%%==============================================================================
\begin{lecture}
\item Canonical quantization of the Klein-Gordon field
\item Heisenberg picture: Time evolution of the quantized Klein-Gordon field
\end{lecture}
%%==============================================================================

%%==============================================================================
\begin{problemset}
\item The classical complex Klein-Gordon field
\item The quantized complex Klein-Gordon field
\end{problemset}
%%==============================================================================

%%==============================================================================
\reference{PS:19--26}
%%==============================================================================


\chapter{The Klein-Gordon Field}

\section{Canonical Quantization}

\begin{lot}

    \item \Emph{Theory:}
        %
        \begin{enumerate}
            \item \Emph{Real} field $\phi(x)$ \imp{(\seepset{2} for the complex analog)}
            \item Lagrangian: $\L=\hlf(\pmu\phi)^2-\hlf m^2\phi^2$ \imp{(\termdef{free scalar field})}
            \item EOM: $(\partial^2+m^2)\phi=0$ \imp{(Klein-Gordon equation)}
            \item Hamiltonian: $\H=\hlf\pi^2+\hlf(\nabla\phi)^2+\hlf m^2\phi^2$
        \end{enumerate}

    \item \Emph{Canonical quantization:}
        %
        \begin{eqaligned}[eq:phi_com]
            \com{\phi(\vec x)}{\pi(\vec y)} &=i\delta^{(3)}(\vec x-\vec y)
            \\
            \com{\phi(\vec x)}{\phi(\vec y)} &=0
            \\
            \com{\pi(\vec x)}{\pi(\vec y)} &=0
        \end{eqaligned}
        %
        with $\phi^\dag=\phi$, $\pi^\dag=\pi$ \imp{(``real'' field operators)}
        and $\vec x\in\reals^3$.\\[10pt]
        %
        \imp{This is completely analog to the canonical quantization of
            ``points'' known from undergraduate courses on quantum mechanics if
            Kronecker deltas are replaced by delta distributions:
            $\com{q_i}{p_i}=i\delta_{ij}\,\rightarrow\,\com{\phi(\vec x)}{\pi(\vec y)}
            =i\delta^{(3)}(\vec x-\vec y)$.}\\[10pt]
        %
        \imp{For now, we are in the Schrödinger picture where the fields
        do not depend on time!}

\item \Emph{Goals:} 
    %
    \begin{itemize}
        \item Representation of field operators \imp{\uref \cref{eq:phi_com}}
        \item Spectrum of Hamiltonian
        \item Time evolution of field operators \imp{$\phi(\vec x)\mapsto\phi(x)$}
    \end{itemize}

\item \Emph{Motivation:} 
    
    \imp{(The following facts are used to come up with a tentative
    representation for the field operators; that it is correct can be later
    checked rigorously.)}
    %
    \begin{lot}

        \item Fourier transform of KG equation in space:
            %
            \note{%
            \begin{align}
                \phi(\vec x,t)=\int\ddp{3}{p} e^{i\vec p\vec x}\tilde\phi(\vec p,t)
            \end{align}
            }
            \note{Then}
            %
            \begin{align}
                \left[\pt^2+(|\vec p|^2+m^2)\right]\tilde\phi(\vec p,t)=0 
            \end{align}
            %
            \therefore \Emph{Decoupled harmonic oscillators} with frequency $\omp{p}=\sqrt{|\vec p|^2+m^2}$\\
            %
            and constraint $\tilde \phi^*(\vec p,t)=\tilde \phi(-\vec p,t)$ (since $\phi^*=\phi$).

        \item \consider Hamiltonian $H_\mathrm{SHO}=\hlf P^2+\hlf \omega^2 X^2$\\
            %
            Introduce $X=\frac{1}{\sqrt{2\omega}}(a+a^\dag)$ and
            $P=-i\sqrt{\frac{\omega}{2}}(a-a^\dag)$ with $\com{a}{a^\dag}=1$\\
            %
            \therefore $H_\mathrm{SHO}=\omega (a^\dag a+\hlf)$ \imp{(diagonal!)}\\
            %
            Here: $P\leftrightarrow \tilde \pi(\vec p)$ and $X\leftrightarrow \tilde\phi(\vec p)$

