
%%==============================================================================
\begin{lecture}
\item Construction of the Dirac equation
\item Free-particle solutions of the Dirac equation
\end{lecture}
%%==============================================================================

%%==============================================================================
\begin{problemset}
\item Fock states and coherent states
\item Free-particle solutions of the Dirac equation 
\item Review of the Lorentz group
\end{problemset}
%%==============================================================================

%%==============================================================================
\reference{PS:35--49}
%%==============================================================================

\chapter{The Dirac Field}

\section{The Dirac Equation}
\label{sec:dirac_equation}

\imp{So far: Simplest relativistic field equation \therefore Klein-Gordon
equation}\\
%
\imp{Now: Second simplest relativistic field equation \therefore Dirac
equation}

\begin{lot}

    \item \Emph{Observation 1:} Lorentz symmetry of the KG equation:\\
        %
        \imp{We view Lorentz transformations as \emph{active} transformations,
        mapping solutions to different solutions!  This is equivalent to the
        \emph{passive} viewpoint where the coordinate system is transformed
        instead.}
        %
        \begin{lot}

            \item \consider Coordinate transformation: $x'=\Lambda x$ \&
            Field transformation: $\phi'(x')=\phi(x)$

            \item \consider $\phi$ with $(\partial^2+m^2)\phi(x)=0$ for all $x$

            \item \therefore $\phi'(x)=\phi(\Lambda^{-1}x)$ is a new solution:\\
                %
                \imp{Use the chain rule in the first step twice:}
                %
                \begin{subalign}
                    (g^{\mu\nu}\partial_\mu\partial_\nu+m^2)\phi'(x)
                    &=[g^{\mu\nu}\tensor{(\Lambda^{-1})}{^\sigma_\mu}\partial_\sigma\,
                    \tensor{(\Lambda^{-1})}{^\rho_\nu}\partial_\rho+m^2]\phi(\Lambda^{-1}x)
                    \\
                    &\mimp{Use invariance of the metric}
                    \\
                    &=(g^{\sigma\rho}\partial_\sigma\partial_\rho+m^2)\phi(\Lambda^{-1}x)
                    \\
                    &=(\partial^2+m^2)\phi(\Lambda^{-1}x)\stackrel{\text{$\phi$ solution}}{=}0
                \end{subalign}
                %
                \imp{Here $\partial_\sigma\phi(\Lambda^{-1}x)$ must be read
                as $\left.\partial_\sigma\phi(y)\right|_{y=\Lambda^{-1}x}$,
                i.e., we compute the derivative of the function $\phi$
                with respect to its argument $y$ and then plug in the value
                $\Lambda^{-1}x$.}
                
        \end{lot}

    \item \Emph{Observation 2:} 
    \consider Vector fields under rotations: $\vec \phi'(\vec x)=R\vec\phi(R^{-1}\vec x)$\\
        %
        \therefore In general\note{, a field $\phi(x)\in\complex^n$ can
        transform under $n$-dimensional Lorentz transformations as}
        %
        \begin{align}
            \phi'_a(x)=M_{ab}(\Lambda)\phi_b(\Lambda^{-1}x)    
            \qquad a=1,\dots,n
        \end{align}
        %
        where
        %
        \begin{align}
            M(\Lambda')M(\Lambda)\phi(\Lambda^{-1}\Lambda'^{-1}x)\stackrel{!}{=}M(\Lambda'\Lambda)\phi((\Lambda'\Lambda)^{-1}x)
        \end{align}
        %
        is a $n$-dimensional \Emph{representation} of the Lorentz group $\POLG$.

    \item We want a \Emph{first-order} relativistic field equation:
        %
        \begin{align}
           (\partial^\mu\partial_\mu+\const)\phi=0     
           \quad\Rightarrow\quad
           (i\blacksquare^\mu\partial_\mu+\const)\phi=0     
        \end{align}
        %
        \imp{The $i$ anticipates wave-like solutions for real $\blacksquare$.}

    \item Then \imp{(combine 1 \& 2)}
        %
        \begin{lot}

            \item \consider Coordinate transformation: $x'=\Lambda x$ \&
            Field transformation: $\phi'(x')=M(\Lambda)\phi(x)$

            \item \consider $\phi$ with
            $(i\blacksquare^\mu\partial_\mu+\const)\phi(x)=0$ for all $x$

            \item When is $\phi'(x)=M(\Lambda)\phi(\Lambda^{-1}x)$ is a
            new solution?
                %
                \begin{align}
                    (i\blacksquare^\mu\partial_\mu+\const)\phi'(x)
                    =[i\blacksquare^\mu\tensor{(\Lambda^{-1})}{^\nu_\mu}\partial_\nu+\const]\,M(\Lambda)\phi(\Lambda^{-1}x)
                    \stackrel{!}{=}0
                \end{align}
                %
                \imp{Multiply with $M^{-1}(\Lambda)$:}
                %
                \begin{align}
                    \Leftrightarrow\quad
                    [i\underbrace{M^{-1}(\Lambda)\blacksquare^\mu M(\Lambda)\tensor{(\Lambda^{-1})}{^\nu_\mu}}_{%
                    \stackrel{!}{=}\blacksquare^\nu}\partial_\nu+\const]\,
                    \phi(\Lambda^{-1}x)
                    \stackrel{!}{=}0
                \end{align}
                %
                \therefore $\blacksquare^\mu\equiv\gamma^\mu$ must be $n\times n$-matrices with
                %
                \begin{emphalign}
                    M^{-1}(\Lambda)\gamma^\mu M(\Lambda)=\tensor{\Lambda}{^\mu_\nu}\gamma^\nu
                    \label{eq:g_M}
                \end{emphalign}
                %
                \note{The $\gamma$-matrices ``translate'' the
                ``spinor''-representation $M(\Lambda)$ into the
                ``vector''-representation $\Lambda$ and vice versa.}

        \end{lot}

    \item \imp{How to find $\gamma^\mu$ and $M(\Lambda)$?} $\POLG$ is a
        \Emph{Lie group} \imp{(\seepset{3})}:
        %
        \begin{subalign}
            \Lambda &= \exp\left[-\frac{i}{2}\omega_{\alpha\beta}\J^{\alpha\beta}\right]
            \stackrel{\omega\ll 1}{\approx} \Id-\frac{i}{2}\omega_{\alpha\beta}\J^{\alpha\beta}\label{eq:vec_rep}\\
            M(\Lambda) &= \exp\left[-\frac{i}{2}\omega_{\alpha\beta}S^{\alpha\beta}\right]
            \stackrel{\omega\ll 1}{\approx} \Id-\frac{i}{2}\omega_{\alpha\beta}S^{\alpha\beta}\label{eq:spin_rep}
        \end{subalign}
        %
	    $\omega_{\alpha\beta}$ antisymmetric tensor \therefore 3 rotations
	    (angles) + 3 boosts (rapidities)
        
        \note{It is
        $(\J^{\alpha\beta})_{\mu\nu}=i(\delta^\alpha_\mu\delta^\beta_\nu-\delta^\alpha_\nu\delta^\beta_\mu)$.}
       
	    \imp{The $4\times 4$ matrices $\J^{\alpha\beta}$ generate the
	    vector-representation $\Lambda$, $(\hlf,\hlf)$, the $n\times
	    n$-matrices $S^{\alpha\beta}$ the spinor-representation
	    $M(\Lambda)$, $(\hlf,0)\oplus(0,\hlf)$. The generators are
	    antisymmetric in the spacetime indices.}

        \begin{itemize}

            \item Infinitesimal form of \cref{eq:g_M}:
                %
                \begin{align}
                    \com{\gamma^\mu}{S^{\alpha\beta}}\nte\tensor{(\J^{\alpha\beta})}{^\mu_\nu}\gamma^\nu
                    \nte i(g^{\alpha\mu}\gamma^\beta-g^{\beta\mu}\gamma^\alpha)
                    \label{eq:g_M_L}    
                \end{align}

            \item $\J^{\alpha\beta}$ 
                \therefore \Emph{Lie-algebra} of Lorentz group \imp{($J=S,\J$, \seepset{3})}
                %
                \begin{align}
                    \com{J^{\mu\nu}}{J^{\rho\sigma}}\nte
                    i(g^{\nu\rho}J^{\mu\sigma}-g^{\mu\rho}J^{\nu\sigma}
                    -g^{\nu\sigma}J^{\mu\rho}+g^{\mu\sigma}J^{\nu\rho})
                    \label{eq:LA}    
                \end{align}
                %
                \imp{The Lie algebra defines the structure of the Lie group by
                integration and is therefore the same for all representations.}

        \end{itemize}

    \item \Emph{Solution:} Dirac's trick: \consider $\gamma^\mu$ such that
        %
        \begin{emphalign}
            \acom{\gamma^\mu}{\gamma^\nu}=2 g^{\mu\nu}\,\Id_{n\times n}    
            \quad\text{\termdef{Dirac algebra}}
        \end{emphalign}
        %
        \note{This is the 16-dimensional \emph{Clifford algebra}
        $\mathrm{C}\ell_{1,3}(\complex)$.}\\
        %
        Then
        %
        \begin{emphalign}
            S^{\mu\nu}\equiv \frac{i}{4}\com{\gamma^\mu}{\gamma^\nu} 
            \label{eq:Smunu}
        \end{emphalign}
        %
        satisfies the Lorentz algebra \cref{eq:LA} and \cref{eq:g_M_L}.

