In the last post, we dealt with the hydrogen, while in this post, we will solve the helium.

In the last post, we take considerable length, omit many steps, and neglect several other important factors (the Stark effect, the relativistic corrections, the Zeeman effect, etc.) to arrive at an approximate solution to the hydrogen atom. It’s just a proton and an electron! Don’t be discouraged: discovering the truths about the microscopic world is indeed very hard.

In this post, we will take one step further to deal with the helium atom. Before we go into depth, we need to understand another counter-intuitive concept in quantum mechanics: identical particles. Far different from hydrogen’s single electron, we have two electrons now!

This shouldn’t be much a problem in classical mechanics. Even if the two electrons are truly indistinguishable, we can label them differently at one point in time and space, and calculate the trajectories of each of them. This is impossible in the microscopic level and in quantum mechanics, as each particle described by a wave function has amplitudes to take all paths and possibility to be in all positions. Further, if we dare to make measurements, we may change the state of the system. In short, if two particles are identical (like two electrons), we need to device a new formalism to deal with them.

Identical Particles

Recall the notation of a two particle system:

We define the exchange operator

If the two particles represented by the states $| a \rangle, | b \rangle$ are identical (but not necessarily in the same state), they should be in the eigenstate of the exchange operator.

With the basis states $| a, b \rangle, | b, a \rangle$, we have

So we find the two eigenstates of the exchange operator,

Therefore, two identical particles have only two possible states: $|\Psi_S\rangle$ the symmetric state, and $|\Psi_A\rangle$ the antisymmetric state. In fact, the Nature makes choice for the particles in a rather comprehensive way based on their spin. We mentioned that the spin of particles are either integral numbers or half-integral numbers. Particles with integral spin, $s = 1, 2, 3, \dots$ are found only o be in the symmetric state and are called bosons. Examples are fundamental elementary particles such as photons, gluons, and the graviton, composite particles such as pions, and nuclei such as $^4$He. Particles with half-integral spin, $s = \frac{1}{2}, \frac{3}{2}, \frac{5}{2}, \dots$ are found only to be in the antisymmetric state and are called fermions. Examples are fundamental elementary particles such as electrons, muons, neutrinos, and quarks, composite particles such as protons and neutrons, and nuclei such as $^3$He.

At the level of non-relativistic quantum mechanics, this relationship between the spin and the exchange symmetry is a law of nature and referred to as spin-statistics theorem. This is also the 6th postulate we presented in post 1.

Another important implication of this result is that two fermions cannot be in the same state. If two identical particles are in the same state, we can label it as $|a, a\rangle$, and it is automatically symmetric, which is not allowed for fermions. Therefore, we have the Pauli exclusion principle: two electrons cannot be in the same state.

Time-Independent Perturbation Theory

To find the approximate solution to the helium atom, we need an additional mathematics tool. We first write out the Hamiltonian of the helium atom:

where $Z=2$ for helium and we ignore the kinetic energy of the nucleus. Notice that the first four terms are just the Hamiltonian of the hydrogen atom appearing twice with a different $Z$. We’ve already solved the eigenfunction in the last post. However, we have a nasty 5th term, the interaction between the two electrons, which don’t allow us to apply the results from the hydrogen atom easily. To ease up this problem, we will treat the first four term as the unperturbed Hamiltonian $\hat{H_0}$ and the 5th term as a perturbing Hamiltonian $\hat{H_1}$. This method gives valid approximation when the perturbation is much smaller than the unperturbed energy. In this case, the assumption is actually invalid, but the method will nonetheless gives us sensible result and intuition about the helium atom.

Let’s further investigate the solutions to eigenfunctions and eigenvalues with the perturbation method. Formally, we break a Hamiltonian into

In order to keep track of the effect of the perturbing Hamiltonian, we introduce a parameter $\lambda$ into the Hamiltonian

so that when $\lambda=0$, we turn off the perturbation, and when $\lambda=1$, we recover the full Hamiltonian. We further assume that we can express the eigenvalues and the eigenstates of the Hamiltonian as a power-series of $\lambda$:

where the subscript $n$ indicates the energy level, $|\psi_n^{(d)}\rangle$ is the dth-order correction to the energy eigenstate, and $E_n^{(d)}$ is the dth-order energy shift.

