# 12.4: Dirac Theory of the Hydrogen Atom


The theory of Paul Dirac represents an attempt to unify the theories of quantum mechanics and special relativity. That is, one seeks a formulation of quantum mechanics which is Lorentz invariant, and hence consistent with special relativity. For a free particle, relativity states that the energy is given by

$E + p^2c^2 + m^2c^4$

Associating $$E$$ with a Hamiltonian in quantum mechanics, one has

$H^2 = p^2c^2 + m^2c^4 \label{59}$

If $$H$$ and $$p$$ are associated with the same operators as in Schrödinger theory, then one expects the wave equation

$- \hbar \dfrac{\partial^2}{\partial t^2} \Psi = ( -\hbar^2 \nabla^2 c^2 + m^2 c^4) \Psi \label{60}$

This is known as the Klein-Gordan Equation. Unfortunately, attempts to utilize this equation are not successful, since that which one would wish to interpret as a probability distribution turns out to be not positive definite. To alleviate this problem, the square root may be taken to get

$H = \sqrt{ p^2c^2 + m^2c^4} \label{61}$

However, this creates a new problem. What is meant by the square root of an operator? The approach is to guess the form of the answer, and the correct guess turns out to be

$H = c \alpha \cdot p + \beta mc^2 \label{62}$

With this form of the Hamiltonian, the wave equation can be written

$i \hbar \dfrac{\partial \chi}{\partial t} = (c \alpha \cdot p + \beta mc^2) \chi \label{63}$

In order for this to be valid, one hopes that when it is squared the Klein-Gordan equation is recovered. For this to be true, Equation $$\ref{63}$$ must be interpreted as a matrix equation, where $$\alpha$$ and $$\beta$$ are at least $$4 \times 4$$ matrices and the wavefunction is a four-component column matrix.

It turns out that Equation $$\ref{63}$$ describes only a particle with spin 1/2. This is fine for application to the hydrogen atom, since the electron has spin 1/2, but why should it be so? The answer is that the linearization of the Klein-Gordan equation is not unique. The particular linearization used here is the simplest one, and happens to describe a particle of spin 1/2, but other more complicated Hamiltonians may be constructed to describe particles of spin 0,1,5/2 and so on. The fact that the relativistic Dirac theory automatically includes the effects of spin leads to an interesting conclusion - spin is a relativistic effect. It can be added by hand to the non-relativistic Schödinger theory with satisfactory results, but spin is a natural consequence of treating quantum mechanics in a completely relativistic fashion.

Spin is a relativistic effect

Including the potential now in the Hamiltonian, Equation $$\ref{63}$$ becomes

When the square root was taken to linearize the Klein-Gordan equation, both a positive and a negative energy solution was introduced. One can write the wavefunction

where $$\Psi_+$$ represents the two components of $$\chi$$ associated with the positive energy solution and $$\Psi_-$$ represents the components associated with the negative energy solution. The physical interpretation is that $$\Psi_+$$ is the particle solution, and $$\Psi_-$$ represents an anti-particle. Anti-particles are thus predicted by Dirac theory, and the discovery of anti-particles obviously represents a huge triumph for the theory. In hydrogen, however, the contribution of $$\Psi_-$$ is small compared to $$\Psi_+$$. With enough effort, the equations for $$\Psi_+$$ and $$\Psi_-$$ can be decoupled to whatever order is desired. When this is done, the Hamiltonian to order $$v^2/c^2$$ can be written

$H + H_s + \Delta H _{rel} + \Delta H_{so} + \Delta H_d \label{66}$

where

• $$H_s$$ is the original Schrödinger Hamiltonian,
• $$\Delta H _{rel}$$ is the relativistic correction to the kinetic energy,
• $$\Delta H_{so}$$ is the spin-orbit term, and
• $$\Delta H_d$$ is the previously mentioned Darwin term.

## The Darwin Term

The physical origin of the Darwin term is a phenomenon in Dirac theory called zitterbewegung, whereby the electron does not move smoothly, but instead undergoes extremely rapid small-scale fluctuations, causing the electron to see a smeared-out Coulomb potential of the nucleus.

The Darwin term may be written

$\Delta H_d = -\dfrac{e\hbar^2}{8m^2c^2} \nabla^2 \Psi \label{67}$

For the hydrogenic-atom potential $$V = \dfrac{Ze}{r}$$, Equation $$\ref{67}$$ is

$H_d = -\dfrac{Ze^2\pi \hbar^2}{2 m^2c^2} \delta^3(r) \label{68}$

When first-order perturbation theory is applied, the energy correction depends on $$| \Psi(0)|^2$$. This term will only contribute for s states (i.e., (l=0)), since only these wavefunctions have non-zero probability for finding the electron at the origin. The energy correction for $$l=0$$ can be calculated to be

$\Delta E _d = (Z \alpha)^4 mc^2 \dfrac{1}{2n^3} \label{69}$

Including this term, the fine-structure splitting given by Equation $$\ref{58}$$ can be reproduced for all $$l$$. All the effects that go into fine structure are thus a natural consequence of the Dirac theory.