    \item \Emph{Representations:}

        \begin{itemize}

            \item At least $4$-dimensional\\
                %
                \note{(think of the $\gamma^\mu$ as Majorana modes and
                construct ladder operators \therefore 2 modes)}

            \item All 4-dimensional representations are unitarily equivalent\\
                %
                \note{(actually, they constitute the \emph{unique} irrep of
                the Dirac algebra which is 4-dimensional)}

            \item We use the \Emph{Weyl representation} \note{(sometimes
            called \emph{chiral representation})}:
                %
                \begin{emphalign}
                    \gamma^0=\begin{pmatrix}0 & \Id \\ \Id & 0\end{pmatrix}
                    \quad\text{and}\quad
                    \gamma^i=\begin{pmatrix}0 & \sigma^i \\ -\sigma^i & 0\end{pmatrix}
                \end{emphalign}
                %
            \item Henceforth: $\Lambda_\hlf\equiv M(\Lambda)$\\
            %
            \imp{Two ``copies'' of a spin-$\hlf$ projective representation.}

        \end{itemize}

    \item Setting $\const=-m$, we find:
        %
        \begin{emphalign}
            (i\g^\mu\pmu-m)\Psi=0
            \quad\text{\termdef{Dirac equation}}
            \label{eq:DE}
        \end{emphalign}
        %
        \imp{$\Psi(x)$ is a \emph{bispinor}-field with values in
        $\complex^4=\complex^2\oplus\complex^2$.}

    \item The components of the Dirac spinor field satisfy the KG equation:
        %
        \begin{align}
            0={\color{marked}(-i\gmu\pmu-m)}(i\gamma^\nu\partial_\nu-m)\Psi\nte (\partial^2+m^2)\Psi
            \label{eq:Dirac_KG}
        \end{align}
        %
        \imp{The Dirac differential operator is the ``square root'' of the
            Klein-Gordon differential operator. On the right hand side of
            \cref{eq:Dirac_KG} there is an identity $\Id_{4\times 4}$ that
            we omit.}

    \item \Emph{Dirac adjoint:}
        
        \imp{Goal: Lagrangian (which must be a Lorentz scalar).}\\
        %
        \imp{\therefore How to form Lorentz scalars from spinors?}
        %
        \begin{lot}

            \item First try: $\Psi^\dag\Psi$
                %
                \begin{align}
                    \Psi'^\dag\Psi'
                    =\Psi^\dag \underbrace{\Lambda_\hlf^\dag\Lambda_\hlf}_{\neq\Id}\Psi\neq \Psi^\dag\Psi 
                \end{align}
                %
                \imp{$\Lambda_\hlf$ is not unitary because $S^{\mu\nu}$
                is not Hermitian for boosts ($\mu=0$ and $\nu=1,2,3$).\\
                %
                This is a consequence of the non-compactness of the Lorentz
                group due to boosts!}

            \item Define
                %
                \begin{emphalign}
                    \ol \Psi=\Psi^\dag\gamma^0    
                    \quad\text{\termdef{Dirac adjoint}}
                \end{emphalign}
                %
                \easytoshow $\ol\Psi'\Psi'=\ol\Psi\Lambda_\hlf^{-1}\Lambda_\hlf\Psi=\ol\Psi\Psi$
                \limplies Lorentz scalar\\
                %
                \imp{Use \cref{eq:Smunu} and \cref{eq:spin_rep} and the
                Dirac algebra to show this!}

        \end{lot}

    \item \Emph{Lagrangian:}
        %
        \begin{emphalign}
            \L_\mathrm{Dirac}=\ol\Psi(i\gmu\pmu-m)\Psi
        \end{emphalign}
        %
        \imp{\easytoshow Euler-Lagrange equations yield Dirac equation.}

\end{lot}


\begin{blocknote}
    \begin{itemize}

        \item 
            Let $\sigma^\mu\equiv (1,\vec \sigma)^T$ and $\ol\sigma^\mu\equiv (1,-\vec \sigma)^T$
            and $\g^\mu = \begin{pmatrix}0 & \sigma^\mu \\ \ol\sigma^\mu & 0 \end{pmatrix}$\\
            %
            \therefore Dirac equation:
            %
            \begin{align}
                \begin{pmatrix}-m & i\sigma\partial \\ i\ol\sigma\partial & -m\end{pmatrix}\bispi{\psi_L}{\psi_R}=0
            \end{align}

        \item $\psi_{L}$ and $\psi_{R}$ are called left- and right-handed
        \termdef{Weyl spinors}

        \item They do not mix under Lorentz transformations\\
            %
            \note{They form the $(\hlf,0)$ and $(0,\hlf)$ projective
            irreps of the Lorentz group.  Note that the reducibility of the
            $(\hlf,0)\oplus(0,\hlf)$ bispinor representation is manifest in
            the Weyl basis:}
            %
            \note{%
            \begin{subequations}        
                \label{eq:generators_weyl}
                \begin{align}
                    S^{0i}&=\frac{i}{4}\com{\gamma^0}{\gamma^i}=
                    -\frac{i}{2}\begin{pmatrix}\sigma^i & 0 \\ 0 & -\sigma^i\end{pmatrix}
                    \quad\text{(Boosts, anti-Hermitian)}
                    \\
                    S^{ij}&=\frac{i}{4}\com{\gamma^i}{\gamma^j}
                    =\frac{1}{2}\vep^{ijk}\begin{pmatrix}\sigma^k & 0 \\ 0 & \sigma^k\end{pmatrix}
                    \quad\text{(Rotations, Hermitian)}
                \end{align}
            \end{subequations}
            }

        \item For $m=0$, the Dirac equation decouples into the \termdef{Weyl equations}:
            %
            \begin{align}
                i\ol\sigma\partial\,\psi_L=0    
                \quad\text{and}\quad
                i\sigma\partial\,\psi_R=0    
            \end{align}
            %
            \note{Solutions $\psi_R$ and $\psi_L$ are eigenstates of the
            helicity operator $h=\hat p\frac{\vec\sigma}{2}$ with $h=+\hlf$
            called \emph{right-handed} and $h=-\hlf$ \emph{left-handed}. Here,
            $\hat p =\vec p/E$ is the normalized 3-momentum for a massless
            particle.}

    \end{itemize}
\end{blocknote}


\section{Free-Particle Solutions of the Dirac Equation}

\note{Here we consider the Dirac equation as a wave equation for a \emph{single
particle}, or, equivalently, the classical field equation of a complex bispinor
field; what follows is therefore ``first-quantized'' quantum mechanics. We do
this because we need the eigenfunctions of the Dirac differential operator
to construct the field operators when we quantize the Dirac field (``second
quantization'').}

\imp{Detailed calculations: \seepset{3}.}

\begin{lot}

    \item \cref{eq:DE} \limplies $(\partial^2+m^2)\Psi=0$ \imp{(Klein-Gordon
    equation)}, therefore
        %
        \begin{align}
            \Psi^\pm(x)=\psi^\pm(p)e^{\mp ipx}    
            \quad\text{with}\;
            p^2=m^2\;
            \text{and}\;
            p^0>0
            \label{eq:DE_ansatz}
        \end{align}
        %
        \imp{Here $\psi^{\pm}(p)\in\complex^4$ is a complex-valued four-component bispinor.}\\
        %
        \imp{We set $p^0>0$ for both positive ($+$) and negative ($-$) frequency solutions
        and change the sign of $p$ in the exponent (to simplify the discussion below).}

    \item \cref{eq:DE_ansatz} in \cref{eq:DE} yields 
        %
        \begin{align}
            (\pm\g^\mu p_\mu-m)\psi^\pm(p)
            =\begin{pmatrix}-m & \pm p\sigma \\ \pm p\ol\sigma & -m\end{pmatrix}\bispi{\psi^\pm_L}{\psi^\pm_R}
            =0
            \label{eq:DE_1}
        \end{align}

    \item Note \imp{(\seepset{3})}:
        %
        \begin{itemize}

            \item $(p\sigma)(p\ol\sigma)=p^2=m^2$

            \item Eigenvalues of $p\sigma$ and $p\ol\sigma$: $p^0\pm|\vec p|$
            \therefore for $p^0>0$ and $m>0$ positive spectrum\\
                %
                \imp{In particular, $p\sigma$ and $p\ol\sigma$ are
                invertible and the positive square roots $\sqrt{p\sigma}$
                and $\sqrt{p\ol\sigma}$ are Hermitian.}

