Helium (Basics)
Introduction
Helium \((Z=2)\) provides our first look beyond hydrogen. After mastering the hydrogen atom, you might think that adding just one more electron would be a straightforward extension. However, as we’ll see, even this seemingly small step introduces fundamental challenges that will shape our understanding of all multi-electron atoms.
In this chapter, we’ll start with the helium ion (He\(^+\)), which we can solve exactly, and then tackle neutral helium, where we’ll encounter the limitations of our current methods. This journey will help us appreciate why approximate methods are not just convenient, but absolutely necessary in atomic physics.
1. The He\(^+\) Ion
We begin with the helium ion (He\(^+\)), which has lost one of its two electrons and therefore has only a single electron orbiting the nucleus. This makes it hydrogen-like, meaning we can use everything we learned about hydrogen with just a minor modification.
The key insight is that for ions with only one electron, we can directly apply the hydrogen atom solutions that we have derived previously. The only change needed is to account for the fact that the nucleus now has a charge of \(+2e\) instead of \(+e\). Mathematically, this means replacing \(e^2 \to Ze^2\) throughout all hydrogen equations, where \(Z=2\) for helium.
1.1 Energy levels
Let’s see how this works for the energy levels. Recall that for hydrogen, we derived: \[ E_n = -\frac{m_e}{2\hbar^2}\left(\frac{e^2}{4\pi\epsilon_0}\right)^2 \frac{1}{n^2} \]
For a hydrogenic ion with nuclear charge \(Z\), the stronger nuclear attraction pulls the electron in more tightly, and we get: \[ E_n = -\frac{m_e}{2\hbar^2}\left(\frac{Ze^2}{4\pi\epsilon_0}\right)^2 \frac{1}{n^2} \]
Notice that the energy scales as \(Z^2\). This can be written more compactly as: \[ E_n = -13.6\,\text{eV} \cdot \frac{Z^2}{n^2} \]
For helium (\(Z=2\)), the ground state energy becomes: \[ E_1 = -13.6 \times 4\,\text{eV} = -54.4\,\text{eV} \]
This is four times deeper than hydrogen’s ground state of \(-13.6\) eV. The electron is bound much more tightly because it feels the full attraction of two protons in the nucleus.
The Rydberg is a convenient energy unit in atomic physics: \(1\,\text{Rydberg} = 13.6\,\text{eV}\). In these units, hydrogen’s ground state is exactly \(-1\) Rydberg, and He\(^+\) is \(-4\) Rydberg. This makes comparing hydrogenic systems straightforward.
1.2 Bohr radius
The stronger nuclear attraction also affects the size of the electron’s orbit. The Bohr radius, which characterizes the typical distance of the electron from the nucleus, also scales with nuclear charge.
For hydrogen, recall: \[ a_0 = \frac{4\pi\epsilon_0\hbar^2}{m_e e^2} \approx 0.529\,\text{Å} \]
For a hydrogenic ion with nuclear charge \(Z\): \[ a = \frac{4\pi\epsilon_0\hbar^2}{m_e Ze^2} = \frac{a_0}{Z} \]
For helium (\(Z=2\)), the electron orbit is compressed to half the hydrogen size. This makes physical sense: the stronger nuclear charge pulls the electron closer. The electron cloud in He\(^+\) is much more compact than in hydrogen.
1.3 Wave functions
The wave functions for hydrogenic ions follow the same functional form as hydrogen, but with the scaled Bohr radius \(a = a_0/Z\). This reflects both the tighter binding and the smaller orbital size.
For the hydrogen ground state, we had: \[ \psi_{100} = \frac{1}{\sqrt{\pi a_0^3}} e^{-r/a_0} \]
For a general hydrogen-like ion: \[ \psi_{100} = \sqrt{\frac{Z^3}{\pi a_0^3}} e^{-Zr/a_0} \]
Notice two changes: the normalization constant includes a factor of \(Z^{3/2}\) (because the orbital is smaller, the wave function is more concentrated), and the exponential decay is faster by a factor of \(Z\).
For helium (He\(^+\)) specifically: \[ \psi_{100}^{\text{He}} = \frac{2\sqrt{2}}{\sqrt{\pi a_0^3}} e^{-2r/a_0} \]
So far, everything has been straightforward. The He\(^+\) ion is completely solved, and we understand it just as well as we understand hydrogen. But now comes the interesting part: what happens when we add back the second electron?
2. The Neutral Helium Atom
Now we consider neutral helium with two electrons. You might think we could just put two electrons into the ground state and be done with it. However, there’s a complication: the two electrons repel each other. This electron-electron repulsion changes everything and makes the problem impossible to solve exactly.
This is a watershed moment in atomic physics. With hydrogen and hydrogen-like ions, we could write down exact solutions. With helium, we encounter the first system that forces us to approximate.