    \end{lot}

\item \imp{This motivates the} \Emph{Field operators}
    %
    \begin{eqaligned}[eq:phi_op]
        \phi(\vec x)
        &:=\int\ddp{3}{p}\frac{1}{\sqrt{2\omp{p}}}\left(\ap{p} e^{i\vec p\vec x}+\adp{p} e^{-i\vec p\vec x}\right)
        \\
        &=\int\ddp{3}{p}\underbrace{\frac{1}{\sqrt{2\omp{p}}}\left(\ap{p}+a_{-\vec p}^\dag\right)}_{\tilde \phi(\vec p)}e^{i\vec p\vec x}
        \\
        \pi(\vec x)&:=\int\ddp{3}{p}\underbrace{(-i)\sqrt{\frac{\omp{p}}{2}}\left(\ap{p}-a_{-\vec p}^\dag\right)}_{\tilde \pi(\vec p)}
        e^{i\vec p\vec x}
    \end{eqaligned}
    %
    \aside{(Use colors to skip second line.)}\\
    \imp{The $-\vec p$ is necessary to make the fields ``real'': $\phi^\dag = \phi$ and $\pi^\dag=\pi$.}\\
    with \Emph{momentum modes}
    %
    \begin{align}
        \com{\ap{p}}{\adp{q}}=(2\pi)^3\delta^{(3)}(\vec p-\vec q)    
        \label{eq:ap_com}
    \end{align}
    %
    \imp{(All other commutators vanish.)}\\[5pt]
    %
    \imp{\easytoshow\cref{eq:ap_com}\land\cref{eq:phi_op}\limplies\cref{eq:phi_com} (Check this at home!)}

\item \Emph{Hamiltonian:}
    %
    \begin{emphalign}
        H\nte \int\ddp{3}{p}\omp{p}\Big (\adp{p}\ap{p}+\hlf \underbrace{\com{\ap{p}}{\adp{p}}}_{\propto\,\delta(0)=\infty}\Big )
    \end{emphalign}
    %
    \imp{Ignore the infinite term since only relative energies are physical!\\This
    infinity accounts for the zero-point energies of all harmonic oscillator
    modes.}\\
    \note{Dropping this infinity is called \emph{normal ordering} (\fref later).}

\item \Emph{Eigenstates \& Spectrum:}
    %
    \begin{itemize}

        \item \easytoshow $\com{H}{\adp{p}}=\omp{p}\adp{p}$
        
        \therefore \consider Vacuum $\ket{0}$ with $\ap{p}\ket{0}=0$
        
        \therefore Eigenstates $\adp{p}\adp{q}\dots\ket{0}$ \imp{(span complete Hilbert space)}
        
        \imp{\therefore Irreducible representation of momentum mode algebra \cref{eq:ap_com}}

        \item Energy: $E_{\vec p}=\omp{p}=+\sqrt{|\vec p|^2+m^2}$ 
            \imp{(relativistic dispersion, positive energies!)}

        \item (Kinetic) momentum:
            %
            \begin{align}
                P^i=\int\dd{3}{x}\pi(\vec x)(-\partial_i)\phi(\vec x)\nte \int\ddp{3}{p} p^i \adp{p}\ap{p}
                \label{eq:Pop}
            \end{align}
            %
            \imp{This is now an operator!}

        \item Statistics: $\adp{p}\adp{q}\ket{0}=\adp{q}\adp{p}\ket{0}$ 

    \end{itemize}
    %
    \therefore Excitations $\adp{p}$ commute and carry additive energy \&
    momentum\\
    %
    \therefore \Emph{Bosonic particles} \imp{(in momentum space)}

\item \Emph{Normalization:}

    \begin{lot}

        \item \consider $\Lambda=\R'L_3(\beta)\R\in \POLG$ 
            \therefore $p'=(E_{\vec p'},\vec p')=\Lambda p$ with $p=(E_{\vec p},\vec p)$
            
            \imp{Recall that all Lorentz transformations can be generated
            from spatial rotations and a boost $L_3(\beta)$ in $z$-direction!}

        \item Jacobian in space: 
            $\Det{\frac{\partial\vec p'}{\partial\vec p}}\nte\diff{p_3'}{p_3}\nte\frac{E_{\vec p'}}{E_{\vec p}}$\\
            %
            \therefore $\delta^{(3)}(\vec p-\vec q)=\frac{E_{\vec p'}}{E_{\vec p}}\,\delta^{(3)}(\vec p'-\vec q')$\\
            %
            \therefore $\delta^{(3)}(\vec p-\vec q)$ is not Lorentz invariant 
            but $E_{\vec p}\,\delta^{(3)}(\vec p-\vec q)$ is!
           