        \end{itemize}

    \item \consider $\psi^\pm_L=\sqrt{p\sigma}\,\xi^\pm$ with arbitrary, 
        normalized [$(\xi^\pm)^\dag\xi^\pm=1$] \Emph{spinor} $\xi^\pm\in\complex^2$:
        %
        \begin{align}
            &\text{\cref{eq:DE_1}}\;\Rightarrow
            -m\sqrt{p\sigma}\,\xi^\pm \pm p\sigma\,\psi^\pm_R=0
            \\
            &\stackrel{\sqrt{p\sigma}\sqrt{p\ol\sigma}=m}{\Leftrightarrow}\quad
            \psi^\pm_R=\pm\frac{m}{\sqrt{p\sigma}}\xi^\pm=\pm\sqrt{p\ol\sigma}\xi^\pm
        \end{align}
        %
        \imp{The second equation in \cref{eq:DE_1} yields the same solution.}

    \item \Emph{Solutions:}
        
        \imp{Conventional notation: $\xi^+\mapsto \xi$, $\xi^-\mapsto \eta$
        and $\psi^+\mapsto u$, $\psi^-\mapsto v$}\\
        %
        \imp{Basis states: $\xi^s$ with $\xi^1=\bispi{1}{0}$ and
        $\xi^2=\bispi{0}{1}$ (same for $\eta^s$)}
        %
        \begin{emphalign}
            \Psi^+(x)&=\underbrace{\bispi{\sqrt{p\sigma}\xi^s}{\sqrt{p\ol\sigma}\xi^s}}_{u^s(p)}e^{-ipx}
            \quad\text{(positive frequency solutions)}
            \\
            \Psi^-(x)&=\underbrace{\bispi{\sqrt{p\sigma}\eta^s}{-\sqrt{p\ol\sigma}\eta^s}}_{v^s(p)}e^{+ipx}
            \quad\text{(negative frequency solutions)}
        \end{emphalign}
        %
        with $p^2=m^2$, $p^0>0$ and $s=1,2$
        
        \imp{\therefore \emph{Four} linearly independent solutions for each
        3-momentum $\vec p$ ($\pm$ and $s=1,2$).}

    \item Some relations \imp{(\seepset{3})}:
        %
        \begin{itemize}

            \item \Emph{Orthonormality:}
                
                Let $\ol u^s\equiv (u^s)^\dag\gamma^0$ and $\ol v^s\equiv
                (v^s)^\dag\gamma^0$, then
                %
                \begin{eqaligned}[eq:uv]
                    \ol u^r u^s&=2m\,\delta^{rs}\quad\text{and}\quad(u^r)^\dag u^s=2\Ep{p}\,\delta^{rs}\\
                    \ol v^r v^s&=-2m\,\delta^{rs}\quad\text{and}\quad(v^r)^\dag v^s=2\Ep{p}\,\delta^{rs}\\
                    \ol v^r u^s&=\ol u^r v^s=0\\
                    u^{r\dag}(\vec p)v^s(-\vec p)&=v^{r\dag}(-\vec p)u^s(\vec p)=0
                \end{eqaligned}
                %
                \imp{Note that $\ol u u$ is Lorentz invariant whereas $u^\dag
                u\propto \Ep{p}$ is not!\\
                %
                Note that $(u^r)^\dag v^s\neq 0$ and $(v^r)^\dag u^s\neq 0$!\\
                %
                For massless particles, the normalization condition is given
                by $(u^r)^\dag u^s=2\Ep{p}\,\delta^{rs}$.}

            \item \Emph{Spin sums:}
                
                Let $\slashed p\equiv \gamma^\mu p_\mu$ (\termdef{Feynman
                slash notation}), then
                %
                \begin{eqaligned}[eq:uvsum]
                    \sum_s u^s(p)\ol u^s(p)&=\slashed p + m\Id\\
                    \sum_s v^s(p)\ol v^s(p)&=\slashed p - m\Id   
                \end{eqaligned}
                %
                \note{Useful if one wants to sum over spin-polarizations of
                fermions (\fref later).}

        \end{itemize}

\end{lot}


%%==============================================================================
\begin{lecture}
\item Dirac field bilinears
\item Fermi statistics and the quantization of the Dirac field
\end{lecture}
%%==============================================================================

%%==============================================================================
\reference{PS:49--62}
%%==============================================================================

\section{Dirac Field Bilinears}

\begin{lot}

    \item \Emph{Definition:}
        %
        \begin{emphalign}
            \gamma^5\equiv i\gamma^0\gamma^1\gamma^2\gamma^3
            \note{= -\frac{i}{4!}\vep^{\mu\nu\rho\sigma}\gamma_\mu\gamma_\nu\gamma_\rho\gamma_\sigma}
            \stackrel{\text{Weyl basis}}{=} \begin{pmatrix}-\mathds{1} & 0 \\ 0 & \mathds{1} \end{pmatrix}
        \end{emphalign}
        %
        with
        %
        \begin{align}
            (\gamma^5)^\dag=\gamma^5,    
            \quad
            (\gamma^5)^2=\mathds{1},
            \quad
            \acom{\gamma^5}{\gamma^\mu}=0
        \end{align}
        %
        \imp{The $\gamma^5$ matrix is labeled with ``5'' instead of
        ``4'' for historical reasons ($\gamma^0$ was sometimes called
        $\gamma^4$).  Important: $\gamma^5$ does \emph{not} belong to
        $\gamma^\mu=(\gamma^0,\gamma^1,\gamma^2,\gamma^3)$ but is just a
        name. Sums like $\gamma^\mu \partial_\mu$ only run over $\mu=0,1,2,3$.}

	    \note{The last relation implies $\com{\gamma^5}{S^{\mu\nu}}=0$,
	    i.e., the Dirac bispinor representation must be reducible
	    according to Schur's lemma: $(\hlf,0)\oplus(0,\hlf)$}

    \item The following \Emph{bilinears} $\ol\Psi\Gamma\Psi$ transform under
    the Lorentz group as \ldots
        %
        \begin{emphalign}
            \begin{array}{rlll}
                \Gamma = & 1 & \text{scalar} & \times 1\\
                & \gamma^\mu & \text{vector} & \times 4\\
                & \sigma^{\mu\nu}\equiv\frac{i}{2}\com{\gamma^\mu}{\gamma^\nu}=i\gamma^{[\mu}\gamma^{\nu]} & \text{tensor} & \times 6\\
                & \gamma^\mu\gamma^5& \text{pseudo-vector} & \times 4\\
                & \gamma^5& \text{pseudo-scalar} & \times 1
            \end{array}
            \label{eq:bilinears}
        \end{emphalign}
        %
        \imp{The notation $\gamma^{[\mu}\dots\gamma^{\nu]}$ denotes the
        completely antisymmetrized product.}
        
        \imp{Any $4\times 4$ matrix $\Gamma$ can be decomposed into these
        16 matrices with definite transformation properties under Lorentz
        transformations.}
        
        \imp{The prefix \emph{pseudo-} marks quantities that transform under
        continuous Lorentz transformations $\Lambda\in\POLG$ as usual but
        pick up an additional sign under parity transformations.}
        
        \note{This is similar to the cross product $a\times b$ in three
        dimensions which produces a \emph{pseudo-vector} from the two vectors
        $a$ and $b$ with respect to the Euclidean group (= isometries of
        Euclidean space). E.g., angular momentum $\vec L=\vec r\times\vec p$
        is not a vector but a pseudo-vector.}
        
        For example,
        %
        \begin{align}
            (j^\mu)'=\ol\Psi'\gamma^\mu\Psi'
            =\ol\Psi \Lambda_{\hlf}^{-1}\gamma^\mu \Lambda_{\hlf}\Psi
            =\tensor{\Lambda}{^\mu_\nu}\ol\Psi \gamma^\nu \Psi
            =\tensor{\Lambda}{^\mu_\nu}j^\nu
            \label{eq:current_transform}
        \end{align}
        %
        transforms as a Lorentz 4-vector.

