Helium Atom
- Quantum Mechanics Unclassified
· 10 min 
from 「Modern Quantum Mechanics」: Sakurai, J. J.
Understanding the Helium atom by variational method.
Ground state of the helium atom
Let’s approximate the ground state energy of the helium atom using the variational method. The Hamiltonian of the system is; ignoring the motion of the nucleus (Born–Oppenheimer approximation):
$ \global\def\Zef{Z\rq} $
\[ H = -\frac{\hbar^2}{2m_e}(\laplacian_1+\laplacian_2) -\frac{e^2}{4\pi\vpmt}\left[ \frac{2}{r_1}+\frac{2}{r_2}-\frac{1}{\abs{\b{r}_1-\b{r}_2}} \right] \]
Let’s set the trial wavefunction as the product of the ground state wavefunction of the hydrogenic atom system, ignoring the interaction of electrons, which are regarded as identical. Here we add a parameter $\Zef$ for the variation.
\[ \begin{align*} \psi(\b{r}_1,\b{r}_2) &= \psi _{100}(\b{r_1})\psi _{100}(\b{r}_2) \nl &= \frac{\Zef^3}{\pi a_0^3} \exp\left[ -\frac{\Zef}{a_0}(r_1+r_2) \right] \end{align*} \]
Then, let’s change the Hamiltonian to the following form for simple calculations.
\[ H = \left( -\frac{\hbar^2}{2m_e}\laplacian_1-\frac{\Zef e^2}{4\pi\vpmt r_1} \right) + \left( -\frac{\hbar^2}{2m_e}\laplacian_2-\frac{\Zef e^2}{4\pi\vpmt r_2} \right) +\frac{e^2}{4\pi\vpmt}\left[ \frac{\Zef-2}{r_1}+\frac{\Zef-2}{r_2}+\frac{1}{\abs{\b{r}_1-\b{r}_2}} \right] \]
We see that the terms inside parentheses are the Hamiltonian of hydrogen-like atoms of atomic number $\Zef$, respectively. However, here $\Zef$ doesn’t have to be an integer since it’s a parameter for the variational method; it reflects the shielding effect (or screening effect) of the electrons. Therefore, $\Zef$ is an effective nuclear charge $Z_\text{eff}$, and we’ll calculate it later.
Recall that $\Expct{\dfrac{1}{r}}=\dfrac{\Zef}{a_0}$ is satisfied at the ground state of the hydrogen-like atom. The expectation value of the Hamiltonian is, where $E_1$ is the ground state energy of the hydrogen:
\[ \begin{align*} \expct{\hat{H}} &= \Zef^2E_1 + \Zef^2E_1 + \frac{(\Zef-2)e^2}{4\pi\vpmt}\left[ \Expct{\frac{1}{r_1}}+\Expct{\frac{1}{r_2}} \right] + \frac{e^2}{4\pi\vpmt}\Expct{\frac{1}{\abs{\b{r}_1-\b{r}_2}}} \nl &= 2\Zef^2E_1-4\Zef(\Zef-2)E_1 + \frac{e^2}{4\pi\vpmt}\Expct{\frac{1}{\abs{\b{r}_1-\b{r}_2}}} \end{align*} \]
Let’s calculate the expectation value:
\[ \Expct{\frac{1}{\abs{\b{r}_1-\b{r}_2}}} = \left(\frac{Z^3}{\pi a_0^3}\right)^2 \underbrace{ \int _{\R^3}\int _{\R^3} \frac{1}{\abs{\b{r}_1-\b{r}_2}} \exp\left[ -\frac{2\Zef}{a_0}(r_1+r_2) \right] \d^3\b{r}_1 \d^3\b{r}_2 } _{\dps I} \]
Let’s use the multipole expansion using spherical harmonics, where $ Y_0^0 \equiv \dfrac{1}{\sqrt{4\pi}} $. By the orthogonality of the spherical harmonic functions, we get:
\[ \begin{align*} I &= \int_0^\infty\int_0^\infty \exp\left[ -\frac{2\Zef}{a_0}(r_1+r_2) \right] \sum_{l=0}^\infty\sum_{m=-l}^l \frac{4\pi}{2l+1}\frac{r_<^l}{r_>^{l+1}} r_1^2r_2^2 \,\d r_1\d r_2 \oint_{\Omega_1}\oint_{\Omega_2} Y_l^{m\ast}(\theta_1,\varphi_1) Y_l^m(\theta_2,\varphi_2) \,\d\Omega_2\,\d\Omega_1 \nl &= \int_0^\infty\int_0^\infty \exp\left[ -\frac{2\Zef}{a_0}(r_1+r_2) \right] \sum_{l=0}^\infty\sum_{m=-l}^l \frac{4\pi}{2l+1}\frac{r_<^l}{r_>^{l+1}} r_1^2r_2^2 \,\d r_1 \d r_2 \cdot 4\pi \oint_{\Omega_1} Y_l^{m\ast}(\theta_1,\varphi_1)Y_0^0(\theta_1,\varphi_1)\d\Omega_1 \oint_{\Omega_2} Y_0^{0\ast}(\theta_2,\varphi_2)Y_l^m(\theta_2,\varphi_2)\d\Omega_2 \nl &= \int_0^\infty\int_0^\infty \exp\left[ -\frac{2\Zef}{a_0}(r_1+r_2) \right] \sum_{l=0}^\infty\sum_{m=-l}^l \delta_{l0}\delta_{m0} \frac{(4\pi)^2}{2l+1}\frac{r_<^l}{r_>^{l+1}} r_1^2r_2^2 \,\d r_1 \d r_2 \nl &= (4\pi)^2 \int_0^\infty\int_0^\infty \exp\left[ -\frac{2\Zef}{a_0}(r_1+r_2) \right] \frac{r_1^2r_2^2}{r_>} \,\d r_1 \d r_2 \nl &= \frac{5\pi^2}{8\Zef^5}a_0^5 \end{align*} \]
Therefore,
\[ \begin{align*} \frac{e^2}{4\pi\vpmt}\Expct{\frac{1}{\abs{\b{r}_1-\b{r}_2}}} &= \frac{e^2}{4\pi\vpmt} \left(\frac{\Zef^3}{\pi a_0^3}\right)^2 \frac{5\pi^2}{8\Zef^5}a_0^5 \nl &= -\frac{5}{4}\Zef E_1 \end{align*} \]
Then we finally get the expectation of the Hamiltonian:
\[ \begin{align*} \expct{\hat{H}} &= \left( -2\Zef^2+\frac{27}{4}\Zef \right) E_1 \nl &= \left( 2\Zef^2-\frac{27}{4}\Zef \right) \RydE \end{align*} \]
This value is minimized when $\Zef$ is $\dfrac{27}{16}$.
\[ \begin{align*} \expct{\hat{H}} &\ge \frac{729}{128}E_1 \nl &\approx -77.489 \,\eV \end{align*} \]
An experimental value is $-78.975 \,\eV$, which shows that our approximation is pretty accurate. (error of $ 1.88\,\% $).
We can also calculate the ionization energy of helium. Since the first ionization energy of helium is trivially (and also approximately) $ 4\RydE=54.423\,\eV $, we get the second ionization energy naturally, of which value is $23.066\,\eV$. Although this isn’t accurate enough (over $5\,\%$ of the error to experiments) to use, it still provides us the rough value.
Summarizing,
- $ Z_\text{eff} \approx 1.6875 $
- $ E_\text{gs} \approx -77.489\,\eV $
- The wavefunction of the ground state helium can be thought of as the product of the wavefunctions of $\ce{He+}$.
Helium-like atom
A two-electron atom or helium-like ion is a quantum mechanical system consisting of one nucleus with a charge of $Ze$ and just two electrons. ($\ce{H-}$, $\ce{He}$, $\ce{Li+}$, $\ce{Be^2+}$, …)
The expectation of Hamiltonian can be computed in the same way,
\[ \expct{\hat{H}} = \left( -2\Zef^2+\frac{16Z-5}{4}\Zef \right)E_1 \]
Therefore,
- $ Z_\text{eff} \approx Z-\dfrac{5}{16} $
- $ E_\text{gs} \approx 2\left(Z-\dfrac{5}{16}\right)^2 E_1 $
This is approximation is quite good, surprisingly! Compare with experimental values in the tables here.
Better approximations
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Excited states of the helium atom
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