Relativistic Quantum Mechanics: Klein-Gordon & Dirac Formulations

1. The Incompatibility of Relativity and the Schrödinger Equation

The standard Schrödinger equation $i\hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m}\nabla^2\psi + V\psi$ is fundamentally asymmetric: it contains a first-order derivative in time, but second-order derivatives in space. In special relativity, space and time are inextricably unified into spacetime coordinates $x^\mu = (ct, \vec{x})$. Any physical wave equation invariant under Lorentz transformations must treat space and time derivatives on an equal footing.

2. Four-Vector Kinematics and the Einstein Energy-Momentum Relation

Using natural covariant tensor notation with the Minkowski metric signature $\eta_{\mu\nu} = \text{diag}(+1, -1, -1, -1)$:

$$x^\mu = (x^0, x^1, x^2, x^3) = (ct, x, y, z), \quad x_\mu = \eta_{\mu\nu} x^\nu = (ct, -x, -y, -z)$$ $$p^\mu = (E/c, \vec{p}), \quad p_\mu = (E/c, -\vec{p})$$

The relativistic invariant scalar product of four-momentum is the rest mass shell constraint:

$$p^\mu p_\mu = \frac{E^2}{c^2} - \vec{p}^2 = m^2 c^2 \iff E^2 = \vec{p}^2 c^2 + m^2 c^4$$

Canonical quantum operator replacement maps four-momentum to four-gradient:

$$E \to i\hbar \frac{\partial}{\partial t}, \quad \vec{p} \to -i\hbar\nabla \iff \hat{p}^\mu = i\hbar\partial^\mu = i\hbar \left(\frac{1}{c}\frac{\partial}{\partial t}, -\nabla\right)$$

1. Derivation of the Klein-Gordon Equation

Applying canonical operator replacement directly to the relativistic energy-momentum invariant $E^2 - \vec{p}^2 c^2 = m^2 c^4$:

$$\left( -\hbar^2 \frac{\partial^2}{\partial t^2} + \hbar^2 c^2 \nabla^2 \right)\phi(\vec{r}, t) = m^2 c^4 \phi(\vec{r}, t)$$ $$\left( \frac{1}{c^2}\frac{\partial^2}{\partial t^2} - \nabla^2 + \frac{m^2 c^2}{\hbar^2} \right)\phi(\vec{r}, t) = 0 \iff \left(\Box + \frac{m^2 c^2}{\hbar^2}\right)\phi = 0$$

where $\Box \equiv \partial^\mu \partial_\mu = \frac{1}{c^2}\frac{\partial^2}{\partial t^2} - \nabla^2$ is the d'Alembertian operator. This is the Klein-Gordon equation (1926).

2. The Continuity Equation and Negative Probability Density

Multiplying the Klein-Gordon equation by $\phi^*$ and subtracting the complex conjugate equation multiplied by $\phi$:

$$\phi^* \frac{\partial^2\phi}{\partial t^2} - \phi \frac{\partial^2\phi^*}{\partial t^2} - c^2\left(\phi^* \nabla^2\phi - \phi \nabla^2\phi^*\right) = 0$$ $$\frac{\partial}{\partial t}\left[ \frac{i\hbar}{2mc^2}\left(\phi^* \frac{\partial\phi}{\partial t} - \phi \frac{\partial\phi^*}{\partial t}\right) \right] + \nabla\cdot\left[ -\frac{i\hbar}{2m}\left(\phi^* \nabla\phi - \phi \nabla\phi^*\right) \right] = 0$$

This defines the conserved four-current $\partial_\mu j^\mu = 0$ with probability density $\rho$:

$$\rho = \frac{i\hbar}{2mc^2}\left(\phi^* \frac{\partial\phi}{\partial t} - \phi \frac{\partial\phi^*}{\partial t}\right)$$

The Crisis: Because the Klein-Gordon equation is second order in time, both $\phi$ and $\frac{\partial\phi}{\partial t}$ can be chosen independently as initial Cauchy conditions. Consequently, $\rho$ can be negative! In 1928, this was viewed as a fatal defect for a single-particle probability density. (Later, Pauli and Weisskopf showed that the Klein-Gordon equation correctly describes spin-0 bosons like pions, with $\rho$ reinterpreted as electric charge density $j^0 = \rho_q$).