            \aside{Use colors to shorten this!}\\
            %
            \imp{3D volumes are \emph{not} invariant under boosts due to
            Lorentz contraction!}

        \item \Emph{Single-particle eigenstates:}
            %
            \begin{align}
                \ket{\vec p}:={\color{marked}\sqrt{2E_{\vec p}}}\;\adp{p}\ket{0}    
                \quad\Rightarrow\quad
                \braket{\vec p}{\vec q}
                =(2\pi)^3\,\underbrace{{\color{marked}2E_{\vec p}}\,\delta^{(3)}(\vec p-\vec q)}_{
                    \text{Lorentz invariant}}
                \label{eq:sp_eigenstates}
            \end{align}
            %
            \imp{This follows directly from the commutation relations.\\The
            $2$ is just convention.}

    \end{lot}

\item \Emph{Lorentz transformations} $\Lambda\in\POLG$:\\
    %
    \imp{We need a unitary \emph{representation} of the Lorentz group $\POLG$ on the Hilbert space!}
    %
    \begin{align}
        U(\Lambda)\ket{\vec p}:=\ket{\Lambda\vec p}
        \quad\Leftrightarrow\quad
        U(\Lambda)\adp{p}U^{-1}(\Lambda)=\sqrt{\frac{E_{\Lambda\vec p}}{E_{\vec p}}}\,a^\dag_{\Lambda\vec p}
    \end{align}
    %
    \imp{It is $(\Lambda\vec p)^i \equiv \Lambda^{i}_{\mu}p^\mu$ (i.e.,
    the spatial projection).
    
    Note that the ``boost part'' of $\Lambda$ is hidden in the normalization
    of the state!}

    \note{You can check that $U^\dag(\Lambda)U(\Lambda)=\Id$ using the
        single-particle states and our normalization of these states
        \cref{eq:sp_eigenstates}. The representation $U(\Lambda)$ is
        \emph{infinite-dimensional} and can therefore be \emph{unitary}
        despite the non-compactness of the Lorentz group.}


\item \Emph{Interpretation of $\phi(\vec x)$:}
    %
    \begin{align}
        \phi(\vec x)\ket{0}=\int\ddp{3}{p}{\color{marked}\frac{1}{2E_{\vec p}}}e^{-i\vec p\vec x}\ket{\vec p}
    \end{align}
    %
    For non-relativistic $|\vec p|\ll m\,\Rightarrow\,E_{\vec p}\approx\const$\\
    %
    \note{\therefore State $\ket{\vec x}$ of particle at position $\vec x$}\\
    %
    \therefore $\phi(\vec x)$ creates particle at position $\vec x$
    %
    \note{%
    \begin{itemize}

        \item This interpretation is also consistent with the ``position-space
        representation'' $\bra{0}\phi(\vec x)\ket{\vec p}\nte e^{i\vec p\vec x}$.

        \item The factor $\frac{1}{2E_{\vec p}}$ suppresses large momenta
        and ``smears out'' the position of the particle on length scales of
        its Compton wavelength $\lambda_c=1/m$ (\fref space-like two-point
        correlation function $\bra{0}\phi(\vec x)\phi(\vec y)\ket{0}\sim
        e^{-m|\vec x-\vec y|}$).

    \end{itemize}}

\end{lot}

\begin{blocknote}
    \label{note:LI_measure}
    \begin{itemize}

        \item Projector on single-particle sector:
        $\mathds{1}_1=\int\ddp{3}{p}\ket{\vec p}\frac{1}{2E_{\vec p}}\bra{\vec
        p}$

        \item \consider $f(p)$ Lorentz invariant
        \therefore $\int\ddp{3}{p}\frac{f(p)}{2E_{\vec p}}$ is Lorentz invariant

    \end{itemize}
\end{blocknote}


\section{The Klein-Gordon Field in Space-Time}

\imp{So far: Schrödinger picture}\\
%
\imp{Now: Heisenberg picture}

\begin{lot}

    \item \Emph{Heisenberg operators:} 
        $\phi(x)=\phi(\vec x,t)=e^{iHt}\phi(\vec x)e^{-iHt}$ \imp{(similar for $\pi(x)$)}

