0
$\begingroup$

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 $$ H = \frac{p^{2}}{2m_e} - \frac{Z*e^2}{|r|}~?$$

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

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}

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.