August 27, 2024

Lectures on Introduction to Quantum Field Theory
Ghanashyam Date<EMAIL_ADDRESS><EMAIL_ADDRESS>
Chennai Mathematical Institute
H1, SIPCOT IT Park,
Siruseri, Kelambakkam 603103, INDIA.

These are lecture notes of the QFT-I course I gave in an online mode at
Chennai Mathematical Institute. The course focussed on the free
relativistic quantum fields, their interactions in the perturbative
scattering framework, standard computations of QED processes, radiative
corrections at 1-loop with renormalization and an introduction to the
toolbox of path integrals.

§ WHAT IS QUANTUM FIELD THEORY?

There is of course no short answer to this question. There are different
facets of quantum fields. An aspect that is used extensively in
condensed matter physics is the `QFT as a framework for computing
processes in interacting many body systems'. Here, each of the many
bodies typically has its internal energy states eg atoms/molecules at
various lattice sites, localized states in periodic potentials etc. The
interactions take place by making transitions among these internal
states. Since the internal states are discrete, the interactions also
involve discrete exchanges of energy/momentum/spin etc which is
described by a quantum field eg., $[\Psi(\vec{r}),
\Psi^{\dagger}(\vec{r}')] \sim \delta^3(\vec{r} - \vec{r}')$. With large
number of bodies involved, a quantum field capable of indefinite number
of transitions is a well suited framework.

There is another aspect which makes quantized fields essential
when we want to describe interactions of even few bodies in a
relativistically covariant manner. Special relativity imposes two
crucial modifications: (a) the notion of causality is modified by the
finite upper speed limit and (b) equivalence of mass and energy demands
that the rest masses be also included in the energy conservation. The
latter one allows annihilation and creation of `particles' at the
expense of energy. The framework of non-relativistic quantum mechanics
is not capable of accommodating these possibilities. Why is it so?

For this, we need to be sharper about what is meant by “relativistic

§.§ Relativistic Covariance

Special relativity posits that the space-time that is appropriate arena
for describing physical processes is the Minkowski space-time -
$\mathbb{R}^4$ with the flat metric. This allows the space-time to be
described in terms of coordinates $x^{\mu} \leftrightarrow (t, x^i)$ and
metric $\eta^{\mu\nu} = diag( -1, 1, 1, 1) = \eta_{\mu\nu}$. All
inertial observers, use such a space-time and relate their descriptions
of the experiments by using the coordinate transformations: $x^{'\mu} =
\Lambda^{\mu}_{~\nu}x^{\nu} + a^{\mu}$. The $\Lambda^{\mu}_{~\nu}$
denote the Lorentz transformations. These transformations, termed
Poincare transformations, preserve the space-time metric. Under
composition of transformations, these form the Poincare group.

In general, covariance of a `structure' with respect to some group
means that there is a homomorphism of that group, onto the invariance
group of the structure. Here, invariance group of the structure means
the group of transformations of the structure which preserves the
structure. For instance, if the `structure' is a vector space, then its
invariance group must be the group of linear transformations. For the
quantum framework, the structure is a projective Hilbert space (Hilbert
space modulo non-zero scaling) and the basic observable quantities are
the transition probabilities: $|\langle\psi|\phi\rangle|^2/(\|\psi\|
\|\phi\|)$. It is a theorem due to Wigner that the transformations of
physical states (elements of the projective Hilbert space) preserving
the transition probabilities can be represented by linear or anti-linear
transformation of the Hilbert space, $\psi' = A\psi$, satisfying
$\langle\psi'|\phi'\rangle = \langle\psi|\phi\rangle ~ \mbox{or} ~
\langle\psi'|\phi'\rangle = \langle\phi|\psi\rangle.$ The former defines
$A$ to be a unitary operator while the latter defines an anti-unitary operator. The anti-unitary operator is necessarily
anti-linear as well. The only anti-unitary operator one encounters is
the time reversal operator (more on this later). Thus
covariance requires the Hilbert space of the quantum system to carry a
unitary representation of the Poincare group.

If $g \in G \to R(g)$ denotes a group action which satisfies the
composition rule: $R(g_1\cdot g_2) = R(g_1)\cdot R(g_2)$, $R(g)$
constitutes a representation of the group $G$. On a Hilbert space, we
denote a group action as $|\psi'\rangle := g\cdot|\psi\rangle :=
U(g)|\psi\rangle$. It follows that $g_1\cdot(g_2\cdot|\psi\rangle) =
U(g_1)( U(g_2)|\psi\rangle) = U(g_1\cdot g_2)|\psi\rangle = (g_1\cdot
g_2)\cdot|\psi\rangle$. This action also induces a natural action on the
operators via $\langle\psi|g\cdot A|\psi\rangle := \langle
g^{-1}\cdot\psi|A|g^{-1}\cdot\psi\rangle \leftrightarrow A' := g\cdot A
:= U(g)AU(g)^{\dagger}$. Check that (a) this is indeed a homomorphism:
$(g_1\cdot g_2)\cdot A = g_1\cdot(g_2\cdot A)$ and (b)
$\langle\psi'|A'|\phi'\rangle = \langle\psi|A|\phi\rangle$. All
algebraic relations among the operators are preserved under this group
action.  Incidentally, this is exactly how one defines the action of
diffeomorphisms on functions on a manifold: $[g\cdot f](p) :=
f(g^{-1}\cdot p)$ (this is not the pull back definition). All this
essentially says that covariance is implemented through a representation
of the group. Which representation?

§.§ Compatibility with quantum conditions

Any particular quantum system is however distinguished by what
Dirac called quantum conditions eg. $[q^i, p_j] = i\hbar
\delta^i_{~j}$. It is on the Hilbert space carrying a representation of
the quantum conditions that the covariance group must be represented.
To appreciate what this means, let us take the specific examples of the
Galilean group and the Poincare group and try to seek unitary
representations on the Hilbert space of a single particle with the above
quantum conditions with, $i, j = 1,2,3$. Taking the generators of the
infinitesimal transformations to be represented by self-adjoint
operators, the Lie algebras of the two groups have the following
commutation relations:

\begin{eqnarray*}
\mbox{Galilean transformations} & : & x'^i = R^{i}_{~j}x^j + v^i t +
a^i~~,~~ t' = t + b ~~~,~~~  R^i_{~m}R^j_{~n}\delta^{mn} = \delta^{ij}
\mbox{Generators} & : & P_0, P_i, C_i, M_{ij}~ ; ~~~ \mbox{with
non-vanishing commutators:}\\
\big[M_{ij}, M_{kl}\big] & = & i\bigg(\delta_{ik}M_{jl} -
(i\leftrightarrow j) - (k\leftrightarrow l) + (i,k \leftrightarrow
j,l)\bigg) ; \\
\big[M_{ij}, P_k\big] & = & i\bigg(\delta_{ik}P_j - \delta_{jk}P_i\bigg)
~~ ; ~~
\big[M_{ij}, C_k\big] ~ = ~ i\bigg(\delta_{ik}C_j - \delta_{jk}C_i\bigg)
; \\
\big[C_i, P_j\big] & = & i M\delta_{ij} ~~,~~ \big[C_i, P_0\big] ~ = ~ i
P_i \ . \\
\mbox{Poincare transformations} & : &  x'^{\mu} :=
\Lambda^{\mu}_{~\nu}x^{\nu} + a^{\mu} ~~~ ,  ~~~
\Lambda^{\mu}_{~\alpha}\Lambda^{\nu}_{~\beta}\eta^{\alpha\beta} =
\eta^{\mu\nu} , \eta = diag(-1, 1, 1, 1) ;  \\
\mbox{Generators} & : & P_0, P_i, K_i, M_{ij}~ ; ~~~ \mbox{with
non-vanishing commutators:}\\
\big[M_{ij}, M_{kl}\big] & = & i\bigg(\delta_{ik}M_{jl} -
(i\leftrightarrow j) - (k\leftrightarrow l) + (i,k \leftrightarrow
j,l)\bigg) ; \\
\big[M_{ij}, P_k\big] & = & i\bigg(\delta_{ik}P_j - \delta_{jk}P_i\bigg)
~~ ; ~~
\big[M_{ij}, K_k\big] ~ = ~ i\bigg(\delta_{ik}K_j - \delta_{jk}K_i\bigg)
; \\
\big[K_i, P_j\big] & = & i \delta_{ij}P_0 ~~,~~ \big[K_i, P_0\big] ~ = ~
i P_i  ~~~ \mbox{and,} \\
\big[K_i, K_j\big] & = & i M_{ij} \ .
\end{eqnarray*}

Both have the same number of generators, 10, and almost the same Lie
algebra except that the Galilean algebra has a `central extension'
denoted by the (mass) parameter, $M$ and the `Galilean boost' generators
$C_i$ commute among themselves unlike the `Lorentz boost' generators
$K_i$ .

These can be expressed in terms of the basic operators $q^i, p_j$ as
follows[Thomas Jordan, 0810.4637 and 1st reference from it. The
reference 1 shows that if $q^i$ is taken to represent position
and transform under Galilean group accordingly, then the
Galilean brackets suffice to fix the remaining generators,
including $P$ up to a constant. This also shows that there are
no functions on the phase space which are invariant under the
Galilean group apart form constant functions.].

\begin{eqnarray*}
\mbox{Galilean} & ~~,~~ & \mbox{Poincare} \\
M_{ij} := q_ip_j - q_j p_i ~,~ P_i := p_i & , & M_{ij} := q_ip_j - q_j
p_i ~,~ P_i := p_i ~ ; \\
C_i := Mq_i ~~,~~ P_0 := \frac{\vec{p}\cdot\vec{p}}{2M} & , & K_i :=
q_iP_0 ~~,~~ P_0 := \sqrt{\vec{p}\cdot\vec{p} + M^2} ~ .
\end{eqnarray*}

The parameter $M$ is naturally identified with the mass of our particle
system. Notice that only the boost generators and the $P_0$ generators
are different in the two groups.

Ex: Verify that the above definitions of the generators, satisfy
the respective algebras, up to factors of $\hbar$.

Ex: Under infinitesimal group action, the infinitesimal change in
any operator is given by $\delta_{\vec{\epsilon}}A := i[A,
\vec{\epsilon}\cdot\vec{T}]$, where $\vec{\epsilon}$ denotes the
infinitesimal parameters and $\vec{T}$ denotes the generators. Show that
the infinitesimal changes, $\delta_{\vec{\epsilon}}q^i,
\delta_{\vec{\epsilon}}p_i$ are at the most linear in $q^i, p_i$
for the Galilean case but not so for the Poincare case for the $K_i,

In particular, the Poincare group action is non-linear on the
basic operators. This is problematic for the following reason. At the
classical level, the observables are sufficiently smooth functions on
the phase space and the set of all observables can be given the
structure of an algebra: i.e. a vector space with a multiplication
(product of two functions) defined. A similar feature holds in the
quantum case. Observables are self-adjoint operators. The set of
observables too can be given the structure of an algebra (product of two
operators). In both cases, the algebras can be thought of as being
generated by the basic observables. The action of a group on the basic
variables induces a corresponding action on all operators. There are
domain issues in the quantum case, but much more importantly, the
multiplication is non-commutative which leads to the ordering issues.

§.§ Manifest Covariance

When the basic observables transform non-linearly, it is much harder to
deduce the transformation properties of the other operators. This is
compounded further by the non-commutativity in the quantum case. When
the transformations of basic variables are linear, the non-commutativity
issue exists but is usually easier to manage. With this in mind we now
put the additional requirement that the basic observables transform linearly i.e. in effect tensorially (modulo inhomogeneous action of
translations). This is sometimes called the requirement of manifest

Invoking manifest covariance discards Poincare group action on the phase
space of a particle. The action of the Galilean group is however

This could of course be generalized to $N-$particles, i,j = 1,…3N.
We see that non-relativistic quantum theory of $N-$particles is not
capable of implementing manifest Poincare covariance. and we can expect
potential hazards in incorporating the relativistic causality and the
creation/annihilation processes.

Could we not extend the quantum conditions to a relativistically
covariant form?

For instance, define a quantum system by postulating quantum
conditions as $[q^{\mu}, p_{\nu}] = i\hbar\delta^{\mu}_{~\nu}$. Then the
Poincare generators can be taken to be $M_{\mu\nu} := q_{\mu}p_{\nu} -
q_{\nu}p_{\mu} $ and $P_{\mu} := p_{\mu}$. It is trivial to check the
Poincare Lie algebra. These also show that the basic observables are
tensors of ranks as indicated by their index positions. This could be a
model for a relativistic particle. However, now there is another problem
in quantum theory.

From the basic commutators, we can define the creation/annihilation
operators: $a^{\mu} := \Case{1}{\sqrt{2\hbar}} (q^{\mu} + i
\eta^{\mu\nu}p_{\nu})$ and its adjoint $(a^{\mu})^{\dagger}$, satisfying
the commutation relations $[a^{\mu}, (a^{\nu})^{\dagger}] =
\eta^{\mu\nu}$. As usual, the states are labeled by the eigenvalues of
the number operators and the vacuum state is defined by
$a^{\mu}|0\rangle = 0\ \forall \ \mu$. This is a Lorentz covariant form
of defining the vacuum state. It follows immediately that $\langle
0|[a^0, (a^0)^{\dagger}]|0\rangle = \|(a^0)^{\dagger}|0\rangle\|^2 =
\eta^{00} = -1$.

Additionally, if we were to interpret $x^0$ as a “time” coordinate so
that $p_0$ is interpreted as energy, then the energy eigenvalues
must necessarily be unbounded below. The Stone-von Neumann theorem
implies that if the canonical commutation relations arise from the
infinitesimal form of the Weyl relations, then the spectra of both the
conjugate operators consists of all real numbers.

Thus, the manifestly Lorentz covariant quantum conditions cannot be realized on a Hilbert space!

Let us turn attention to systems with infinitely many degrees of
freedom. The best known example is classical electrodynamics. This is
also relativistically covariant (in fact, led to special relativity!).
Its consequence, the electromagnetic waves, surprisingly show
particulate behavior.  While classically, an accelerated charge radiates
waves at the frequency of its mechanical motion, in atomic systems the
electrons only radiate at frequencies related by energy differences
between levels.  When photographic films are seen at low intensity
light, the `light marks' are localized. In interference experiments, the
interference pattern builds up grain by grain. The photo-electric effect
also shows a threshold behavior. All these are very particulate
manifestations. The observation of intensity correlations, the $(g-2)$
measurement and the Lamb shift all point to going beyond classical
electrodynamics. The validation by these high precision measurements
support the quantum electrodynamical theory - a quantum field theory.
The most recent example is of course the prediction of the Higgs
particle and its discovery.  These are positive arguments in favor of
quantum field theoretic framework being well suited for a quantum theory
in relativistic regimes.

