5.3: Two-Particle Systems

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 .

Contributors and Attributions

• Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)

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