Hydrogen atom, what's the wave equation for the atom's nucleus? 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: 


*

*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?)

*In the case when hydrogen was traveling, say along $x$ axis, would there be an extra influence/interaction towards the electron's wave equation?
 A: I write this to supplement the correct answer of @BenCrowell and the, in my view, incomplete answer of @EmilioPisanty. In my opinion, and it seems also to be Ben Crowell's, the question of the OP clearly aimed at the QM wave function description of the proton in the hydrogen model. 
The usual approach to include the effect of the proton in the hydrogen problem is to decouple the Hamiltonian into the Hamiltonian for translational motion of the center of mass and the Hamiltonian for the relative motion of electron and proton, which have the distance $\vec r=\vec r_\text{e} - \vec r_\text{p}$, which is a generalized coordinate. (See Emilio Pisanty's answer.) This Hamiltonian describing the relative motion is for a single fictitious particle with electron charge with the reduced mass $\mu=\left(1/m_\text{e}+1/m_\text{p}\right)^{-1}$ in the central Coulomb potential $\frac {e}{4\pi \epsilon_0 |\vec r|}$ with a distance $\vec r$ from the origin. In the center of mass frame, this is the only Hamiltonian necessary to describe the hydrogen atom.  For this the time-independent Schrödinger equation reads:$$H\psi\left(\vec r\right)=\left(\frac {\vec p^2}{2\mu} - \frac{e^2}{4\pi \epsilon_0 |\vec r|}\right)\psi\left(\vec r\right)=E\psi\left(\vec r\right) \tag1 $$ By solving this Schrödinger equation you get all the energy eigenvalues of the hydrogen atom including the motion effect of the proton. However, you have to keep in mind that the wave solutions $\psi\left(\vec r\right) $ (eigenfunctions) obtained are for this fictitious particle of reduced mass $\mu$ describing the combined proton-electron system, not for the electron or for the proton itself. 
Thus the question arises, whether and how the electron and proton can be described separately with wave functions giving, e.g., their spatial probability distribution. Ben Crowell has already given a correct short answer for this without a derivation. I try to show how this can be obtained from the wave functions $\psi\left(\vec r\right)$ of the fictitious particle system.
In the center of mass reference frame the center of mass position vector is zero yielding $$m_\text{e}\vec r_\text{e}+m_\text{p}\vec r_\text{p}=0 \tag 2$$ and $$\vec r_\text{e}=-\frac {m_\text{p}}{m_\text{e}}\vec r_\text{p} \tag 3$$The distance vector $\vec r$ can be expressed by the electron or the proton position vector $$\vec r=\vec r_\text{e} -\vec r_\text{p}=\vec r_\text{e}\left(\frac {m_\text{e}+m_\text{p}}{m_\text{p}}\right)=-\vec r_\text{p}\left(\frac {m_\text{e}+m_\text{p}}{m_\text{e}}\right) \tag 4$$ Thus, the wave solution of eq. (1) yields $$\psi \left(\vec r\right)=\psi\left(\vec r_\text{e}\frac {m_\text{e}+m_\text{p}}{m_\text{p}}\right)=\psi\left(-\vec r_\text{p}\left(\frac {m_\text{e}+m_\text{p}}{m_\text{e}}\right)\right)$$ Therefore, the wave functions for the electron and for the proton, $\psi_\text{e}\left(\vec r_\text{e}\right)$ and $\psi_\text{p}\left(\vec r_\text{p}\right)$,  are obtained from the wave function $\psi\left(\vec r\right)$ by simple coordinate scalings. And the wave function of the proton is related to the one of the electron by the coordinate scaling [eq. 2] $$\psi_\text{p}\left(\vec r_\text{p}\right) =\psi_\text{e}\left(-\vec r_\text{e} \frac {m_\text{e}}{m_\text{p}}\right) \tag 5$$ 
This shows that the electron and the proton wave functions can be derived from the reduced mass system wave function and that they are perfectly correlated and centered around the center of mass, as Ben Crowell has shown in his answer. The proton wave function is simply a scaled version of the electron wave function. This means, e.g., that in the ground s-state of the atom the maximum position probability density of the proton lies on a spherical shell around the center of gravity with radius $$r_\text{p}=\frac {m_\text{e}}{m_\text{p}} r_\text{e} \approx \frac {m_\text{e}}{m_\text{p}}r_{\text{Bohr}}\ \tag 6$$  which is much smaller than the Bohr radius.
I would be grateful if you could correct me or give an explanation in case you find something wrong in this supplemental derivation.
A: 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.
A: By conservation of momentum, the center of mass of the atom is what actually stays fixed. This implies that there is a perfect correlation between the wavefunctions $\Psi$ of the electron and $\Phi$ of the proton:
$$\Phi(x)=\Psi(-(M/m)x),$$
where $M$ is the mass of the proton and $m$ is the mass of the electron.
The effect on the energy levels is to replace the electron's mass with the reduced mass.