    \item \Emph{Heisenberg equation:} 
        $i\pt\O=\com{\O}{H}$ for $\O=\phi,\pi$ yields
        %
        \begin{subalign}
            i\pt\phi(x)
            &=
            \note{%
            \com{\p(x)}{\int\dd{3}{y}
            \left\{\hlf\pi^2(\vec y,t)+\hlf(\nabla\phi(\vec y,t))^2+\hlf m^2\phi^2(\vec y,t)\right\}}
            \nonumber
            }
            \\
            &=
            \note{%
            \int\dd{3}{y}i\delta^{(3)}(\vec x-\vec y)\pi(\vec y,t)
            \nonumber
            }
            \\
            &=i\pi(x)
            \\
            i\pt\pi(x)&\nte -i(-\nabla^2+m^2)\phi(x)
        \end{subalign}
        %
        \begin{emphalign}
            \Rightarrow\quad(\pt^2-\nabla^2+m^2)\phi(x)=0\quad\text{(Klein-Gordon equation)}
        \end{emphalign}

    \item Time-evolution of modes:
        %
        \begin{subalign}
            e^{iHt}\ap{p}e^{-iHt}&=\ap{p}e^{-iE_{\vec p}t}\\
            e^{iHt}\adp{p}e^{-iHt}&=\adp{p}e^{+iE_{\vec p}t}
        \end{subalign}
        %
	    \aside{Use colors to skip last row.}\\ 
        %
        \imp{This can be shown informally by using $H_{\vec p}=E_{\vec
        p}\,\adp{p}\ap{p}$ and counting excitations (i.e., on the number
        basis).}

    \item \Emph{Field operators:}
        %
        \begin{emphalign}
        \phi(x)
        &=\int\ddp{3}{p}\frac{1}{\sqrt{2\Ep{p}}}\left.
        \left(\ap{p} e^{-ipx}+\adp{p} e^{ipx}\right)\right|_{p^0=\Ep{p}}\\
        \pi(x)&=\pt\p(x)
        \end{emphalign}
        %
        \imp{Here, $px=p^\mu x_\mu=\Ep{p}t-\vec p\vec x$; note that $p^0=\Ep{p}$.\\
        %
        In the following, $\ap{p}$ and $\adp{p}$ always denote the
        \emph{time-independent} Schrödinger operators!
        }

\end{lot}

\begin{blocknote}
    \begin{lot}

        \item Hamiltonian generates time translations:\\
            %
            $\phi(\vec x,t)=e^{iHt}\underbrace{\phi(\vec x,0)}_{\phi(\vec x)}e^{-iHt}$

        \item Total momentum operator generates space translations:\\
            %
            $\phi(\vec x)=e^{-i\vec P\vec x}\phi(\vec 0)e^{i\vec P\vec x}$

        \item Four-momentum operator generates space-time translations:\\
            %
            $\phi(x)=e^{iPx}\phi(0)e^{-iPx}$\\
            %
            \imp{Here, $P^\mu=(H,\vec P)$ where $\vec P$ is defined in \cref{eq:Pop}.}

    \end{lot}
\end{blocknote}

\begin{blocknote}

    \imp{Note that $p^0=\Ep{p}$ is always positive:}
    %
    \begin{itemize}

        \item $e^{-ipx}$ 
            $\leftrightarrow$ \Emph{positive-frequency} solution of KG equation 
            $\leftrightarrow$ destruction operator $\ap{p}$

        \item $e^{+ipx}$ 
            $\leftrightarrow$ \Emph{negative-frequency} solution of KG equation 
            $\leftrightarrow$ creation operator $\adp{p}$

    \end{itemize}
    %
    \imp{As \emph{single-particle wavefunctions}, solutions with
    positive/negative frequency correspond to solutions with positive/negative energy. Note that there
	are only excitations with positive energy in the quantized field theory!}

\end{blocknote}


%%==============================================================================
\begin{lecture}
\item Causality
\item Green's functions of the Klein-Gordon theory 
\item The Feynman propagator
\end{lecture}
%%==============================================================================

%%==============================================================================
\reference{PS:27--31}
%%==============================================================================