What is a QFT framework? Let us keep in mind the familiar
classical electromagnetic field. As an example, consider a reflective
cavity in which some classical electromagnetic field exist. This field
satisfies the source free, first order in time,  Maxwell equations which
imply that the electric and magnetic fields satisfy a wave equation. Let
us put some boundary conditions (the cavity is closed, say). We can
write the general solution of the wave equation as a linear combination
of an infinite set of mode functions, which are by themselves
convenient solutions satisfying the boundary condition. Typically, this
selects a set of frequencies/wavelength. The expansion coefficients
encode the relative weightage of the mode functions. One can express the
energy and the momentum contained in the cavity field, in terms of the
expansion coefficients with the disappearance of the mode functions -
their memory remains only in the frequencies/wavelengths. Classically,
the expansion coefficients are complex numbers without any particular
restriction (except that the total energy/momentum should be finite!).
Schematically, we quantize the electromagnetic field by putting hats on
the fields or equivalently on the expansion coefficients. The
quantization procedure requires imposition of quantum conditions which
make the expansion coefficients similar to creation/annihilation
operators - and we have the quanta of the cavity field. Any
exchange of energy/momentum within the cavity or with any outside
environment is done in terms of these quanta, typically referred to as
photons. The many body facet mentioned at the beginning does
essentially this process in reverse.

It is useful to recall two equivalent ways of thinking about quantum
harmonic oscillator. We may think of it as a single particle performing
motion which is somehow “discretized” or as a collection of
“quanta”, each carrying an energy $\hbar\omega$ and the `particle
motion' being understood as consequences of emission/absorption of these
quanta. It is the latter view which is more convenient in the context of
a field.

In nutshell then, A quantum field is a means to hold or supply an
arbitrary collection of quanta and all interactions are transactions of
energy/momentum/charges etc in terms of these quanta.

This view is certainly borne out in perturbative analysis but by no
means the most general one.

§ THE POINCARE GROUP AND ITS REPRESENTATION: MASS AND SPIN/HELICITY

§.§ Poincare group, Lie algebra and Casimir invariants

The Poincare group: The Poincare group or inhomogeneous Lorentz
group is defined by the transformations of space-time coordinates,

\begin{equation}
x'^{\mu} = \Lambda^{\mu}_{~\nu}x^\nu + a^{\mu} ~,~
\Lambda^{\mu}_{~\alpha}\Lambda^{\nu}_{~\beta}\eta^{\alpha\beta}
= \eta^{\mu\nu} ~,~ \eta^{\mu\nu} = diag(-1, 1, 1, 1) =
\eta_{\mu\nu} ~,~ a^{\mu} \in \mathbb{R}^4.
\end{equation}

The defining conditions on $\Lambda^{\mu}_{~\nu}$ imply that,

\begin{eqnarray}
\eta_{\mu\nu}\Lambda^{\mu}_{~\alpha}\Lambda^{\nu}_{~\beta} =
\eta_{\alpha\beta} = \Lambda_{\nu\alpha}\Lambda^{\nu}_{~\beta} &
\Rightarrow & \delta^{\alpha}_{~\beta} =
\Lambda_{\nu}^{~\alpha}\Lambda^{\nu}_{~\beta} \\
\therefore \Lambda_{\nu}^{~\alpha} =
(\Lambda^{-1})^{\alpha}_{~\nu} ~ ~ \leftrightarrow ~ ~
(\Lambda^{-1})_{\nu}^{~\alpha} = \Lambda^{\alpha}_{~\nu} & &
\mbox{Note the index positions}
\end{eqnarray}

It is convenient to view $\Lambda^{\mu}_{~\nu}$ as matrices. It is then
important to adopt a consistent convention regarding the index
positions.  Viewed as matrices, in the last equation,
$\Lambda_{\nu}^{~\alpha}$ is the left inverse of $\Lambda$ while
$\Lambda^{\alpha}_{~\nu}$ is the right inverse of $\Lambda$.

The infinitesimal Transformations are defined by taking
$\Lambda^{\mu}_{~\nu} = \delta^{\mu}_{~\nu} + \omega^{\mu}_{~\nu}$ and
$a^{\mu} = \epsilon^{\mu}$. The defining equation for Lorentz
transformations then imply $\omega^{\mu\nu} = - \omega^{\nu\mu}\ ,
\omega^{\mu\nu} := \omega^{\mu}_{~\alpha}\eta^{\alpha\nu}$. This also
explains the parameter counting of 6 for the Lorentz and 4 for the

Thus, the Poincare group is a 10 parameter continuous group (actually a
Lie group) with the 6 parameters in $\Lambda$ constituting the Lorentz subgroup and the 4 parameters $a^{\mu}$ constituting the translations subgroup.  We denote a generic element of the group as
$(\Lambda, a)$.

Clearly $det \Lambda = \pm 1$. The transformations with positive
determinant are called proper Lorentz transformations while those
with negative one are called improper. Furthermore,
$-(\Lambda^{0}_{~0})^2 + \sum^{3}_{i=1}(\Lambda^0_{~i})^2 = -1
\Rightarrow \Lambda^0_{~0} = \pm \sqrt{1 + \sum_i(\Lambda^0_{~i})^2}$.
Those with positive $\Lambda^0_{~0}$ are called orthochronous
Lorentz transformations.  The identity transformation, being both proper
and orthochronous, the subgroup of proper, orthochronous transformations
is the continuous subgroup connected to the identity. The two improper
transformations with $\Lambda = diag(1, -1, -1, -1)$ and $\Lambda =
diag(-1, 1, 1, 1)$ are called space inversion and time
reversal transformations. It can be shown that any Lorentz
transformation can be obtained from a proper, orthochronous
transformation followed by space inversion and/or time reversal.

Group composition: The composition of two Poincare transformations
is defined as: $(\Lambda_2, a_2)\cdot(\Lambda_1, a_1) =
(\Lambda_2\Lambda_1, \Lambda_2 a_1 + a_2)$. It follows that
$(\mathbb{1}, 0)$ is the identity and $(\Lambda, a)^{-1} =
(\Lambda^{-1}, -\Lambda^{-1}a)$.

Check: $~(\Lambda, a)\cdot(\mathbb{1}, 0)\cdot(\Lambda,
a)^{-1} = (\mathbb{1}, \Lambda b)$   i.e.   $g H g^{-1} \in H ~ \forall~
g$. Such a subgroup $H$ is called an invariant or normal subgroup.
Thus the translations subgroup is a normal subgroup of the Poincare
group and the Poincare group is said to be a semi-direct product of
the Lorentz and the translations groups.

Check: Any group element can be uniquely written as:

\[(\Lambda, a) = (\Lambda, 0)\cdot(\mathbb{1}, \Lambda^{-1}a) =
(\mathbb{1}, a)\cdot(\Lambda, 0)\]

While it is possible to proceed with the abstract identification of the
Lie algebra via the vector fields generating the infinitesimal
transformation, we can simplify by directly considering a unitary
representation of the Poincare group. That is, a set of unitary
operators $U(g): \cal{H} \to \cal{H}$ such that (i) $U^{\dagger}(g)U(g)
= \mathbb{1} = U(g)U(g)^{\dagger}$ and (ii) $U(g_2 g_1) = U(g_2)
U(g_1)~, ~ \forall g's \in G$. To allow for the possibility of infinite
dimensional representations, we use operators on Hilbert space. We also
restrict to irreducible representations i.e. the representation
space has no proper subspace which is invariant under the $U(g)$
operators. It is a result (from Schur's lemma) that if $U(g)$ is a
unitary, irreducible representation, then the only bounded operator that
commutes with all $U(g)$ operators is a multiple of the identity
operator. Its converse also holds, namely, if the only operator that
commutes with all the $U(g)$'s is a multiple of the identity operator,
then the unitary representation is irreducible.  A simple application of
this result is that unitary, irreducible representations of abelian
groups are one dimensional. Hence the translation subgroup has only
1-dimensional irreducible representations.

We define the infinitesimal generators as,

\begin{equation}
U(\Lambda, a) = U(\delta^{\mu}_{~\alpha} +
\omega^{\mu}_{~\alpha}, \epsilon^{\mu}) := \mathbb{1} +
\frac{i}{2}\omega_{\mu\nu}M^{\mu\nu} - i \epsilon_{\mu}P^{\mu} +
o(\omega^2, \epsilon^2) ~ ~,~ ~ M, P \mbox{~are self-adjoint .}
\end{equation}

The commutation relations among the generators are obtained from using
the homomorphism property, $U(\Lambda, a)^{-1} U(\lambda, b) U(\Lambda,
a) = U[ (\Lambda, a)^{-1}\cdot(\lambda, b)\cdot(\Lambda, a) ]$ and
taking $\lambda = \mathbb{1} + \omega, b = \epsilon$. Using the
definition of the generators, it follows,

\begin{eqnarray} \omega_{\mu\nu}U(\Lambda, a)^{-1} M^{\mu\nu} U(\Lambda,
a) & = &
(\Lambda^{-1}\omega\Lambda)_{\alpha\beta}M^{\alpha\beta} - 2
(\Lambda^{-1}\omega a)_{\alpha}P^{\alpha} \\
\epsilon_{\mu}U(\Lambda, a)^{-1}P^{\mu}(\Lambda, a) & = &
(\Lambda^{-1}\epsilon)_{\alpha}P^{\alpha}\ .  ~ ~ \mbox{Using}
\nonumber \\
(\Lambda^{-1}\epsilon)_{\alpha} & = &
\eta_{\alpha\sigma}(\Lambda^{-1})^{\sigma}_{~\mu}\epsilon^{\mu}
= (\Lambda^{-1})_{\alpha}^{~\mu}\epsilon_{\mu} =
\Lambda^{\mu}_{~\alpha}\epsilon^{\alpha} , \\
(\Lambda^{-1}\omega\Lambda)_{\alpha\beta} =
\omega_{\mu\nu}\Lambda^{\mu}_{~\alpha}\Lambda^{\nu}_{~\beta} &
\mbox{and} & (\Lambda^{-1}\omega a)_{\alpha} =
\omega_{\mu\nu}\Lambda^{\mu}_{~\alpha}a^{\nu} \ ; \nonumber \\
U(\Lambda, a)^{-1}M^{\mu\nu}U(\Lambda, a) & = &
\Lambda^{\mu}_{~\alpha}\Lambda^{\nu}_{~\beta}M^{\alpha\beta} -
(\Lambda^{\mu}_{~\alpha}a^{\nu} -
\Lambda^{\nu}_{~\alpha}a^{\mu})P^{\alpha} ,
\label{LorentzTensors} \\
U(\Lambda, a)^{-1} P^{\mu} U(\Lambda, a) & = &
\Lambda^{\mu}_{~\nu}P^{\nu} \nonumber
\end{eqnarray}

The last two equations show that under the `adjoint' action of the
group, the generators transform as Lorentz tensors.  To deduce the
commutation relations, take $\Lambda, a$ to be infinitesimal i.e.
$U(\Lambda, a)^{\pm 1} = \mathbb{1} \pm
\frac{i}{2}\omega_{\alpha\beta}M^{\alpha\beta} \mp i
\epsilon_{\alpha}P^{\alpha}. $ Substitution and reading off coefficients

\begin{eqnarray}
\left[M^{\mu\nu}, M^{\alpha\beta}\right] & = & i
\left(\eta^{\mu\alpha}M^{\nu\beta} - (\mu \leftrightarrow \nu)
- (\alpha \leftrightarrow \beta) + (\mu,\alpha \leftrightarrow
\nu,\beta) \right) \\
\left[M^{\mu\nu}, P^{\alpha}\right] & = & i
\big(\eta^{\mu\alpha}P^{\nu} - \eta^{\nu\alpha}P^{\mu}\big) \\
\left[P^{\mu}, P^{\nu}\right] & = & 0\ .
\end{eqnarray}

It is convenient to introduce the notation, $K^i := M^{i0}$ and $J_i :=
\frac{1}{2}\epsilon_{ijk}M^{jk} \leftrightarrow M^{ij} =
\epsilon^{ijk}J_k$ and also $H := P^0.$ Then the Poincare commutators
take the form (non-zero commutators only),

\begin{eqnarray}\label{PoincareLieAlgebra}
\left[J_i, J_j\right] ~ = ~ i\epsilon_{ij}^{~ ~k}J_k & , &
\left[J_i, K_j\right] ~ = ~ i\epsilon_{ij}^{~ ~ l}K_l \\
\left[K_i, K_j\right] ~ = ~ - i\epsilon_{ij}^{~ ~l}J_l & , &
\left[J_i, P_j\right] ~ = ~ i\epsilon_{ij}^{~ ~ k}P_k \\
\left[K_i, P_j\right] ~ = ~ - iH \delta_{ij} & , & \left[K_i,
H\right] ~ = ~ iP_i
\end{eqnarray}

Define the Pauli-Lubanski vector, $W_{\mu} :=
\frac{1}{2}\epsilon_{\mu\nu\alpha\beta}M^{\nu\alpha}P^{\beta}~ , ~
\epsilon_{0123} := 1$. It follows that $[P^{\mu}, W^{\nu}] = 0\ ,\
[M^{\mu\nu}, W^{\lambda}] = i\big(\eta^{\mu\alpha}W^{\nu} -
\eta^{\nu\alpha}W^{\mu}\big)$ and $[W_{\mu}, W_{\nu}] = -i
\epsilon_{\mu\nu\alpha\beta}W^{\alpha}P^{\beta}$. These relations imply
that $P^2 := \eta_{\mu\nu}P^{\mu}P^{\nu} ~,~ W^2 :=
\eta^{\mu\nu}W_{\mu}W_{\nu}$ commute with all the generators of the
Poincare group and hence also with all the group elements of the proper,
orthochronous Poincare group. These are the two independent Casimir
invariants of the Poincare group and must be multiples of identity
operator in irreducible representations. These multiples serve to label
the unitary, irreducible representations.

§.§ Representations of the proper, orthochronous Poincare group:

Each of its representations induces a representation of its Lie algebra
<ref>. Since the $P^{\mu}$'s commute, their
simultaneous eigenvectors can be taken as a group of labels for a basis.

\begin{equation}
P^{\mu}|k^{\mu}, \xi\rangle ~ = ~ k^{\mu}|k^{\mu}, \xi\rangle ~
~ \Rightarrow  ~  P^2 |k^{\mu}, \xi\rangle ~ = ~ k^2|k^{\mu},
\xi\rangle ~ ~ , ~ ~ k^2 = \eta_{\mu\nu}k^{\mu}k^{\nu} \ .
\end{equation}

Here, $\xi$ collectively labels the degenerate eigenvectors and $k^2$ is
the value of the Casimir $P^2$. Thus, first level of classification is
done by the value of the Lorentz invariant $k^2$. There are four classes
that arise naturally:

(i) $m^2 := - k^2 > 0$ (massive), (ii) $k^2 = 0, k^{\mu} \neq 0$
(massless), (iii) $k^2 > 0$ (tachyonic), and $~ ~ ~ ~ ~ $(iv) $k^{\mu} =
0$ (vacuum). The vacuum representation is just a single vector. The
tachyonic representations have not showed up in experiments yet.

$\bullet$ Since $P^{\mu}$ transforms as a Lorentz vector, so does
$k^{\mu}$ and for $k^2 \le 0$, the $sgn(k^0)$ distinguishes two further
subclasses of the massive and the massless representations.

We restrict to the massive and the massless representations with
$k^0 > 0$ as being physically relevant.

$ \bullet$ Under a Lorentz transformation, $(\Lambda, 0)$, $k^{\mu}$
goes to another point on the hyperboloid $k^2 = $ constant. How does the
collective label $\xi$ transform? We would like to deduce the action of
$U(\Lambda, a)$ on the vectors $|k^{\mu}, \xi\rangle$. Since
$U(\mathbb{1}, a) = exp{ -i a^{\mu}P_{\mu}} $, we already know that
$U(\mathbb{1}, a)|k^{\mu}, \xi\rangle = exp{ -i a^{\mu}k_{\mu}}|k^{\mu},
\xi\rangle$ and  we focus on the Lorentz transformations $U(\Lambda, 0)
=: U(\Lambda)$.

