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} $
\[ \hamiltonian = -\frac{\hbar^2}{2m_e}(\nabla_1^2+\nabla_2^2) -\frac{e^2}{4\pi\epzro}\left[ \frac{2}{r_1}+\frac{2}{r_2}-\frac{1}{\abs{\b{r}_1-\b{r}_2}} \right] \]
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.
\[ \hamiltonian = \left( -\frac{\hbar^2}{2m_e}\nabla_1^2-\frac{\Zef e^2}{4\pi\epzro r_1} \right) + \left( -\frac{\hbar^2}{2m_e}\nabla_2^2-\frac{\Zef e^2}{4\pi\epzro r_2} \right) +\frac{e^2}{4\pi\epzro}\left[ \frac{\Zef-2}{r_1}+\frac{\Zef-2}{r_2}+\frac{1}{\abs{\b{r}_1-\b{r}_2}} \right] \]
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:
\[ \begin{align*} \expct{\hat{\hamiltonian}} &= \Zef^2E_1 + \Zef^2E_1 + \frac{(\Zef-2)e^2}{4\pi\epzro}\left[ \Expct{\frac{1}{r_1}}+\Expct{\frac{1}{r_2}} \right] + \frac{e^2}{4\pi\epzro}\Expct{\frac{1}{\abs{\b{r}_1-\b{r}_2}}} \nl &= 2\Zef^2E_1-4\Zef(\Zef-2)E_1 + \frac{e^2}{4\pi\epzro}\Expct{\frac{1}{\abs{\b{r}_1-\b{r}_2}}} \end{align*} \]
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,
\[ \begin{align*} \frac{e^2}{4\pi\epzro}\Expct{\frac{1}{\abs{\b{r}_1-\b{r}_2}}} &= \frac{e^2}{4\pi\epzro} \left(\frac{Z^3}{\pi a_0^3}\right)^2 \frac{5\pi^2}{8\Zef^5}a_0^5 \nl &= -\frac{5}{4}\Zef E_1 \end{align*} \]
Then we finally get the expectation of the Hamiltonian:
\[ \begin{align*} \expct{\hat{\hamiltonian}} &= \left( -2\Zef^2+\frac{27}{4}\Zef \right) E_1 \nl &= \left( 2\Zef^2-\frac{27}{4}\Zef \right) \Ryd \end{align*} \]
This value is minimized when $\Zef$ is $\dfrac{27}{16}$.
\[ \begin{align*} \expct{\hat{\hamiltonian}} &\ge \frac{729}{128}E_1 \nl &\approx -77.489 \,\eV \end{align*} \]
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\Ryd=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,
\[ \expct{\hat{\hamiltonian}} = \left( -2\Zef^2+\frac{16Z-5}{4}\Zef \right)E_1 \]
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
⚠️Studying⚠️
Excited states of the helium atom
⚠️Studying⚠️