\subsubsection{Causality}

\consider Amplitude for a particle to propagate from $y$ to $x$:
%
\begin{emphalign}
    D(x-y)\equiv \bra{0}\phi(x)\phi(y)\ket{0}
    \nte \int\ddp{3}{p}\frac{1}{2\Ep{p}}e^{-ip(x-y)}
    \label{eq:Dboson}
\end{emphalign}
%
\imp{This expression is Lorentz invariant, i.e., $D(\Lambda(x-y))=D(x-y)$
for all $\Lambda\in\POLG$ [more generally, for all orthochronous Lorentz
transformations $\Lambda\in \mathrm{O}^+(1,3)$].}\\
%
\note{This is \emph{not} true for non-orthochronous Lorentz transformations which
flip the sign of $D(x-y)$ since $\Ep{p}=p^0\mapsto -p^0=-\Ep{p}$!}

\begin{lot}

    \item\consider\Emph{Time-like distance:} $x^0-y^0=t$ and $\vec x-\vec y=0$
        %
        \begin{subalign}
            D(x-y)&=\frac{4\pi}{(2\pi)^3}\int_0^\infty\d{p}\frac{p^2}{2\sqrt{p^2+m^2}}\,e^{-i\sqrt{p^2+m^2} t}
            \\
            &=\frac{1}{4\pi^2}\int_m^\infty\d{E}\sqrt{E^2-m^2}\,e^{-iEt}
            \\
            &\stackrel{t\to\infty}{\neq}0\quad\text{\imp{(actually not convergent)}}
            \\
            &\note{\stackrel{t\to\infty}{\sim} e^{-imt}}\quad\text{\aside{(this is very hand-wavy)}}
        \end{subalign}
        %
        \therefore Does not vanish \therefore Propagation possible

    \item\consider\Emph{Space-like distance:} $x^0-y^0=0$ and $\vec x-\vec y=\vec r$
        %
        \begin{subalign}
            D(x-y)&=\int\ddp{3}{p}\frac{1}{2\Ep{p}}\,e^{i\vec p\vec r}
            \\
            &=\frac{2\pi}{(2\pi)^3}\int_0^\infty\d{p}\frac{p^2}{2\Ep{p}}\frac{e^{ipr}-e^{-ipr}}{ipr}
            \\
            &=\frac{-i}{2(2\pi)^2r}\int_{-\infty}^\infty\d{p}\frac{p\,e^{ipr}}{\sqrt{p^2+m^2}}
            \label{eq:Dsl2}
        \end{subalign}
        %
        Use \Emph{Cauchy's integral theorem} with the following path:
        %
        \begin{center}
            \includegraphics[width=0.6\linewidth]{tikz/cauchy}
        \end{center}
        % 
        \begin{itemize}

                \item \imp{Show that the curved sections vanish for $R\to\infty$
                and $\vep\to 0$, respectively!}\\
                %
                \note{Showing that $B,F\to 0$ for $R\to\infty$ is
                actually tricky and requires some kind of regularization
                (that P\&S are silent about) to exponentially suppress the
                oscillating terms close to the real axis. One way to to fix
                this is to focus on the \emph{asymptotics} $r\to\infty$
                (which is our goal here). The oscillating terms can then be
                exponentially suppressed in the limit $r\to\infty$ so that
                the contributions from the arcs become negligible. Strictly
                speaking, the non-convergent integral \cref{eq:Dsl2} should
                be \emph{defined} by such an appropriately chosen limit.}

                \item \imp{It is $C=E$ since the minus from the opposite direction
                and the branch cut cancel.}

        \end{itemize}
      
        Then
        %
        \begin{align}
            D(x-y)&=-C-E=-2C
            \\
            &=\frac{-i}{(2\pi)^2r}\int_{im}^{\infty}\d{p}\frac{p\,e^{ipr}}{\sqrt{p^2+m^2}}
            \\
            &\stackrel{\rho=-ip}{=}\frac{1}{4\pi^2r}\int_{m}^{\infty}
            \d{\rho}\frac{\rho\,e^{-\rho r}}{\sqrt{\rho^2-m^2}}
            \label{eq:Dsl}
            \\
            &\stackrel{r\to\infty}{\sim}e^{-mr}
        \end{align}
        %
        \therefore Vanishes exponentially \imp{(but non-zero!) \therefore Problem?}\\
        %
        \note{The integral \cref{eq:Dsl} can be evaluated in terms of
        \emph{modified Bessel functions of the second kind}, the asymptotics
        of which is known and yields the given exponential decay. Note that
        simply upper bounding the integrand by $e^{-\rho r}\leq e^{-mr}$
        leaves a diverging integral behind so that one ends up with the
        useless upper bound $e^{-mr}\times\infty$.}


