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\begin{document}

\title{The CPT Theorem}
\date{January 8, 2009}
\author{Nils Kanning}
\maketitle

\begin{abstract}
In the context of classical mechanics and field theory continuous
symmetry transformations and the associated conserved quantities were
discussed. These continuous symmetries were also analyzed in quantum
field theory.

In contrast the CPT theorem is a statement about discrete symmetry
transformations. It can be shown that under few assumptions a
quantum field theory is symmetric with regard to the combination of
the inversion of spatial coordinates (${\rm P}$), the inversion of charge
(${\rm C}$) and the inversion of the direction of time (${\rm T}$).

Before this theorem is discussed and a proof is illustrated, a
brief introduction to symmetries in quantum field theory is given and
discrete symmetries are studied.
\end{abstract}

\section{Symmetries in Quantum Field Theory}

\subsection{What is a Symmetry?}
In quantum field theory a symmetry transformation is a one-to-one
mapping $\mathcal{U}$ of states
\begin{align}
\label{eq:utrafo}
  |\alpha\rangle \to |\alpha'\rangle = \mathcal{U} |\alpha\rangle,
\end{align}
given that two conditions hold.

First the transition probabilities between two states $|\alpha\rangle$
and $|\beta\rangle$ have to be conserved:
\begin{align*}
  |\langle\alpha|\beta\rangle| = |\langle\alpha'|\beta'\rangle|
\end{align*}
A theorem of Wigner then implies that the mapping $\mathcal{U}$ is
either unitary or antiunitary.

Second the equality of transition probabilities should also hold after
the time evolution of the states:
\begin{align}\label{eq:prob}
  &|\langle\alpha|e^{iH(t_\alpha-t_\beta)}|\beta\rangle|\nonumber\\
  &=|\langle\alpha'|e^{iH(t_\alpha-t_\beta)}|\beta'\rangle|
\end{align}
Therefore it is imposed that the mapping commutes with the Hamiltonian
$[\mathcal{U},H]=0$.\footnote{Eq. \ref{eq:prob} does not hold for
  transformations involving the time variable -- especially time
  reversal. But $[\mathcal{U},H]=0$ is imposed in this case too.}

\subsection{How are Quantum Fields transformed by a Symmetry?}
A symmetry transformation in classical field theory is given by the
transformation of the spacetime points
\begin{align*}
  x \to x' = \Lambda x
\end{align*}
and the according transformation of the field:
\begin{align}
\label{eq:trafo_field_class}
  \phi(x) \to \phi'(x)=U(\Lambda)\phi(\Lambda^{-1}x)
\end{align}
The expression $U(\Lambda)$ has the structure of a matrix, possibly
mixing the components of the field -- e.g. vector components of vector
fields or spinor components of spinor fields. In case of a scalar
field it is a number.

It is assumed that in quantum field theory this symmetry
transformation is implemented by a unitary operator $\mathcal{U}$
transforming the states as introduced in Eq. \ref{eq:utrafo}.  To
arrive at the transformation of the quantum field operators, a set of
states is applied to Eq. \ref{eq:trafo_field_class}, now interpreted
as quantum fields:
\begin{align*}
  \langle\alpha'|U(\Lambda)\phi(\Lambda^{-1}x)|\beta'\rangle\\
  =\langle\alpha'|\phi'(x)|\beta'\rangle
  &=\langle\alpha|\phi(x)|\beta\rangle
\end{align*}
The last equality is demanded from the physical point of view.  From
this equation the desired transformation law can be extracted using
the unitarity of $\mathcal{U}$:
\begin{align}
\label{eq:trafo_field_quant}
  \mathcal{U}\phi(x)\mathcal{U^\dagger} = U(\Lambda)\phi(\Lambda^{-1}x)
\end{align}
This formula is of importance for the further analysis.

\section{Discrete Symmetries}
Discrete symmetry transformations are studied exemplary using the
spatial inversion (${\rm P}$) in the case of the charged free scalar
field. The ${\rm C}$ and ${\rm T}$ symmetries are only briefly
discussed for this field.