	    \note{\easytoshow $j^\mu$ is the conserved Noether current
	    corresponding to the continuous symmetry $\Psi\to e^{i\alpha}\Psi$
	    of the Dirac Lagrangian.}

        \imp{Note that $\gamma^\mu$ does \emph{not} transform under Lorentz
        transformations [it has no prime in \cref{eq:current_transform}, recall
        our derivation of the gamma matrices in \cref{sec:dirac_equation}]. The
        tuple of gamma matrices $\gamma^\mu$ is not a Lorentz 4-vector (despite
        its Greek upper index!) but a fixed set of basis vectors in the Dirac
        algebra; $\gamma_\mu\equiv g_{\mu\nu}\gamma^\nu$ is defined as usual.}

\end{lot}


\section{Quantization of the Dirac Field}

\begin{lot}

    \item Lagrangian: $\L=\ol\Psi(i\gmu\pmu-m)\Psi$

    \item Canonical momentum: $\Pi_a=\pdiff{\L}{\dot\Psi_a}=i\Psi_a^*$

    \item Hamiltonian: $H=\int\dd{3}{x}\Psi^\dag\underbrace{[-i\vec\alpha\nabla+m\beta]}_{=H_D}\Psi$
        with $\vec\alpha=\gamma^0\vec\gamma$ and $\beta=\gamma^0$\\
        %
        \imp{$H_D$ is the Dirac Hamiltonian of single-particle quantum
        mechanics.}\\
        %
        \therefore Expand $\Psi$ in eigenmodes of $H_D$ to diagonalize $H$

    \item Eigenmodes: $H_Du^s(\vec p)e^{i\vec p \vec x}=\Ep{p}\rep$ and
    $H_Dv^s(\vec p)e^{-i\vec p \vec x}=-\Ep{p}\rep$\\
        %
        \note{This can be seen from
        $[i\gamma^0\partial_0+i\vec\gamma\nabla-m]\Psi=0$ (\bref last
        lecture).}

    \item Mode expansion:
        %
        \begin{align}
            \Psi(\vec x)=\sum_s\int\ddp{3}{p}\frac{1}{\sqrt{2\Ep{p}}}
            \,\left[a_{\vec p}^su^s(\vec p)e^{i\vec p\vec x}+b_{\vec p}^sv^s(\vec p)e^{-i\vec p\vec x}\right]
            \label{eq:D_me}
        \end{align}
        %
        \imp{$a^s_{\vec p}$ and $b^s_{\vec p}$ are operator-valued expansion
        coefficients. We do not yet fix their algebra!}

    \item Use
        %
        \begin{align}
            H_D\Psi(\vec x)=\sum_s\int\ddp{3}{p}\sqrt{\frac{\Ep{p}}{2}}
            \,\left[a_{\vec p}^su^s(\vec p)e^{i\vec p\vec x}-b_{\vec p}^sv^s(\vec p)e^{-i\vec p\vec x}\right]
        \end{align}
        %
        then \imp{(using the orthonormality relations \cref{eq:uv})}
        %
        \begin{align}
            H=\int\dd{3}{x}\Psi^\dag H_D\Psi\nte\sum_s\int\ddp{3}{p}\,\Ep{p}(
            a_{\vec p}^{s\dag}a_{\vec p}^{s}-b_{\vec p}^{s\dag}b_{\vec p}^{s})
            \label{eq:D_H}
        \end{align}
        %
        \imp{You do not need reordering of operators to show this. The
        algebra is still undefined!}

\end{lot}

\subsubsection{First try: Commutator}

\begin{lot}

    \setcounter{enumi}{6}

    \item Canonical quantization with equal-time \Emph{commutators}:
        %
        \begin{eqaligned}
            \com{\Psi_a(\vec x)}{\Pi_b(\vec y)} &= i\delta_{ab}\df{3}{\vec x-\vec y}
            \quad\Leftrightarrow\quad\com{\Psi_a(\vec x)}{\Psi_b^\dag(\vec y)} = \delta_{ab}\df{3}{\vec x-\vec y}\\
            \com{\Psi_a(\vec x)}{\Psi_b(\vec y)} &= 0
            \label{eq:cq_boson}
        \end{eqaligned}

    \item \easytoshow Mode algebra
        %
        \begin{align}
            \com{a^r_{\vec p}}{a^{s\dag}_{\vec q}} &= \com{b^r_{\vec p}}{b^{s\dag}_{\vec q}} 
            = (2\pi)^3\delta^{rs}\df{3}{\vec p-\vec q}\\
            \com{a^r_{\vec p}}{b^{s(\dag)}_{\vec q}} &= 0
        \end{align}
        %
        \imp{Show (using the mode expansion \cref{eq:D_me} and the spin
        sums \cref{eq:uvsum}) that this is equivalent to the commutators of
        the fields.}\\
        %
        \note{Beware: Eq. (3.89) of P\&S is mathematically ill-defined since
        $\Psi\Psi^\dag$ is a matrix but $\Psi^\dag\Psi$
            is not (it's just sloppy math that doesn't belong in a textbook
            for students).  Do it right, i.e., componentwise: $\com{\Psi_a(\vec
            x)}{\Psi^\dag_b(\vec y)}=\delta_{ab}\df{3}{\vec x-\vec y}$.}
        
        \therefore Irreducible Representation = \Emph{Bosonic} Fock space

    \item \Emph{Problem:} $\left(b^{s\dag}_{\vec p}\right)^n\ket{0}$ has
    energy $-n\Ep{p}\xrightarrow{n\to\infty} -\infty$\\
    %
    \therefore No stable vacuum state \note{(the spectrum of $H$ is unbounded
    below)}

    \item Fix (?): $b\leftrightarrow b^\dag$ \aside{(Use colors to modify
    the previous derivation.)}
        %
        \begin{lot}

            \item $\Psi(\vec x)=\dots [a_{\vec p}^s\dots+b_{\vec p}^{s\dag}\dots]$

            \item $H=\dots (a_{\vec p}^{s\dag}a_{\vec p}^{s}-b_{\vec p}^{s}b_{\vec p}^{s\dag})$

            \item $\com{b^r_{\vec p}}{b^{s\dag}_{\vec q}} = -(2\pi)^3\delta^{rs}\df{3}{\vec p-\vec q}$

            \item $H=\dots (a_{\vec p}^{s\dag}a_{\vec p}^{s}-b_{\vec p}^{s\dag}b_{\vec p}^{s})+\const$

            \item $\com{H}{b^{s\dag}_{\vec p}}=\Ep{p}b^{s\dag}_{\vec
            p}$ \therefore $b^{s\dag}_{\vec p}$ creates a particle with
            \emph{positive} energy!  \therefore $H\geq 0$\\
            %
            \imp{It \Emph{seems} that we solved the problem: The spectrum
            of the Hamiltonian is now bounded from below.}

            \item \Emph{But:}
                %
                \begin{align}
                    \|b^{s\dag}_{\vec p}\ket{0}\|^2
                    =\bra{0}\com{b^s_{\vec p}}{b^{s\dag}_{\vec p}}\ket{0}
                    =-(2\pi)^3\df{3}{0}<0
                \end{align}
                %
            \therefore Negative norm states \imp{(i.e., the constructed
            representation is not a Hilbert space)}

        \end{lot}

    \item \Emph{Conclusion:} \cref{eq:cq_boson} implies
        %
        \begin{itemize}

            \item either an instability of the vacuum

            \item or a loss of unitarity 

        \end{itemize}
        %
        \therefore No consistent quantization possible!

\end{lot}

\subsubsection{Second try: Anticommutator}

\begin{lot}

    \setcounter{enumi}{6}

    \item Canonical quantization with equal-time \Emph{anti}commutators:
        %
        \begin{emphalign}
            \acom{\Psi_a(\vec x)}{\Psi_b^\dag(\vec y)} = \delta_{ab}\df{3}{\vec x-\vec y}
            \quad\text{and}\quad
            \acom{\Psi_a(\vec x)}{\Psi_b(\vec y)} = 0
            \label{eq:cq_fermion}
        \end{emphalign}
        %
        \imp{Note that these are \emph{equal-time} anticommutators!}

    \item \easytoshow Mode algebra
        %
        \begin{emphalign}
            \acom{a^r_{\vec p}}{a^{s\dag}_{\vec q}} = \acom{b^r_{\vec p}}{b^{s\dag}_{\vec q}} 
            = (2\pi)^3\delta^{rs}\df{3}{\vec p-\vec q}
            \quad\text{and}\quad
            \acom{a^r_{\vec p}}{b^{s(\dag)}_{\vec q}} = 0
            \label{eq:fma}
        \end{emphalign}
        %
        \imp{The proof is similar to the bosonic case above.}
        
        \therefore Irreducible Representation = \Emph{Fermionic} Fock space

    \item \Emph{Problem:} $b^{s\dag}_{\vec p}\ket{0}$ has energy $-\Ep{p}$
    \& infinite sum over momenta\\
    %
    \therefore Still no stable vacuum state\\
    %
    \note{(The spectrum of $H$ is still unbounded below due to the sum
    over momenta.)}

    \item Fix (?): $b\leftrightarrow b^\dag$ \imp{(we saw above that it changes the sign of the excitation energies)}
        %
        \begin{enumerate}

            \item \Emph{Hamiltonian:}
                %
                \begin{emphalign}
                    H&=\sum_s\int\ddp{3}{p}\,\Ep{p}(a_{\vec p}^{s\dag}a_{\vec p}^{s}-b_{\vec p}^{s}b_{\vec p}^{s\dag})\\
                    &=\sum_s\int\ddp{3}{p}\,\Ep{p}(a_{\vec p}^{s\dag}a_{\vec p}^{s}+b_{\vec p}^{s\dag}b_{\vec p}^{s})-\infty
                    \label{eq:D_H2}
                \end{emphalign}
                %
                \imp{We will drop the infinite constant henceforth.}
                \aside{(Cross the $-\infty$.)}