2.1 The Hamiltonian
Let’s write down the full Hamiltonian for neutral helium and see what we’re up against. We need to include:
- The kinetic energy of both electrons
- The potential energy of each electron due to the nucleus
- The repulsion between the two electrons
This gives us: \[ \hat{H} = \underbrace{-\frac{\hbar^2}{2m_e}\nabla_1^2}_{\text{KE}_1} + \underbrace{-\frac{\hbar^2}{2m_e}\nabla_2^2}_{\text{KE}_2} - \underbrace{\frac{Ze^2}{4\pi\epsilon_0 r_1}}_{\text{PE}_1} - \underbrace{\frac{Ze^2}{4\pi\epsilon_0 r_2}}_{\text{PE}_2} + \underbrace{\frac{e^2}{4\pi\epsilon_0|\vec{r}_1 - \vec{r}_2|}}_{\text{electron repulsion}} \]
where:
- Subscripts 1 and 2 refer to the two electrons
- \(r_1 = |\vec{r}_1|\) and \(r_2 = |\vec{r}_2|\) are distances from the nucleus
- The last term represents the electron-electron repulsion
The problem is that last term: \(|\vec{r}_1 - \vec{r}_2|\) is the distance between the two electrons, and it couples the motion of the two particles. This term prevents us from separating the Schrödinger equation into two independent one-electron problems. The electrons are entangled—where one electron is affects where the other can be, and vice versa.
It is important to recognize that no exact analytical solution exists for the helium atom. This is not due to insufficient mathematical sophistication, but rather reflects a fundamental property of systems with multiple interacting particles. If the two-electron problem defies exact solution, the situation for heavier atoms—with their many-body interactions—is considerably more challenging. The need for systematic approximation methods thus emerges as a central theme in atomic and molecular physics.
2.2 Zeroth-order approximation
Let’s make the simplest possible approximation: what if we just ignore the electron-electron repulsion? This is obviously unphysical, but it will give us a baseline to compare against.
If we ignore the electron repulsion term, the Hamiltonian separates into two independent hydrogen-like problems, one for each electron: \[ \psi(\vec{r}_1, \vec{r}_2) = \psi_{\text{H-like}}(\vec{r}_1) \psi_{\text{H-like}}(\vec{r}_2) \]
For the ground state, both electrons occupy the 1s orbital (we’ll deal with the Pauli exclusion principle later when we discuss spin): \[ \psi_{100}(\vec{r}_1) \psi_{100}(\vec{r}_2) = \frac{8}{\pi a_0^3} e^{-2(r_1 + r_2)/a_0} \]
Each electron contributes \(-54.4\) eV (as we calculated for He\(^+\)), so the total ground state energy in this approximation is: \[ E_{100} = 2 \times (-54.4\,\text{eV}) = -108.8\,\text{eV} \approx -109\,\text{eV} \]
2.3 Comparison with experiment
Now for the moment of truth: how does our approximation compare to reality? The experimental value for the helium ground state is: \[ E_{\text{exp}} \approx -79\,\text{eV} \]
We’re off by about \(30\) eV! That’s a huge discrepancy—roughly 40% of the experimental value. This tells us something important: the interaction term contributes significantly and cannot be neglected. Our zeroth-order approximation, while instructive, is far too crude.
2.4 Physical interpretation
Why is our calculated energy so much more negative (lower) than the experimental value? The answer lies in what we ignored: electron-electron repulsion.
In our approximation, each electron feels the full \(Z=2\) nuclear attraction and nothing else. In reality, the two electrons repel each other. This repulsion raises the energy (makes it less negative) compared to the independent-particle picture.
Think of it this way: the nuclear attraction tries to pull both electrons close to the nucleus, but the electrons push each other away. The actual ground state is a compromise between these competing effects. The electrons can’t get as close to the nucleus as they would if they were alone, so the binding energy isn’t as deep as our simple model predicts.
The \(30\) eV difference is a direct measure of the average electron-electron repulsion energy in the helium ground state. This is a substantial fraction of the total binding energy, showing that electron correlations are crucial in determining atomic structure.
3 First-Order Perturbation Theory
To improve upon our crude zeroth-order approximation, we now treat the electron-electron repulsion term as a small perturbation (\(H'\)) to the main, solvable Hamiltonian (\(H_0\)).