\subsection{Scalar Fields}

\subsubsection{Parity (P)}
In a classical theory ${\rm P}$ is a symmetry if, given a trajectory,
the trajectory with inverted spatial coordinates also solves the
equations of motion (cf. Fig. \ref{img:p}).

\begin{figure}[ht]
  \centering
  \begin{tikzpicture}[domain=-1.7:1.7]
    % the coordinate frame
    \draw[-latex] (0,0) -- (2.5,0) node[right] {$\vec x$}; 
    \draw[-] (0,0) -- (-0.1,0); 
    \draw[-latex] (0,0) -- (90:2.5) node[above] {$t$};
    \draw[-] (0,0) -- (90:-0.1);
    
    % some coordinates
    \coordinate (traj_start) at (.5,0.4);
    \coordinate (traj_end) at (2.1,2);
    \coordinate (traj2_start) at (-.5,0.4);
    \coordinate (traj2_end) at (-2.1,2);

    % trajectory
    \draw[-] 
    (traj_start)
    .. controls (0.8,1) and (1.7,1) .. 
    node[right=4pt] {$\vec x(t)$}
    (traj_end);

    % transformed trajectory
    \draw[-, densely dashed] 
    (traj2_start) 
    .. controls (-0.8,1) and (-1.7,1) .. 
    node[left=2pt, text width = 1.3cm] {$\vec x'(t) $ \newline$=-\vec x(t)$}
    (traj2_end);

    %all dots
    \fill [] (traj_start) circle (1pt);
    \fill [] (traj_end) circle (1pt);
    \fill [] (traj2_start) circle (1pt);
    \fill [] (traj2_end) circle (1pt);
  \end{tikzpicture}
  \caption{Parity transformation $\vec x'(t)$ of a trajectory $\vec x(t)$ in
    classical mechanics}
  \label{img:p}
\end{figure}

In the context of classical field theory the scalar field $\varphi(x)$
transforms according to:
\begin{align*}
  \varphi(x) \to \varphi'(x)=\eta_\text{P}\varphi(t,-\vec{x})
\end{align*}
Here the phase $\eta_\text{P}$ remains a free parameter.
It is easy to check that the new field solves the Klein-Gordon
equation and thus is a symmetry of the classical theory:
\begin{align*}
  (\square+m^2)\varphi'(x)=\varphi(t,-\vec{x})=0
\end{align*}

The unitary operator implementing this transformation in quantum
theory is denoted by $\mathcal{P}$.
Then Eq. \ref{eq:trafo_field_quant} is employed to get the
transformation of the quantum field:
\begin{align*}
  \mathcal{P}\varphi(x)\mathcal{P}^\dagger = \eta_\text{P}\varphi(t,-\vec{x})
\end{align*}

Using a plane wave decomposition
\begin{align*}
  \varphi(x) = \int \widetilde{{\rm d}k}\;a(k) e^{-ikx}+b^\dagger(k) e^{ikx}
\end{align*}
the transformation laws for the creation and annihilation operators
$a^\dagger(k)$, $b^\dagger(k)$, $a(k)$ and $b(k)$ can be
derived. The operator $\mathcal{P}$ can then be explicitly constructed
in terms of creation and annihilation operators and it can be verified
that it commutes with the Hamiltonian. So the parity transformation is
a symmetry of the free scalar field theory.

\subsubsection{Charge Conjugation (C)}
For the classical charged free scalar field the charge conjugation is
achieved by a complex conjugation of the solution
(cf. Fig. \ref{img:c} for a mechanical system). It is then shown
that the conjugated solution also solves the Klein-Gordon equation.

The unitary operator implementing this symmetry is labeled $\mathcal{C}$.

\begin{figure}[ht]
  \centering
  \begin{tikzpicture}[domain=-1.7:1.7]
    % the coordinate frame
    \draw[-latex] (0,0) -- (2.5,0) node[right] {$\vec x$}; 
    \draw[-] (0,0) -- (-0.1,0); 
    \draw[-latex] (0,0) -- (90:2.5) node[above] {$t$};
    \draw[-] (0,0) -- (90:-0.1);

    % some coordinates
    \coordinate (traj_start) at (.5,0.4);
    \coordinate (traj2_start) at (2.,0.4);