            \item The mode algebra \cref{eq:fma} is \Emph{invariant} under
            $b\leftrightarrow b^\dag$!\\
                %
                \therefore Unitarity is preserved \Emph{and} Hamiltonian is
                bounded from below\\
                %
                \imp{\therefore With anticommutation relations, quantization
                is consistently possible!}

        \end{enumerate}

    \item \Emph{Heisenberg picture:}
    
	\imp{Now that we have a representation where the Hamiltonian generates
	a unitary time evolution, we can switch to the Heisenberg picture:}

    With
    %
    \begin{align}
        e^{iHt}a_{\vec p}^s e^{-iHt}\nte a_{\vec p}^s\,e^{-i\Ep{p}t}
        \quad\text{and}\quad
        e^{iHt}b_{\vec p}^s e^{-iHt}\nte b_{\vec p}^s\,e^{-i\Ep{p}t}
    \end{align}
    %
    and $\Psi(x)=e^{iHt}\Psi(\vec x)e^{-iHt}$ we find
    %
    \begin{emphalign}
        \Psi(x)
        &=\sum_s\int\ddp{3}{p}\frac{1}{\sqrt{2\Ep{p}}}\,
        \left[a_{\vec p}^su^s(p)e^{-ipx}+b_{\vec p}^{s\dag}v^s(p)e^{ipx}\right]
        \\
        \ol\Psi(x)
        &=\sum_s\int\ddp{3}{p}\frac{1}{\sqrt{2\Ep{p}}}\,
        \left[a_{\vec p}^{s\dag}\ol u^s(p)e^{ipx}+b_{\vec p}^{s}\ol v^s(p)e^{-ipx}\right]
    \end{emphalign}
    %
    \imp{These are operator-valued spinor fields, i.e., functions (more
    precisely: distributions) on Minkowski spacetime that assign to an event $x$
    a tuple (``spinor'') of operators that act on the fermionic Fock space where
    the states of the quantized theory live.}

\end{lot}

\subsubsection{Continuous symmetries \& Conserved charges}

\begin{itemize}

    \item Time translation \therefore Hamiltonian \imp{(see above)}

    \item Spatial translations \therefore Momentum operator
        %
        \begin{emphalign}
            \vec P\nte\int\dd{3}{x}\Psi^\dag(-i\nabla)\Psi
            \nte\sum_s\int\ddp{3}{p}\,\vec p\,(a_{\vec p}^{s\dag}a_{\vec p}^{s}+b_{\vec p}^{s\dag}b_{\vec p}^{s})
        \end{emphalign}

    \item Rotations \therefore Angular momentum operator $\vec J$
        \note{%
            \begin{align}
                \vec J=\int\dd{3}{x}\Psi^\dag\left[\vec x\times (-i\nabla)+\hlf\vec\Sigma\right]\Psi
                \quad\text{with}\quad \vec \Sigma = \begin{pmatrix}\vec\sigma & 0 \\ 0 & \vec\sigma\end{pmatrix}
                \label{eq:J}
            \end{align}
        }

    \item Global phase rotations $e^{i\alpha}\Psi$\\
    %
    \easytoshow Conserved current $j^\mu=\ol\Psi\gmu\Psi$\\
    %
    \therefore Conserved charge:
        %
        \begin{emphalign}
            Q=\int\dd{3}{x}\Psi^\dag\Psi
            &\nte\sum_s\int\ddp{3}{p}(a_{\vec p}^{s\dag}a_{\vec p}^{s}+b_{-\vec p}^{s}b_{-\vec p}^{s\dag})
            \\
            &=\sum_s\int\ddp{3}{p}(a_{\vec p}^{s\dag}a_{\vec p}^{s}-b_{\vec p}^{s\dag}b_{\vec p}^{s})+\infty
        \end{emphalign}
        %
        \imp{In QED we will couple the fermions to the EM field; then, $Q$
        is the total EM charge of the fermion field.}
        
        \note{Recall that in \emph{single-particle quantum mechanics} the
        global phase rotation symmetry gives rise to a positive density
        and a current that can be interpreted as probability current; the
        conserved charge corresponds then to the total probability to find
        the single particle somewhere. Because of the normal ordering (=
        dropping the infinite constant) this interpretation does no longer
        apply as $Q$ can become negative.}

\end{itemize}
%
\note{The operators of conserved charges generate symmetry transformations
of the Hamiltonian.}

\subsubsection{Excitations = Particles}

\begin{emphalign}
    \begin{array}{rll}
        a_{\vec p}^{s\dag}\ket{0}: &\text{\Emph{Fermion} with} 
        &\text{energy $\Ep{p}$,}\\
        &&\text{momentum $\vec p$,} \\
        &&\text{spin $J=\hlf$ (polarization $s$),}\\
        &&\text{and charge $Q=+1$}
        \\[10pt]
        b_{\vec p}^{s\dag}\ket{0}: &\text{\Emph{Antifermion} with} 
        &\text{energy $\Ep{p}$,}\\
        &&\text{momentum $\vec p$,} \\
        &&\text{spin $J=\hlf$ (polarization opposite to $s$),}\\
        &&\text{and charge $Q=-1$}
    \end{array}
\end{emphalign}
%
\imp{In QED, the fermions will be \emph{electrons} and the antifermions
\emph{positrons}.}

\begin{blocknote}
\note{%
    \begin{itemize}

        \item The two states for $s=1,2$ \emph{suggest} a spin-$\hlf$
        representation

        \item To show this, the action of $\vec J$ (see \cref{eq:J}) on
        one-particle states must be studied

        \item One finds for particles at rest:
            %
            \begin{align}
                J_z\,a_{\vec 0}^{s\dag}\ket{0}=\pm\hlf a_{\vec 0}^{s\dag}\ket{0}
                \quad\text{and}\quad
                J_z\,b_{\vec 0}^{s\dag}\ket{0}=\mp\hlf b_{\vec 0}^{s\dag}\ket{0}
            \end{align}
            %
            with $\xi^{s=1}=\bispi{1}{0}$ and $\xi^{s=2}=\bispi{0}{1}$.

    \end{itemize}
}
\end{blocknote}

\subsubsection{Lorentz transformations}

\begin{lot}

    \item \consider Lorentz transformation $\Lambda\in\POLG$ on single
    particle state $\ket{\vec p,s}_a\equiv \sqrt{2\Ep{p}}\,a_{\vec p}^{s\dag}\ket{0}$:\\
    %
    \imp{(The subscript $a$ ($b$) denotes the state of a(n) (anti-)fermion;
    we omit it, $\ket{\vec p,s}\equiv\ket{\vec p,s}_a$, when the distinction
    is not important.)}
        %
        \begin{align}
            \ket{\vec p,s}\mapsto U(\Lambda)\ket{\vec p,s}    
        \end{align}
        %
        $U(\Lambda)$: representation of $\POLG$ on Fock space\\
        %
        \imp{For generic rotations/boosts, this mixes the two spin components!}

    \item \consider \Emph{Special case:} quantization axis parallel to boost and/or
    rotation axis
        
        \therefore Spin polarizations do not mix:
        %
        \begin{align}
            U(\Lambda)\,a_{\vec p}^s\,U^{-1}(\Lambda)=\sqrt{\frac{E_{\Lambda\vec p}}{\Ep{p}}}\,a_{\Lambda \vec p}^s
        \end{align}
        %
	    \note{Note that spins mix under generic Lorentz transformations:
	    $a_{\vec p}^1\leftrightarrow a_{\vec q}^2$.}

	% \note{Note that particles and antiparticles never mix under Lorentz
	%     transformations, $a_{\vec p}^s\nleftrightarrow b_{\vec q}^r$, but
	%     spins generally do, $a_{\vec p}^1\leftrightarrow a_{\vec q}^2$.}

    \item \imp{Consider this special case, then:}
        %
        \begin{align}
            \braket{\vec p,s}{\vec q,r}
            =\underbrace{2\Ep{p}(2\pi)^3\df{3}{\vec p-\vec q}\delta^{rs}}_{\text{Lorentz invariant}}
            =\bra{\vec p,s}U^\dag(\Lambda)U(\Lambda)\ket{\vec q,r}
        \end{align}
        %
        \therefore $U(\Lambda)$ is \Emph{unitary}

    \item Now we have 3 representations:
        %
        \begin{emphalign}
            \begin{array}{rlll}
                \Lambda & \text{acts on 4-vectors in $\reals^{1,3}$} & D=4 & \text{not unitary}\\
                \Lambda_{\hlf} & \text{acts on bispinors in $\complex^{2}\oplus\complex^{2}$} & D=4 & \text{not unitary}\\
                U(\Lambda) & \text{acts on states in fermionic Fock space} & D=\infty & \text{unitary}
            \end{array}
        \end{emphalign}