The Lorentz transformation relation (<ref>) imply,

\begin{eqnarray}
P^{\mu}\big( U(\Lambda)|k, \xi\rangle\big) & = &
\Lambda^{\mu}_{~\nu}U(\Lambda)P^{\nu}|k, \xi\rangle =
\big(\Lambda^{\mu}_{~\nu}k^{\nu}\big) \big(U(\Lambda)|k,
\xi\rangle\big) ~ ~ ~ ~ \Rightarrow \nonumber \\
U(\Lambda)|k, \xi\rangle & = & \sum_{\xi'}C_{\xi, \xi'}|\Lambda
k, \xi'\rangle \ .
\end{eqnarray}

We want to find the coefficients $C_{\xi, \xi'}$ corresponding to
irreducible representations of the Lorentz group. In principle,
these coefficients could (and do) depend on $k$ label, however, their
number (the degree of degeneracy) must be the same over a given
hyperboloid. This is essentially a continuity argument - the label $k$
changes continuously while the degree of degeneracy is integer valued.

Let $\hat{k}$ be some convenient, fixed vector on the $k^2 \leq 0$
hyperboloid and let $L(k)$ be some arbitrarily chosen Lorentz
transformation such that $k^{\mu} = L(k)^{\mu}_{~\nu}\hat{k}^{\nu}$. The
Lorentz transformation $L$ depends explicitly on $k$ (and also
implicitly on $\hat{k}$) and also is not unique. For instance, for
$\hat{k} = (M, \vec{0})$, any $k$ can be obtained by a combination of a
rotation and a boost transformation. Having made a choice of $L(k)$,

\begin{equation} \label{NkDefn}
|k, \xi\rangle := N(k)U(L(k))|\hat{k}, \xi\rangle
\end{equation}

Since this is a definition, the same label $\xi$ appears on both sides.
It follows,

\begin{equation*}
U(\Lambda)|k, \xi\rangle = N(k) U(\Lambda
L(k))|\hat{k},\xi\rangle ~ = ~ N(k)\left[ U(L(\Lambda k))\cdot
U( (L(\Lambda k))^{-1} )\right] U(\Lambda)\cdot U(L(k))|\hat{k},
\xi\rangle \ .
\end{equation*}

The last three factors of $U$ combine to give $U(L^{-1}(\Lambda
k))\cdot\Lambda\cdot L(k)$, which acting on $\hat{k}$ takes $\hat{k} \to
k \to \Lambda k \to \hat{k}$ since $(\Lambda k)^{\mu} =[L(\Lambda
k)]^{\mu}_{~\nu}\hat{k}^{\nu}$.  Denote: $W(\Lambda, k) :=
L^{-1}(\Lambda k)\cdot\Lambda\cdot L(k)$. It follows that $W(\Lambda,
k)\hat{k} = \hat{k}\ \forall \Lambda$.

The Lorentz transformations, $W(\Lambda, k)$ leaving the $\hat{k}$
invariant, form a subgroup of the Lorentz transformations. It is known
as the little group of $\hat{k}$ or stability subgroup of
$\hat{k}$. Defining a representation $D(W)$ by, $U(W(\Lambda,
k))|\hat{k}, \xi\rangle  := \sum_{\xi'}D_{\xi'\xi}(W)|\hat{k},
\xi'\rangle$ , we get,

\begin{eqnarray}
U(\Lambda)|k,\xi\rangle & = & N(k)U(L(\Lambda
k))\sum_{\xi'}D_{\xi' \xi}(W)|\hat{k}, \xi'\rangle \nonumber \\
& = & N(k)\sum_{\xi'}D_{\xi' \xi}(W)\left[U(L(\Lambda
k))|\hat{k}, \xi'\rangle.\right] \nonumber \\
\therefore U(\Lambda)|k,\xi\rangle & = & \frac{N(k)}{N(\Lambda
k)}\sum_{\xi'}D_{\xi' \xi}(W)|\Lambda k, \xi'\rangle
\end{eqnarray}

In the last equation, we have used the definition (<ref>).  The
irreducible representations of the little group can thus be used to
characterize the irreducible representations of the Lorentz group. We
are not done yet, the normalization factors $N(k)$ need to be

Determination of $N(k)$:

For different $\hat{k}, \hat{k}'$, the corresponding little groups are
different. The unitarity of the Poincare (and hence of the Lorentz
group) representation implies the representations of the little groups
be also unitary. We can choose $\langle\hat{k}', \xi'|\hat{k},
\xi\rangle = \delta_{\xi, \xi'}\delta^3(\hat{k} - \hat{k}')$ and we
would like this to be preserved under Lorentz transformations i.e. with
the hatted $k$'s being replaced by their Lorentz transforms. This is
possible only for a specific choice of $N(k)$ which we determine now.

Note that this requirement associates the ortho-normalization with the
hyperboloids determined by $\hat{k}^2, (\hat{k}')^2$. Since a delta
function is defined in the context of an integration, we are implicitly
envisaging an integration over the hyperboloids and not over
$\mathbb{R}^4$. The integration is to be Lorentz invariant. This can be
inferred as follows.

\begin{eqnarray}
\int_{\mathrm{mass shell}}\theta(k^0)f(k) & := &
\int_{\mathbb{R}^4}d^4k \theta(k^0)\delta(k^2 + m^2)f(k) ~ = ~
\int d^3k dk^0 \theta(k^0)\delta(- (k^0)^2 + \vec{k}\cdot\vec{k}
+ m^2)f(k) \nonumber \\
& = & \int d^3k \frac{f(\sqrt{\vec{k}\ ^2 + m^2}, \vec{k})}{2
\sqrt{\vec{k}\ ^2 + m^2}}  ~ ~ ~ \because \delta(f(x)) = \sum_i
\frac{\delta(x - x_i)}{|f'(x_i)|} \ .
\end{eqnarray}

Although there are two roots from the delta function of the mass shell,
the $\theta(k^0)$ picks out only one term with $\omega_k := k^0 := +
\sqrt{\vec{k}\cdot\vec{k} + m^2}$. The $\frac{d^3k}{2\omega_k}$ is the
Lorentz invariant volume element (measure) on the mass shell.
Since we have,

\begin{equation}
f(\vec{k}) = \int_{\mathbb{R}^3}d^3k'
f(\vec{k}')\delta^3(\vec{k} - \vec{k}') = \int_{\mathbb{R}^3}
\frac{d^3k'}{2\omega_k} f(\vec{k})\left[2\omega_k
\delta^3(\vec{k} - \vec{k}')\right] ,
\end{equation}

we identify the invariant delta function on the mass shell as:

\[
\delta_{inv}^3(\vec{k} - \vec{k}') := 2\omega_k\delta^3(\vec{k} -
\vec{k}') = 2\sqrt{\vec{k}^2 + m^2}\delta^3(\vec{k} - \vec{k}')  .
\]

Let $k = L(k)\hat{k}$ and define $k' := L(k)\hat{k}'$ with
the same $L(k)$.

\[
\langle k', \xi'|k, \xi\rangle = N^*(k')N(k)\langle \hat{k}',
\xi'|U^{\dagger}(L(k))\ U(L(k))|\hat{k}, \xi\rangle =
|N(k)|^2\delta_{\xi',\xi}\delta^3(\hat{k}' - \hat{k}) .
\]

We have used unitarity of $U(L(k))$.  But we also know that
$\delta^3(\vec{k}' - \vec{k})$ and $\delta(\hat{k}' - \hat{k})$ are
related by the same Lorentz transformation. Therefore using invariant
delta function we have,

\begin{eqnarray}
\omega_k\delta^3(\vec{k}' - \vec{k}) & = &
\omega_{\hat{k}}\delta^3(\hat{k}' - \hat{k})  \\
\therefore \langle k', \xi'|k, \xi\rangle & = &
\delta^3(\vec{k}' - \vec{k}) ~ ~ \mbox{and the choice,} \\
|N(k)| & := & \sqrt{\frac{\hat{k}^0}{k^0}} ~ ~ \Rightarrow ~ ~
\\
\langle k', \xi'|k, \xi\rangle & = &
\delta_{\xi',\xi}\delta^3(\vec{k} - \vec{k}') ~ ~ \mbox{and}~
U(\Lambda)|k, \xi\rangle & = & \sqrt{\frac{(\Lambda
k)^0}{k^0}}\sum_{\xi'}D_{\xi'\xi}(W(\Lambda, k))|\Lambda k,
\xi'\rangle
\end{eqnarray}

§.§ Little groups

It remains to determine the little groups for the various cases.

The vacuum representation ($\mathbf{k^{\mu} = 0}$) :

The little group is all of the Lorentz group.  However the
representation is the trivial one: $U(\Lambda) = \mathbb{1}, \ \forall ~
\Lambda$.

Massive representation ($\mathbf{k^2 = - m^2}$) :
Choose $\hat{k}^{\mu} = (M, \vec{0})$. The little group is defined by

$\Lambda^{\mu}_{~\nu}\hat{k}^{\nu} = \hat{k}^{\mu} \Rightarrow
\Lambda^0_{~0} = 1, \Lambda^i_{~0} = 0$. The defining condition on
$\Lambda, \eta^{00} = \Lambda^0_{~0}\Lambda^{0}_{~0}\eta^{00} +
(\Lambda^0_{~i})^2\delta^{ii},$ gives $\Lambda^i_{~0} = 0$. Hence
$\Lambda$ is block diagonal, with $\delta^{ij} = \Lambda^{i}_{~k}
\Lambda^j_{~l}\delta^{kl}$ and the little group is $SO(3)$, the group of
rotations. Its unitary representations are all finite dimensional and
labeled by its Casimir, $J^2 = j(j+1), j \in \mathbb{N}/2$. The half
integer ones come from the double cover $SU(2)$. The $D_{\xi'\xi}$ are
the usual $(2s+1)$ dimensional representations.

The representations of this class are thus labeled by `mass', m and
`spin' s.

Massless representation ($\mathbf{k^2 = 0}$) :

Choose $\hat{k}^{\mu} = (k, 0, 0, k), k > 0$. It is more convenient to
determine the generators of the little group. These are some linear
combinations of the Lorentz generators $M^{\alpha\beta}$. Consider the

\begin{eqnarray*}
\left[\epsilon_{\alpha\beta}M^{\alpha\beta}, P^{\lambda}\right] & = &
i\big(\epsilon^{\lambda}_{~\beta}P^{\beta} -
\epsilon_{\alpha}^{~\lambda}P^{\alpha}\big) ~ ~ ~ \Rightarrow \\
\left[\epsilon_{\alpha\beta}M^{\alpha\beta}, P^{\lambda}\right]|\hat{k},
\xi\rangle & = & i\big(\epsilon^{\lambda}_{~\beta}\hat{k}^{\beta} -
\epsilon_{\alpha}^{~\lambda}\hat{k}^{\alpha}\big)|\hat{k}, \xi\rangle ~
= ~ 2i\epsilon^{\lambda}_{~\alpha}\hat{k}^{\alpha}.
\end{eqnarray*}

The l.h.s. of the above equation vanishes because, the generators of the
little group, $\epsilon\cdot M$ leave the eigenstate of $\hat{k}$
invariant. The r.h.s. Then gives the conditions on $\epsilon$'s.
Explicitly, $0 = \epsilon^{\lambda}_{~\alpha}\hat{k}^{\alpha} =
(\epsilon^{\lambda}_{~0} + \epsilon^{\lambda}_{~3})k) \Rightarrow
\epsilon_{\alpha 0} + \epsilon_{\alpha 3} = 0$. This in turn gives
$\epsilon_{03} = 0, \epsilon_{01} = \epsilon_{13}, \epsilon_{02} =
\epsilon_{23}$. Hence,

\begin{eqnarray}
\frac{1}{2}\epsilon_{\alpha\beta}M^{\alpha\beta}|_{little} & = &
\epsilon_{12}M^{12} + \epsilon{13}(M^{13} - M^{01}) +
\epsilon_{23}(M^{23} - M^{02}) \nonumber \\
& := & \epsilon_1(J_1 - K_2) + \epsilon_2(-J_2 - K_2) +
\epsilon_3J_3
\end{eqnarray}

It is conventional to denote: $A := J_2 + K_1, B := -J_1 + K_2$ and
write the little algebra of $\hat{k} = k(1, 0, 0, 1)$ as,

\begin{equation}
[A, B] = 0 ~,~ [J_3, A] = iB ~,~ [J_3, B] = -iA \ .
\end{equation}

This is the Euclidean group in 2 dimensions: two translations and one
rotation. Its representations are not as familiar. To study these,
define $R(\theta) := exp{\ i\theta J_3}$. It is easy to check that
$R(\theta)\ A\ R^{-1}(\theta) = A cos(\theta) - B sin(\theta)$ and
$R(\theta)\ B\ R^{-1}(\theta) = A sin(\theta) + B cos(\theta)$.

Consider a representation $D$ in which $A|\hat{k}, a, b\rangle =
a|\hat{k}, a, b\rangle$ and $B|\hat{k}, a, b\rangle = b|\hat{k}, a,
b\rangle$. Define $|\hat{k}, a, b, \theta\rangle := R(\theta)|\hat{k},
a, b\rangle$. It follows that,

\[
A|\hat{k}, a, b, \theta\rangle = (a cos\theta - b
sin\theta)|\hat{k}, a, b, \theta\rangle  ~ , ~ ~ B|\hat{k}, a, b,
\theta\rangle = (a sin\theta + b cos\theta)|\hat{k}, a, b,
\theta\rangle  ~ .
\]

Hence, if $a, b$ are non-zero, then for each $\theta$, the states
$|\hat{k}, a, b, \theta\rangle$ are simultaneous eigenstates of $A$ and
$B$. Such a representation has the
continuous parameter $\theta$ and no such additional parameter is seen
in observations. To avoid such a parameter, we restrict the
representation of the Little group to those for which $a = b = 0$ or $A,
B$ vanish in the representation. The little group then effectively
reduces to just $U(1)$ generated by $J_3$. Its irreducible
representations are labeled by half integers. We denote these
representations as, $J_3|\hat{k}, \sigma\rangle = \sigma|\hat{k},
\sigma\rangle, \sigma \in \frac{1}{2} \mathbb{Z}$. The label $\sigma$ is
called helicity. The one dimensional representation is denoted as
$D_{\sigma'\sigma}$. The representations with $a, b$ nonzero, are called
“continuous spin representations”.

In summary:

\begin{eqnarray} \label{ReprnSummary}
|k, \sigma\rangle & := &
\sqrt{\frac{\hat{k}^0}{k^0}}U(L(k))|\hat{k}, \sigma\rangle \\
\mbox{\underline{For $m > 0$:}} & & \nonumber \\
U(\Lambda)|k, \sigma\rangle & = & \sqrt{\frac{(\Lambda
k)^0}{k^0}} \sum_{\sigma'}
D^j_{\sigma\sigma'}\left[W(\Lambda,k)\right] |\Lambda k,
\sigma'\rangle  ~ ~ ~ where, \nonumber \\
W(\Lambda, k) & := & L^{-1}(\Lambda k)\cdot\Lambda\cdot L(k) ~ ~
~ , ~ ~ ~ L(k)\hat{k} := k ~ ~ ~ \sigma ~\mbox{is the $j_3$
eigenvalue} \\
\mbox{\underline{For $m = 0$:}} & & \nonumber \\
U(\Lambda)|k, \sigma\rangle & = & \sqrt{\frac{(\Lambda
k)^0}{k^0}} exp\left\{i \sigma\theta(\Lambda, k)\right\}|\Lambda
k, \sigma\rangle  ~ ~ ~ where, \mbox{$\sigma$ is the helicity
label,}\nonumber \\
W(\Lambda, k) & := & L^{-1}(\Lambda k)\cdot\Lambda\cdot L(k) ~
:= ~ S(\alpha(\Lambda, k), \beta(\Lambda, k))\ R(\theta(\Lambda,
\end{eqnarray}

Let us evaluate the second Casimir invariant of the Poincare group,
$W^2$. For this we evaluate $W_{\mu}|\hat{k}, \sigma\rangle$.

