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 $\chi$ 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

$$i \hbar \frac{\partial \chi}{\partial t}=\left(c \alpha \cdot \mathbf{p}+\beta m c^2-\frac{Z e^2}{r}\right) \chi .$$

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

$$\chi=\binom{\Psi_{\text {+}}}{\Psi_{-}},$$

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 can be reproduced for all $l$. All the effects that go into fine structure are thus a natural consequence of the Dirac theory.

The hydrogen atom can be solved exactly in Dirac theory, where the states found are simultaneous eigenstates of $H, J^2$, and $J_z$, since these operators can be shown to mutually commute. The exact energy levels in Dirac theory are

$$E_{n j}=m c^2\left[1+\left(\frac{Z \alpha}{n-\left(j+\frac{1}{2}\right)+\sqrt{\left(j+\frac{1}{2}\right)^2-(Z \alpha)^2}}\right)^2\right]^{-1 / 2} .$$

This can be expanded in powers of $Z \alpha$, yielding

$$E_{n j}=m c^2\left\{1-\frac{1}{2} \frac{(Z \alpha)^2}{n^2}\left[1+\frac{(Z \alpha)^2}{n}\left(\frac{1}{j+\frac{1}{2}}-\frac{3}{4 n}\right)\right]+\ldots\right\}$$

This includes an amount $m c^2$ due to the relativistic energy associated with the rest mass of the electron, along with the principle energy levels and fine structure, in exact agreement to order $(Z \alpha)^4$ with what was previously calculated. However, even this exact solution in Dirac theory is not a complete description of the hydrogen atom, and so the the next section describes further effects not yet discussed.

Footnote

See Bjorken and Drell Chapter 4 for a thorough discussion of the transfer

Contributors and Attributions

• Randal Telfer (JWST Astronomical Optics Scientist, Space Telescope Science Institute)