    \item \imp{Action by conjugation on field operators} \easytoshow
        %
        \begin{emphalign}
            U(\Lambda)\Psi(x)U^{-1}(\Lambda)=\Lambda^{-1}_\hlf\Psi(\Lambda x)    
            \label{eq:UPsiU}
        \end{emphalign}

\end{lot}

%%==============================================================================
\begin{lecture}
\item The spin-statistics theorem
\item The Dirac propagator
\item Causality
\item Discrete symmetries of the Dirac theory
\end{lecture}
%%==============================================================================

%%==============================================================================
\begin{problemset}
\item The relativistic hydrogen atom
\item Parity transformation of Dirac spinors
\end{problemset}
%%==============================================================================

%%==============================================================================
\reference{PS:62--71}
%%==============================================================================

\subsubsection{Spin-statistics theorem}

\begin{itemize}

    \item \Emph{Observation:}
        %
        \begin{align}
            \begin{array}{rlll}
                \text{Klein-Gordon field $\phi$:} &\text{Spin 0 (scalar)} 
                &\text{\therefore commutator} &\text{\therefore bosonic excitations}\\
                \text{Dirac field $\Psi$:} &\text{Spin $\hlf$ (spinor)} 
                &\text{\therefore anticommutator} &\text{\therefore fermionic excitations}
            \end{array}
        \end{align}
        %
        \imp{This is no coincidence but hints at a more fundamental
        connection:}

    \item \Emph{Spin-statistics theorem:}
        %
        \begin{emphalign}
            \left.
            \begin{array}{r}
                \text{Lorentz invariance}\\
                \text{Causality}\\
                \text{Positive energies}\\
                \text{Positive norms}
            \end{array}
            \right\}
            \Rightarrow
            \left\{
            \begin{array}{rcl}
                \text{Integer spin} &\leftrightarrow &\text{Bosons}\\
                \text{Half-integer spin} &\leftrightarrow &\text{Fermions} 
            \end{array}
            \right.
        \end{emphalign}
        %
        \imp{This means, whenever you quantize a relativistic field that
        transforms under a (projective) half-integer spin representation,
            the Poisson bracket must be replaced by \emph{anti}commutators.
            Otherwise unitarity is lost or the vacuum becomes unstable.}

        \item %
        \begin{samepage}
        \Emph{``Proof by picture'':}%
        %
        \begin{center}
            \includegraphics[width=0.9\linewidth]{spin-statistics}
        \end{center}
        %
        \imp{Rigorous proofs are elaborate and quite technical.}
        
        \note{\uref\url{http://math.ucr.edu/home/baez/spin_stat.html}}
        \end{samepage}
   
\end{itemize}


\subsubsection{Dirac Propagator}

\imp{All that follows is very similar to our discussion of the Klein-Gordon propagator.\\
For details, we refer the student to the corresponding notes.}

\begin{lot}

    \item Propagation amplitudes \aside{(use colors to skip this calculation)}:
        %
        \begin{align}
            \bra{0}\Psi_a(x)\ol\Psi_b(y)\ket{0}
            &=\int\ddp{3}{p}\frac{1}{2\Ep{p}}e^{-ip(x-y)}\underbrace{\sum_s u^s_a(p) \ol u^s_b(p)}_{(\slashed{p}+m)_{ab}}
            \\
            &=\note{(i\slashed{\partial}_x+m)_{ab}D(x-y)}
            \\
            &\stackrel{x^0>y^0}{=}\int\ddp{4}{p}\frac{i(\slashed{p}+m)_{ab}}{p^2-m^2+i\vep}e^{-ip(x-y)}
        \end{align}
        %
        \begin{align}
            \bra{0}\ol\Psi_b(y)\Psi_a(x)\ket{0}
            &=\int\ddp{3}{p}\frac{1}{2\Ep{p}}e^{-ip(y-x)}\underbrace{\sum_s v^s_a(p) \ol v^s_b(p)}_{(\slashed{p}-m)_{ab}}
            \\
            &=\note{-(i\slashed{\partial}_x+m)_{ab}D(y-x)}
            \\
            &\stackrel{x^0<y^0}{=}-\int\ddp{4}{p}\frac{i(\slashed{p}+m)_{ab}}{p^2-m^2+i\vep}e^{-ip(x-y)}
        \end{align}
        %
        \note{%
        Remember:
        %
        \begin{center}
            \includegraphics[width=0.75\linewidth]{tikz/feynman_contour}
        \end{center}
        %
        \begin{itemize}
            \item $x^0>y^0$: close contour below
            \item $x^0<y^0$: close contour above
        \end{itemize}
        }

    \item \imp{Therefore we define the} \Emph{Feynman propagator} \imp{of the Dirac field}:
        %
        \begin{emphalign}
            S^{ab}_F(x-y)&=\int\ddp{4}{p}\frac{i(\slashed{p}+m)_{ab}}{p^2-m^2+i\vep}e^{-ip(x-y)}
            \\
            &=
            \begin{cases}
                \phantom{-}\bra{0}\Psi_a(x)\ol\Psi_b(y)\ket{0} & \text{for}\;x^0>y^0\\ 
                -\bra{0}\ol\Psi_b(y)\Psi_a(x)\ket{0} & \text{for}\;x^0<y^0
            \end{cases}
            \\
            &\equiv \bra{0}\T\Psi_a(x)\ol\Psi_b(y)\ket{0}
        \end{emphalign}
        %
        \Emph{Note:} For $t_1>t_2$ it is
        $\T\Psi(t_2)\Psi(t_1)\equiv-\Psi(t_1)\Psi(t_2)$ for fermionic fields!
        
        \imp{The Feynman propagator $S_F(x-y)$ of the Dirac field is a
        $4\times 4$ matrix.}

    \note{%
    \item Similarly, one can derive the \emph{Retarded Green's function}:
        %
        \begin{align}
            S_R^{ab}(x-y)\equiv \theta(x^0-y^0)\,\bra{0}\acom{\Psi_a(x)}{\ol\Psi_b(y)}\ket{0}    
            \nte (i\slashed{\partial}_x+m)_{ab}D_R(x-y)
        \end{align}
        %
	    Here, $D_R(x-y)$ is the retarded Green's function of the
	    Klein-Gordon field;\\
        %
	    $\slashed\partial_x$ denotes derivatives with respect to the
	    variables $x^\mu$ for $\mu=0,1,2,3$ and generates the $\slashed{p}$
	    in the integral.
    }

\end{lot}

\subsubsection{Causality}

\begin{lot}

    \item \Emph{Measurable operators}: 
        %
        $\hat O(x)=\sum\prod_{i=1}^{\text{even}\,N}(\Psi_i^{(\dag)}(x)\;\vee\;
        \partial\Psi_i^{(\dag)}(x)\;\vee\;\partial^2\Psi_i^{(\dag)}(x)\;\dots)$
        
        Example: $j^\mu=\ol\Psi\gmu\Psi$ \imp{(check that this is Hermitian!)} 
        (but \Emph{not} $\Psi_a+\Psi_a^\dag$!)
        
        \imp{Restricting observables to field polynomials of even degree
            ensures that space-like separated observables \emph{commute}
            if space-like separated fields \emph{anticommute} (which is the
            best we can hope for given our quantization conditions).}
        
        \note{The answer to the question \emph{``Why restrict observables
        to even degree expressions in the fields?''} is therefore: Because
        these are the only observables that do not violate causality in a
        theory built from fermionic fields. (There are also more rigorous
        arguments for this: \uref Superselection.)}

    \item Causality \imp{for fermionic fields} $\Leftrightarrow$
    $\acom{\Psi_a(x)}{\ol\Psi_b(y)}=0$ for $(x-y)^2<0$\\
        %
        \imp{All other anticommutators vanish trivially.
            Note that here $x=(t,\vec x)$ and $y=(t',\vec y)$, i.e., we
            consider the anticommutator at \emph{different} times.}
            
            We find \imp{(using results from above)}
            %
            \begin{align}
                \acom{\Psi_a(x)}{\ol\Psi_b(y)}
                &\nte(i\slashed{\partial}_x+m)_{ab}[D(x-y)-D(y-x)]\\
                &\mimp{$(x-y)^2<0$}\\
                &=(i\slashed{\partial}_x+m)_{ab}[D(x-y)-D(x-y)]=0
            \end{align}
            %
            \imp{The argument is the same as for the Klein-Gordon field.}\\
            %
            \note{Recall: $D(x-y)=\int\ddp{3}{p}\frac{1}{2\Ep{p}}e^{-ip(x-y)}$}
            
\end{lot}


\section{Discrete Symmetries of the Dirac Theory}

\subsubsection{Review of the Lorentz group}

\imp{Details: \seepset{3}}

\begin{itemize}

    \item \imp{Lorentz group $\LG$ = Lie group with four disconnected components}

    \item \imp{Continuous Lorentz transformations = \emph{Proper orthochronous
        Lorentz group} $\POLG$}

