Ze Chen

Dirac Equation

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The Equation and the Solution

We demand that $$\vb*{\alpha}$$ and $$\beta$$ satisfy the commutation relations \begin{align*} \beta^2 &= \mathbb{1}; \\ \qty{\alpha_i,\beta} &= 0; \\ \qty{\alpha_i,\alpha_j} &= \mathbb{1}\delta_{ij}. \end{align*} A conventional representation is given by $\vb*{\alpha} = \begin{pmatrix} 0 & \vb*{\sigma} \\ \vb*{\sigma} & 0 \end{pmatrix};\quad \beta = \begin{pmatrix} \mathbb{1} & 0 \\ 0 & -\mathbb{1} \end{pmatrix}.$ Now we have the Dirac equation $i\hbar\pdv{{\psi}}{t} = (c\vb*{\alpha}\cdot \vb*{p} + \beta mc^2){\psi}.$

Probability Current

Under the interpretation $\rho = \psi^\dagger \psi;\quad \vb*{j} = \psi^\dagger \vb*{\alpha} \psi;$ the continuity equation $\pdv{\rho}{t} + \div \vb*{j} = 0$ holds.

Electromagnetic Interaction

The Dirac equation of a particle in a electromagnetic field is written as $i\hbar\pdv{\psi}{t} = \qty[c\vb*{\alpha}\cdot\qty(\vb*{p} - q\vb*{A}/c)+\beta mc^2 + q\phi]\psi.$

In Pure Magnetic Field

In the following we set $$\phi = 0$$. Writing $$\psi(t) = \psi e^{-iEt/\hbar}$$ we get $(c\vb*{\alpha}\cdot \vb*{\pi} + \beta mc^2)\psi = E\psi,$ where $$\vb*{\pi} = \vb*{p} - q\vb*{A}/c$$ is the kinetic momentum.

Let $$\displaystyle \psi = \begin{pmatrix}\chi \\ \Phi\end{pmatrix}{}$$ where $$\chi$$ is called the large component and $$\Phi$$ the small component. The Dirac equation is rewritten as

\begin{align*} (\vb*{\sigma}\cdot \vb*{A})(\vb*{\sigma}\cdot \vb*{B}) &= \vb*{A}\cdot \vb*{B} + i\vb*{\sigma}\cdot \vb*{A}\times \vb*{B};\\ \vb*{\pi} \times \vb*{\pi} &= \frac{iq\hbar}{c}\vb*{B}. \end{align*}

\begin{align*} \chi &= \qty(\frac{c\vb*{\sigma}\cdot \vb*{\pi}}{E-mc^2})\Phi; \\ \Phi &= \qty(\frac{c\vb*{\sigma}\cdot \vb*{\pi}}{E+mc^2})\chi. \end{align*}

Let $$E_S = E - mc^2$$ denote the energy that appeared in the Schrödinger equation. In the nonrelativistic limit we get $\qty[\frac{(\vb*{p} - q\vb*{A}/c)^2}{2m} - \frac{q\hbar}{2mc}\vb*{\sigma}\cdot \vb*{B}] \chi = E_S \chi.$

Hydrogen Fine Structure

Let $$V$$ denote the Coulomb potential in the hydrogen atom. The Dirac equation is rewritten as \begin{align*} \Phi &= \qty(\frac{c\vb*{\sigma}\cdot \vb*{p}}{E-V+mc^2})\chi; \\ (E-V-mc^2) &= c\vb*{\sigma}\cdot \vb*{p}\qty[\frac{1}{E-V+mc^2}]c\vb*{\sigma}\cdot \vb*{p}\chi. \end{align*} In the nonrelativistic limit we keep the $$(E_S - V)$$ on the denominator on the RHS to the first order and find $E_S \chi = H\chi = \qty[\frac{\vb*{p}^2}{2m} + V - \frac{\vb*{\sigma}\cdot \vb*{P}(E_S -V)\vb*{\sigma}\cdot \vb*{P}}{4m^2c^2}]\chi.$ To retain the conservation of probability and to modify $$H$$ into a Hermitian operator, we solve for $\chi_S = \qty(1+\frac{\vb*{p}^2}{8m^2c^2})\chi$ instead of $$\chi$$. The equation then becomes (working to order $$\alpha^4$$) $E_S \chi_S = H_S \chi_S = \qty(H + \qty[\frac{\vb*{p}^2}{8m^2c^2},V]).$ Approximate to order $$\alpha^4$$ we find \begin{align*} H_S &= \frac{\vb*{p}^2}{2m} + V - \frac{\vb*{p}^4}{8m^3c^2} - \frac{i\vb*{\sigma}\cdot \vb*{p}\times \qty[\vb*{p}, V]}{4m^2c^2} + \qty{-\frac{\vb*{p}\cdot \qty[\vb*{p},V]}{4m^2c^2}+\frac{\qty[\vb*{p}\cdot \vb*{p},V]}{8m^2c^2}} \\ &= \frac{\vb*{p}^2}{2m} + V - \frac{\vb*{p}^4}{8m^3c^2} + \frac{e^2}{2m^2c^2r^2}\vb*{S}\cdot \vb*{L} + \frac{e^2\hbar^2 \pi}{2m^2 c^2}\delta^3(\vb*{r}) \end{align*} These items accounts for $$T$$, $$V$$, relativistic $$T$$, spin-orbit interaction, and the Darwin term sequentially.

2021/2/8 23:43:12

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