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Physics LibreTexts

5.3: Two-Particle Systems

  • Page ID
    15752
  • 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

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

      \( \newcommand {\ltapp} {\stackrel {_{\normalsize<}}{_{\normalsize \sim}}}\) \(\newcommand {\gtapp} {\stackrel {_{\normalsize>}}{_{\normalsize \sim}}}\) \(\newcommand {\btau}{\mbox{\boldmath$\tau$}}\) \(\newcommand {\bmu}{\mbox{\boldmath$\mu$}}\) \(\newcommand {\bsigma}{\mbox{\boldmath$\sigma$}}\) \(\newcommand {\bOmega}{\mbox{\boldmath$\Omega$}}\) \(\newcommand {\bomega}{\mbox{\boldmath$\omega$}}\) \(\newcommand {\bepsilon}{\mbox{\boldmath$\epsilon$}}\)