6.4: Two-Particle Systems

Consider a system consisting of two particles, mass \(\begin{equation}m_{1} \text { and } m_{2}\end{equation}\), interacting via the potential \(\begin{equation}V\left(x_{1}-x_{2}\right)\end{equation}\) which only depends on the relative positions of the particles. According to Eqs. (419) and (426), the Hamiltonian of the system is written

\begin{equation}H\left(x_{1}, x_{2}\right)=-\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\left(x_{1}-x_{2}\right)\end{equation}

Let

\begin{equation}x^{\prime}=x_{1}-x_{2}\end{equation}

 

be the particles' relative position, and
\begin{equation}X=\frac{m_{1} x_{1}+m_{2} x_{2}}{m_{1}+m_{2}}\end{equation}

 

the position of the center of mass. It is easily demonstrated that
\begin{equation}\frac{\partial}{\partial x_{1}}=\frac{m_{1}}{m_{1}+m_{2}} \frac{\partial}{\partial X}+\frac{\partial}{\partial x^{\prime}}\end{equation}

\begin{equation}\frac{\partial}{\partial x_{2}}=\frac{m_{2}}{m_{1}+m_{2}} \frac{\partial}{\partial X}-\frac{\partial}{\partial x^{\prime}}\end{equation}

Hence, when expressed in terms of the new variables, \(\begin{equation}x^{\prime} \text { and } X\end{equation}\), the Hamiltonian becomes

\begin{equation}H\left(x^{\prime}, X\right)=-\frac{\hbar^{2}}{2 M} \frac{\partial^{2}}{\partial X^{2}}-\frac{\hbar^{2}}{2 \mu} \frac{\partial^{2}}{\partial x^{\prime 2}}+V\left(x^{\prime}\right)\end{equation}

where

\begin{equation}M=m_{1}+m_{2}\end{equation}

 

is the total mass of the system, and
\begin{equation}\mu=\frac{m_{1} m_{2}}{m_{1}+m_{2}}\end{equation}

 

the so-called reduced mass. Note that the total momentum of the system can be written
\begin{equation}P=-i \hbar\left(\frac{\partial}{\partial x_{1}}+\frac{\partial}{\partial x_{2}}\right)=-i \hbar \frac{\partial}{\partial X}\end{equation}

The fact that the Hamiltonian (442) is separable when expressed in terms of the new coordinates [i.e., \(\begin{equation}\left.H\left(x^{\prime}, X\right)=H_{x^{\prime}}\left(x^{\prime}\right)+H_{X}(X)\right]\end{equation}\) suggests, by analogy with the analysis in the previous section, that the wavefunction can be factorized: i.e.,

\begin{equation}\psi\left(x_{1}, x_{2}, t\right)=\psi_{z^{\prime}}\left(x^{\prime}, t\right) \psi_{X}(X, t)\end{equation}

 

Hence, the time-dependent Schrödinger equation (423) also factorizes to give
\begin{equation}i \hbar \frac{\partial \psi_{x^{\prime}}}{\partial t}=-\frac{\hbar^{2}}{2 \mu} \frac{\partial^{2} \psi_{x^{\prime}}}{\partial x^{\prime 2}}+V\left(x^{\prime}\right) \psi_{x^{\prime}}\end{equation}

and

\begin{equation}\text { i } \hbar \frac{\partial \psi_{X}}{\partial t}=-\frac{\hbar^{2}}{2 M} \frac{\partial^{2} \psi_{X}}{\partial X^{2}}\end{equation}

 

The above equation can be solved to give
\begin{equation}\psi_{X}(X, t)=\psi_{0} \mathrm{e}^{\mathrm{i}\left(P^{\prime} X / \hbar-E^{\prime} t / \hbar\right)}\end{equation}

where \(\begin{equation}\psi_{\mathrm{b}}, P^{\prime}, \text { and } E^{\prime}=P^{\prime 2} / 2 M\end{equation}\) are constants. It is clear, from Eqs. (445), (446), and (449), that the total momentum of the system takes the constant value \(\begin{equation}P^{\prime}\end{equation}\) : i.e., momentum is conserved.

Suppose that we work in the centre of mass frame of the system, which is characterized by \(\begin{equation}P^{\prime}=0\end{equation}\). It follows that \(\begin{equation}\psi_{X}=\psi_{0}\end{equation}\). In this case, we can write the wavefunction of the system in the form

\begin{equation}\psi\left(x_{1}, x_{2}, t\right)=\psi_{x^{\prime}}\left(x^{\prime}, t\right) \psi_{\mathrm{b}} \equiv \psi\left(x_{1}-x_{2}, t\right), \text { where }\end{equation}

\begin{equation}\mathrm{i} \hbar \frac{\partial \psi}{\partial t}=-\frac{\hbar^{2}}{2 \mu} \frac{\partial^{2} \psi}{\partial x^{2}}+V(x) \psi\end{equation}

In other words, in the center of mass frame, two particles of mass \(\begin{equation}m_{1} \text { and } m_{2}\end{equation}\), moving in the potential \(\begin{equation}V\left(x_{1}-x_{2}\right)\end{equation}\), are equivalent to a single particle of mass \(\begin{equation}\mu, \text { moving in the potential } V(x), \text { where } x=x_{1}-x_{2}\end{equation}\). This is a familiar result from classical dynamics.

Contributors

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

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