In my lectures on quantum mechanics, it was said that the Hamiltonian for a hydrogenic atom could be split into relative motion and translational motion, as follows:

$$\hat{H} = - \frac{\hbar^2}{2m}\nabla^2_{cm} - \frac{\hbar^2}{2\mu}\nabla^2 - \frac{Ze^2}{4\pi\epsilon_0r}$$

and that one only need consider the relative motion:

$$\hat{H} = - \frac{\hbar^2}{2\mu}\nabla^2 - \frac{Ze^2}{4\pi\epsilon_0r}$$

This statement is reiterated in Atkins' Physical Chemistry (10th Edition), as well as his Molecular Quantum Mechanics (4th Edition), and various other online resources, but I am yet to find an explanation as to why.

Why can we, and why do we, neglect the translational motion of the atom?


1 Answer 1


We can do that because the variables have separated. The Hamiltonian of relative motion depends on the relative coordinates. The translational part depends on the coordinates of the center of mass. The two never mix. They don't interact; they don't even care about each other.

Separation of variables in various guises is an immensely powerful technique of solving PDE. You will encounter it time and again, including this very chapter of whatever textbook you are using, just a few lines below, when you'll start to actually solve the equation for hydrogen atom by splitting it into radial and angular parts.

So it goes.

  • $\begingroup$ I understand separability but the author goes on to write $- \frac{\hbar^2}{2\mu}\nabla^2 - \frac{Ze^2}{4\pi\epsilon_0r} = E\Psi$, declaring it the Schrödinger equation for a hydrogenic atom. What is the physical intuition behind seemingly totally discarding the translational motion? (i.e. how can you disregard translational motion and still set the result equal to $E$?) $\endgroup$
    – Jacob
    Nov 14, 2017 at 12:11
  • 2
    $\begingroup$ This is $E$ of relative motion. Translational motion is associated with its own kinetic energy; it can be anything, so why bother. $\endgroup$ Nov 14, 2017 at 12:37

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