    % trajectory
    \draw[-] 
    (traj_start)
    .. controls +(0.3,0.6) and +(-.2,-0.7) .. 
    node[left=6pt] {$\vec x(t)$}
    +(1.5,1.5);
    
    % current vector and charge
    \draw[-latex] 
    (traj_start)+(1.5,1.5) 
    node[above right] {$\vec j$} 
    node[left] {$Q$} 
    -- 
    +(2.0,1.5);

    % transformed trajectory
    \draw[-, densely dashed] 
    (traj2_start) 
    .. controls +(0.3,0.6) and +(-.2,-0.7) .. 
    node[below right] {$\vec x'(t)$\newline$=\vec x(t)$} 
    +(1.5,1.5);
    
    % transformed current vector and charge
    \draw[-latex] (traj2_start)+(1.5,1.5)    
    node[above left] {$-\vec j$} 
    node[right] {$-Q$} 
    -- 
    +(1.0,1.5);

    % dots
    \fill [] (traj_start) circle (1pt);
    \fill [] (traj_start)+(1.5,1.5) circle (1pt);
    \fill [] (traj2_start) circle (1pt);
    \fill [] (traj2_start)+(1.5,1.5) circle (1pt);

  \end{tikzpicture}
  \caption{Charge conjugation in classical mechanics: inversion of the charge $Q$
    and the current $\vec j$}
  \label{img:c}
\end{figure}


\subsubsection{Time Reversal (T)}
In case of the Klein-Gordon equation the time reversed solution is
obtained by inverting the sign of the time argument. To get
the right current vector also a complex conjugation is needed for
charged fields (cf. Fig. \ref{img:t} for the mechanical problem).

\begin{figure}[ht]
  \centering
  \begin{tikzpicture}[domain=-1.7:1.7]
    % the coordinate frame
    \draw[-latex] (0,0) -- (2.5,0) node[right] {$x_1$}; 
    \draw[-] (0,0) -- (-0.1,0); 
    \draw[-latex] (0,0) -- (90:2.5) node[above] {$x_2$};
    \draw[-] (0,0) -- (90:-0.1);

    % some coordinates
    \coordinate (traj_start) at (.5,0.4);
    \coordinate (traj2_start) at (1.5,0.4);

    % trajectory
    \draw[-, 
    decoration={markings, mark=at position .55 with {\arrow{latex};}}, 
    postaction={decorate}] 
    (traj_start) 
    .. controls +(0.3,0.6) and +(-.2,-0.7) .. 
    node[left=6pt] {$\vec x(t)$}
    +(1.5,1.5);

    % current vector and charge
    \draw[-latex] 
    (traj_start)+(1.5,1.5)
    node[left] {$Q$} 
    -- 
    +(1.5+.2,1.5+0.7)
    node[below right] {$\vec j$};

    % transformed trajectory
    \draw[-, densely dashed, 
    decoration={ markings, mark=at position .55 with {\arrowreversed{latex};}}, 
    postaction={decorate}] (traj2_start)
    .. controls +(0.3,0.6) and +(-.2,-0.7) .. 
    node[below right] {$\vec x'(t)=\vec x(-t)$} 
    +(1.5,1.5);
    
    % transformed current vector and charge
    \draw[-latex] 
    (traj2_start)+(1.5,1.5) 
    node[above right] {$Q$}
    -- 
    +(1.5-.2,1.5-0.7)
    node[above right] {$-\vec j$};

    % dots
    \fill [] (traj_start) circle (1pt);
    \fill [] (traj_start)+(1.5,1.5) circle (1pt);
    \fill [] (traj2_start) circle (1pt);
    \fill [] (traj2_start)+(1.5,1.5) circle (1pt);

  \end{tikzpicture}
  \caption{Time reversal of a mechanical system: same charge $Q$
    and inversion of the current $\vec j$}
  \label{img:t}
\end{figure}

The implementation of this transformation in quantum theory differs
from the two symmetries discussed above.

