15.5: Relativistic Quantum Mechanics

The Schrödinger equation itself it clearly inconsistent with relativity; It has second derivatives of space, and first derivatives of time. If we use the relativistic expression for energy \(E^2 = |{\bf p}|^2 c^2 + m^2 c^4\) we obtain

\[-\hbar^{2} \frac{\partial^{2}}{\partial t^{2}} \phi(\mathbf{r}, t) = -\hbar^{2} c^{2} \nabla^{2} \phi(\mathbf{r}, t)+m^{2} c^{4} \phi(\mathbf{r}, t) \nonumber\]

which is called the Klein Gordon equation. It has solutions describing a relativistic quantum particle, but others which describe particles of negative total energy, together with negative probabilities for finding them! Applied to hydrogen it gets the relativistic kinetic energy correction correct, but it doesn’t account for other observed relativistic effects, such as the spin-orbit correction or the Darwin term (see Atomic and molecular physics).

Dirac tried keeping time and space on an equal footing using a linear equation

\[i \hbar \frac{\partial}{\partial t} \psi(\mathbf{r}, t) = \left\{c \underline{\alpha} \cdot \hat{\underline{p}}+\beta m c^{2} \right\} \psi(\mathbf{r}, t) = \hat{H} \psi(\mathbf{r}, t) \quad \text { where } \underline{\alpha} \cdot \hat{p} = -i \hbar \left(\alpha_{x} \frac{\partial}{\partial x}+\alpha_{y} \frac{\partial}{\partial y}+\alpha_{z} \frac{\partial}{\partial z} \right) \nonumber\]

Consider a free particle, no terms in the Hamiltonian \(\hat{H}\) should depend on \({\bf r}\) or \(t\) as these would describe forces. Dirac assumed that \(\alpha_i\) and \(\beta\) are independent of position, time, momentum and energy, so \(\underline{\alpha}\) and \(\beta\) commute with \({\bf r}\), \(t\), \(\underline{\hat{p}}\) and \(E\) but not necessarily with each other.

Since relativistic invariance must be maintained, i.e., \(E^2 = |\underline{p}|^2 c^2 + m^2 c^4\),

\[\begin {align*} \hat{H}^{2} \psi(\mathbf{r}, t) &= \left(c^{2}|\underline{p}|^{2}+m^{2} c^{4} \right) \psi(\mathbf{r}, t) \\ &= \left\{c \underline{\alpha} \cdot \hat{\underline{p}}+\beta m c^{2} \right\} \left\{c \underline{\alpha} \cdot \hat{\underline{p}}+\beta m c^{2} \right\} \psi(\mathbf{r}, t) \end{align*} \nonumber\]

Expand the RHS of this equation, being very careful about the ordering of \(\alpha_i\) and \(\beta\)

\[\begin {align*} \hat{H}^{2} \psi(\mathbf{r}, t) & = \left\{ c^{2} \left[ \left(\alpha_{x} \right)^{2} \left(\hat{p}_{x} \right)^{2}+ \left(\alpha_{y} \right)^{2} \left(\hat{p}_{y} \right)^{2}+ \left(\alpha_{z} \right)^{2} \left(\hat{p}_{z} \right)^{2} \right]+m^{2} c^{4} \beta^{2} \right\} \psi(\mathbf{r}, t) \\& +c^{2} \left\{ \left(\alpha_{x} \alpha_{y}+\alpha_{y} \alpha_{x} \right) \hat{p}_{x} \hat{p}_{y}+ \left(\alpha_{y} \alpha_{z}+\alpha_{z} \alpha_{y} \right) \hat{p}_{y} \hat{p}_{z}+ \left(\alpha_{z} \alpha_{x}+\alpha_{x} \alpha_{z} \right) \hat{p}_{x} \hat{p}_{z} \right\} \psi(\mathbf{r}, t) \\ &+ m c^{3} \left\{ \left(\alpha_{x} \beta+\beta \alpha_{x} \right) \hat{p}_{z}+ \left(\alpha_{y} \beta+\beta \alpha_{y} \right) \hat{p}_{z}+ \left(\alpha_{z} \beta+\beta \alpha_{z} \right) \hat{p}_{z} \right\} \psi(\mathbf{r}, t) \end {align*} \nonumber\]

relativistic invariance for the free particle requires that the second and third term are zero, and so

\[ \begin{array} {rcl} \left(\alpha_{x} \right)^{2} = \left(\alpha_{y} \right)^{2} = \left(\alpha_{z} \right)^{2} = \beta^{2} = 1 \\ \alpha_{i} \alpha_{j}+\alpha_{j} \alpha_{i} = 0 & (i \neq j) \\ \alpha_{x} \beta+\beta \alpha_{x} = 0 & \text { (and similarly for } y, z) \end{array} \nonumber\]

Thus \(\alpha_i\) and \(\beta\) cannot be just numbers. The simplest representation for \(\alpha\) and \(\beta\) are 4x4 matrices, meaning that the wavevector is a 4-component vector. When we work this through, there are no negative probabilities, but two of the components turn out to have negative energy.

It turns out that the four components accurately describe the two spin states of the electron and the positron. More remarkably, Dirac solved the equation before the positron had even been discovered!