1. Dirac's Linearization Postulate

To ensure a positive-definite probability density $\rho \ge 0$, Paul Dirac (1928) sought a relativistic wave equation that is strictly first order in time:

$$i\hbar \frac{\partial \psi}{\partial t} = \hat{H}_D \psi$$

For relativistic covariance, it must also be first order in spatial gradients:

$$\hat{H}_D = c(\vec{\alpha}\cdot\hat{\vec{p}}) + \beta m c^2 = -i\hbar c \sum_{k=1}^3 \alpha_k \frac{\partial}{\partial x_k} + \beta m c^2$$

Squaring the Hamiltonian $\hat{H}_D^2 \psi = -\hbar^2 \frac{\partial^2\psi}{\partial t^2}$ and requiring that every component of $\psi$ identically satisfies the relativistic dispersion relation $E^2 = p^2 c^2 + m^2 c^4$:

$$\hat{H}_D^2 = c^2 \sum_{j,k} \frac{1}{2}\{\alpha_j, \alpha_k\} p_j p_k + m c^3 \sum_k \{\alpha_k, \beta\} p_k + \beta^2 m^2 c^4 \equiv c^2 \vec{p}^2 + m^2 c^4$$

2. The Clifford Algebra

Equating coefficients yields the required algebraic conditions for the coefficients $\alpha_1, \alpha_2, \alpha_3, \beta$:

$$\{\alpha_j, \alpha_k\} = \alpha_j \alpha_k + \alpha_k \alpha_j = 2\delta_{jk}\hat{I}$$ $$\{\alpha_k, \beta\} = \alpha_k \beta + \beta \alpha_k = 0$$ $$\alpha_k^2 = \hat{I}, \quad \beta^2 = \hat{I}$$

Since $\alpha_k$ and $\beta$ anticommute, their eigenvalues must be $\pm 1$ and their traces must vanish ($\text{Tr}(\alpha_k) = \text{Tr}(\beta) = 0$). This requires their matrix dimension $N$ to be even. In $N = 2$, only three anticommuting matrices exist (the Pauli matrices). Thus, the minimal dimensionality is $N = 4$!

3. The Dirac-Pauli Representation and Gamma Matrices

In the standard Dirac-Pauli representation:

$$\vec{\alpha} = \begin{pmatrix} 0 & \vec{\sigma} \\ \vec{\sigma} & 0 \end{pmatrix}, \quad \beta = \begin{pmatrix} \hat{I}_{2\times 2} & 0 \\ 0 & -\hat{I}_{2\times 2} \end{pmatrix}$$

Multiplying the Dirac equation by $\beta / c$ and defining the covariant Dirac gamma matrices:

$$\gamma^0 \equiv \beta, \quad \vec{\gamma} \equiv \beta\vec{\alpha} \implies \gamma^\mu = (\beta, \beta\vec{\alpha})$$ $$\gamma^0 = \begin{pmatrix} \hat{I} & 0 \\ 0 & -\hat{I} \end{pmatrix}, \quad \gamma^k = \begin{pmatrix} 0 & \sigma_k \\ -\sigma_k & 0 \end{pmatrix}$$

The gamma matrices satisfy the fundamental Clifford Algebra:

$$\{\gamma^\mu, \gamma^\nu\} = \gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = 2\eta^{\mu\nu}\hat{I}_{4\times 4}$$

The Dirac equation takes the elegant covariant form:

$$\left(i\hbar \gamma^\mu \partial_\mu - mc\right)\psi = 0 \iff (i\hbar \gamma^\mu \partial_\mu - mc)\psi = 0$$

