Dirac field

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1 Mass term
2 Canonical Variables
3 Equations of Motion
4 Symmetries and Conserved Quantities
5 Classical Solutions
6 Canonical Quantization
7 Path integral quantization
8 Exercises

Mass term

Define the left- and right-handed spinors

$\Psi_L = \frac{1}{2}\left(1 - \gamma^5\right) \Psi\,$ ,

$\Psi_R = \frac{1}{2}\left(1 + \gamma^5\right) \Psi\,$ .

In the Weyl basis, we may write $\Psi\,$ in terms of two Weyl spinors

$\Psi = \begin{pmatrix} \psi_L \\ \psi_R \end{pmatrix}\,$ .

Then, note that

$\Psi_L^\dagger \gamma^0 \Psi_L\,$	$=\frac{1}{4} \Psi^\dagger \left(1 - \gamma^5 \right)^\dagger \gamma^0 \left(1 - \gamma^5 \right)\Psi\,$ ,
	$=\frac{1}{4} \Psi^\dagger \left(\gamma^0 - \left\{\gamma^5, \gamma^0 \right\}+ \gamma^5 \gamma^0 \gamma^5 \right)^\dagger \Psi\,$ ,
	$=0\,$ ,

Then, e.g.,

$\mathcal{L}_m\,$	$= -m(\bar\Psi \Psi)\,$
	$= -m( \bar\Psi_L \Psi_R + \bar\Psi_R \Psi_L) \,$ ,
	$= -m(\psi_L^\dagger \psi_R + \psi_R^\dagger \psi_L)\,$ .

Including a Dirac mass therefore breaks chiral symmetry.

Canonical Variables

Equations of Motion

Symmetries and Conserved Quantities

Classical Solutions

Canonical Quantization

We start by expanding the field in terms of modes, treating $a^s_{\mathbf{p}}$ and $b^s_{\mathbf{p}}$ as operators:

$\psi(\mathbf{x}) = \int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_{\mathbf{p}}}} e^{-i \mathbf{p\cdot x}} \sum_{s=1,2} \left[a^s_\mathbf{p} u^s(\mathbf{p}) + b^s_\mathbf{-p} v^s(\mathbf{-p}) \right]$

We then postulate the following anticommutation relations,

$\left\{ a^r_\mathbf{p}, a^{s\dagger}_\mathbf{q} \right\} = \left\{ b^r_\mathbf{p}, b^{s\dagger}_\mathbf{q} \right\} = (2\pi)^3 \delta^{(3)}(\mathbf{p-q})\delta^{rs}$

and take all other anticommutators to vanish. Inserting this into the Hamiltonian gives:

$H\,$	$= \int d^3x\,\bar{\psi} \left(i \gamma^i \partial_i + m \right)\psi$
	$= {\int \!\!\!\int \!\!\!\int} \frac{d^3x\,d^3p\,d^3q}{(2\pi)^6\sqrt{4E_{\mathbf{p}}E_{\mathbf{q}} }}\sum_{rs} \left[ a^{r\dagger}_\mathbf{p} \bar{u}^{r}(\mathbf{p}) + b^{r\dagger}_\mathbf{-p} \bar{v}^{r}(\mathbf{-p}) \right] (\gamma^i q_i +m) \left[ a^s_\mathbf{q} u^s(\mathbf{q}) + b^s_\mathbf{-q} v^s(\mathbf{-q}) \right] e^{-i(\mathbf{q-p})\cdot\mathbf{x}}$

Using the Fourier transform of the delta function, this can be simplified to:

$H\,$	$= \int\!\!\frac{d^3p}{(2\pi)^3 2E_{\mathbf{p} }} \sum_{rs} \left[ a^{r\dagger}_\mathbf{p} \bar{u}^{r}(\mathbf{p}) + b^{r\dagger}_\mathbf{-p} \bar{v}^{r}(\mathbf{-p}) \right] (\gamma^i p_i +m) \left[ a^s_\mathbf{p} u^s(\mathbf{p}) + b^s_\mathbf{-p} v^s(\mathbf{-p}) \right]$
	$=\int\!\!\frac{d^3p}{(2\pi)^3 2E_{\mathbf{p} }} \sum_{rs} \left[ a^{r\dagger}_\mathbf{p} \bar{u}^{r}(\mathbf{p}) + b^{r\dagger}_\mathbf{-p} \bar{v}^{r}(\mathbf{-p}) \right] \left[ a^s_\mathbf{p} (\gamma^0p_0)u^s(\mathbf{p}) + b^s_\mathbf{-p} (-\gamma^0p_0)v^s(\mathbf{-p}) \right]$
	$=\int\!\!\frac{d^3p\,p_0}{(2\pi)^3 2E_{\mathbf{p} }} \sum_{rs} \left[a^{r\dagger}_\mathbf{p} a^s_\mathbf{p} u^{r\dagger}(\mathbf{p}) u^s(\mathbf{p}) + b^{r\dagger}_\mathbf{-p} a^s_\mathbf{p} v^{r\dagger}(\mathbf{-p})u^s(\mathbf{p}) - a^{r\dagger}_\mathbf{p} b^s_\mathbf{-p} u^{r\dagger}(\mathbf{p}) v^s(\mathbf{-p}) - b^{r\dagger}_\mathbf{-p} b^s_\mathbf{-p} v^{r\dagger}(\mathbf{-p}) v^s(\mathbf{-p}) \right]$