It is assumed that time reversal is implemented by an operator
$\mathcal{T}$ such that $\mathcal{T}|\alpha\rangle=|\alpha'\rangle$
and $[\mathcal{T},H]=0$.
The time evolution of the states is generated by the Hamiltonian:
\begin{align*}
  |\alpha(t)\rangle &= e^{-iHt}|\alpha\rangle\\
  |\alpha'(t)\rangle &= e^{-iHt}\mathcal{T}|\alpha\rangle
\end{align*}
Next using the physical interpretation of $\mathcal{T}$ one can
identify $\mathcal{T}|\alpha(t)\rangle = |\alpha(-t)\rangle$. Thus if
$\mathcal{T}$ would be unitary:
\begin{align*}
  |\alpha'(-t)\rangle = \mathcal{T}e^{-iHt}|\alpha\rangle &\\
  =e^{-iHt}\mathcal{T}|\alpha\rangle &= |\alpha'(t)\rangle
\end{align*}
Consequently $\mathcal{T}$ cannot be unitary and has to be an antiunitary
operator, which in this case causes a complex conjugation in the
exponential and leads to the correct result $|\alpha'(-t)\rangle$.

\subsection{Summary of Transformation Laws}
\label{sec:summary}
Because for the discussion of the CPT theorem the transformations laws
for scalar, vector and spinor fields are needed, a summary of the
results is given here.\footnote{The computations are worked out in
  some detail in \cite{Greiner1993}. The results for the spinor field
are also derived in \cite{Itzykson1980}.}

It can be shown that the phases $\eta_\text{P}$, $\eta_\text{C}$ and
$\eta_\text{T}$ are arbitrary for charged fields and real for neutral
fields.\footnote{This remark is verified in \cite{Greiner1993}.}

\subsubsection[Scalar Field (Spin 0)]{Scalar Field (Spin $0$)}
\begin{align*}
  \mathcal{P} \varphi(x) \mathcal{P}^\dagger &=
  \eta_\text{P}\varphi(t,-\vec{x})\\
  \mathcal{C} \varphi(x) \mathcal{C}^\dagger &=
  \eta_\text{C}\varphi^\dagger(t,\vec{x})\\
  \mathcal{T} \varphi(x) \mathcal{T}^\dagger &=
  \eta_\text{T}\varphi(-t,\vec{x})
\end{align*}

\subsubsection[Vector Field (Spin 1)]{Vector Field (Spin $1$)}
\begin{align*}
  \mathcal{P} A^\mu(x) \mathcal{P}^\dagger &=
  \eta_\text{P}A_\mu(t,-\vec{x})\\
  \mathcal{C} A^\mu(x) \mathcal{C}^\dagger &=
  -\eta_\text{C}{A^\mu}^\dagger(t,\vec{x})\\
  \mathcal{T} A^\mu(x) \mathcal{T}^\dagger &=
  \eta_\text{T}A^\mu(-t,\vec{x})
\end{align*}

\subsubsection[Spinor Field (Spin 1/2)]{Spinor Field (Spin $1/2$)}
\begin{align*}
  \mathcal{P} \psi(x) \mathcal{P}^\dagger &=
  \eta_\text{P}\gamma^0\psi(t,-\vec{x})\\
  \mathcal{C} \psi(x) \mathcal{C}^\dagger &=
  \eta_\text{C}C\overline{\psi}^{T}(t,\vec{x})\\
  \mathcal{T} \psi(x) \mathcal{T}^\dagger &= 
  \eta_\text{T}A\psi(-t,\vec{x})
\end{align*}

The convenient notation $\overline{\psi}=\psi^\dagger\gamma^0$ and
the convention $\gamma^5 = i \gamma^0\gamma^1\gamma^2\gamma^3$
are used. Furthermore $C$ and $A$ are matrices acting on the spinor
components with the properties:
\begin{align*}
  C\gamma_\mu C^{-1} &= -\gamma_\mu^T\\
  C^{-1} = C^T &= -C\\
  A &= -i\gamma^5C
\end{align*}
In the Dirac representation $C$ becomes:
\begin{align*}
  C = i\gamma^0\gamma^2
\end{align*}

\subsection{Consequences and Outlook}
As illustrated using the ${\rm P}$ transformation for the scalar field,
it can be shown that ${\rm C}$, ${\rm P}$ and ${\rm T}$ are
symmetries of the free quantum fields. This means that the
corresponding unitary or antiunitary operators commute with the
Hamiltonian of these systems.