1. The Dirac Adjoint Spinor and Conserved Current

The Dirac wavefunction $\psi(x)$ is a 4-component column vector termed a Dirac bispinor. Taking the Hermitian conjugate of $(i\hbar\gamma^0\partial_0 + i\hbar\vec{\gamma}\cdot\nabla - mc)\psi = 0$:

$$-i\hbar \partial_0\psi^\dagger (\gamma^0)^\dagger - i\hbar \nabla\psi^\dagger\cdot(\vec{\gamma})^\dagger - mc\psi^\dagger = 0$$

Since $(\gamma^0)^\dagger = \gamma^0$ and $(\gamma^k)^\dagger = -\gamma^k$, multiplying from the right by $\gamma^0$ and using $\gamma^k\gamma^0 = -\gamma^0\gamma^k$:

$$i\hbar \partial_\mu \bar{\psi} \gamma^\mu + mc\bar{\psi} = 0$$

where the Dirac adjoint spinor is defined as:

$$\bar{\psi} \equiv \psi^\dagger \gamma^0$$

Combining the two equations:

$$\bar{\psi}(i\hbar\gamma^\mu\partial_\mu\psi) + (i\hbar\partial_\mu\bar{\psi}\gamma^\mu)\psi = i\hbar\partial_\mu(\bar{\psi}\gamma^\mu\psi) = 0$$

This defines the conserved probability four-current:

$$j^\mu \equiv c\bar{\psi}\gamma^\mu\psi, \quad \partial_\mu j^\mu = 0$$

The time component is the probability density:

$$\rho = \frac{j^0}{c} = \bar{\psi}\gamma^0\psi = \psi^\dagger (\gamma^0)^2 \psi = \psi^\dagger \psi = \sum_{a=1}^4 |\psi_a|^2 \ge 0$$

Dirac's formulation triumphantly resolves the negative probability crisis: $\rho$ is strictly positive-definite!

1. Plane Wave Ansatz and Spinor Decomposition

For a free particle with four-momentum $p^\mu = (E/c, \vec{p})$, we seek plane wave solutions:

$$\psi(x) = u(\vec{p}) e^{-ip\cdot x/\hbar} = \begin{pmatrix} \phi \\ \chi \end{pmatrix} e^{-i(Et - \vec{p}\cdot\vec{r})/\hbar}$$

where $\phi, \chi$ are two-component Pauli spinors. Substituting into the Dirac equation:

$$\begin{pmatrix} mc^2 & c\vec{\sigma}\cdot\vec{p} \\ c\vec{\sigma}\cdot\vec{p} & -mc^2 \end{pmatrix} \begin{pmatrix} \phi \\ \chi \end{pmatrix} = E \begin{pmatrix} \phi \\ \chi \end{pmatrix}$$ $$\implies (E - mc^2)\phi = c(\vec{\sigma}\cdot\vec{p})\chi$$ $$\implies (E + mc^2)\chi = c(\vec{\sigma}\cdot\vec{p})\phi$$

2. Positive Energy Solutions ($E = +E_p = +\sqrt{p^2 c^2 + m^2 c^4}$)

Expressing the lower spinor $\chi$ in terms of the upper spinor $\phi$:

$$\chi = \frac{c(\vec{\sigma}\cdot\vec{p})}{E_p + mc^2}\phi$$

For an electron at rest ($\vec{p} = 0$), $\chi = 0$, meaning $\phi$ represents the two familiar non-relativistic spin states (spin-up $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and spin-down $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$). The normalized positive-energy four-spinors are:

$$u^{(s)}(\vec{p}) = \sqrt{\frac{E_p + mc^2}{2mc^2}} \begin{pmatrix} \chi_s \\ \frac{c(\vec{\sigma}\cdot\vec{p})}{E_p + mc^2}\chi_s \end{pmatrix} \quad (s = 1, 2)$$

3. Negative Energy Solutions ($E = -E_p = -\sqrt{p^2 c^2 + m^2 c^4}$)

Setting $\psi(x) = v(\vec{p}) e^{+ip\cdot x/\hbar}$, the upper component $\phi$ becomes smaller than $\chi$:

$$\phi = \frac{c(\vec{\sigma}\cdot\vec{p})}{-E_p - mc^2}\chi = -\frac{c(\vec{\sigma}\cdot\vec{p})}{E_p + mc^2}\chi$$ $$v^{(s)}(\vec{p}) = \sqrt{\frac{E_p + mc^2}{2mc^2}} \begin{pmatrix} \frac{c(\vec{\sigma}\cdot\vec{p})}{E_p + mc^2}\chi_s' \\ \chi_s' \end{pmatrix} \quad (s = 1, 2)$$

There are exactly two positive-energy states and two negative-energy states for every momentum $\vec{p}$, representing the two spin orientations of particle and antiparticle.