    \item Four components connected by discrete transformations:
        %
        \begin{align}
            \text{Parity}\quad P&:\,(t,\vec x)\mapsto (t,-\vec x)\\
            \text{Time reversal}\quad T&:\,(t,\vec x)\mapsto (-t,\vec x) 
        \end{align}
        %
        \begin{center}
            \includegraphics[width=1.0\linewidth]{tikz/structure_lorentz_group}
        \end{center}

\end{itemize}

% \begin{sidewaysfigure}[p]
%     \begin{center}
%         \includegraphics[width=0.9\linewidth]{poincare-group}
%     \end{center}
% \end{sidewaysfigure}


\subsubsection{Parity}

\imp{Details: \seepset{4}}

\begin{lot}

    \item Unitary representation on Fock space:
        %
        \begin{align}
            U(P)\,a_{\vec p}^s\,U^{-1}(P)=\underbrace{\eta_a}_{+1}\,a_{-\vec p}^s    
            \quad\text{and}\quad
            U(P)\,b_{\vec p}^s\,U^{-1}(P)=\underbrace{\eta_b}_{-1}\,b_{-\vec p}^s    
        \end{align}
        %
        \imp{Note that we do \emph{not} want spin to change under $P$ because
        angular momentum $\vec L=\vec r\times\vec p$ also does not pick up
        a sign under inversion (it is a pseudo-vector).}\\[5pt]
        %
        \note{Note that often $U(P)$ is simply written $P$.}

    \item Equivalent to
        %
        \begin{emphalign}
            U(P)\Psi(t,\vec x)U^{-1}(P)=\underbrace{\gamma^0}_{P_\hlf}\Psi(\underbrace{t,-\vec x}_{Px})
            \label{eq:parity_dirac}
        \end{emphalign}
        %
        \note{The $\gamma^0$-matrix exchanges the left- and right-handed Weyl
        sectors of the bispinor; this makes sense as a parity transformation
        of space should switch chirality.}

    \item \note{%
        Dirac field bilinears (examples):
        %
        \begin{alignat}{2}
            U(P)\ol\Psi\Psi U^{-1}(P)&\nte +\ol\Psi\Psi(t,-\vec x) &&\quad\rightarrow\text{scalar}\\
            U(P)\ol\Psi\gamma^5\Psi U^{-1}(P)&\nte -\ol\Psi\gamma^5\Psi(t,-\vec x) &&\quad\rightarrow\text{pseudo-scalar}
        \end{alignat}
        %
        % \item Dirac Lagrangian $\L_D=\ol\f(i\slashed{\partial}-m)\f$ is parity-symmetric: $U(P)\L_D U^{-1}(P)=\L_D$
        }

\end{lot}


\subsubsection{Time Reversal}

\begin{lot}

    \item Time reversal should \dots
        %
        \begin{itemize}

            \item $U(T)\Psi(t,\vec x)U^{-1}(T)= T_\hlf\Psi(-t,\vec x)$,

            \item $U(T)a^s_{\vec p}U^{-1}(T)=a^?_{-\vec p}$,

            \item flip spins \imp{(motivated by $\vec L=\vec r\times\vec p\mapsto -\vec L$)},

            \item be a symmetry of the Dirac theory: $\com{U(T)}{H}=0$,

            \item obey $U^{-1}(T)=U^\dag(T)$.\\
                %
                \imp{This is required for any symmetry to preserve overlaps: \uref Wigner's theorem.}

        \end{itemize}
        %
        \note{Note that often $U(T)$ is simply written $T$.}

    \item \Emph{Problem:}
        %
        \begin{alignat}{2}
            &&\Psi(t,\vec x) &= e^{iHt}\Psi(\vec x) e^{-iHt}\\    
            &\Rightarrow\quad
            &U(T)\Psi(t,\vec x)U^{-1}(T) &= e^{iHt}U(T)\Psi(\vec x)U^{-1}(T)e^{-iHt}\\    
            &\Rightarrow\quad
            &T_\hlf\Psi(-t,\vec x)\ket{0} &= e^{iHt}T_\hlf\Psi(\vec x)\ket{0}\\
            &\Rightarrow\quad
            &T_\hlf e^{-iHt}\Psi(\vec x)\ket{0} &= e^{iHt}T_\hlf\Psi(\vec x)\ket{0}\\
            &\Rightarrow\quad
            &\underbrace{e^{-2iHt}}_{\text{time-dependent!}}T_\hlf\Psi(\vec x)\ket{0} &= T_\hlf\Psi(\vec x)\ket{0}
        \end{alignat}
        %
        \imp{Here we used that $\com{U(T)}{H}=0$ and $H\ket{0}=0$.}
        
        \therefore Not possible \imp{(for invertible $T_\hlf$ and arbitrary
        times $t$)}!

    \item \Emph{Solution:} $U(T)$ must be \Emph{antiunitary/antilinear}:
        %
        \begin{emphalign}
            U(T)c=c^*U(T)    
            \quad\text{for}\;c\in\complex
            \label{eq:antiu}
        \end{emphalign}
        %
        \note{The relation \cref{eq:antiu} makes $U$
        \emph{antilinear}; anti\emph{unitarity} means that in addition
        $\braket{U\psi}{U\phi}=\braket{\psi}{\phi}^\ast$ for all states $\psi$
        and $\phi$. Antiunitary operators can be written as $U=V\mathcal{K}$
        where $V$ is a unitary operator and $\mathcal{K}$ denotes complex
        conjugation.}

        \aside{Highlight the differences with colors in the derivation above:
        %
        \begin{alignat}{2}
            &&\Psi(t,\vec x) &= e^{iHt}\Psi(\vec x) e^{-iHt}\\    
            &\Rightarrow\quad
            &U(T)\Psi(t,\vec x)U^{-1}(T) &= e^{-iHt}U(T)\Psi(\vec x)U^{-1}(T)e^{iHt}\\    
            &\Rightarrow\quad
            &T_\hlf\Psi(-t,\vec x)\ket{0} &= e^{-iHt}T_\hlf\Psi(\vec x)\ket{0}\\
            &\Rightarrow\quad
            &T_\hlf e^{-iHt}\Psi(\vec x)\ket{0} &= e^{-iHt}T_\hlf\Psi(\vec x)\ket{0}\\
            &\Rightarrow\quad
            &\underbrace{\Id}_{\text{time-independent!}}T_\hlf\Psi(\vec x)\ket{0} &= T_\hlf\Psi(\vec x)\ket{0}
        \end{alignat}
        }

    \item Transformation of \Emph{spin}:
        %
        \begin{lot}

            \item Spinors: \consider Spin basis $\xi^s$ ($s=1,2$) along arbitrary axis $\vec n$:
                %
                \begin{align}
                    \xi^1=\bispi{\cos\frac{\theta}{2}}{e^{i\phi}\sin\frac{\theta}{2}} 
                    \quad\text{and}\quad
                    \xi^2=\bispi{-e^{-i\phi}\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}
                \end{align}
                %
                \note{That is, $\xi^1=\ket{\uparrow}$ and $\xi^2=\ket{\downarrow}$.}\\
                %
                ``Time-reversed'' (=flipped) spinors:
                %
                \begin{align}
                    \ol{\xi^s}\equiv -i\sigma^2(\xi^s)^*    
                    \quad\Rightarrow\quad
                    \bicase{\ol{\xi^{1}}}{\ol{\xi^{2}}}
                    =\bicase{\xi^{2}}{-\xi^{1}}
                    \label{eq:xims}
                \end{align}
                %
                \note{%
                    Indeed, if $\vec n\cdot\vec\sigma\,\xi=+\xi$, we have
                    %
                    \begin{align}
                       \vec n\cdot\vec\sigma\,(-i\sigma^2\xi^*)
                       =-i\sigma^2(-\vec n\cdot\vec\sigma)^*\xi^*
                       =i\sigma^2(\xi^*)
                       =-(-i\sigma^2\xi^*)
                    \end{align}
                    %
                    where we used $\vec\sigma\sigma^2=\sigma^2(-\vec\sigma^*)$.\\[10pt]
                    %
                    Note that $T=-i\sigma^2K$ (where $K$ denotes complex
                    conjugation) is the conventional representation of
                    time-reversal symmetry for spinful fermions that you
                    might know from condensed matter physics (e.g., to
                    classify symmetry-protected topological phases).
                }

            \item Bispinors:
                %
                \begin{eqaligned}[eq:us]
                    u^s(p)\equiv\bispi{\sqrt{p\sigma}\,\xi^s}{\sqrt{p\ol\sigma}\,\xi^s}
                    \quad&\text{and}\quad
                    v^s(p)\equiv\bispi{\sqrt{p\sigma}\,\ol{\xi^{s}}}{-\sqrt{p\ol\sigma}\,\ol{\xi^{s}}}
                    \\
                    \ol{u^s}(p)\equiv\bispi{\sqrt{p\sigma}\,\ol{\xi^s}}{\sqrt{p\ol\sigma}\,\ol{\xi^s}}
                    \quad&\text{and}\quad
                    \ol{v^s}(p)\equiv\bispi{\sqrt{p\sigma}\,\ol{\ol{\xi^{s}}}}{-\sqrt{p\ol\sigma}\,\ol{\ol{\xi^{s}}}}
                \end{eqaligned}
                %
                \aside{Use colors to skip the second row.}
                