It also turns out that these three transformations are symmetries of
quantum electrodynamics. So the operators also commute with the
Hamiltonian of quantum electrodynamics including the interaction
between the electron and the photon field.

But one can imagine interaction terms causing a violation of these
symmetries. This effect is also realized in nature. For example
the weak interaction violates the ${\rm P}$ symmetry.

However, the combination of all three transformations, called ${\rm
  CPT}$, remains a symmetry of ``almost'' every quantum field theory. This
result is discussed in the following section.


\section{CPT Theorem}
The theorem states that the ${\rm CPT}$ transformation is a symmetry
of a quantum field theory if two conditions are fulfilled.

First the theory has to be described by a Lagrange density which is
Lorentz invariant\footnote{Here Lorentz invariant means invariance
  under proper orthochronous Lorentz transformations.},
local\footnote{A Lagrange density is local if the fields are all
  evaluated at the same spacetime point and only polynomials of the
  fields appear.}, hermitian and normal ordered.

Second the theory is quantized according to the spin-statistics
theorem. Integer spin fields commute and half integer spin fields anticommute.


\subsection{Proof}
The proof illustrated here\footnote{This way of proofing the theorem
  is also chosen in \cite{Itzykson1980} and \cite{Greiner1993}.} only
includes scalar (spin 0), vector (spin 1) and spinor fields (spin
1/2), but the statement of the theorem is true for higher spins
too.\footnote{A more general proof is presented in
  \cite{Grawert1959}.}

The ${\rm CPT}$ transformation is represented by the operator $\Theta
= \mathcal{CPT}$. This operator is antiunitary since $\mathcal{C}$ and
$\mathcal{P}$ are unitary and $\mathcal{T}$ is antiunitary.

The overall structure of the proof is to show that the transformation
of the Lagrange density $\mathcal{L}$ is given by $\Theta
\mathcal{L}(x)\Theta^\dagger=\mathcal{L}(-x)$. Then it can be
concluded that $\Theta$ commutes with the Hamiltonian $H$ and thus
$\Theta$ is a symmetry.

\subsubsection{Transformation of the Lagrange density}
Because the Lagrange density is Lorentz invariant, it has to be a
Lorentz scalar. So all Lorentz indices are contracted and no spinor
index remains either.

The idea is then to show that every quantity\footnote{The colons
  indicate normal ordering.}  $:T^{\mu_1\ldots\mu_m}(x):$ in the
Lagrange density involving $m$ Lorentz indices transforms under
$\Theta$ to:
\begin{align}
  \label{eq:trafo_law}
  &\Theta\ : T^{\mu_1\ldots\mu_m}(x) : \Theta^\dagger \nonumber\\
&=(-1)^m : {T^{\mu_1\ldots\mu_m}}^\dagger(-x) : 
\end{align}
Since in the Lagrange density all indices are contracted, the total
number of indices is even and using Eq. \ref{eq:trafo_law} follows:
\begin{align*}
  \mathcal{L}(x)\Theta^\dagger=\mathcal{L}^\dagger(-x)= \mathcal{L}(-x)
\end{align*}
Where the last equality holds because $\mathcal{L}$ is assumed to be
hermitian.

Next the behaviour according to Eq. \ref{eq:trafo_law} is checked for
the quantities appearing in the Lagrange density.

\paragraph{Transformation of Fields}
Using the results summarized in Sec. \ref{sec:summary} the
transformation of the fields can be computed.

For the scalar fields one gets:
\begin{align*}
  \Theta \varphi(x) \Theta^\dagger
  &=\mathcal{C}\mathcal{P}\mathcal{T}\varphi(x)\mathcal{T}^\dagger\mathcal{P}^\dagger\mathcal{C}^\dagger\\
  &=\eta_\text{C}\eta_\text{P}\eta_\text{T}\varphi^\dagger(-x)\\
  &=\varphi^\dagger(-x)
\end{align*}
Remembering the remark about the phases in Sec. \ref{sec:summary}, all
phases are chosen to be $1$. The result agrees with
Eq. \ref{eq:trafo_law} since there is no Lorentz index and accordingly
no overall minus sign.