1. Minimal Electromagnetic Coupling in the Dirac Equation

In the presence of an electromagnetic four-potential $A^\mu = (\Phi/c, \vec{A})$, canonical momentum is replaced by kinetic momentum:

$$p^\mu \to \pi^\mu = p^\mu - q A^\mu = p^\mu + e A^\mu \quad (q = -e)$$

The Dirac equation becomes:

$$i\hbar \frac{\partial \psi}{\partial t} = \left[ c\vec{\alpha}\cdot(\hat{\vec{p}} + e\vec{A}) + \beta m c^2 - e\Phi \right]\psi$$

2. The Non-Relativistic Limit: Recovery of the Pauli Equation

Writing $\psi = \begin{pmatrix} \phi \\ \chi \end{pmatrix} e^{-imc^2 t/\hbar}$ to separate out the fast rest-mass oscillation, the coupled equations for the large component $\phi$ and small component $\chi$ are:

$$i\hbar \frac{\partial \phi}{\partial t} = c\vec{\sigma}\cdot\vec{\Pi}\chi - e\Phi\phi$$ $$i\hbar \frac{\partial \chi}{\partial t} + 2mc^2\chi = c\vec{\sigma}\cdot\vec{\Pi}\phi - e\Phi\chi$$

In the non-relativistic limit ($|i\hbar\partial_t \chi| \ll 2mc^2\chi$ and $|e\Phi| \ll mc^2$):

$$\chi \approx \frac{\vec{\sigma}\cdot\vec{\Pi}}{2mc}\phi$$

Substituting $\chi$ back into the equation for $\phi$:

$$i\hbar \frac{\partial \phi}{\partial t} = \left[ \frac{(\vec{\sigma}\cdot\vec{\Pi})^2}{2m} - e\Phi \right]\phi$$

Using the Pauli identity $(\vec{\sigma}\cdot\vec{A})(\vec{\sigma}\cdot\vec{B}) = \vec{A}\cdot\vec{B} + i\vec{\sigma}\cdot(\vec{A}\times\vec{B})$:

$$(\vec{\sigma}\cdot\vec{\Pi})^2 = \vec{\Pi}^2 + i\vec{\sigma}\cdot(\vec{\Pi}\times\vec{\Pi}) = (\vec{p} + e\vec{A})^2 + i\vec{\sigma}\cdot(-i\hbar e\nabla \times \vec{A}) = (\vec{p} + e\vec{A})^2 + e\hbar(\vec{\sigma}\cdot\vec{B})$$

Substituting this result yields the famous Pauli Equation:

$$i\hbar \frac{\partial \phi}{\partial t} = \left[ \frac{(\vec{p} + e\vec{A})^2}{2m} + \frac{e\hbar}{2m}(\vec{\sigma}\cdot\vec{B}) - e\Phi \right]\phi$$

3. The Natural Gyromagnetic Factor $g = 2$

Notice the interaction term with the magnetic field:

$$\hat{H}_{\text{mag}} = \frac{e\hbar}{2m}(\vec{\sigma}\cdot\vec{B}) = \frac{e}{m}\left(\frac{\hbar}{2}\vec{\sigma}\right)\cdot\vec{B} = \frac{e}{m}(\vec{S}\cdot\vec{B}) = 2\left(\frac{e}{2m}\right)\vec{S}\cdot\vec{B} = -\vec{\mu}_s \cdot \vec{B}$$

where $\vec{\mu}_s = -g_s \frac{e}{2m}\vec{S}$. Dirac's theory proves that:

$$g_s = 2$$

In non-relativistic physics, $g=2$ had to be inserted as an ad-hoc empirical postulate. In Dirac's relativistic equation, **electron spin ($s=1/2$) and the anomalous gyromagnetic factor ($g=2$) emerge automatically and inevitably from the requirement of relativistic spacetime covariance!**

1. Systematic Expansion Beyond Leading Order

To systematically decouple the positive and negative energy states to order $(v/c)^2$, one applies a canonical unitary transformation $\psi' = e^{i\hat{S}}\psi$, termed the Foldy-Wouthuysen (FW) transformation.