The full Hamiltonian is: \[ \hat{H} = \hat{H}_0 + \hat{H}' \]
where the unperturbed Hamiltonian is: \[ \hat{H}_0 = -\frac{\hbar^2}{2m_e}\nabla_1^2 -\frac{Ze^2}{4\pi\epsilon_0 r_1} - \frac{\hbar^2}{2m_e}\nabla_2^2 -\frac{Ze^2}{4\pi\epsilon_0 r_2} \]
and the perturbation is the electron-electron repulsion term: \[ \hat{H}' = \frac{e^2}{4\pi\epsilon_0 r_{12}} \]
3.1 The First-Order Energy Correction
The first-order correction to the ground state energy, \(E^{(1)}\), is the expectation value of the perturbation over the unperturbed ground state wavefunction (\(\psi^{(0)}\)): \[ E^{(1)} = \langle \psi^{(0)} | \hat{H}' | \psi^{(0)} \rangle \]
Our zeroth-order ground state wavefunction is the product of two independent hydrogen-like wavefunctions for \(Z=2\): \[ \psi^{(0)}(\vec{r}_1, \vec{r}_2) = \psi_{100}(\vec{r}_1) \psi_{100}(\vec{r}_2) \]
The integral for the energy correction becomes: \[ E^{(1)} = \int\int |\psi_{100}(\vec{r}_1)|^2 |\psi_{100}(\vec{r}_2)|^2 \frac{e^2}{4\pi\epsilon_0 |\vec{r}_1 - \vec{r}_2|} d^3r_1 d^3r_2 \]
This integral is non-trivial but can be solved. Its value is found to be \(\frac{5}{8} \frac{Ze^2}{4\pi\epsilon_0 a_0}\), where \(a_0\) is the Bohr radius. Given that the unperturbed ground state energy of a hydrogenic atom is \(-13.6 \text{ eV} \cdot Z^2\), we can re-express the correction in terms of this familiar unit.
For helium (\(Z=2\)): \[ E^{(1)} = \frac{5}{8} (2) \left( \frac{e^2}{4\pi\epsilon_0 a_0} \right) = \frac{5}{4} \left( 13.6\,\text{eV} \right) = 34.0\,\text{eV} \]
3.2 Result and Physical Interpretation
The new, first-order approximation for the helium ground state energy is the sum of the zeroth-order energy and the first-order correction: \[ E^{(1)}_{\text{total}} = E^{(0)} + E^{(1)} = -108.8\,\text{eV} + 34.0\,\text{eV} = -74.8\,\text{eV} \]
This result of \(-74.8\,\text{eV}\) is a significant improvement over the zeroth-order value of \(-108.8\,\text{eV}\). It is also much closer to the experimental value of approximately \(-79\,\text{eV}\).
While much better, our calculated energy is still too high (i.e., less negative). This is because first-order perturbation theory calculates the average electron-electron repulsion using the unperturbed wavefunctions, which do not account for the screening effect. In reality, each electron’s wavefunction is distorted by the presence of the other, effectively reducing the nuclear charge they experience. Our calculation overestimates the energy by about 4.2 eV because it fails to capture this subtle, but important, quantum mechanical correlation between the electrons.
To get even closer to the experimental value, we would need to go to higher orders in perturbation theory or, more effectively, use the variational method.
4. Summary
Let’s recap what we’ve learned:
He\(^+\) (one electron): This is exactly solvable as a hydrogen-like ion with \(Z=2\). The energy levels scale as \(Z^2\), giving a ground state of \(E_1 = -54.4\,\text{eV}\). The electron orbits at half the Bohr radius of hydrogen due to the stronger nuclear attraction.
Neutral He (two electrons): The electron-electron repulsion makes an exact solution impossible. We have explored two levels of approximation:
- Our zeroth-order approximation (ignoring repulsion) gives a ground state energy of \(E^{(0)} \approx -109\,\text{eV}\).
- Using first-order perturbation theory (treating repulsion as a correction), the result improves significantly to \(E^{(1)}_{\text{total}} \approx -74.8\,\text{eV}\).
Comparison with Experiment: The experimental value is \(\sim -79\,\text{eV}\).
- The zeroth-order model has a large discrepancy of \(\sim 30\,\text{eV}\), showing the repulsion cannot be ignored.
- The first-order perturbation theory result is much closer, with a smaller error of \(\sim 4.2\,\text{eV}\).
The Gap and its Physical Significance:
- The remaining difference between our best approximation and the experimental value highlights the importance of electron correlation effects, particularly the screening effect. Our first-order perturbation theory overestimates the repulsive energy because it doesn’t account for how the electrons’ positions affect each other.
We do not yet have a method that fully captures the electron correlation effects responsible for the remaining discrepancy. While perturbation theory can be continued to higher orders, a more systematic approach is needed. This will be introduced when we discuss the Hartree-Fock approximation, which accounts for electron-electron interactions more effectively in multi-electron systems by incorporating the idea of an average electric field.
This helium example illustrates a fundamental challenge in atomic physics: even the simplest multi-electron atom requires approximation methods beyond the exact solutions available for hydrogen-like systems. As we move to heavier atoms, the situation only becomes more complex, with many electrons all interacting with each other. The methods we develop to handle helium will form the foundation for understanding all of chemistry and atomic physics. The good news is that while we can’t solve the many-electron problem exactly, we can develop systematic approximations that work remarkably well. The journey from helium to heavy atoms is one of the great success stories of 20th-century physics.