The calculation for the vector field reveals as desired:
\begin{align*}
  \Theta A^\mu(x) \Theta^\dagger = -{A^\mu}^\dagger(-x)
\end{align*}

Computing the transformation of the spinor field one
gets:\footnote{The $^*$ denotes complex conjugation.}
\begin{align}
\label{eq:spinor}
  \Theta \psi(x) \Theta^\dagger = -i\gamma^5\psi^{*}(-x)
\end{align}
And for the adjoint field:
\begin{align}
\label{eq:spinorad}
  \Theta \psi^\dagger(x) \Theta^\dagger = i\psi^T(-x)\gamma^{5\,\dagger}
\end{align}
These two results seem to disagree with Eq. \ref{eq:trafo_law}, but
since $\psi(x)$ is a spinor field there are still spinor indices
remaining.

Because only Lorentz invariant bilinear products of spinors without
remaining spinor indices can appear in the Lagrange density, their
transformation behaviour is studied next.

\paragraph{Transformation of Spinor Bilinear Products}
A spinor bilinear product is of the structure:
\begin{align*}
  : \overline{\psi}^A(x)\Gamma\psi^B(x) :
\end{align*}
The new indices $A$ and $B$ are different spinor fields.\footnote{It
  has to be assumed that different half integer spin fields also
  anticommute. Furthermore different integer spin fields commute and
  integer spin fields commute with half integer spin fields.} The
matrix $\Gamma$ is a $4\times 4$ complex valued matrix acting on the
spinor components.

The transformation of the spinor bilinear product is displayed in
detail here because it is an example of why normal ordering and the
spin-statistics theorem are needed for the proof.
\begin{align*}
  &\Theta:\overline{\psi}^A(x)\Gamma\psi^B(x):\Theta^\dagger\\
  &=\;:\Theta{\psi^A}^\dagger(x)\Theta^\dagger\Theta\gamma^0\Gamma\Theta^\dagger\Theta\psi^B(x)\Theta^\dagger:\\
  &=\;:i{\psi^A}^T(-x){\gamma^5}^\dagger{\gamma^0}^*\Gamma^*(-i)\gamma^5{\psi^B}^*(-x):
\end{align*}
For this first step Eq. \ref{eq:spinor}, Eq. \ref{eq:spinorad} and the
antiunitarity of $\Theta$ are needed.

Next the Dirac representation of $\gamma^5$ is used to arrive
at:\footnote{In the Dirac representation the only non-vanishing
  entries of $\gamma^5$ are the secondary diagonal elements, which are
  equal to $1$.}
\begin{align*}
  &\Theta:\overline{\psi}^A(x)\Gamma\psi^B(x):\Theta^\dagger\\
  &=\;:{\psi^A}^T(-x)({\gamma^5}^\dagger{\gamma^0}\Gamma\gamma^5)^*{\psi^B}^*(-x):
\end{align*}
The expression in parentheses has to be analyzed in detail. It can be
shown that a basis of the $4\times 4$ complex matrices can be
constructed using Dirac's $\gamma$-matrices and products of these. So
$\Gamma$ can essentially be thought of as product of $n$ matrices
$\gamma^\mu$ where $\mu=0, \ldots, 3$. By making use of the
property\footnote{The braces denote the anticommutator.}
$\{\gamma^5,\gamma^\mu\} = 0$ and ${\gamma^5}^\dagger\gamma^5=1$, the
expression in parenthesis can be rewritten:
\begin{align*}
   {\gamma^5}^\dagger{\gamma^0}\Gamma\gamma^5 &=
   {\gamma^5}^\dagger\gamma^5{\gamma^0}\Gamma(-1)^{n+1}\\
   &= {\gamma^0}\Gamma(-1)^{n+1}
\end{align*}
If $m$ is the number of remaining Lorentz indices in $\Gamma$ it can
be concluded that $(-1)^m=(-1)^n$ since a contraction involves an even
number of indices. Which means:
\begin{align*}
   {\gamma^5}^\dagger{\gamma^0}\Gamma\gamma^5
   &= {\gamma^0}\Gamma(-1)^{m+1}
\end{align*}