(i) $\mathbf{-\hat{k}^2 = m^2 \neq 0}$: For $\hat{k} = (m, 0, 0, 0)$,

$W_{\mu}|\hat{k}, \sigma\rangle = \frac{1}{2}\epsilon_{\mu\nu\alpha
0}M^{\nu\alpha} m |\hat{k}, \sigma\rangle$. This vanishes for $\mu = 0$
and for $\mu = i$, we get $W_{i}|\hat{k}, \sigma\rangle =
-\frac{1}{2}\epsilon_{0ijk}M^{jk} m |\hat{k}, \sigma\rangle = - m
J_i|\hat{k}, \sigma\rangle$. Therefore,

\[
W^2|\hat{k}, \sigma\rangle = m^2\ j(j+1)|\hat{k}, \sigma\rangle.
\]

(ii) $\mathbf{-\hat{k}^2 = 0}$: For $\hat{k} = k(1,0,0,1)$,

$W_{\mu}|\hat{k}, \sigma\rangle = \frac{1}{2}(\epsilon_{\mu\nu\alpha 0}
+ \epsilon_{\mu\nu\alpha 3}M^{\nu\alpha})\ k |\hat{k}, \sigma\rangle$.
This implies,

\begin{eqnarray}
W_0|\hat{k}, \sigma\rangle ~ = ~ k\sigma |\hat{k}, \sigma\rangle & , &
W_1|\hat{k}, \sigma\rangle ~ = ~ k(-J_1 + K_2) |\hat{k}, \sigma\rangle =
0 \\
W_3|\hat{k}, \sigma\rangle ~ = ~ - k\sigma |\hat{k}, \sigma\rangle  & ,
& W_2|\hat{k}, \sigma\rangle ~ = ~ k(-J_2 - K_1) |\hat{k}, \sigma\rangle
= 0 .  \end{eqnarray}

Hence $W^2|\hat{k}, \sigma\rangle = [- (k\sigma)^2 + (-k\sigma)^2]
|\hat{k}, \sigma\rangle = 0$, for our choice of representations.

Note that the action of the translations subgroup on the Poincare
representation is given by, $U(\mathbb{1},a)|k, \sigma\rangle =
exp{-ia^{\mu} k_{\mu}}|k, \sigma\rangle$ while that of the Lorentz group
is given in the box. Since any element of the Poincare group can be
uniquely written as a product of a Lorentz transformation times a
translation, we have specified the full action. Since this action is
determined by the choice of the unitary, irreducible representation of
the little group, the Poincare representation is said to be induced by a representation of the little group.

These representations are taken to identify elementary

§.§ The Discrete Subgroups and their actions

Recall that we have two improper Lorentz transformations, the space
inversion $\cal{P}: \Lambda^{\mu}_{~\nu} = diag(1, -1, -1, -1)$ and the
time reversal $\cal{T}: \Lambda^{\mu}_{~\nu} = diag(-1, 1, 1, 1)$. All
Lorentz transformations can be generated from the proper, orthochronous
transformations combined with either or both of these.  Clearly these
transformations are order 2 i.e. $\cal{P}^{-1} = \cal{P}$ and
$\cal{T}^{-1} = \cal{T}$. Let us assume that these have an action on the
physical state space preserving probabilities. Then by Wigner's theorem,
these are represented either as linear and unitary or anti-linear and
anti-unitary operators on the Hilbert space. Let us use the same
symbols to denote the corresponding operators which should be clear from
the context.

The homomorphism property,

\begin{equation}\label{HomomorphismProp}
U(\Lambda, a)^{-1}U(\lambda, b)U(\Lambda, a) = U\left(
(\Lambda,a)^{-1}\cdot(\lambda,b)\cdot (\lambda,a)\right)
\end{equation}

gives us the action of space inversion and time reversal on the proper,
orthochronous transformations. For $a = 0$ and $\Lambda = \cal{P}
~\mathrm{or}~ \cal{T}$ leads to:

\begin{equation} \label{DiscreteOnPoincare}
\cal{P}^{-1}U(\lambda, b)\cal{P} ~ = ~
U(\cal{P}^{-1}\lambda\cal{P}, \cal{P}^{-1}b) ~ ~ , ~ ~
\cal{T}^{-1}U(\lambda, b)\cal{T} ~ = ~
U(\cal{T}^{-1}\lambda\cal{T}, \cal{T}^{-1}b) .
\end{equation}

The same homomorphism (<ref>), together with the
infinitesimal form, $U(\mathbb{1} + \omega, \epsilon) = \mathbb{1} +
\Case{i}{2}\omega_{\alpha\beta}M^{\alpha\beta} -i
\epsilon_{\alpha}P^{\alpha}$ for $(U(\lambda,b)$, gives us the

\[
U^{-1}(\Lambda) M^{\mu\nu}U(\Lambda) =
\Lambda^{\mu}_{~\alpha}\Lambda^{\nu}_{~\beta}M^{\alpha\beta} ~
~,~ ~ U^{-1}(\Lambda) P^{\mu} U(\Lambda) =
\Lambda^{\mu}_{~\nu}P^{\nu} .
\]

Following the same steps as before but not canceling the factors
of $i$ to allow for anti-linearity, we get,

\begin{eqnarray}
\cal{P}^{-1}(iP^0)\cal{P} = iP^0  & , &
\cal{P}^{-1}(iP^i)\cal{P} = -iP^i  \label{InversionDefn}\\
\cal{T}^{-1}(iP^0)\cal{T} = -iP^0  & , &
\cal{T}^{-1}(iP^i)\cal{T} = iP^i \label{TimeReversalDefn}
\end{eqnarray}

Let if possible $\cal{P}$ be anti-linear. Then it anti-commutes
with $P^0$ which represents the energy. This in turn means that for
every positive energy state, there is a negative energy state. This
conflicts with the energy being bounded below. Identical implication
follows if $\cal{T}$ is linear!  Hence, the space inversion
operator must be linear and unitary while the time reversal operator
must be anti-linear and anti-unitary.

With this understood, the action of these discrete operators on the
Poincare generators is given by,

\begin{eqnarray}\label{DiscreteOnGenerators}
\cal{P}^{-1}P^0\cal{P} = P^0 ~,~ \cal{P}^{-1}P^i\cal{P} = - P^i & , &
\cal{P}^{-1}J_i\cal{P} = J_i ~,~ \cal{P}^{-1}K_i\cal{P} = - K_i
\label{SpaceInversionAction}\\
\cal{T}^{-1}P^0\cal{T} = P^0 ~,~ \cal{T}^{-1}P^i\cal{T} = - P^i & , &
\cal{T}^{-1}J_i\cal{T} = - J_i ~,~ \cal{T}^{-1}K_i\cal{T} =  K_i
\label{TimeReversalAction}
\end{eqnarray}

From these defining actions, we can determine how these operators act on
the unitary, irreducible representations of the Poincare group.

It is easy to see that under space inversion and time reversal, the two
Casimir invariants are invariant: $\cal{P}^{-1}P^2\cal{P} = P^2$ and
$\cal{P}^{-1}W^2\cal{P} = W^2$. This is obvious from the action of these
operators on the generators which transform as their tensor indices
indicate. The Casimir invariants are Lorentz scalars (not
pseudo-scalars) and hence invariant under inversions and time reversal.
Thus both parity and time reversal will not mix different irreducible
representations. To evaluate their actions on individual vectors, we
note that the irreducible representations are characterized in terms of
$\hat{k}$ vector and an eigenvalue(s) of $J_3$ generator.

Action of $\cal{P}:$

$m > 0$:

Here, $\hat{k} = (m, 0, 0, 0)$ and $\sigma = -s, -s+1, \ldots s-1, s$,
$s$ being the spin. Since $\cal{P}$ commutes with $P^0$, anti-commutes
with $P^i$ and commutes with $J_3$, we see that $\cal{P}|\hat{k},
\sigma\rangle$ also has the same eigenvalues of the $P^{\mu}, J_3$.
Hence, $\boxed{\cal{P}|\hat{k}, \sigma\rangle = \eta_{\sigma}|\hat{k},
\sigma\rangle}$, with $\eta_{\sigma}$ being a phase (since the vectors
are non-degenerate). Next, from the usual angular momentum algebra, we

\begin{eqnarray*}
(J_1 \pm iJ_2)|\hat{k}, \sigma\rangle & = & \sqrt{(s\mp\sigma)(s
\pm \sigma + 1)}|\hat{k},\sigma\pm 1\rangle ~ \Rightarrow \\
(J_1 \pm iJ_2)(\cal{P}|\hat{k}, \sigma\rangle) & = &
\sqrt{(s\mp\sigma)(s \pm \sigma + 1)}(\cal{P}|\hat{k},\sigma\pm
1\rangle) ~ \Rightarrow ~ \eta_{\sigma} = \eta_{\sigma\pm 1} \ .
\end{eqnarray*}

Thus $\eta_{\sigma}$ is independent of $\sigma$ and equal $\pm 1$ since
$\cal{P}^2 = \mathbb{1}$. The phase $\eta$ is called the intrinsic
parity of the representation. What about action on the states $|k,
\sigma\rangle$ ? Recall that these states are obtained as

\[
|k, \sigma\rangle =
\sqrt{\frac{\hat{k}^0}{k^0}}U(L(k))|\hat{k},\sigma\rangle ~ ~,~
~ k := L(k)\hat{k} ~ ~ ~ \mbox{for a chosen $L(k)$}.
\]

Acting by the matrix $\cal{P}$ on the defining equation for the Lorentz
transformation $L(k)$, gives $(\cal{P}k) =
(\cal{P}L(k)\cal{P}^{-1})(\cal{P}\hat{k})$ which implies that
$(\cal{P}L(k)\cal{P}^{-1}) = L(\cal{P}k)$. The homomorphism property,
(<ref>), then gives $\cal{P}^{-1}U(L(k))\cal{P} =
U(L(\cal{P}k))$. Acting on $|k, \sigma\rangle$ gives,

\begin{equation}
\cal{P}|k, \sigma\rangle ~ = ~
\sqrt{\frac{\hat{k}^0}{k^0}}U(L(\cal{P}k))\eta|\hat{k},\sigma\rangle
~ ~ = ~ ~ \eta|\cal{P}k, \sigma\rangle .
\end{equation}

Thus $\cal{P}$ acts on all vectors of the representation with the same
$\eta$ and thus, $\eta$ is a property of the irreducible

$m = 0$:

Now $\hat{k} = (k, 0, 0, k)$ and the helicity $\sigma$ is the
eigenvalues of the $J_3$, the angular momentum direction along the
momentum direction. Since ${\cal P}$ changes the sign of the momentum
but not of the angular momentum, a state $|\hat{k},\sigma\rangle \propto
|-\hat{k}, -\sigma\rangle$. Unlike the massive case, the direction of
the reference momentum $\hat{k}$ is changed and this makes it
inconvenient to infer the action of parity on a general state. For this
reason, it is useful to choose a rotation $R_2$ about, say, the 2-axis
to bring back $-\hat{k} \to \hat{k}$.  Hence consider a new operator
${\cal P}' := U_2{\cal P}$, with $U_2 := e^{-i\pi J_2}$. But $U_2$ also
rotates $J_3$ (just as $ \hat{k}$) and hence the helicity is unchanged
by the $U_2$ operation.  Hence, $\boxed{ {\cal P}'|\hat{k},
\sigma\rangle = \eta_{\sigma}|\hat{k}, -\sigma\rangle }$. Here
$\eta_{\sigma}$ is some other phase factor. What about the action of
$\cal{P}$ on a general state $|p, \sigma\rangle$ defined in the equation
(<ref>)? Note that we are using $p := (|p|, \vec{p})$
instead of $k$ to denote the general vector.

Let $R(p, \hat{k})$ be the rotation that rotates the reference $\hat{k}$
in the direction of the general vector $p = (|\vec{p}|, \vec{p})$. Let
$B(p,\hat{k})$ denote the boost along the 3-direction which changes the
reference magnitude $k$ to $|\vec{p}|$. We can thus go from $\hat{k}$ to
$p$ by first using the boost $B$ followed by the rotation $R(p,
\hat{k}):  |p,\sigma\rangle \sim U\big( R(p, \hat{k})B(p,
\hat{k})\big)|\hat{k}, \sigma\rangle$. Noting that $\cal{P}'$ commutes
with boosts along the 3-axis, we have,

\begin{eqnarray*}
U(B(p, \hat{k})) & = & \cal{P}'U(B(p, \hat{k}))(\cal{P}')^{-1} =
U_2\cal{P}B\cal{P}^{-1}U_2^{-1} ~ ~ \Rightarrow ~ ~
\cal{P}U(B(p, \hat{k})) ~ = ~ U_2^{-1}U(B)U_2\cal{P} \\
\therefore \cal{P}|p, \sigma\rangle & \sim & U\big(R(p,\hat{k})\big)
\cal{P} U\big(B(p,\hat{k})\big) |\hat{k},\sigma\rangle ~ = ~ U\big(R(p,
\hat{k})R_2^{-1}B\big) \ \big(\eta_{\sigma}|\hat{k}, -\sigma\rangle\big)
\end{eqnarray*}

Now, although $R(p, \hat{k})R_2^{-1}$ is a rotation that takes the
3-axis along  $-\vec{p}$, its unitary representative is not quite
$U(R(-\vec{p}, \hat{k}))$. It introduces additional phases. I refer you
to the equation (2.6.21) in [1]. The net result is that,
$\boxed{\cal{P}|p,\sigma\rangle = \eta_{\sigma}exp\{\mp
i\pi\sigma\}|\cal{P}(p),-\sigma\rangle.} $ The additional phase $\mp
\pi\sigma$ correlates with the sign of the $2-$component of the momentum

Action of $\cal{T}$:

$m > 0$:

Now $\cal{T}$ commutes with $P^0$, anti commutes with $P^i$ and $J_i$.

\[
\therefore P^i\cal{T}|\hat{k}, \sigma \rangle = 0 ~ , ~
P^0\cal{T}|\hat{k}, \sigma \rangle = m |\hat{k}, \sigma \rangle
~, ~ J_3|\hat{k}, \sigma \rangle = - \sigma |\hat{k}, \sigma
\rangle .
\]

The last one implies that $\cal{T}|\hat{k}, \sigma \rangle =
\zeta_{\sigma}|\hat{k}, -\sigma \rangle$, where $\zeta$ is a phase
factor. From the angular momentum algebra we get,

\begin{eqnarray*}
(J_1 \pm iJ_2)|\hat{k}, \sigma\rangle & = & \sqrt{(s\mp\sigma)(s
\pm \sigma + 1)}|\hat{k},\sigma\pm 1\rangle ~ \Rightarrow \\
(- J_1 \pm iJ_2)(\cal{T}|\hat{k}, \sigma\rangle) & = &
\sqrt{(s\mp\sigma)(s \pm \sigma + 1)}(\cal{T}|\hat{k},\sigma\pm
1\rangle) \nonumber \\
& = & \sqrt{(s\mp\sigma)(s \pm \sigma + 1)}\zeta_{\sigma\pm
1}|\hat{k}, - \sigma\mp 1\rangle)
~ \Rightarrow -\zeta_{\sigma} = \zeta_{\sigma\pm 1} \ .
\end{eqnarray*}

$\zeta_{\sigma + 1} = \zeta_{\sigma -1}$. Choosing the phase for say
$\zeta := \zeta_{\sigma = j}$, the phase for $j-1, j-2, \dots$ alternate
between $\pm\zeta$. Thus we can write the phase as $\zeta_{\sigma} =
\zeta (-1)^{j - \sigma}$, $\zeta$ is an arbitrarily chosen phase.  The
phase $\zeta$ can actually be absorbed away by putting $|\hat{k}, \sigma
\rangle' := \sqrt{\zeta}|\hat{k}, \sigma \rangle$. The action of
$\cal{T}$ on the new vector gives,

\[
\cal{T}|\hat{k},\sigma \rangle' = \sqrt{\zeta^*}\cal{T}|\hat{k},
\sigma \rangle = |\zeta|(-1)^{j- \sigma}|\hat{k},-\sigma \rangle
\ .
\]

Note: A similar manipulation with the $\eta$ phase will
retain the phase in the redefined vectors and thus the intrinsic parity
cannot be absorbed away. The anti-linearity of $\cal{T}$ is responsible
for this feature. We may retain the irrelevant phase $\zeta$.