    \item \consider \Emph{Measurements} $A$ and $B$: can affect each other
        iff $\com{A}{B}\neq 0$\\
        %
        Simplest choice: $A=\phi(x)$ and $B=\phi(y)$\\
        %
        \imp{Causality is preserved if all observables commute at space-like
        separations!}\\
        %
        \imp{Since $\pi=\pt\phi$, it is sufficient for $\com{\phi(x)}{\phi(y)}$
        to vanish for $(x-y)^2<0$.}
        %
        \begin{subalign}
            \com{\p(x)}{\p(y)}&=    
            \begin{aligned}[t]
                &\int\ddp{3}{p}\frac{1}{\sqrt{2\Ep{p}}}
                \int\ddp{3}{q}\frac{1}{\sqrt{2\Ep{q}}}
                \\
                &\times
                \com{
                    \left(\ap{p} e^{-ipx}+\adp{p} e^{ipx}\right)
                }{
                    \left(\ap{q} e^{-iqy}+\adp{q} e^{iqy}\right)
                }
            \end{aligned}
            \\
            &=\int\ddp{3}{p}\frac{1}{2\Ep{p}}\left(e^{-ip(x-y)}-e^{ip(x-y)}\right)
            \\
            &=D(x-y)-D(y-x)
        \end{subalign}
        %
        Let $(x-y)^2<0$ space-like \therefore $\exists\Lambda^*\in\POLG\;:\;\Lambda^*(x-y)=-(x-y)$:\\
        \imp{The \emph{proper orthochronous Lorentz group} $\POLG$ is a
        connected subgroup of the Lorentz group $\LG$, the elements of which connect continuously to the identity.}
        %
        \begin{center}
            \includegraphics[width=0.6\linewidth]{tikz/lorentz_trafo}
        \end{center}
        %
        \imp{Continuous transformations (rotations in space and boosts) allow for $(x-y)\mapsto -(x-y)$ only
        if $(x-y)^2<0$. For time-like distances, this requires discontinuous transformations (time-reversal).}
        \\[10pt]
        Then
        %
        \begin{emphalign}
            \com{\p(x)}{\p(y)}
            &=D(x-y)-D(\Lambda^*(y-x))\nonumber\\
            &\note{(x-y)^2<0}\nonumber\\
            &=D(x-y)-D(x-y)\equiv 0
            \quad\text{(Causality)}
        \end{emphalign}
        %
        \imp{For time-like separation, $(x-y)^2>0$, there is no such continuous
        transformation and the argument breaks down.} \\[10pt]
        %
        \note{The first line follows from the Lorentz invariant integral
        measure in Note~\ref{note:LI_measure} and the definition of the
        propagator in \cref{eq:Dboson}. Remember that $D$ is only invariant
        under \emph{orthochronous} Lorentz transformations but picks up a
        minus sign under time inversion!}

\end{lot}

\subsubsection{The Propagator}

\begin{lot}

    \item Since $\com{\p(x)}{\p(y)}\propto\mathds{1}$ \imp{(the commutator
    is a c-number)}, we can write ($x^0>y^0$ for now)\\
    %
    \note{``c-number'' historically denotes scalar multiples of the identity,
    i.e. classical/commuting/complex ``numbers''.}
    
