Section 22.2 Reduced Mass
It is helpful to consider briefly how the quantum two-body problem separates into an equation governing the center of mass and an equation describing the system around the center of mass, comparing this process to the classical problem. The quantum two-body problem in three dimensions is very messy, but all the essential features of the calculation show up in a simple onedimensional model. So, for simplicity, let’s consider a system of two particles, \(m_1\) and \(m_2\text{,}\) lying on a line at positions \(x_1\) and \(x_2\text{,}\) and let the interaction between the particles be represented by a potential energy \(U\) that depends only on \(x=x_1-x_2\text{,}\) the separation distance between the particles. Don’t worry about how the particles can get past each other on the line—this is a simple toy model; just imagine that they can pass right through each other.
Our first job, as always, is to identify the Hamiltonian \(\Hop\) for the system. Because energies are additive, the kinetic part of the Hamiltonian is just the sum of the kinetic parts for two individual particles and the potential \(U(x)\) describes the interaction between them. Therefore the Hamiltonian is
\begin{equation}
\Hop
= -\frac{\hbar^2}{2m_1}\frac{\partial^2}{\partial x_1^2}
- \frac{\hbar^2}{2m_1}\frac{\partial^2}{\partial x_1^2} + U(x)\tag{22.2.1}
\end{equation}
and the wave function \(\Psi\) is a function of the positions of both particles (and of course time) \(\Psi=\Psi(x_1,x_2,t)\text{.}\)
Inspired by our experience with classical two-body systems, we will try rewriting the Hamiltonian (22.2.1) in terms of the center-of-mass coordinate \(X\text{,}\) given by
\begin{equation*}
X = \frac{m_1x_1+m_2x_2}{m_1+m_2}
\end{equation*}
and the relative coordinate \(x\text{.}\) We will use the chain rule of calculus to transform the partial derivatives in (22.2.1) to derivatives with respect to \(x\) and \(X\text{.}\) (Please see Appendix [A], especially the worked example on plane polar coordinates.) The transformations for first derivatives are:
\begin{align}
\frac{\partial}{\partial x_1}
\amp= \frac{\partial x}{\partial x_1}\frac{\partial}{\partial x}
+ \frac{\partial X}{\partial x_1}\frac{\partial}{\partial X}
= \frac{\partial}{\partial x}
+ \frac{m_1}{m_1+m_2}\frac{\partial}{\partial X}\tag{22.2.2}\\
\frac{\partial}{\partial x_2}
\amp= \frac{\partial x}{\partial x_2}\frac{\partial}{\partial x}
+ \frac{\partial X}{\partial x_2}\frac{\partial}{\partial X}
= -\frac{\partial}{\partial x}
+\frac{m_2}{m_1+m_2}\frac{\partial}{\partial X}\tag{22.2.3}
\end{align}
It is important to note that we cannot simply write equations (22.2.2)–(22.2.3) for the second derivative, which is what we need for the Hamiltonian (22.2.1). To find the second derivative, we must apply the first derivative rules (22.2.2)–(22.2.3) twice:
\begin{align}
\frac{\partial^2}{\partial x_1^2}
\amp= \frac{\partial}{\partial x_1}\frac{\partial}{\partial x_1}\Psi\notag\\
\amp= \left(
\frac{\partial}{\partial x}
+ \frac{m_1}{m_1+m_2}\frac{\partial}{\partial X}
\right) \left(
\frac{\partial}{\partial x}
+ \frac{m_1}{m_1+m_2}\frac{\partial}{\partial X}
\right) \Psi\notag\\
\amp= \frac{\partial^2}{\partial x^2}\Psi
+ \frac{2m_1}{m_1+m_2} \frac{\partial^2}{\partial x\partial X}\Psi
+ \left(\frac{m_1}{m_1+m_2}\right)^2\frac{\partial^2}{\partial X^2}\Psi\tag{22.2.4}\\
\frac{\partial^2}{\partial x_2^2}
\amp= \frac{\partial}{\partial x_2}\frac{\partial}{\partial x_2}\Psi\notag\\
\amp= \left(
-\frac{\partial}{\partial x}
+ \frac{m_2}{m_1+m_2}\frac{\partial}{\partial X}
\right) \left(
-\frac{\partial}{\partial x}
+ \frac{m_2}{m_1+m_2}\frac{\partial}{\partial X}
\right) \Psi\notag\\
\amp= \frac{\partial^2}{\partial x^2}\Psi
- \frac{2m_2}{m_1+m_2} \frac{\partial^2}{\partial x\partial X}\Psi
+ \left(\frac{m_1}{m_1+m_2}\right)^2\frac{\partial^2}{\partial X^2}\Psi\tag{22.2.5}
\end{align}
Substituting into the Hamiltonian (22.2.1), we obtain for Schrödinger’s equation
\begin{equation}
\left(
-\frac{\hbar^2}{2\mu}\frac{\partial^2}{\partial x}^2
-\frac{\hbar^2}{2(m_1+m^2)}\frac{\partial^2}{\partial X^2} + U(x)
\right) \Psi(X,x,t)
= i\hbar \frac{\partial}{\partial t} \Psi(X,x,t)\tag{22.2.6}
\end{equation}
By transforming to these coordinates, the middle terms in equations (22.2.4) and (22.2.5) have canceled, enabling us to separate the dependence on \(x\) from the dependence on \(X\text{.}\) We can now write
\begin{equation}
\Psi(x,X,t) = \Psi_M(X)\Psi_\mu(x)T(t)\tag{22.2.7}
\end{equation}
After a separation of variables procedure (see Appendix [B]) on equation (22.2.7), we find that the ordinary differential equation governing the variable \(X\) has a simple, recognizable form (see Problem [ProbsC].[14.3b]). The solution has the same form as the free-particle solution to the Schrödinger equation (also called the plane-wave solution to the equation)
\begin{equation}
\Psi_M(X) = e^{iP_XX/\hbar}\tag{22.2.8}
\end{equation}
where \(P_X\) represents the momentum associated with the motion of the center of mass. All observables in quantum mechanics involve the probability density, i.e. terms of the form \(\Psi^*\Psi\text{,}\) so if we are evaluating observables associated with the relative motion, the pure phase contribution from the center-of-mass has no effect. We can therefore ignore the center-of-mass motion and concentrate only on the relative motion.
We have arrived at a conclusion in the quantum analysis of the two-body problem that is similar to our analysis of the classical problem (but for different reasons). We have again replaced the more complicated two-body system with a fictitious one-body system, involving the relative coordinate and the reduced mass. Once we have solved the problem and found \(\Psi_\mu(x)\) and \(T(t)\text{,}\) we can then reverse the procedure in this section to find the wave function \(\Psi(x_1,x2,t)\) describing the original two-body system. The analysis in three dimensions is the same, except that we must do the calculation three times, once for each of the rectangular coordinates.