The action of time reversal operator on the general vectors $|k, \sigma
\rangle$ proceeds in exactly the same manner as above leading to,
$\cal{T}|k,\sigma \rangle = \zeta(-1)^{j-\sigma}|k,-\sigma \rangle$. It
follows immediately that $\cal{T}^2|k,\sigma \rangle =
(-1)^{2j}|k,\sigma \rangle$ which distinguishes different spins.

$m = 0$:

The analysis here proceeds in an exactly same manner as that for the
parity and we just note the final result as:
$\boxed{\cal{T}|p,\sigma\rangle = \xi_{\sigma}{\sigma}exp\{\mp
i\pi\sigma\}|\cal{T}(p),\sigma\rangle.} $ The additional phase $\mp
\pi\sigma$ correlates with the sign of the $2-$component of the momentum

This completes the basic definitions related to the Poincare group and
its unitary, irreducible representations.

These representations specify the attributes that we may assign to
elementary quanta. However, these are not suitable for manifest
covariance. Firstly, the label $\sigma$ transforms covariantly only
under the little group and not the full Lorentz group. Secondly, the
label $k$ is restricted to the positive hyperboloid and not the full
$\mathbb{R}^4$. This makes it inconvenient to take Fourier transform to
link them with space-time coordinates. For this we need to construct
representations on vector valued functions on the space-time. This is
done next.

§ REPRESENTATIONS SUITABLE FOR MANIFEST COVARIANCE: “FIELD

As noted earlier, the abstractly classified unitary, irreducible
representations of the Poincare group are not suitable for manifest
Lorentz covariance. For this, we begin by defining vector space of
suitably chosen function and define an action (homomorphism) of the
Poincare group. These will be reducible in general and irreducibility
will be imposed by partial differential equation (“field equations”).
To make the action unitary, requires an inner product to be defined.
This will further show a `doubling of representations' leading to the
particle/anti-particle interpretation.

Let $\Psi^A(x)$ be a finite dimensional column vector of a suitable
class of complex/real valued functions of the space-time coordinates
$x^{\mu}$. The defining action of the Poincare group is: $x \to gx :=
x_{(\Lambda,a)} = \Lambda x + a$. It induces a corresponding action on
the functions which we denote as: $\Psi \to (g\Psi) := \Psi_g$,

\[
\Psi_g(x) = D(h(g))\Psi(g^{-1}x) ~ ~ \leftrightarrow ~ ~
\Psi'_{(\Lambda,a)}(x) = D^A_{\ B}(h(\lambda,a))\Psi^B(
\]

Here, $D$ is a finite dimensional representation of the Lorentz
group, $h(\Lambda,a)$ is a map from the Poincare group to the Lorentz
group. The above action is required to be a homomorphism, i.e.
$(g\Psi)(x) = D(h(g))\Psi(g^{-1}x)$ such that

\begin{eqnarray}
\left(g'\left(g\Psi\right)\right)(x) & = & (g'g\Psi)(x) ~
\forall~ g,g' \in \mathrm{Poincare \ and}~ \forall ~ x .  \\
l.h.s. & = & D(h(g'))\ (g\Psi)(g'^{-1}x) = D(h(g')) D(h(g))
\Psi(g^{-1}g'^{-1}x) \nonumber \\
& = & D(h(g')h(g)) \Psi( (g'g)^{-1}x ) . \\
r.h.s. & = & D(h(g'g))\Psi(g'g)^{-1}x) ~ ~ \therefore ~ h(g)'s ~
\mbox{must satisfy,} \\
D(h(g'))\ D(h(g)) & = & D(h(g'g)) ~ \forall ~ g, g' \in
~\mbox{Poincare. Equivalently,} \\
D( h(g')h(g) ) & = & D (h(g'g) )  ~ ~ \mbox{which implies,}
\nonumber \\
h(g')h(g) & = & h(g'g) ~ , ~ \mbox{or $h(g)$ must be a
\end{eqnarray}

Since our primary focus is on the Lorentz group, and $h(g) =
h( (\mathbb{1},a)\cdot(\Lambda,0)) = h( (\mathbb{1},a)) h( (\Lambda,0)
)$, we may choose $h( (\mathbb{1}, a) ) = \mathbb{1}$. This
reduces the homomorphism from Poincare into Lorentz, to a homomorphism
from Lorentz to Lorentz. We can and do take, this homomorphism to the
identity homomorphism. With this we write,

\begin{equation}
\Psi_{(\Lambda,a)}(x) ~ := ~ D(\Lambda) \Psi( \Lambda^{-1}(x -
a) ).
\end{equation}

§.§ Induced representation of Poincare from inducing
representation of Lorentz

We have thus got an induced representation of the Poincare group
from an inducing representation $D(\Lambda)$ of the Lorentz group
and we take $D$ to be an irreducible representation. There is no
requirement of unitarity as $\Psi^A(x)$ is not a quantum
mechanical wave function. Additionally, it is a fact that the Lorentz
group has no non-trivial finite dimensional, unitary
representations. Our task now reduces to finding out all possible,
irreducible, finite dimensional representations of the Lorentz group.

To appreciate irreducibility, consider the infinitesimal action:
$(\Lambda, a) = (\mathbb{1} + \omega, \epsilon), \ D(\Lambda) =
D(\mathbb{1} + \frac{i}{2}\omega_{\alpha\beta}T^{\alpha\beta})$ where,
$T$ represent the Lorentz generators in the $D$ representation.  Then,

\begin{equation}
\Psi_{\mathbb{1} + \omega, \epsilon}(x) = (\mathbb{1} +
\frac{i}{2}\omega\cdot T)(\Psi(x - \omega x - \epsilon)) ~ = ~
\Psi(x) + \frac{i}{2}\omega\cdot T\Psi(x) -
(\omega^{\alpha}_{~\beta}x^{\beta} +
\epsilon^{\alpha})\partial_{\alpha}\Psi|_x
\end{equation}

This defines the Poincare generators acting on the $\Psi(x)$.

\begin{eqnarray}
M^{\alpha\beta}\Psi & := & (T^{\alpha\beta})^A_{~B}\Psi^B(x) +i
(\eta^{\gamma\alpha}x^{\beta} -
\eta^{\gamma\beta}x^{\alpha})\partial_{\gamma}\Psi^A(x) \\
P_{\mu}\Psi & := & -i \partial_{\mu}\Psi^A(x)
\end{eqnarray}

The two Casimir invariants, $P^2, W^2$ acting on $\Psi$ should evaluate
to some constants to satisfy the necessary condition for irreducibility
of the Poincare representation. The $P^2$ Casimir is easy to evaluate
and gives,

\begin{equation} \label{Casimir1onPsi}
(P^2\Psi)^A = \eta^{\mu\nu}(-i)^2\partial_{\mu}\partial_{\nu}\Psi^A ~ =
~ - \Box\Psi^A ~ ~ ~ ~ ~ ~ \Box := -\partial_0^2 + \nabla^2 \ .
\end{equation}

For the second Casimir we have,

\begin{eqnarray}
(W_{\mu}\Psi)^A & = & \frac{1}{2} \epsilon_{\mu\nu\alpha\beta}
(M^{\nu\alpha}P^{\beta}\Psi)^A \nonumber \\
& = & \frac{1}{2} \epsilon_{\mu\nu\alpha\beta}
\left[(T^{\nu\alpha})^A_{~B} + i\delta^A_{~B} \left(\eta^{\gamma\nu}
x^{\alpha} - \eta^{\gamma\alpha} x^{\nu}\right) \partial_{\gamma}\right]
\left[-i\delta^B_{~C} \partial^{\beta}\right]\Psi^C \nonumber \\
& = & -\frac{i}{2} \epsilon_{\mu\nu\alpha\beta} (T^{\nu\alpha})^A_{~B}
\partial^{\beta}\Psi^B + \epsilon_{\mu\nu\alpha\beta} x^{\alpha}
\partial^{\nu}\partial^{\beta}\Psi^A
= -\frac{i}{2} \epsilon_{\mu\nu\alpha\beta} (T^{\nu\alpha})^A_{~B}
\partial^{\beta}\Psi^B + 0
\end{eqnarray}

Note that the 2-derivative term has canceled. Then,

\begin{eqnarray}
(W^2\Psi)^A & = & \frac{1}{2}\epsilon^{\mu\nu'\alpha'\beta'}
(M_{\nu'\alpha'}P_{\beta'})^A_{~B}(W_{\mu}\Psi)^B  \nonumber \\
& = & \frac{1}{2} \epsilon^{\mu}_{~\nu'\alpha'\beta'}
\left[(T^{\nu'\alpha'})^A_{~B} + i\delta^A_{~B}
(\eta^{\gamma\nu'}x^{\alpha'} - \eta^{\gamma\alpha'}x^{\nu'})
\partial_{\gamma}\right] \left[-i\delta^{B}_{~C}\partial^{\beta'}
\right] \left(W_{\mu}\Psi\right)^C \nonumber \\
& = & -\frac{i}{2} \epsilon^{\mu}_{~\nu'\alpha'\beta'}
(T^{\nu'\alpha'})^A_{~B} \partial^{\beta'}(W_{\mu}\Psi)^B +
\epsilon^{\mu}_{~\nu'\alpha'\beta'} x^{\alpha'} \partial^{\nu'}
\partial^{\beta'}(W_{\mu}\Psi)^A \nonumber \\
& = & - \frac{1}{4}\epsilon^{\mu}_{~\nu'\alpha'\beta'}
\epsilon_{\mu\nu\alpha\beta} (T^{\nu'\alpha'} T^{\nu\alpha})^A_{~B}
\partial^{\beta'}\partial^{\beta}\Psi^B
\end{eqnarray}

Notice that $W^2\Psi = 0$ for non-one dimensional $D$ representation.

As illustrations consider two examples: (i) $D$ is the trivial representation
of the Lorentz group and (ii) $D$ is the defining representation of the
Lorentz group.

$D(\Lambda) = 1$:  Now $W^2\Psi = 0$ while $P^2\Psi =
-\Box\Psi = -m^2\Psi$. Thus, irreducibility requires that $\Psi$ must
satisfy $(\Box - m^2)\Psi(x) = 0$ and we recognize this as the
Klein-Gordon equation. The $\Psi$ transforms as: $\Psi_{(\Lambda, a)}(x)
= \Psi(\Lambda^{-1}(x - a))$.

$D(\Lambda)$ is the defining representation: This means
$\Lambda^{\mu}_{~\nu} = \delta^{\mu}_{~\nu} + \omega^{\mu}_{~\nu} :=
\delta^{\mu}_{~\nu} + \frac{i}{2} (\omega_{\alpha\beta}
T^{\alpha\beta})^{\mu}_{~\nu}$ and hence

\[
(T^{\alpha\beta})^{\mu}_{~\nu} := -i (\eta^{\alpha\mu}
\delta^{\beta}_{~\nu} - \eta^{\beta\mu} \delta^{\alpha}_{~\nu})
\]

This gives,
\[
(T^{\nu'\alpha'}T^{\nu\alpha})^{\rho}_{~\sigma} = -\left[
\eta^{\nu'\rho}\eta^{\nu\alpha'}\delta^{\alpha}_{~\sigma}
\eta^{\alpha'\rho}\eta^{\nu\nu'}\delta^{\alpha}_{~\sigma}
\eta^{\nu'\rho}\eta^{\alpha\alpha'}\delta^{\nu}_{~\sigma}
\eta^{\alpha'\rho}\eta^{\alpha\nu'}\delta^{\nu}_{~\sigma}\right]
\]

This is manifestly antisymmetric in $(\nu'\alpha')$ and $(\nu\alpha)$.
Contraction with the epsilons give equal contributions and cancel the
factor of 4, leading to $(W^2\Psi)^{\rho} =
\epsilon^{\mu\rho\nu}_{~~~\beta'} \epsilon_{\mu\nu\sigma\beta}
\partial^{\beta\beta'}\Psi^{\sigma}$.