                \imp{Note that here $\ol{u^s}$ is \emph{not} the Dirac
                adjoint $\ol{u}^s$!}\\
                %
                \imp{Recall that the basis $\eta^s$ in the definition of
                $v^s(p)$ was arbitrary.}

            \item Define the modes:
                %
                \begin{align}
                    \bicase{\ol{\ap{p}^{1}}}{\ol{\ap{p}^{2}}}
                    \equiv
                    \bicase{\ap{p}^{2}}{-\ap{p}^{1}}
                    \quad\text{and}\quad
                    \bicase{\ol{\bp{p}^{1}}}{\ol{\bp{p}^{2}}}
                    \equiv
                    \bicase{\bp{p}^{2}}{-\bp{p}^{1}}
                    \label{eq:ams}
                \end{align}
                %
                \aside{Skip the second part.}
                
                \note{Note that this is analog to \cref{eq:xims}!}

            \item Let $\tilde p\equiv (p^0,-\vec p)$ and show
                %
                \begin{eqaligned}[eq:ums]
                    \ol{u^{s}}(\tilde p)&\nte -\gamma^1\gamma^3[u^s(p)]^*\\
                    \ol{v^{s}}(\tilde p)&\nte -\gamma^1\gamma^3[v^s(p)]^*
                \end{eqaligned}
                %
                Note: $\ol{\ol{\xi^{s}}}=-\xi^s$ used in $\ol{v^s}$
                
                \imp{Use \cref{eq:us} and $\sqrt{\tilde
                p\sigma}\,\sigma^2=\sigma^2\,\sqrt{p\sigma^*}$ to show this!}
            
        \end{lot}

    \item \Emph{Definition:}
        %
        \begin{emphalign}
            \left.
            \begin{array}{r}
            \text{Antilinearity \cref{eq:antiu}}\\[5pt]
            U(T)\,a_{\vec p}^s U^{-1}(T)\equiv \ol{a^{s}_{-\vec p}}\\[5pt]
            U(T)\,b_{\vec p}^s U^{-1}(T)\equiv \ol{b^{s}_{-\vec p}}    
            \end{array}
            \right\}
            \Rightarrow
            \begin{aligned}[c]
                &U(T)\Psi(t,\vec x)U^{-1}(T)\\
                \nte &\underbrace{(\gamma^1\gamma^3)}_{T_\hlf}\,\Psi(-t,\vec x)
            \end{aligned}
        \end{emphalign}
        %
        \imp{Use \cref{eq:ums} and \cref{eq:ams} and
        $\ol{\ap{p}^2}\,\ol{u^2}(p)=\ap{p}^1 u^1(p)$ etc.\ to show this!}
        
        \note{%
            Note that in Weyl representation
            %
            \begin{align}
                T_{\hlf}=\begin{pmatrix}i\sigma^y & 0 \\ 0 & i\sigma^y\end{pmatrix}
            \end{align}
            %
            i.e., time-reversal acts on spins but does not mix chiralities (as
            parity did). This makes sense, because (for massless particles)
            chirality = helicity and helicity is the projection of spin on
            momentum: $\vec S\cdot\vec p$. Since both spin (angular momentum)
            $\vec S$ \emph{and} linear momentum $\vec p$ change sign under
            time-reversal, helicity does not.
        }

    \note{%
    \item Dirac field bilinears (example: $j^\mu=\ol\Psi\gmu\Psi$):
        %
        \begin{align}
            U(T)j^\mu(t,\vec x)U^{-1}(T)&\nte 
            \begin{cases}
                +j^\mu(-t,\vec x) &\text{for}\;\mu=0\\
                -j^\mu(-t,\vec x) &\text{for}\;\mu=1,2,3
            \end{cases}
        \end{align}
        %
        \therefore As expected for density ($\mu=0$) and 3-current ($\mu=1,2,3$)
    }

\end{lot}


\subsubsection{Charge Conjugation}

\begin{lot}

    \item Discrete, \Emph{non-spacetime} symmetry that exchanges particle and antiparticle:
        %
        \begin{emphalign}
            U(C)\aps{p}{s}U^{-1}(C)=\bps{p}{s}
            \quad\text{and}\quad
            U(C)\bps{p}{s}U^{-1}(C)=\aps{p}{s}
        \end{emphalign}
        %
        \imp{Note that there is no representation on Minkowski space as this
        is an ``internal'' symmetry.}
        
        \note{Often $U(C)$ is simply written $C$.}

    \item \imp{Use \cref{eq:us} to show:}
        %
        \begin{align}
            u^s(p)\nte-i\g^2(v^s(p))^*    
            \quad\text{and}\quad
            v^s(p)\nte-i\g^2(u^s(p))^*    
        \end{align}

    \item Then
        %
        \begin{align}
            &U(C)\Psi(x)U^{-1}(C)
            \\
            &\color{notecolor}=\sum_s\int\ddp{3}{p}\frac{1}{\sqrt{2\Ep{p}}}
            \left[
                -i\g^2(v^s(p))^*\bps{p}{s} e^{-ipx}
                -i\g^2(u^s(p))^*\adps{p}{s} e^{ipx}
            \right]
            \\
            &\nte-i\g^2(\Psi^\dag)^T=-i(\ol\Psi\g^0\g^2)^T
        \end{align}

    \item Therefore:
        %
        \begin{emphalign}
            U(C)\Psi U^{-1}(C)&=-i(\ol\Psi\g^0\g^2)^T\note{(=\underbrace{-i\gamma^2}_{\equiv C_\hlf}\f^*)}
            \label{eq:charge_conjugation_def}\\
            \quad\text{and}\quad
            U(C)\ol\Psi U^{-1}(C)&\nte-i(\g^0\g^2\Psi)^T
        \end{emphalign}
        %
        \begin{itemize}

        \item \imp{Note that $C$ essentially exchanges $\Psi\leftrightarrow\ol\Psi$
        but is not antiunitary!}
        
        \item \note{To show this, recall that $\g^0$ and $\g^2$ are symmetric
        matrices.}
        
        \item \note{It is $C_\hlf^\dag=C_\hlf$, $C_\hlf^2=\Id$ and $C_\hlf\gmu
        C_\hlf=-{\gamma^\mu}^*$.}
        
        \item \note{%
        Note that the expression in parantheses is only true for the
        transformation of \emph{classical} (i.e. ``first quantized'') Dirac
        fields and can be used to show the symmetry of the classical Dirac
        equation. However, if you take the $^\ast$ to conjugate complex numbers
        \emph{and} Hilbert space operators, $\Psi^*\equiv(\Psi^\dag)^T$,
        it is valid for the quantized field as well.
        }

        \end{itemize}

    \note{%
    \item Dirac field bilinears (examples):
        %
        \begin{alignat}{2}
            U(C)\ol\Psi\Psi U^{-1}(C)&\nte \ol\Psi\Psi &&\quad\text{(Scalar)}\\
            U(C)\ol\Psi\gmu\Psi U^{-1}(C)&\nte -\ol\Psi\gmu\Psi &&\quad\text{(Vector)}
        \end{alignat}
    }

\end{lot}

\begin{blocknote}
    \begin{itemize}

        \item Any relativistic QFT must be invariant under $\POLG$ \note{($=L^\uparrow_+$)}

        % \item The free Dirac Lagrangian $\L_D=\ol\Psi(i\gmu\pmu-m)\Psi$ is $\{C,P,T\}$-invariant

        \item The (classical) Dirac equation $(i\gmu\pmu-m)\Psi=0$ is
        $\{C,P,T\}$-invariant

        \item The (quantized) Dirac theory is $\{C,P,T\}$-invariant:\\
        %
        $\com{H}{U(X)}=0$ for $X=P,T,C$

        \item Weak interactions \imp{(of the standard model)} violate $C$
        and $P$ but preserve $CP$ and $T$ \imp{(\uref Wu experiment)}

        \item Rare processes \imp{(decay of neutral kaons)} violate $CP$
        and $T$ but preserve $CPT$

        \item $CPT$ seems to be a perfect symmetry of nature

        \item \Emph{$CPT$ theorem}:
        %
        \begin{align}
            \left.
            \begin{array}{r}
                \text{$\POLG$ invariance}\\
                \text{Causality}\\
                \text{Locality}\\
                \text{Stable vacuum}
            \end{array}
            \right\}
            \Rightarrow
            \text{$CPT$ symmetry}
        \end{align}

    \end{itemize}
\end{blocknote}


