I am wondering if the Hamilton operator of an atom always is calculated by the same formular $$ H = \frac{p_e^{2}}{2m} + \frac{P_N^{2}}{2M} -\frac{Z*e^2}{|r|}$$ with index e stands for electron and $N$ for Nucleus. I have also seen some diferences in this formular. Is this the same as $$ H = \frac{p^{2}}{2m_e} - \frac{Z*e^2}{|r|}~?$$

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    $\begingroup$ What about an atom with more than 1 electron? $\endgroup$ – Qmechanic Jan 9 '18 at 20:00

The Hamiltonian for any number of electrons and any number of nuclei is:

\begin{align} H=&-\sum_i \frac{\hbar^2}{2M_i}\nabla_{R_i}^2 -\sum_i \frac{\hbar^2}{2m_i}\nabla_{r_i}^2 + \\ &+\frac{e^2}{4\pi\epsilon_0}\sum_{i<j}\frac{1}{|r_i-r_j|} - \frac{e^2}{4\pi\epsilon_0}\sum_{i,j}\frac{Z_i}{|R_i-r_j|} + \\ &+\frac{e^2}{4\pi\epsilon_0}\sum_{i<j}\frac{Z_iZ_j}{|R_i-R_j|} \end{align}

where $M_i$ ($m_i$) are the nuclear (electronic) masses, $R_i$ ($r_i$) are the nuclear (electronic) positions, and $Z_i$ are the atomic numbers.

The first two terms are the KE of the nuclei and electrons, followed by the electrostatic electron-electron repulsion, the electron-nucleus attraction, and the nucleus-nucleus repulsion.

To answer your question: for hydrogen you set $Z=1$, and you have a single electron and a single nucleus, so the Hamiltonian reduces to:

\begin{align} H=&-\frac{\hbar^2}{2M}\nabla_{R}^2 -\frac{\hbar^2}{2m}\nabla_{r}^2 - \frac{e^2}{4\pi\epsilon_0}\frac{1}{|R-r|} \end{align}

You may then take the nucleus as your centre of mass (i.e. fix $R=0$), in which case it further simplifies to

\begin{align} H=&-\frac{\hbar^2}{2m}\nabla_{r}^2 - \frac{e^2}{4\pi\epsilon_0}\frac{1}{|r|} \end{align}

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  • $\begingroup$ If you include nucleus nucleus repulsion then there are no bound solutions. Also what about spin? Relativistic corrections? $\endgroup$ – lalala Feb 16 '19 at 2:32

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