Recall the general identity:

\begin{equation} \label{EpsilonIdentity}
\epsilon^{a_1\cdots a_j\ a_{j+1}\cdots a_n} \epsilon_{a_1\cdots
a_j\ b_{j+1}\cdots b_n} ~ = ~ (-1)^s\ (n-j)!\ j! \
\delta^{[a_{j+1}}_{~ ~b_{j+1}}\ \cdots\ \delta^{a_n]}_{~ ~b_n} \
\end{equation}

where $s$ is the index of the metric i.e. number of negative eigenvalues
of the metric. In our case the metric is the Minkowski metric and $s =
1$. This gives,

\[
\epsilon^{\mu\rho\nu}_{~ ~ ~ \beta'}\epsilon_{\mu\nu\sigma\beta}
~ = ~ - \epsilon^{\mu\nu\rho}_{~ ~ ~
\beta'}\epsilon_{\mu\nu\sigma\beta} ~ = ~ -(-1)\ 2!\ 2!\
\frac{1}{2}\ (\delta^{\rho}_{\sigma}\eta_{\beta'\beta} -
\eta_{\beta'\sigma}\delta^{\rho}_{\beta}) ,
\]

leading to,

\begin{eqnarray}\label{CasimirsOnVector}
(W^2\Psi)^{\rho} & = & 2(\delta^{\rho}_{\sigma}\eta^{\beta'\beta} -
\eta^{\beta\rho}\delta^{\beta'}_{\sigma}) \partial^2_{~
\beta'\beta}\Psi^{\sigma} ~ = ~ 2 \left[ \Box \Psi^{\rho} -
\partial^{\rho}(\partial_{\sigma}\Psi^{\sigma}\right] \ , ~
\mbox{and} \\
(P^2\Psi)^{\rho} & = & - \Box\Psi^{\rho}
\end{eqnarray}

Note: We have specified the action of the Poincare group on
$\Psi^{A}$. How does $\partial_{\mu}\Psi^A$ transform? We can think of
the derivative as a new quantity with two indices. The index $A$ will
transform by $(T^{\alpha\beta})^A_{~B}$ as before while the index $\mu$
will transform by the defining representation of the Lorentz
group, $(T^{\alpha\beta})_{\mu}^{~\nu} = +i(\eta^{\alpha\nu}
\delta^{\beta}_{\mu} - \eta^{\beta\nu}\delta^{\alpha}_{\mu})$.  In the
above calculation of the second $W^{\mu}$, this was missed. However it
may be checked that this extra contribution vanishes in the $W^2$

§.§ Irreducibility and field equations

The Poincare representation will be irreducible provided the Casimir invariants
have fixed values. Let $P^2\Psi^{\rho} = -m^2\Psi{\rho}$ and
$W^2\Psi^{\rho} = C\Psi^{\rho}$. For non-zero mass, with spin $j$, we know that $W^2$
equals $m^2\ j(j+1)$ and hence $C = m^2j(j+1)$. Thus $\Psi^{\rho}$

\[
\Box \psi^{\rho} = m^2\Psi^{\rho} ~ ~,~ ~ 2(m^2\Psi^{\rho} -
\partial^{\rho}\partial\cdot\Psi) = m^2 j(j+1)\Psi^{\rho}
\]

We have two cases to consider. The defining vector representation of the
Lorentz group, splits as $1 \oplus 0$ under the rotation subgroup. Hence
$j = 1 \ or\ 0$. For $j = 1$, taking the divergence of the second
equation implies $\partial\cdot\Psi = 0$. For $j = 0$, this argument
gives an identity. However, vanishing of $W^2$ gives
$\partial_{\rho}\partial\cdot\Psi = m^2\psi_{\rho}$. Differentiating by
$\partial_{\sigma}$ and anti-symmetrising in the indices gives,
$\partial_{\rho}\Psi_{\sigma} = \partial_`{\sigma}\Psi_{\rho}
\Rightarrow \Psi_{\rho} = \partial_{\rho}\Phi$ for some scalar $\Phi$
which is defined up to a constant. This scalar satisfies $(\Box -
m^2)\Phi = $constant which can be taken to be zero. Thus,

if the defining representation is to give the massive, spin one
irreducible representation, $\Psi^{\rho}$ must satisfy: $(\Box -
m^2)\Psi^{\rho} = 0 = \partial\cdot\Psi$. If it is to give the
massive, spin zero irreducible representation, then $\Psi_{\rho}
= \partial_{\rho}\Phi$ with $(\Box - m^2)\Phi = 0$.

For $m = 0$ and our restriction on the representations of the little
group, the Casimir conditions give, $\Box \Psi_{\mu} = 0$ and
$\partial_{\mu}\partial_{\nu}\Psi^{\nu} = 0$. Introduce a new field,
$A_{\mu} := \Psi_{\mu} + \partial_{\mu}\Lambda$. It follows that,

\[
\Box A_{\mu} = 0 + \partial_{\mu}\Box\Lambda ~ ~,~
\partial_{\mu}\partial^{\nu}A_{\nu} = 0 +
\partial_{\mu}\Box\Lambda  ~ ~ ~ \Rightarrow \Box A_{\mu} -
\partial_{\mu}\partial^{\nu}A_{\nu} = 0 ~ =
\partial^{\nu}\big(\partial_{\nu}A_{\mu} -
\partial_{\mu}A_{\nu}\big)
\]

We recognize the last equality as the source free Maxwell equation, including its gauge invariance: $A_{\mu} \rightarrow A_{\mu} +
\partial_{\mu}\Lambda$!

These examples show that the induced representation $\Psi^{A}$ of the
Poincare group defined through the inducing representation of the
Lorentz subgroup, is in general reducible. Requiring that
the Poincare Casimir invariants, evaluated on $\Psi$ gives the values
corresponding to the unitary, irreducible representation of the Poincare
group, imposes differential conditions on $\Psi$ eg the
Klein-Gordon equation and the subsidiary conditions. Usually, these
equations are proposed as the field equations. Here we see them
arising as irreducibility conditions.

Let us return to the classification of the finite dimensional,
representations of the Lorentz group.

We already know that the defining representation acts as: $v'^{\mu} =
\Lambda^{\mu}_{~\nu}v^{\nu} \leftrightarrow v'_{\mu} =
\Lambda_{\mu}^{~\nu} v_{\nu} = (\Lambda^{-1})^{\nu}_{~\mu} v_{\nu}$.
From these we can trivially construct other irreducible representations
by taking tensor products and subtracting suitable `traces'. For
instance, a rank-2 tensor $V^{\mu\nu}$ transforms as $V'^{\mu\nu} =
\Lambda^{\mu}_{~\alpha}\Lambda^{\nu}_{~\beta}V^{\alpha\beta}$. This is a
reducible representation though. Why? Because its symmetrised and
anti-symmetrised parts transform among themselves and thus form
invariant subspaces. Furthermore, tensors of the form $\eta^{\mu\nu}C$
also transforms into the same form. Hence the symmetric combination
splits further into traceless and trace part. Thus we get the space
$\cal{V}$ of the second rank tensors decomposes as:

\[
\cal{V} = \cal{V}_{anti-sym} \oplus \cal{V}_{sym, traceless} \oplus
\cal{V}_{trace} , ~ ~ \mbox{each being an irreducible representation.}
\]

Analogous construction can be carried out for higher rank tensors. These
however do not exhaust all the finite dimensional irreducible
representations. There are the spinor representations which are missed.
It is a result that all finite dimensional representations of
pseudo-orthogonal groups, $SO(p,q), p+q \ge 2$ can be constructed from
the irreducible representations of the corresponding Clifford algebra.

§.§ Clifford Algebras and Spinor representations

Clifford algebras are algebras generated by elements, $\{\mathbb{1},
\gamma^{\mu}, \mu = 0, \ldots, (p+q-1)\}$ satisfying $\gamma^{\mu}
\gamma^{\nu} + \gamma^{\nu} \gamma^{\mu} = 2
\bar{\eta}^{\mu\nu}\mathbb{1}, \bar{\eta}^{\mu\nu} = diag(-1, \dots ,
-1, +1, \dots +1)$. There are $p$ negative and $q$ positive signs.  We
have put a bar on the metric since we will relate it to the $\eta$
defining the Lorentz group for which $p = 1$ and $q = 3$. Note that for
$\mu \neq \nu$ the gamma's anti-commute, $(\gamma^0)^2 =
\bar{\eta}^{00}$ and $(\gamma^i)^2 = \bar{\eta}^{ii}$. Since the gamma's
anti-commute, their bilinear satisfy commutation relations.

Let $\Sigma^{\mu\nu} := a(\gamma^{\mu}\gamma^{\nu} -
\gamma^{\nu}\gamma^{\mu})$. We will choose the proportionality constant
$a$ suitably. Consider,

\begin{eqnarray}
\left[\Sigma^{\mu\nu}, \Sigma^{\alpha\beta}\right] & = & a^2\left\{
\left[\gamma^{\mu}\gamma^{\nu},
\gamma^{\alpha}\gamma^{\beta}\right] - (\mu \leftrightarrow \nu)
- (\alpha \leftrightarrow \beta) + (\mu,\alpha \leftrightarrow
\nu,\beta)\right\} \\
\gamma^{\mu}\gamma^{\nu}\gamma^{\alpha}\gamma^{\beta} & = &
2\bar{\eta}^{\nu\alpha}\gamma^{\mu}\gamma^{\beta} -
2\bar{\eta}^{\mu\alpha}\gamma^{\nu}\gamma^{\beta} +
2\bar{\eta}^{\nu\beta}\gamma^{\alpha}\gamma^{\mu} -
2\bar{\eta}^{\mu\beta}\gamma^{\alpha}\gamma^{\nu}  \nonumber \\
\therefore \frac{1}{2}\left[\gamma^{\mu}\gamma^{\nu},
\gamma^{\alpha}\gamma^{\beta}\right] & = & \left[ -
\bar{\eta}^{\mu\alpha}\gamma^{\nu}\gamma^{\beta} +
\bar{\eta}^{\nu\alpha}\gamma^{\mu}\gamma^{\beta} -
\bar{\eta}^{\mu\beta}\gamma^{\alpha}\gamma^{\nu} +
\bar{\eta}^{\nu\beta}\gamma^{\alpha}\gamma^{\mu} \right]\nonumber \\
\therefore \left[\Sigma^{\mu\nu}, \Sigma^{\alpha\beta}\right] & = &
(4a)\left\{ - \bar{\eta}^{\mu\alpha}\Sigma^{\nu\beta} + (\mu
\leftrightarrow \nu) + (\alpha \leftrightarrow \beta) - (\mu,\alpha
\leftrightarrow \nu,\beta)\right\}
\end{eqnarray}

This has the same form as the algebra satisfied by the Lorentz
generators: $\left[M^{\mu\nu}, M^{\alpha\beta}\right] =
i(\eta^{\mu\alpha}M^{\nu\beta} - - +)$, the $M$'s were defined through
$U(\Lambda = 1 + \omega) = \mathbb{1} + \Case{i}{2} \omega_{\mu\nu}
M^{\mu\nu}$. Thus we need $-4a\bar{\eta}^{\mu\nu} = i\eta^{\mu\nu}$.
There are two ways to satisfy this and we choose our conventions
as: $\bar{\eta} = -\eta$ and $a = \Case{i}{4}$. Thus,

\begin{equation}\label{CliffordDefns}
\boxed{ \gamma^{\mu}\gamma^{\nu} + \gamma^{\nu}\gamma^{\mu} = -
2\eta^{\mu\nu} ~ ~ ~ , ~ ~ ~ \Sigma^{\mu\nu} :=
\frac{i}{4}(\gamma^{\mu}\gamma^{\nu} - \gamma^{\nu}\gamma^{\mu})
~ ~=~  ~ \frac{i}{4}\left[\gamma^{\mu}, \gamma^{\nu}\right] .  }
\end{equation}

The $\Sigma$'s satisfy the Lorenz Lie algebra.  Furthermore, the
definitions imply that $\boxed{[\Sigma^{\mu\nu}, \gamma^{\lambda}] =
i(\eta^{\mu\lambda}\gamma^{\nu} - \eta^{\nu\lambda}\gamma^{\mu})}$ i.e.
the $\gamma$'s transform as Lorentz vectors.

It is useful to define $\boxed{\gamma_5 :=
+i\gamma^0\gamma^1\gamma^2\gamma^3}$ Just from the Clifford algebra and
definitions it follows that $\gamma^2_5 = + \mathbb{1}$,
$\gamma_5\gamma^{\mu} = - \gamma^{\mu}\gamma_5$, and $[\gamma_5,
\Sigma^{\mu\nu}] = 0$.

Claim: It is always possible to choose the $\gamma$'s to be
finite dimensional unitary matrices.

This is based on the following facts: (a) The set of elements
$ G := \{\pm\mathbb{1}, \pm \gamma^{\mu}, \pm\gamma^{\mu}\gamma{\nu},
\pm\gamma^{\mu}\gamma^{\nu}\gamma^{\lambda}, \dots, \pm
\gamma_5\}$ with all indices distinct in the products, forms a finite
group of 32 elements (in our 4 dimensional case); (b) All representations of
finite groups can be made unitary; (c) representation theory of finite
groups applied to the group $G$ gives that there is exactly one,
non-trivial, unitary, irreducible representation of G and hence of the
Clifford algebra and that its dimension is $2^{[4/2]} = 4$. The results
also extends to other dimensions.

Unitarity of the $\gamma$'s and their squares being $\pm\mathbb{1}$,
imply further that $(\gamma^0)^{\dagger} = \gamma^0$ and
$(\gamma^i)^{\dagger} = - \gamma^i$. This in turn gives
$(\Sigma^{0i})^{\dagger} = - \Sigma^{0i}\ , \ (\Sigma^{ij})^{\dagger} =
\Sigma^{ij}$ and $\gamma_5^{\dagger} = \gamma_5$.

Hence, the representation $D$ of the Lorentz group provided by
$\Sigma$'s is (i) non-unitary and (ii) reducible, since
$\gamma_5$ is Hermitian and commutes with the generators.

Since $\gamma$'s are finite dimensional, their traces are defined and
all of them including $\gamma_5$ are traceless. In particular, this
implies that $\gamma_5$ has two eigenvalues equal to +1 and other two
equal to -1. Hence $\Case{1\pm\gamma_5}{2}$ are projection matrices and
reduce the 4 dimensional representation of the Lorentz group into two
irreducible representations of dimensions 2.

By convention, $\Psi_L := \frac{1 - \gamma_5}{2}\Psi , \leftrightarrow
\gamma_5\Psi_L = -\Psi_L$ is  called a left handed Weyl spinor
while $\Psi_R := \frac{1 + \gamma_5}{2}\Psi , \leftrightarrow
\gamma_5\Psi_R = + \Psi_R$ is  called a right handed Weyl spinor.

The 4-component $\Psi^A$'s are called Dirac spinors while the
2-component projections, $\Psi_{\pm} := \Case{1\pm\gamma_5}{2}\Psi$ are
called Weyl spinors. Choosing $D$ to be generated by the
$\Sigma_{\pm} := \Case{1\pm\gamma_5}{2}\Sigma$ give irreducible
representations of the Lorentz group and in turn an irreducible
representation of the Poincare group .

Returning to Poincare representation, consider the combination
$\gamma^{\mu}P_{\mu}$, apparently a Lorentz scaler. Indeed, recalling
that $[M^{\mu\nu}, P_{\lambda}] = +i(\delta^{\mu}_{\lambda}P^{\nu} -
\delta^{\nu}_{\lambda}P^{\mu})$, it follows that,

\[
\left[T^{\mu\nu}, \gamma^{\lambda}P_{\lambda}\right] =
\left[\Sigma^{\mu\nu}, \gamma^{\lambda}\right]P_{\lambda} +
\gamma^{\lambda}\left[M^{\mu\nu}, P_{\lambda}\right] = 0\ .
\]

We have used $T$ to denote a general representation, $M$ to denote the
defining representation and $\Sigma$ to denote the reducible spinor
representation of the Lorentz group. We use the conventional
abbreviation $a\!\!\!/ := \gamma^{\mu}a_{\mu}$ for all 4-vectors

Consider the Poincare Casimir $P^2$. Irreducibility forces $P^2\Psi =
-m^2\Psi$. Let $m \neq 0$. The combinations, $m\mathbb{1} \pm p\!\!\!/ $
commutes with the Lorentz generators and also satisfy, $(m \pm
p\!\!\!/)^2 = m^2 + (p\!\!\!/)^2 \pm 2m p\!\!\!/ = 2(m \pm p\!\!\!/)$.
Therefore, $\pi_{\pm} := \Case{m \pm p\!\!\!/}{2m}$ is a projection
matrix operator (since $P$ is an operator) that also commutes with the
Lorentz generators. It trivially commutes with the Poincare generators
as well. Hence the Poincare representation induced by the Dirac spinor
is reducible in yet another way, the irreducible subspaces being
provided by $(\pm m + p\!\!\!/)\Psi = 0 = (- i\gamma^{\mu}\partial_{\mu}
\pm m)\Psi = 0$. This is just the Dirac equation! Note that the
projection property holds only for non-zero mass.