This is used to continue the transformation of the spinor bilinear product:
\begin{align*}
  &\Theta:\overline{\psi}^A(x)\Gamma\psi^B(x):\Theta^\dagger\\
  &= (-1)^{m+1}:{\psi^A}^T(-x)(\gamma^0\Gamma)^*{\psi^B}^*(-x):\\
  &= (-1)^{m+1}:{\psi^A_\alpha}^T(-x)(\gamma^0\Gamma)^*_{\alpha\beta}{\psi^B}^*_{\beta}(-x):\\
  &= (-1)^{m+1}:{\psi^A_\alpha}(-x)(\gamma^0\Gamma)^\dagger_{\beta\alpha}{\psi^B}^\dagger_{\beta}(-x):
\end{align*}
The lower indices are the spinor components and summation with respect
to these indices is to be understood.

Finally the two fields are exchanged. Because of the quantization
according to anticommutation relations an additional factor of $-1$
appears. Possible infinities due to $\delta$-functions in the
anticommutation relations are eliminated by the normal ordering
prescription.
\begin{align*}
  &\Theta:\overline{\psi}^A(x)\Gamma\psi^B(x):\Theta^\dagger\\
  &= (-1)^m:{\psi^B}^\dagger_{\beta}(-x)(\gamma^0\Gamma)^\dagger_{\beta\alpha}{\psi^A_\alpha}(-x):\\
  &= (-1)^m:\left(\overline{\psi}^A(-x)\Gamma\psi^B(-x)\right)^\dagger:
\end{align*}
Consequently also this term transform as desired in
Eq. \ref{eq:trafo_law}.

\paragraph{Transformation of Complex Numbers}

Because $\Theta$ is antiunitary a complex number is complex
conjugated. This agrees with Eq. \ref{eq:trafo_law} too.

\paragraph{Transformation of Differential Operators}

By writing the action of a differential operator $\partial_\mu$ on a
field as a limit one can show that the operator $\partial_\mu$ is
transformed to $-\partial_\mu$, which is in agreement with
Eq. \ref{eq:trafo_law}.

This finally establishes the desired transformation behaviour for the
whole Lagrange density.

\subsubsection{Commutation with the Hamiltonian}

Proceeding from the Lagrange density $\mathcal{L}$ to the Hamiltonian
density $\mathcal{H}$, one finds that the transformation behaviour
stays the same:
\begin{align*}
  \Theta\mathcal{H}(x)\Theta^\dagger= \mathcal{H}(-x)
\end{align*}

The result for the Hamiltonian is now obtained via integration over the
spatial dimensions:
\begin{align*}
  \Theta H(t)\Theta^\dagger
  &=\int {\rm d}^3x\; \Theta\mathcal{H}(x)\Theta^\dagger\\
  &= \int {\rm d}^3x\; \mathcal{H}(-t,-\vec{x})\\
  &= H(-t) = H
\end{align*}
Where the last equality holds because $H$ is conserved. Restating the
above means $[\Theta,H]=0$ and thus $\Theta$ is a symmetry of the
theory.

\subsection{Remarks and Consequences}

The CPT theorem is a deep result of quantum field theory because only
few assumptions about the theory are needed.

While the above proof is only formal, there also exists a proof of the
CPT theorem in the framework of axiomatic quantum field
theory.\footnote{This proof can be found in \cite{Streater1964} and
  \cite{Greenberg2003}.}

Looking at the consequences of the theorem, it can be shown that the
mass of a particle and the corresponding antiparticle are equal. The
same holds true for the lifetimes. This points out possibilities to
test the CPT theorem experimentally.

A strong consequence of a possible CPT violation would be the
violation of Lorentz invariance.\footnote{This implication is proven
  in \cite{Greenberg2002}.} But so far no CPT violation has been
observed.

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\bibitem{Streater1964} Streater, R. F., Wightman, A. S.:
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\bibitem{Greenberg2003} Greenberg, O. W.: 
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\bibitem{Greenberg2002} Greenberg, O. W.: 
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\bibitem{Grawert1959} Grawert, G., Lüders, G., Rollnik, H.: 
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\bibitem{Yamamoto2002} Yamamoto, H.: Discrete symmetries (Lecture
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\end{document}