This post contains a summary of my lecture notes for the second half of last Wednesday’s QFT-I lecture.

[Click here for a PDF with the full notes for this portion of the lecture.]

### DISCLAIMER: Very rough notes from class, with some additional side notes.

These are notes for the UofT course PHY2403H, Quantum Field Theory, taught by Prof. Erich Poppitz, fall 2018.

### Summary:

- We found that the Dirac Hamiltonian is

\begin{equation*}

H

=

\int d^3 x

\Psi^\dagger

\lr{

– i \gamma^0 \gamma^j \partial_j \Psi + m \gamma^0

}

\Psi.

\end{equation*} - We found that our plane wave solutions

\(\Psi_u = u(p) e^{-i p \cdot x}\), and \( \Psi_v = v(p) e^{i p \cdot x} \), were eigenvectors of the operator portion of the Hamiltonian

\begin{equation*}

\begin{aligned}

-\gamma^0 \lr{ i \gamma^j \partial_j – m } \Psi_u &= p_0 \Psi_u \\

-\gamma^0 \lr{ i \gamma^j \partial_j – m } \Psi_v &= -p_0 \Psi_v.

\end{aligned}

\end{equation*} - We formed a linear superposition of our plane wave solutions

\begin{equation}\label{eqn:qftLecture21b:800}

\Psi(\Bx, t)

=

\sum_{s = 1}^2

\int \frac{d^3 p}{(2 \pi)^3 \sqrt{ 2 \omega_\Bp } }

\lr{

e^{-i p \cdot x} u^s_\Bp a_\Bp^s

+

e^{i p \cdot x} v^s_\Bp b_\Bp^s

}.

\end{equation} - and expressed the Dirac Hamiltonian in terms of creation and anhillation operators

\begin{equation*}

H_{\text{Dirac}}

=

\sum_{r = 1}^2

\int \frac{d^3 p }{(2\pi)^3 }

\omega_\Bp

\lr{

a^{r \dagger}_\Bp

a^r_\Bp

–

b^{r \dagger}_{-\Bp}

b^r_{-\Bp}

}.

\end{equation*} - Finally, we interpreted this using the Dirac Sea argument
- It was claimed that the \( a, b\)’s satisfied anticommutator relationships

\begin{equation}\label{eqn:qftLecture21b:940}

\begin{aligned}

\symmetric{a^s_\Bp}{a^{r \dagger}_\Bq} &= \delta^{sr} \delta^{(3)}e(\Bp – \Bq) \\

\symmetric{b^s_\Bp}{b^{r \dagger}_\Bq} &= \delta^{sr} \delta^{(3)}(\Bp – \Bq),

\end{aligned}

\end{equation}

where all other anticommutators are zero

\begin{equation}\label{eqn:qftLecture21b:960}

\symmetric{a^r}{b^s} =

\symmetric{a^r}{b^{s\dagger}} =

\symmetric{a^{r\dagger}}{b^s} =

\symmetric{a^{r\dagger}}{b^{s\dagger}} = 0.

\end{equation}

and used these to algebraically remove the negative energy states of the Hamiltonian.