Note: The projectors $\frac{1 \pm \gamma_5}{2}$ reduce the
$D$ representation of the Lorentz group and each of these induces an
irreducible representation of the Poincare group. By contrast, $
\frac{m \pm \dsl{p}}{2m}$, do not reduce the $D$ representation, but
nevertheless reduces the Poincare representations. Are either of these
Poincare representations further reduced by the other projector? The
answer is `No' because $\gamma_5$ anti-commutes with $\dsl{p}$ and thus
exchanges the two projectors.

Note: Consider the action of $\dsl{p}$ on the left(right)
handed Weyl spinors. It follows immediately that $\gamma_5(\dsl{p})
\Psi_{L/R} = \pm \dsl{p} \Psi_{R/L}$. Therefore, if we restrict to any
one irreducible Weyl representation, then $\dsl{p}$ must annihilate it.
That is Weyl spinors satisfy the massless Dirac equation - also
known as the Weyl equation. Conversely, if we have a Dirac spinor
$\Psi$ that satisfies the Weyl equation, then its decomposition into its
Weyl spinors also satisfy the Weyl equation individually: $\dsl{p}(\Psi_L
+ \Psi_R) = 0 \Rightarrow \dsl{p}\Psi_{L/R} = 0$.

* Form the anti-symmetrized products of the
$\gamma$-matrices, satisfying the Clifford algebra
$\{\gamma^{\mu}, \gamma^{\nu}\} = -2\eta^{\mu\nu}$,

$\Gamma^{\mu} := \gamma^{\mu}\\
\Gamma^{\mu\nu} := \gamma^{\mu}\gamma^{\nu} -
\gamma^{\nu}\gamma^{\mu} \\
\Gamma^{\mu\nu\lambda} :=
\gamma^{\mu}\gamma^{\nu}\gamma^{\lambda} +
\gamma^{\nu}\gamma^{\lambda}\gamma^{\mu} +
\gamma^{\lambda}\gamma^{\mu}\gamma^{\nu} -
\gamma^{\nu}\gamma^{\mu}\gamma^{\lambda} -
\gamma^{\mu}\gamma^{\lambda}\gamma^{\nu} -
\gamma^{\lambda}\gamma^{\nu}\gamma^{\mu} \\
\Gamma^{\mu\nu\alpha\beta} :=
\gamma^{[\mu}\gamma^{\nu}\gamma^{\alpha}\gamma^{\beta]}$
with no numerical overall factors. Since the indices
take 4 values only, we cannot have any more antisymmetric
products. Together with $\mathbb{1}$, this set comprises
of 16 matrices.

Show that products of any number of $\gamma$'s can be
expressed  in terms of the  these 16 matrices together
with products of $\eta$'s. Conclude that these 16
matrices constitute a basis for arbitrary products of
$\gamma$'s. Since they have different Lorentz
transformation properties, they are all independent. The
vector space of $k\times k$ matrices has dimension
$k^2$, hence the minimum matrix order for the $\gamma$'s
is 4 and they are necessarily irreducible representation
of the Clifford algebra. Hence our $\gamma$'s are
$4\times 4$. Analogous arguments hold for Clifford
algebra of $n$ dimensions (indices taking $n$ values).

* For spinors satisfying the Dirac equation with $M\neq
0$, evaluate the Pauli-Lubanski scalar,

* The Lorentz $J_i, K_i$ satisfy the algebras $[J_i, J_j]
= i\epsilon_{ij}^{~l} J_l; ~ [K_i, K_j] = -
i\epsilon_{ij}^{~l} K_l; ~ [J_i, K_j] =
i\epsilon_{ij}^{~l} J_l.$ Define: $A_i :=
\Case{1}{2}(J_i + i K_i), B_i := \Case{1}{2}(J_i -
iK_i)$. Check that the $[A_i, B_j] =
i\epsilon_{ij}^{~l}A_l, \ , [B_i, B_j] =
i\epsilon_{ij}^{~l}B_l, ~ , [A_i, B_j] = 0.$ This the
$M^{\mu\nu}$ Lie algebra is equivalent to the direct sum
of two, mutually commuting $SU(2)$ algebras. They have
the Casimir invariants $\vec{A}^2, \vec{B}^2$. Evaluate
these for the spinor representation provided by

In summary:

\begin{eqnarray}
\Psi_{\Lambda,a}(x) & := & D(\Lambda)\Psi(\Lambda^{-1}x - \Lambda^{-1}a)
\ \mbox{(Reducible) Poincare action} \\
& & \mbox{Require Poincare Casimir invariants to take constant values} \nonumber \\
(\Box - m^2)\Phi = 0 & : & \mbox{ Klein-Gordon equation (scalar)
} \\
(\Box - m^2)v^{\mu} = 0, \partial\cdot v = 0 & : & \mbox{ Proca
equation (massive vector)} \\
\Box A^{\mu} = 0, \partial\cdot A = 0 & \leftrightarrow &
\partial_{\mu}F^{\mu\nu} = 0\ , \ F_{\mu\nu} := \partial_{\mu}A_{n} -
\partial_{\nu}A_{\mu} ~ \mbox{Maxwell equation)} \\
(-i\dsl{\partial} + m)\Psi = 0 & : & \mbox{Dirac equation, $m = 0$ is
Weyl equation} \\ \nonumber
\end{eqnarray}

§ UNITARY REPRESENTATIONS ON THE SPACE OF SOLUTIONS: ANTI-PARTICLES

At this stage, we have two sets of irreducible representations of the
Poincare group - the “Particle” representations $\{|k, \sigma
\rangle\}$, and the “field” representations $\Psi^{A}(x)$ satisfying
the appropriate field equations (or irreducibility conditions). The
former are unitary while the latter have no such notion defined for
them. We fill this gap now.

The space of vector valued functions, $\Psi^A$ can be made a complex
vector space easily enough, but we need to define an inner product to
define the notion of unitary. Since the irreducible representations are
solutions of the field equations and the equations are linear, we
consider the complex vector space of the solutions of the field
equations and look for an inner product.

Consider the massive scalar field first:
$(\partial^{\mu}\partial_{\mu} - m^2)\Phi(x) = 0, \Phi_{\Lambda,a}(x) =
\Phi(\Lambda^{-1}(x - a))$. Let $u, v$ be two solutions of the
Klein-Gordon equation. Then,

\[
u^*(\Box -m^2)v - v(\Box - m^2)u^* =
\partial_{\mu}\left(u^*\partial^{\mu}v -
v\partial^{\mu}u^*\right) = 0 .
\]

Hence, $\boxed{J^{\mu} := \lambda\left(u^*\partial^{\mu}v -
v\partial^{\mu}u^* \right)}$ is a conserved current. This has an
immediate implication. Consider a 4 dimensional region bounded by two
hypersurfaces of constant value of the time coordinate. More generally,
these are two Cauchy surfaces. Restricting to solutions which vanish
sufficiently rapidly as one approaches asymptotic infinity along the
space-like directions (eg $|\vec{x}| \to \infty$),

\[
\int_{region}d^4x\partial_{\mu}J^{\mu} = 0 = \int_{\Sigma_2 \cup
\Sigma_1}d^3x J^0 ~ ~ \Rightarrow ~ ~ \int_{\Sigma_2}d^3J^0 =
\int_{\Sigma_1}d^3x J^0 \ .
\]

Here we have used that both the Cauchy surfaces have their normals
directed `outward' (future pointing for the later one and past pointing
for the earlier one). This suggests that we define a candidate inner
product between two solutions as,

\begin{equation}
(v, u) ~ := ~ \lambda\int_{\Sigma}d^3x\ J^0(v,u)~ , ~ J^0(v,u)
:= v^*\partial^0u - u\partial^0v^* \ , ~ \Sigma ~ \mbox{a Cauchy
\end{equation}

The conserved current implies that the inner product is independent of
the Cauchy surface.

Note: On a Cauchy surface, the solution and its time
derivative form an initial data and the inner product is really defined
on these data. However the $\Sigma-$independence of the inner product
allows us to think of this as an inner product on the space of

Is $(v, u)$ really an inner product? It is (i) linear in $u$ and
anti-linear in $v$; (ii) $(v, u)^* = (u, v)$ provided $\lambda^* =
- \lambda$ and (iii) $(u, u) \ge 0$ with equality for $u = 0$?

To check the third property, let us write a solution of the field
equation in the form $u(t,\vec{x}) = e^{\pm i\omega
t}u_{\omega}(\vec{x})\ , \omega > 0$. Let us conventionally  call
$e^{-i\omega t}$ as a positive frequency solution. Then, $(u, u) =
\lambda (-2 i\omega) \int_{\Sigma}d^3x |u_{\omega}(\vec{x})|^2$. The
inner product then satisfies the crucial third property provided we
choose $\lambda = +i|\lambda| := i$. The absolute value of $\lambda$ has
been taken to be one as it only affects the normalization of the

Thus, with the convention adopted, the $(v, u)$ with $\lambda = i$ is
indeed an inner product on the subspace of positive frequency
solution. Equally well, the choice $\lambda = -i$ defines an inner
product on the subspace of negative frequency solutions.

Consider a family of solution, the plane wave solution, $u_{\vec{k}}(x)
:= A_{\vec{k}}e^{ik\cdot x}, \ k\cdot x := -k^0t + \vec{k}\cdot\vec{x},
\ k^0 := \sqrt{\vec{k}^2 + m^2} =: \omega_{\vec{k}}$. Substitution

\begin{eqnarray}
(u_{\vec{k}'}, u_{\vec{k}}) & = & i A^*_{\vec{k}'}A_{\vec{k}}
\int_{\Sigma_t}d^3x (-ik^0 - ik^{'0})e^{i(k - k')x} \\
& = & +A^*_{\vec{k}'}A_{\vec{k}}(k^0 + k^{'0})e^{-i(k^0 -
k^{'0})t} \left( (2\pi)^3\delta^3(\vec{k} - \vec{k}')\right) \\
& = &
\left[(2\pi)^3|A_{\vec{k}}|^2\right]\left[(2k^0)\delta^3(\vec{k}
- \vec{k}')\right]
\end{eqnarray}

Since $k^0$ depends only on the magnitude of $\vec{k}$, the delta
function forces the frequencies to be equal. We recognize the second
square bracket as the Lorentz invariant delta function and choose
the normalization constant to $A_{\vec{k}} := (2\pi)^{-3/2}$ so that,

\begin{eqnarray} \label{Orthonormality0}
u_{\vec{k}}(x) & := &
\frac{1}{(2\pi)^{3/2}}e^{ik\cdot x} ~ ~,~ ~ k\cdot
x := - \sqrt{\vec{k}^2 + m^2}\ t + \vec{k}\cdot\vec{x} ~ ~,~ ~
\vec{k} \in \mathbb{R}^3 \\
(u_{\vec{k}'}, u_{\vec{k}}) & = & \delta^3_{inv}(\vec{k} -
\vec{k}') ~ ~ = ~ ~ (2k^0)\delta^3(\vec{k} - \vec{k}') \ .
\end{eqnarray}

The above family of solutions formally form an orthonormal set in the
space of positive frequency solutions. Technically, these do not belong to the space of solutions which have to die off suitably at
spatial infinity. This is understood as usual by either using `box
normalization' or forming wave-packets. Manipulations done using the
above do not lead to any inconsistency.

Now we are ready to check if the Poincare action on the above inner
product space is unitary. We will check this by showing that under the
Poincare action, the orthonormality is preserved.

\begin{eqnarray}
[u_{\vec{k}}(x)]_{\Lambda,a} & = & u_{\vec{k}}( \Lambda^{-1}(x - a) ) ~
= ~ \frac{1}{(2\pi)^{3/2}}e^{i k\cdot \Lambda^{-1}(x - a)} \\
& = & \frac{1}{(2\pi)^{3/2}}e^{i (\Lambda k)\cdot(x - a)} ~ ~ \because
k_{\mu}(\Lambda^{-1})^{\mu}_{~\nu}x^{\nu} =
(\Lambda_{\nu}^{~\mu}k_{\mu})x^{\nu}  = (\Lambda k)\cdot x \nonumber \\
\therefore \left[u_{\vec{k}}(x)\right]_{\Lambda,a} & = & e^{-i(\Lambda
k)\cdot a}\ u_{\Lambda k}(x) ~ ~ \mbox{and,} \\
\left( [u_{\vec{k}'}]_{\Lambda,a} \ , \ [u_{\vec{k}}]_{\Lambda,a}\right)
& = & e^{i(\vec{\Lambda(k - k')}) \cdot a} \left( [u_{\vec{\Lambda k}'}]
\ , \ [u_{\vec{\Lambda k}}] \right) ~ = ~ e^{i(\vec{\Lambda(k - k')})
\cdot a}\ \delta_{inv}^3(\vec{\Lambda(k - k')}) \nonumber \\
& = & \left( [u_{\vec{k}'}]\ , \ [u_{\vec{k}}]\right) ~ ~ ~ \mbox{Hence,
\end{eqnarray}

Note:  The orthochronous Lorentz group preserves the sign of
the frequency, $k^0 > 0 => (\Lambda k)^0 > 0$, and hence maps positive
(negative) frequency solutions into positive (negative) frequency
solutions. Thus we see that the space of solutions of the
irreducibility condition (field equations) itself decomposes into two Lorentz invariant subspaces of positive/negative frequency
solutions. There is no contradiction with the Casimir being constant -
we just happen to have two unitary representations which have the
same values of the Poincare (orthochronous, proper) Casimir invariants.
These in fact represent `particle' and `anti-particle' representations

The plane wave orthonormal basis, $\{ u_{\vec{k}}(x)\}$ is in one-to-one
and onto correspondence with the `particle basis' $\{|k \rangle\}$.

The same features are exhibited by the other solution spaces, it remains
to identify a conserved current on the space of solutions, define an
inner product, obtain an orthonormal set and show unitarity. Since all
field equations imply that the fields always satisfy the Klein-Gordon
equation, we will always have the positive/negative frequency subspaces
and the particle/anti-particle identification.