Then use $\hat H |\psi_n\rangle = E_n |\psi_n\rangle$,

By equating the first-order terms of $\lambda$ and apply the ket $\langle \psi_n^{(0)}|$, we have

The cancelling the 1st and 3rd term, and presuming $\psi_k^{(0)}|\psi_n^{(0)}\rangle = \delta_{kn}$, we obtain the first-order energy shift:

In this post, we will purely deal with the energy-levels of the helium atom up to the first-order shift, so I will not carry out further derivations to obtain corrections in energy eigenstates.

The Helium Atom

Finally, we proceed to the main course of this post. We’ve already written out the Hamiltonian for the helium atom,

The Ground State

From the last post, we obtain that for the energy eigenvalue of the Hydrogen atom is $E_n = \frac{E_0}{n^2}$ where $E_0 = -13.6 eV$. We can make the result more general by extending hydrogen’s single proton to $Z$ electrons, and obtain is $E_n = \frac{Z^2E_0}{n^2}$.

Therefore, for the ground state of the helium atom which has two ground state electrons, the eigenvalue of $\hat{H_0}$ is

What about the eigenstates of $\hat{H_0}$ for the ground state? The two electrons are in the same energy states 1s, and we have a symmetrical spacial state:

But we know that the helium atom must be in an antisymmetric state, so we must have an antisymmetric spin state. Recall the 4 two-particle states for electrons we introduced in post 4,

The $|0,0\rangle$ state is the only anti-symmetric state,

Together, we have the eigenstate

Now we can calculate the 1st-order energy shift of the system.

where to obtain the final result, we need to substitute in the wave function $\langle r|1,0,0\rangle$ which we obtain in post 7 with the case of $Z=2$ plus much algebra.

Using the first-order shift, we can get our first approximation of the ground state energy of $^4$He,

which is good enough for the experimental result of $-79.0eV$. Note that the “perturbation” is on the order of the unperturbed energy, so it’s normal to get a inprecise result.

The First Excited States

Let’s also consider the first excited states of $^4$He. For the first excites states, we have one electron at the ground state, and one electron at the first excited state.

There are two possible spacial two-particle states, one symmetric and the other antisymmetric:

We’ve already recapped the four two-particle states of the intrinsic spin, where three possible states are symmetric and the fourth is antisymmetric. To make the overall state antisymmetric, we have the following four possible states:

where $m_s = -1, 0, 1$.

The unperturbed energy for each of these states is

We can also express the first-order shift in energy,

We express the first-order shift in the form of $J_{12} \pm K_{12}$, where the first term is similar to the first-order energy shift of the ground state energy. However, the second-term arises purely quantum mechanical and due to the identical nature of the particles.

Referencing Townsend’s [2], we can argue positivity of $K$. Note that if we put $r_1 = r_2$ in the wave functions in the first line of the calculation of the first-order energy shift, we find that the antisymmetric wave functions vanish (due to the minus sign), while the symmetric wave functions add together constructively (due to the plus sign). Thus the electrons in the antisymmetric spatial state tend to avoid each other in space, which should lower their energy due to Coulomb repulsion relative to the electrons in the symmetric spatial state, in which the electrons prefer to be close together.

Thus the existence of such an exchange term produces an energy shift of the total-spin states: the energy of the triplet of spin-1 states (symmetric spin states) is shifted by $J - K$ , while the singlet spin-0 state (the antisymmetric spin state) is shifted in energy by $J + K$. Provided $K$ is positive, the energy of the spin-1 states will be lower than the spin-0 state, as we have argued physically should be true.

By carrying out the calculations of

we can find the first-order corrections to the energies of the first-excited states of the helium atom.

References

[1] Zajc, William. PHYS GU4021-4022 Quantum Mechanics I-II, Department of Physics, Columbia University. 2019-2020.

[2] Townsend, John. A Modern Approach to Quantum Mechanics. 2nd Edition.