}{2\,\mu} \frac{\partial^{\,2}\psi_{x'}}{\partial x'^{\,2}} + V(x')\,\psi_{x'},$ and ${\rm i}\,\hbar\,\frac{\partial\psi_X}{\partial t} = -\frac{\hbar^{\,2}}{2\,M}\frac{\partial^{\,2}\psi_X}{\partial X^{\,2}}.$ The previous equation can be solved to give $\label{ex33} \psi_X(X,t) = \psi_{0}\,{\rm e}^{\,{\rm i}\,(P'\,X/\hbar - E'\,t/\hbar)},$ where $$\psi_0$$, $$P'$$, and $$E' = P'^{\,2}/2\,M$$ are constants. It is clear, from Equations ([exa]), ([exb]), and ([ex33]), that the total momentum of the system takes the constant value $$P'$$. In other words, momentum is conserved.

Suppose that we work in the centre of mass frame of the system, which is characterized by $$P'=0$$ . It follows that $$\psi_X=\psi_0$$. In this case, we can write the wavefunction of the system in the form $$\psi(x_1,x_2,t) = \psi_{x'}(x',t)\,\psi_0\equiv \psi(x_1-x_2,t)$$, where ${\rm i}\,\hbar\,\frac{\partial\psi}{\partial t} = -\frac{\hbar^{\,2}}{2\,\mu} \frac{\partial^{\,2}\psi}{\partial x^{\,2}} + V(x)\,\psi.$ In other words, in the center of mass frame, two particles of mass $$m_1$$ and $$m_2$$, moving in the potential $$V(x_1-x_2)$$, are equivalent to a single particle of mass $$\mu$$, moving in the potential $$V(x)$$, where $$x=x_1-x_2$$. This is a familiar result from classical dynamics .

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