    \begin{subalign}
        \bra{0}\com{\p(x)}{\p(y)}\ket{0}
        &=\int\ddp{3}{p}\frac{1}{2\Ep{p}}\left(e^{-ip(x-y)}-e^{ip(x-y)}\right)
        \\
        &\mimp{Substitute $\vec p\to-\vec p$ to obtain the second term:}
        \\
        &=\int\ddp{3}{p}\left\{
            \frac{e^{-ip(x-y)}|_{p^0=\Ep{p}}}{2\Ep{p}}
            +\frac{e^{-ip(x-y)}|_{p^0=-\Ep{p}}}{-2\Ep{p}}
        \right\}
        \\
        &\mimp{Residue theorem with clockwise orientation (therefore the $-1$):}
        \\
        &\stackrel{x^0>y^0}{=}\int\ddp{3}{p}\int_{\gamma=\gamma_R}\frac{\d{p^0}}{2\pi i}
        \frac{-1}{\underbrace{p^2-m^2}_{(p^0-\Ep{p})(p^0+\Ep{p})}}\,e^{-ip(x-y)}
        \\
        &=\int_{\gamma=\gamma_R}\ddp{4}{p}\frac{i}{p^2-m^2}\,e^{-ip(x-y)}
        \label{eq:DR_cont_int}
    \end{subalign}
    %
    with contour $\gamma_R$
    %
    \begin{center}
        \includegraphics[width=0.6\linewidth]{tikz/contour_retarded}
    \end{center}
    %
    \imp{The arc vanishes in the lower/upper-half pane for $x^0>y^0$ and $x^0<y^0$, respectively.}
    \\[10pt]
    %
    Therefore
    %
    \begin{emphalign}
        D_R(x-y)
        \equiv \theta(x^0-y^0)\bra{0}\com{\p(x)}{\p(y)}\ket{0}
        =\left[\text{\cref{eq:DR_cont_int}}\right]_{\gamma=\gamma_R}
    \end{emphalign}

    \item \Emph{Interpretation:}
    %
    \begin{align}
        (\partial^2+m^2)D_R(x-y)\nte -i\delta^{(4)}(x-y)    
        \label{eq:KG_GF}
    \end{align}
    %
    \therefore \Emph{Retarded Green's function} of Klein-Gordon operator\\
    %
    \imp{``R'' for ``retarded'' since it vanishes for $x^0<y^0$.}\\
    %
    \note{We could have found \cref{eq:DR_cont_int} directly from \cref{eq:KG_GF} by Fourier transformation.}

    \item Alternative contour $\gamma_F$:
    %
    \begin{center}
        \includegraphics[width=1.0\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}
    %
    \begin{emphalign}
        D_F(x-y)
        \equiv\left[\text{\cref{eq:DR_cont_int}}\right]_{\gamma=\gamma_F}
        =&\int\ddp{4}{p}\frac{i}{p^2-m^2+i\vep}e^{-ip(x-y)}    
        \\
        &\text{(Feynman propagator)}\nonumber
    \end{emphalign}
    %
    \imp{The infinitesimal $i\vep$ shifts the poles to $p^0\approx\pm
    (\Ep{p}-i\vep/2\Ep{p})=\pm (\Ep{p}-i\epsilon)$ and yields an equivalent
    prescription of the Feynman propagator without the need to specify a
    contour. Note that $\vep/2\Ep{p}\equiv\epsilon$ are both infinitesimals.}
    \\[10pt]
    %
    We find \note{(using \cref{eq:DR_cont_int} and \cref{eq:Dboson})}
    %
    \begin{emphalign}
        D_F(x-y)&=
        \begin{cases}
            D(x-y) & \text{for}\;x^0>y^0\\ 
            D(y-x) & \text{for}\;x^0<y^0
        \end{cases}
        \\
        &=
        \begin{aligned}[t]
        &\theta(x^0-y^0)\bra{0}\p(x)\p(y)\ket{0}\\
        +&\theta(y^0-x^0)\bra{0}\p(y)\p(x)\ket{0}
        \end{aligned}
        \\
        &\equiv \bra{0}\T\p(x)\p(y)\ket{0}
    \end{emphalign}
    %
    with the \Emph{time-ordering (meta-)operator} $\T$\\ \imp{$\T$ orders
    products of operators by time with the latest to the left.}\\
    %
    \note{It is a \emph{meta-operator} as it operates on \emph{descriptions}
    of operators.  Note that this is different from \emph{super-operators}
    (such as the Lindbladian) which operate on operators.}\\[10pt]
    %
    \note{The Feynman propagator is a Green's function of the KG equation
    (with different boundary conditions than the retarded/advanced Green's
    functions).}\\[10pt]
    %
    Later: Feynman propagator \& Interactions \leadsto \emph{Feynman rules}\\
    %
    \imp{However, so far we only studied the \emph{free} KG field (\therefore
    linear field equation).  Without interactions, however, there is no
    scattering so that there are no characteristic observations possible.}

\end{lot}