At this point the orthogonality relations can be used:

$H\,$

$=\int\!\!\frac{d^3p}{(2\pi)^3 2E_{\mathbf{p} }} \sum_{rs} \left[ 2 p_0 E_{\mathbf{p}} ( a^{r\dagger}_\mathbf{p} a^s_\mathbf{p} \delta^{rs} - b^{r\dagger}_\mathbf{-p} b^s_\mathbf{-p} \delta^{rs}) \right]$

Now $p_0 = E_\mathbf{p}$ and

$H\,$

$=\int\!\!\frac{d^3p}{(2\pi)^3 E_{\mathbf{p} }} \sum_{r} \left[ E_{\mathbf{p}}^2 ( a^{r\dagger}_\mathbf{p} a^r_\mathbf{p} - b^{r\dagger}_\mathbf{-p} b^r_\mathbf{-p}) \right]$

We can change our variable of integration for the second term to give the standard form:

$H\,$

$=\int\!\!\frac{d^3p\, E_{\mathbf{p}}}{(2\pi)^3 } \sum_{r} \left[ a^{r\dagger}_\mathbf{p} a^r_\mathbf{p} - b^{r\dagger}_\mathbf{p} b^r_\mathbf{p} \right]$

We see therefore that the field describes two types of non-interacting particles: one type reated by $a^{r\dagger}_\mathbf{p}$ and one type created by $b^{r\dagger}_\mathbf{p}$ . However, the particles of the second type have negative energy, and so the field can achieve an arbitrarily large negative energy by creating more and more particles and is therefore unstable. One approach to dealing with this problem is to use the experimental observation that fermions obey the so-called Pauli exclusion principle to postulate a ground state in which all negative energy states are already filled and measure all energies (and charges) relative to this ground state. This is the Dirac sea point of view. Another option is to reinterpret the operators $b^{r\dagger}_\mathbf{p}$ and $b^r_\mathbf{p}$ as operators that respectively destroy and create positive energy particles. In this point of view, we observe that for given values of $\mathbf{p}$ and $r\,$ there are only two possible states: $b^{r}_\mathbf{p}\left|1\right\rangle = \left|0\right \rangle$ and $\left|1\right \rangle = b^{r\dagger}_\mathbf{p}\left|0\right\rangle$ . Since $\left|1\right\rangle$ is the ground state, we can repeat the above calculations and let $b^{r}_\mathbf{p} \to b^{r\dagger}_\mathbf{p}$ and $b^{r\dagger}_\mathbf{p} \to b^{r}_\mathbf{p}$ . We then obtain

$H =\int\!\!\frac{d^3p\, E_{\mathbf{p}}}{(2\pi)^3 } \sum_{r} \left[ a^{r\dagger}_\mathbf{p} a^r_\mathbf{p} - b^{r}_\mathbf{p} b^{r\dagger}_\mathbf{p} \right]$ .

Making use of the anticommutation relations, this can be changed to

$H =\int\!\!\frac{d^3p\, E_{\mathbf{p}}}{(2\pi)^3 } \sum_{r} \left[ a^{r\dagger}_\mathbf{p} a^r_\mathbf{p} + b^{r\dagger}_\mathbf{p} b^{r}_\mathbf{p} - (2\pi)^3 \delta^{(3)}(0)\right]$ .

If we measure all our energies relative to the state with no particles in it, then this becomes

$H =\int\!\!\frac{d^3p\, E_{\mathbf{p}}}{(2\pi)^3 } \sum_{r} \left[ a^{r\dagger}_\mathbf{p} a^r_\mathbf{p} + b^{r\dagger}_\mathbf{p} b^{r}_\mathbf{p}\right]$ .

Note that in order to achieve the result that particles of the second type are created with positive energy, we had to make use of an anticommutation relation. This would not have been possible had we used a commutation relation instead. This is a manifestation of the spin statistics theorem.

Path integral quantization

Exercises

∗∗∗ By making use of $(\gamma^5)^2 = 1\,$ and $\{\gamma^\mu, \gamma^5\} = 0\,$ , what can you say about the eigenvalues of $i\gamma^\mu \partial_\mu - m\,$ ?

∗∗∗ Show that $\ln \det\, (i\gamma^\mu \partial_\mu - m) = \tfrac{1}{2}\operatorname{tr} \ln\, (\partial^2 - m^2)\,$ . Here the arguments of $\det\,$ and $\operatorname{tr}\,$ should be thought of as operators acting on spinor fields. Optional: complete the calculation in momentum space.