In the past few weeks, I tried to watch an series of anime each weekend; so far, I have watched Gurren Lagann, Initial D, and re-watched Another. This week the player is Magi: The Labyrinth of Magic. All of these are very good watches, I've watched more anime than all past 3 years combines – ironically.


This week's particle physics finally have entered the realm of quantum mechanics, specifically, relativistic wave equations.

Klein-Gordon Equations

Following Schrodinger's legacy, people recognized that:

$$ \hat{\mathcal{H}}\Psi(t) = i\hbar \frac{\partial}{\partial t} \Psi(t) $$

By the correspondence principle, we recognize that for a free moving particle, Hamiltonian is just kinetic energy $\frac{p^2}{2}$, and replacing $p$ with its coordinate space operator brings us:

$$ -\frac{\hbar^2}{2m}\nabla^2 \Psi(\vec{r}, t) = i\hbar \frac{\partial}{\partial t} \Psi(t) $$

For each energy eigen states at hand, this also equals to $E\Psi$, but wait a second, what energy are we talking about here? It looks like some naive energy pre Einstein's time. If you think that, you're stepping into K-G equation already!

Fundamentally, just like any problem, the Hamiltonian is just a "general" operators, in the sense that, we are expected to put in the correct terms when actually solving the equation. But if the "speed" of the particle in question is too fast, clearly $E\neq \frac{p^2}{2m}$, so we need to use the theory of special relativity. This relation between energy and momentum is often called "dispersion relation".

Special relativity says invariant mass squared:

$$ m^2 = p^2 = E^2 - \vec{p}^2 $$

By making the same identification of $E\rightarrow i\hbar \frac{\partial}{\partial t}$ we get (in n.u):

$$ \left[-\frac{\partial^2}{\partial t^2} + \nabla^2\right]\phi(x) = m^2 \phi(x) $$

where $\phi(x)$ is a 4-vector wave equation so to speak. You can check 2 things from this point on:

  1. Act on a supposedly energy eigen state $N e^{-i(Et-\vec{p}\cdot\vec{r})/\hbar}$ and see if you get back the dispersion relation.
  2. Make the approximation $ E = \sqrt{\vec{p}^ 2 c^ 2 + (mc^ 2)^ 2} \approx mc ^2$ and see if you get the Schrodinger's equation back as a non-relativistic limit.

Dirac equation

Now Paul Dirac was a curious man, he did not like the second order equation. He wanted it to be 1-st order, whatever the cost is. Know we know what the equation has – basically 3 parts:

$$ \begin{cases} \nabla \\ m\\ i\frac{\partial}{\partial t} \end{cases} $$

I'm pretty sure the internal activities goes some thing along:

Hell, what if we just, stick some coefficients and force the equation obey special relativity dispersion relation and figure out what the coefficients need to be???

And, off we go:

$$ \left[\vec{\alpha} \cdot (-i\nabla) + \beta m\right] \Psi(x) = i \frac{\partial}{\partial t}\Psi (x) $$

The notation already suggests that the coefficients are not numbers, rather something monstrous. Recall, in the operator form, our constrain looks like this (add $i$ to $\nabla$):

$$ -\frac{\partial^2}{\partial t^2} - (-i\nabla)^2 = m^2 $$

Re-arrange and compare to Dirac equation, we see that by construction:

$$ \left[\vec{\alpha} \cdot (-i\nabla) + \beta m\right]^2 = - (\frac{\partial}{\partial t})^2 = -\frac{\partial^2}{\partial t^2} - (-i\nabla)^2 $$

We need the two ends equal, obviously we can't have cross terms of $\nabla$ and $m$, upon inspection, we conclude the following constrains ({} is anti commutator):

$$ \begin{cases} \alpha_i^ 2 = 1 \\ \{\alpha_i, \alpha_j\} = \delta_{ij} \\ \{\alpha_i, \beta\} = 0 \\ \beta^2 =1 \end{cases} $$

It is fairly clear that numbers cannot satisfy this.


The we do some trick to make life later easier, multiply both sides of Dirac equation by $\beta$ and make new variables:

$$ \left[\vec{\gamma} \cdot (-i\nabla) + m\right] \Psi(x) = \gamma^0 i \frac{\partial}{\partial t}\Psi (x) $$

So $\gamma^ 0 = \beta$ and $\vec{\gamma} = \beta \vec{\alpha}$. By using Feynman's slash notation (this is disgusting to say the least):

$$ i\partial\!\!\!/ \Psi(x)= m\Psi(x) $$

Where $\partial\!\!\!/ = \gamma^ \mu \partial_\nu$.


Come back to the constrains, we have additional constrains from quantum mechanics. First, Hamiltonian is Hermitian means:

$$ H = \vec{\alpha}\cdot\vec{p} + \beta m = H^\dagger $$

so each $\alpha_i$ has to be hermitian and so does $beta$.

Second, $\alpha_i \beta + \beta \alpha_i =0$ means:

$$ \begin{align} \alpha_i \beta^2 + \beta \alpha_i \beta &= 0 \\ Tr(\alpha_i) = -Tr(\beta \alpha_i \beta) &= -Tr(\alpha_i)\\ \Rightarrow Tr(\alpha_i) &= 0 \end{align} $$

Similarly we can also show that $\beta$ is traceless.

Most importantly, we can obtain the size of these matrix by the following neat trick:

$$ \begin{align} \text{det}(\alpha_i \beta) &= \text{det}(-\beta \alpha_i) \\ \text{det}(\alpha_i) \text{det}(\beta) &= \text{det}(-1I) \text{det}(\beta) \text{det}(\alpha_i) \\ &\Rightarrow \text{det}(-1I) = 1 \end{align} $$

This some how forces the dimension of the identity matrix to be even! So maybe they are 2x2 matrices? Unfortunately, a set of basis for 2x2 hermitian matrix has to contain the identity matrix, which is NOT traceless. So we have to go to 4x4.

In 4x4, we have plenty of choice. By some conversion, the common choice looks like this:

$$ \gamma^0 = \begin{pmatrix} I_2 & 0 \\ 0 & -I_2 \end{pmatrix}, \gamma^1 = \left(\begin{array}{cccc} 0 & \sigma_x \\ -\sigma_x & 0 \end{array}\right), \gamma^2 = \left(\begin{array}{cccc} 0 & \sigma_y \\ -\sigma_y & 0 \end{array}\right), \gamma^3 = \left(\begin{array}{cccc} 0 & \sigma_z \\ -\sigma_z & 0 \end{array}\right). $$

Where each $\sigma$ is a Pauli matrix – painful, I know. Thus our wave equation is expand into a vector of wave equations:

$$ \begin{align} \Psi(x) &= \left(\begin{matrix} \Psi_1(x) \\ \Psi_2(x) \\ \Psi_3(x) \\ \Psi_4(x) \end{matrix}\right) \end{align} $$

This is a 4-component spinor, as a spoiler, the basis is spin-up electron, spin-down electron, spin-down positron, spin-up positron. I know right, mind==blown.