Reduced masses

A molecular vibration normally involves all the atoms moving simultaneously. This is clearly very different from the simple harmonic motion of a mass attached to a spring that is attached to an immovable object. Nevertheless, it is convenient to visualize a molecular vibration as consisting of a single mass, M, on the end of a spring of force constant k. For such a system, the period of vibration, T, is given by:

$\begin{displaymath}T=2\pi\sqrt{\frac{M}{k}}. \end{displaymath}$

How, then, do we relate the complicated motion of a molecular vibration to the mass and spring model?

During a molecular vibration, each atom follows a simple harmonic motion. So the problem is, to what extent does each atom contribute to the mass, and to what extent does each atom contribute to the spring?

In order to answer this, first consider some simple systems. In the system H-X, where X has a very large mass, compared to that of the H, the effective mass is obviously that of H. In H₂, the effective mass is half that of a single H. Why is this so? In H-X, particle X is stationary, and particle H contributes 100% of the energy to the vibration. In H₂, each particle obviously contributes 50%, but now the center of mass is half way between the two particles. If the force constants are the same in H-X and in H-H, then the period of vibration of H-X will be 2^1/2 times that of H-H. This is the same period as for a system of two particles, each of which having a mass twice that of a H particle. For a system of two particles, A and B, having masses M_A and M_B, the vibrational wavefunction, $\psi_v$ , is:

$\begin{displaymath}\psi_v=\sqrt{\frac{M_B}{M_A+M_B}}\psi_A-\sqrt{\frac{M_A}{M_A+M_B}}\psi_B. \end{displaymath}$

This can be interpreted as particle A moves $(\sqrt{\frac{M_B}{M_A+M_B}})^2$ in the time particle B moves $(\sqrt{\frac{M_A}{M_A+M_B}})^2$ . The center of mass, $\rho$ , stays constant:

$\begin{displaymath}\rho=\sum_iM_i\delta x_i = M_A\frac{M_A}{M_A+M_B}- M_B\frac{M_A}{M_A+M_B} = 0. \end{displaymath}$

The square of the coefficients of the wavefunction represent the contribution to the motion. The effective mass, $\mu$ , for this system is:

$\begin{displaymath}\mu = \frac{M_A\times M_B}{M_A+M_B}. \end{displaymath}$

What fraction is due to A and what fraction is due to B? From the wavefunction, the intensity of A is $\frac{M_B}{M_A+M_B}$ , and the relative rate of motion is also $\frac{M_B}{M_A+M_B}$ , so the contribution to the effective mass due to A is:

$\begin{displaymath}(\frac{M_B}{M_A+M_B})M_A; \end{displaymath}$

likewise, for B:

$\begin{displaymath}(\frac{M_A}{M_A+M_B})M_B. \end{displaymath}$

Consider two particles, A and B, of mass 1 and 4, respectively. The wavefunction for the vibration is:

$\begin{displaymath}\psi_v = \sqrt{\frac{4}{5}}\psi_A-\sqrt{\frac{1}{5}}\psi_B, \end{displaymath}$

where A contributes

$\begin{displaymath}\frac{16}{25}\times 1 = 0.64 \end{displaymath}$

and B contributes

$\begin{displaymath}\frac{1}{25}\times 4 = 0.16 \end{displaymath}$

to the effective mass of $\frac{4}{5}$ .

In other words, the contribution to the effective mass is equal to the intensity of the wavefunction on each atom, times the mass of the atom, times the intensity of the wavefunction. This is intuitively correct: the total vibration is composed of contributions from each particle, and the amount each particle contributes is proportional to its intensity in the wavefunction. The mass of each particle is also proportional to its intensity in the wavefunction.

Extension to polyatomic molecules is now trivial. The effective mass is given by:

$\begin{displaymath}\mu = \sum_A <\!\psi_AM_A\psi_A\!>\times <\!\psi_A\vert\psi_A\!>. \end{displaymath}$

When written in this way, the quantum nature of the expression is obvious. However, for computational convenience, the effective mass is rewritten as:

$\begin{displaymath}\mu = \sum_A(\psi_A^2)^2\times M_A \end{displaymath}$

$\begin{displaymath}\mu = \sum_A(\sum_{i=1}^3c_{A_i}^2)^2M_A. \end{displaymath}$

This expression is suitable for most systems. However, it is not a well-defined quantity. Under certain circumstances involving degenerate vibrations, the quantity μ can become ill-defined. This phenomenon can be attributed to the fact that the reduced mass is not an observable.