Quantum Mechanics – What’s the Wave Equation for a Hydrogen Atom’s Nucleus?

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I learned from the class about the equation for hydrogen atom's electron where textbook assumed that the center/nuclei of hydrogen atom was fixed at origin.

However, since every particle was a wave, the nuclei of the hydrogen atom (say only contain one proton) could be seen a wave as well.

My question was that:

  1. What's the wave equation for the proton in hydrogen atom? Was it simply a traveling wave when the atom was moving, and a Dirac Delta function when it was "fixed"? (Further, what if there was a neutron?)

  2. In the case when hydrogen was traveling, say along $x$ axis, would there be an extra influence/interaction towards the electron's wave equation?

Best Answer

Basically, it's the Schrödinger equation for a free particle, but it's important to note that that particle isn't the proton - it's the entire atom's center of mass.

This is covered in reasonable detail in suitably rigorous textbooks in quantum mechanics (though I can't think of a specific example at the moment), and the basic idea goes like this:

  • You start with the Schrödinger equation for the electronic and nuclear coordinates, i.e. with the hamiltonian $$ H = \frac{1}{2M}\mathbf p_N^2 + \frac{1}{2m}\mathbf p_e^2 - \frac{Ze^2}{|\mathbf r_N-\mathbf r_e|}. $$
  • You then transform your system to a new set of coordinates: one for the relative motion, and one for the center of mass, \begin{align} \mathbf r &= \mathbf r_e-\mathbf r_N &&& \mathbf R &= \frac{1}{M+m}\left(M\mathbf r_N+m\mathbf r_e\right) \\ \mathbf p &= \frac{M}{M+m}\mathbf p_e-\frac{m}{M+m}\mathbf p_N &&& \mathbf P &= \mathbf p_N+\mathbf p_e. \end{align}
  • You verify that the new coordinates satisfy the correct (canonical) commutation relations, i.e. that $[x_j,p_k] = [X_j,P_k] = i\hbar \delta_{jk}$ and $[x_j,P_k]=0 = [X_j,p_k]$.
  • You express the nuclear and electronic coordinates as functions of the transformed coordinates, put them into your hamiltonian, and work away at the transformation, to get $$ H = \frac{1}{2(M+m)}\mathbf P^2 + \frac{1}{2\mu}\mathbf p^2 - \frac{Ze^2}{|\mathbf r|}, $$ where $\mu = \left(\frac1m+\frac1M\right)^{-1}$ is the reduced mass (itself very close to $m$ in the limit where $m\ll M$).

This decomposition completely separates out your (initially coupled) dynamical problem into two separate and quite distinct sub-problems, the usual electronic hamiltonian, $$ H_\mathrm{el} = \frac{1}{2\mu}\mathbf p^2 - \frac{Ze^2}{|\mathbf r|}, $$ and a center-of-mass hamiltonian given by just the free-particle kinetic term, $$ H_\mathrm{COM} = \frac{1}{2(M+m)}\mathbf P^2. $$ That can then be used to get the explicit wave equation for the "nuclear" (actually center-of-mass) motion. In the simplest case this is indeed just the free particle, but it's easy to see how it can be modified to, say,

  • include an explicit potential that specifically addresses the nuclear motion,
  • add the potential for a dipole trap, which works by adding an $\mathbf R$-dependent external potential that couples off-resonance to the electronic motion, which then 'freezes' that degree of freedom to an $\mathbf R$-dependent ground state with an $\mathbf R$-dependent ground-state energy that acts as a trap for the center of mass, or also
  • account for the momentum of a photon that's absorbed by the electronic degrees of freedom,

among many possible applications.

Oh, and also: nothing in my initial procedure is specific to quantum mechanics, and that separation of variables is also present in essentially identical form (i.e. you only need to swap out the canonical commutators for an identical preservation of the Poisson brackets) within classical hamiltonian mechanics.

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