slug: particle-physics-4 date*published: 2019-04-27T06:06:19 date*updated: 2019-04-27T06:06:19 tags: English Posts, Acedemic Notes –-

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](https://en.wikipedia.org/wiki/Another_(novel). 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:

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.

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:

$\begin{cases} \alpha_i \beta^2 + \beta \alpha_i \beta &= 0 \\ Tr(\alpha_i) = -Tr(\beta \alpha_i \beta) &= -Tr(\alpha_i)\\ \Rightarrow Tr(\alpha_i) &= 0 \end{cases}$Similarly we can also show that $\beta$ is traceless.

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

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:

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