For an electron moving in a central electrostatic potential $V(r) = -e\Phi(r)$:

$$\hat{H}_{\text{FW}} = \beta\left(mc^2 + \frac{\vec{p}^2}{2m} - \frac{\vec{p}^4}{8m^3 c^2}\right) + V(r) + \frac{1}{2m^2 c^2}\frac{1}{r}\frac{dV}{dr}(\vec{L}\cdot\vec{S}) + \frac{\hbar^2}{8m^2 c^2}\nabla^2 V(r)$$

2. Rigorous Derivation of Fine Structure Terms

Restricting to the upper two-component spinor ($\beta \to +1$), the effective Hamiltonian reproduces the exact fine structure of hydrogen derived perturbatively in Chapter 3:

1. Relativistic Mass-Kinetic Correction: $\hat{H}_{\text{rel}} = -\frac{\vec{p}^4}{8m^3 c^2}$.
2. Spin-Orbit Interaction with Exact Thomas Precession: $$\hat{H}_{\text{SO}} = \frac{1}{2m^2 c^2}\frac{1}{r}\frac{dV}{dr}(\vec{L}\cdot\vec{S})$$ The kinematic Thomas factor $1/2$ appears automatically without needing non-inertial frame arguments!
3. Darwin Term: $$\hat{H}_{\text{Darwin}} = \frac{\hbar^2}{8m^2 c^2}\nabla^2 V(r) = \frac{\hbar^2}{8m^2 c^2}\left(\frac{e^2}{\varepsilon_0}\delta^3(\vec{r})\right) = \frac{\pi\hbar^2 e^2}{2m^2 c^2(4\pi\varepsilon_0)}\delta^3(\vec{r})$$

1. Dirac's Hole Theory and the Prediction of the Positron

Because relativistic quantum mechanics permits negative energy eigenstates with $E \le -mc^2$, an unconstrained electron would cascade down into infinitely negative energies by radiative emission.

To resolve this instability, Dirac invoked the Pauli exclusion principle:

"All negative-energy states in the universe with $E \le -mc^2$ are completely filled with an invisible, uniform sea of electrons—the Dirac Sea."

If a photon of energy $\hbar\omega \ge 2mc^2 \approx 1.022\text{ MeV}$ is absorbed by an electron in the negative-energy sea, the electron is kicked into a positive-energy state ($E \ge +mc^2$). It leaves behind a vacancy or hole in the sea.

A hole in a sea of charge $-e$ and energy $-E$ behaves physically as a particle of:

$$q_{\text{hole}} = -(-e) = +e, \quad E_{\text{hole}} = -(-E) = +E > 0$$

Dirac thus predicted the positron (the anti-electron), discovered experimentally by Carl Anderson in 1932.

2. The Klein Paradox: Relativistic Tunneling

Consider a relativistic electron of energy $E < V_0$ incident upon an electrostatic step potential of height $V_0$.

In non-relativistic physics, when $V_0 > E$, the wave is exponentially damped inside the barrier with zero transmission. In the Dirac equation, when the potential step exceeds the pair-creation threshold:

$$V_0 > E + mc^2$$

the incident positive-energy state couples directly to the continuum of negative-energy states inside the barrier! The transmission coefficient does not decay to zero; rather, particles penetrate through the barrier with substantial probability. Physically, the gigantic electric field $\mathcal{E} \sim V_0 / \lambda_c$ sparks the vacuum, creating electron-positron pairs: electrons are reflected, while positrons are transmitted into the barrier.