Consider the Proca equation with divergence condition: $(\Box -
m^2)v^{\mu} = 0 = \partial\cdot v$. Let $V$ denote the space of
solutions of these equations. As before, let $v^{\mu}$ and $u^{\mu}$ be
two solutions. Then it follows that $\boxed{J^{\mu} :=
\lambda(v^{*\nu}\partial^{\mu}u_{\nu} -
u^{\nu}\partial^{\mu}v^*_{\nu})}$ is conserved exactly as before. The
divergence condition just selects a subspace of the space of solution of
the Klein-Gordon equation. The inner product is defined as (with the
same convention of positive frequency solutions),

\[
(v, u) ~ := ~ i\int_{\Sigma_t}d^3x \left[v^{*\nu}
\partial^{\mu}u_{\nu} - u^{\nu}\partial^{\mu}v^*_{\nu}\right]
\]

Consider the family of solutions,

\begin{eqnarray}
u^{\mu}_{\vec{k}}(x) & := &
\frac{1}{(2\pi)^{3/2}}\varepsilon^{\mu}(k) e^{ik\cdot x} ~ , ~
k_{\mu}\varepsilon^{\mu}(k) = 0 ~,~ k\cdot x := -k^0\ t +
\vec{k}\cdot\vec{x} \\ %, \ k^0 := \sqrt{\vec{k}^2 + m^2} \\
\left(u_{\vec{k}'}, u_{\vec{k}}\right) & = & \frac{1}{(2\pi)^3}(k^0 +
k^{'0})\left[ \varepsilon^{*\mu} (k') (\varepsilon_{\mu} (k)
\right]e^{-i(k^0 - k^{'0})t} \left[(2\pi)^3\delta^3(\vec{k} -
\vec{k}')\right] \nonumber \\
& = & \left[ \varepsilon^{*\mu} (k') (\varepsilon_{\mu} (k)
\right] \delta^3_{inv}(\vec{k}' - \vec{k}) ~ ~ , ~ ~
\mbox{here} ~ k^0 := \sqrt{\vec{k}^2 + m^2} \ .
\end{eqnarray}

The polarizations $\varepsilon(k)$ satisfying the transversality
condition, $k\cdot\varepsilon(k) = 0$ selects 3 independent vectors for
$m \neq 0$ and 2 independent vectors for $m = 0$ thanks to the
equivalence of $A_{\mu}$ and $A_{\mu} + \partial_{\mu}\Lambda$. To
distinguish the different polarization vectors, we introduce an
additional label, $a$ taking 3 and 2 values respectively for the massive
and massless cases. We choose them to satisfy the orthonormality
relations, $\boxed{\varepsilon^{*}_{\mu}(k,a)\varepsilon^{\mu}(k,b) =
\delta_{ab}}$ and of course $\varepsilon(k, a)\cdot k = 0\ \forall\ a =
1,2,3$.  For massless case, $a$ is usually denoted by $\lambda$ and
takes only two values. The plane wave family of solutions so defined,
constitute an orthonormal set.

Under the Poincare action,

\begin{eqnarray}
\left[u^{\mu}_{\vec{k},a}(x)\right]_{\Lambda,b} & = &
\Lambda^{\mu}_{~\nu}u^{\nu}_{\vec{k},a}(\Lambda^{-1}(x - b)) ~ = ~
\frac{1}{(2\pi)^{3/2}}\Lambda^{\mu}_{~\nu}\varepsilon^{\nu}
(\vec{k},a)e^{ik\Lambda^{-1}(x -b)} \nonumber \\
& = & e^{-i(\Lambda k)\cdot
b}(\Lambda^{\mu}_{~\nu}\varepsilon^{\nu}(k,a))e^{i(\Lambda k)\cdot x}  ~
~ \mbox{But,}\nonumber \\
(\Lambda k)_{\mu}(\Lambda^{\mu}_{~\nu}\varepsilon^{\nu}(k,a)) & = &
k_{\mu}\varepsilon^{\mu}(k,a) ~ = ~ 0 ~ \therefore ~ \mbox{r.h.s. = }~
e^{i(\Lambda k)\cdot b)}\varepsilon^{\mu}(\Lambda k), a)e^{i(\Lambda
k)\cdot x } \nonumber \\
\left[u^{\mu}_{\vec{k},a}(x)\right]_{\Lambda,b} & = & e^{i(\Lambda
k)\cdot b)}u^{\mu}_{\vec{\Lambda k},a}(x) \ .
\end{eqnarray}

Noting further that $(\Lambda\varepsilon(k',a))^* \cdot
(\Lambda\varepsilon(k,b)) = \varepsilon^*(\Lambda k', a) \cdot
\varepsilon(\Lambda k, b)$, unitarity of the Poincare action now follows
in exactly the same manner as for the scalar. As in the case of the
scalar, here too we have two unitary, irreducible representations of the
(orthochronous, proper) Poincare group corresponding to the `particle'
and `anti-particle' tags. For the zero mass case, the only difference is
that the polarizations are restricted to the two spatially transverse

The case of the spinors is more interesting. We follow the same strategy
of looking for a conserved current and a corresponding inner product, but
we will do this for the Dirac equation which is a first order
differential equation. We have:

\[
\Psi^A_{\Lambda,a}(x) = D^A_{~B}(\Lambda)\Psi^B(\Lambda^{-1}(x -
a)) ~ ~,~ ~ (-i\gamma^{\mu}\partial_{\mu} \pm m)\Psi = 0\ .
\]

Note that there are two equations.

To look for a conserved current, we need to consider the (matrix)
adjoint of the Dirac equation. The adjoint will involve
$\gamma^{\dagger}$ which are inconvenient for manipulations. However, it
follows readily that $\boxed{(\gamma^{\mu})^{\dagger} =
\gamma^0\gamma^{\mu}\gamma^0}$. This suggests that we use the Dirac
conjugate defined as: $\bar{\Psi} := \Psi^{\dagger}\gamma^0$. It
follows that

\[
(-i\gamma^{\mu}\partial_{\mu} \pm m)\Psi = 0 ~ ~ \leftrightarrow ~ ~
i\partial_{\mu}\bar{\Psi}\gamma^{\mu} \pm m \bar{\Psi} = 0\ .
\]

Using these equations for two solutions, $\Psi_1, \Psi_2$ and their
Dirac conjugates, it follows that $\boxed{J^{\mu}(\Psi_2, \Psi_1) :=
\lambda(\bar{\Psi}_2\gamma^{\mu}\Psi_1)}$ is the conserved current for
both $\pm\ m$ equations. With the choice $\lambda = +1$, the inner
product on the solutions of Dirac equations is defined as:

\[
(\Psi_2, \Psi_1) := \int_{\Sigma_t}d^3x \bar{\Psi}_2\gamma^0\Psi_1 ~ ~
\leftrightarrow ~ ~ (\Psi, \Psi) = \int_{\Sigma_t}d^3x\Psi^{\dagger}\Psi
~ \ge ~ 0 \ .
\]

Caution: For $\gamma_0$ in the above equation, $(\Psi, \Psi)
\le 0$.

To construct a family of orthonormal solutions, consider the `plain wave
ansatz', $\Phi^A(\vec{k},\sigma,x) := \frac{1}{(2\pi)^{3/2}}
\varphi^A(k,\sigma)e^{ik\cdot x}$. The Dirac equations then require
$(\dsl{k} \pm m)\varphi(k,\sigma) = 0$ or $\dsl{k}\varphi = \mp
m\varphi$. Multiplying by $\dsl{k}$ and using $\dsl{k}\dsl{k} = - k^2$
leads to $k^2 + m^2 = 0 \leftrightarrow k^0 := \pm \sqrt{\vec{k}^2 +
m^2}$. We fix $k^0$ to be positive and call a solution with $e^{ik\cdot
x}$ as having positive frequency and a solution with $e^{-ik\cdot x}$ as
having negative frequency. The two Dirac equations can now be
taken as a single Dirac equation with $+m$, admitting positive and
negative frequency solutions. Explicitly, we refine the ansatz as
$\Phi^A_+(\vec{k},\sigma,x) := \frac{1}{(2\pi)^{3/2}}
u^A(k,\sigma)e^{ik\cdot x}$ and $\Phi^A_-(\vec{k},\sigma,x) :=
\frac{1}{(2\pi)^{3/2}} v^A(k,\sigma)e^{-ik\cdot x}$. Both satisfying
$(-i \dsl{\partial} + m)\Phi_{\pm} = 0$. This implies that $(\dsl{k} +
m)u(k,\sigma) = 0 = (\dsl{k} - m)v(k,\sigma)$. The $u, v$ spinors get
their $k-$dependence through these defining equations. These spinors are
eigen-spinors of $\dsl{k}$ with eigenvalues $\pm\ m$. Since the trace of
the $\gamma$'s vanishes, $\dsl{k}$ is traceless too and hence each
eigenvalue is doubly degenerate i.e. the `$\sigma$' label in the
eigen-spinors takes two values. These will be linked to the
helicities later on.

Substitution of these solutions in the inner product leads to,

\begin{eqnarray}
(\Phi(k',\sigma',), \Phi(k,\sigma)) & = & \int_{\Sigma_t}d^3x
\frac{1}{(2\pi)^3} u^{\dagger}(k',\sigma')u(k,\sigma)e^{-i(k'
-k)\cdot x} \nonumber \\
& = & u^{\dagger}(k',\sigma')u(k,\sigma)e^{-i(k^0 -
k^{'0})t}\delta^3(\vec{k} - \vec{k}') \\
\therefore (\Phi(k',\sigma',), \Phi(k,\sigma)) & = &
\left[\frac{u^{\dagger}(k',\sigma')u(k,\sigma)}{2k^0}\right]
\delta^3_{inv}(\vec{k} - \vec{k}') ~ ~ := ~ ~
\delta_{\sigma',\sigma}\delta^3_{inv}(\vec{k} - \vec{k}') \ .
\end{eqnarray}

We have chosen a normalization of the $u-$spinors. Identical
orthonormality relations follow for the $v-$spinors.

Under Lorentz transformations, the solutions transform as,

\[
\left[\Phi(k,\sigma)\right]_{\Lambda}(x) ~ = ~
\left[D(\Lambda)u(k,\sigma)\right] \frac{e^{i(\Lambda k)\cdot
x)}}{(2\pi)^{3/2}}  ~ ~ ~ \stackrel{?}{=} ~ \Phi(\Lambda k, \sigma)(x)
\]

The last equality will hold if $D(\Lambda)u(k,\sigma) = u(\Lambda k,
\sigma)$ i.e. if $(\dsl{(\Lambda k)} + m) \left[D(\Lambda) u(k,\sigma)
\right] = 0$.

We already have $[\Sigma^{\mu\nu}, \gamma^{\lambda}] =
i(\eta^{\mu\lambda}\gamma^{n} - \eta^{\nu\lambda}\gamma^{\mu})$. This
implies (may be checked by taking the infinitesimal form of

\[
D^{-1}(\Lambda)\gamma^{\mu}D(\Lambda) = \Lambda^{\mu}_{~\nu}\gamma^{\nu}
~ \Rightarrow ~ \gamma^{\mu}D(\Lambda) =
\Lambda^{\mu}_{~\nu}D(\Lambda)\gamma^{n} \ .
\]

It follows,

\begin{eqnarray}
\left[\dsl{(\Lambda k)} + m\right] \left(D(\Lambda) u(k,\sigma) \right)
& = & \left[(\Lambda k)_{\mu}\gamma^{\mu}D(\Lambda) + D(\Lambda)
m\right]u(k,\sigma) \\
& = & \left[(\Lambda k)_{\mu}\ \Lambda^{\mu}_{~\nu}
D(\Lambda)\gamma^{\nu} + D(\Lambda) m\right] u(k,\sigma) \nonumber \\
& = & D(\Lambda) \left[\Lambda_{\mu}^{~\alpha}
k_{\alpha}\Lambda^{\mu}_{~\nu} \gamma^{\nu} + m\right]u(k,\sigma) ~ ~
\mbox{but, $\Lambda_{\mu}^{~\alpha} \Lambda^{\mu}_{~\nu} =
\delta^{\alpha}_{~\nu}$ } \nonumber \\
\therefore \left[\dsl{(\Lambda k)} + m\right] \left(D(\Lambda)
u(k,\sigma) \right) & = & D(\Lambda)\left[k_{\nu}\gamma^{\nu} +
m\right]u(k,\sigma) ~ = ~ 0. ~ \mbox{(Covariance Result)} ~ ~
\end{eqnarray}

Thus, indeed we have $\boxed{D(\Lambda)u(k,\sigma) = u(\Lambda k,
\sigma)}$ and hence,

\[
\left[\Phi(k,\sigma)\right]_{\Lambda,a}(x) ~ = ~ e^{-i(\Lambda k)\cdot
a}\Phi(\Lambda k, \sigma)(x) \ .
\]

Under Lorentz transformations then the orthonormality relation
transforms as,

\begin{eqnarray}
(\Phi(k',\sigma')_{\Lambda,a}, \Phi(k,\sigma)_{\Lambda,a}) & = &
e^{i\Lambda(k' - k)\cdot a} \left(\Phi(\Lambda k',\sigma'), \Phi(\Lambda
k, \sigma)\right) \nonumber \\
& = & \left[ \frac{u^{\dagger} (\Lambda k', \sigma') u(\Lambda k,
\sigma)}{2(\Lambda k)^0} \right] \delta^3_{inv} (\vec{k}' - \vec{k}) ~ =
~ \delta_{\sigma', \sigma}\delta^3_{inv} (\vec{k}' - \vec{k})\ .
\end{eqnarray}

In the last equality we have used the normalization of the $u$ spinors.
This proves the unitarity of the Poincare representation on the positive
frequency solutions. As before, Lorentz transformations do not mix the
subspaces of the positive/negative frequency solutions.

The normalization definition does not look Lorentz invariant, but it is.
Both $u^{\dagger}u = \bar{u}\gamma^0u$ and $k^0$ transform the same way.
A more convenient form will be displayed later on.

Note: The covariance result allows us to define the $u, v$
spinors from their definition in the rest frame (for massive case)i,
$\hat{k} = (m, \vec{0})$. In this frame, the spinors satisfy the
equations: $-m\gamma^0 u( \hat{k}, \sigma) = -m u( \hat{k}, \sigma)\ ,$
and $-m\gamma^0 v( \hat{k}, \sigma) = +m v( \hat{k}, \sigma)$. Thus
these spinors at $\hat{k}$ are eigen-spinors of $\gamma^0$ with
eigenvalues $\pm 1$ respectively. $\gamma^0$ being Hermitian, these
spinors are orthogonal: $u^{\dagger}(\hat{k},\sigma)v(\hat{k}, \sigma) =
0$. As noted before, these eigenvalues are doubly degenerate and
$\sigma$ labels these. For the massless case, $\hat{k} = (k, 0, 0, k)$
and $u, v$ are eigen-spinors of $J_3$ generator of the Little group.

Note: Incidentally, covariance result and the identification
of $u(\hat{k},\sigma), v(\hat{k},\sigma)$ as eigen-spinors of $\gamma^0$
also leads to a completeness relation as follows. Given that
$\gamma^0 u(\hat{k},\sigma) = u(\hat{k}, \sigma)\ , \gamma^0 v(\hat{k},
\sigma) = - v(\hat{k},\sigma)$ and each eigen-space being two
dimensional, $\Rightarrow$ completeness relations (`spectral
representation' - $A = \sum_n \lambda_n |n \rangle \langle n|$ ) takes
the form

\begin{eqnarray*}
\sum_{\sigma}u(\hat{k},\sigma)u^{\dagger}(\hat{k},\sigma) =
\frac{\mathbb{1} + \gamma^0}{2} & ~,~ &
\sum_{\sigma}v(\hat{k},\sigma)v^{\dagger}(\hat{k},\sigma) =
\frac{\mathbb{1} - \gamma^0}{2} \\
\sum_{\sigma}u(\hat{k},\sigma)\bar{u}(\hat{k},\sigma) = \frac{\gamma^0 +
\mathbb{1}}{2} ~ = ~ \frac{-\dsl{\hat{k}} + m}{2m} & ~,~ &
\sum_{\sigma}v(\hat{k},\sigma)\bar{v}(\hat{k},\sigma) = \frac{\gamma^0 -
\mathbb{1}}{2} ~ = ~ - \frac{\dsl{\hat{k}} + m}{2m} \\
\sum_{\sigma}u(\vec{k},\sigma)\bar{u}(\vec{k},\sigma) = \frac{-\dsl{{k}}
+ m}{2m} & ~,~ & \sum_{\sigma}v(\vec{k},\sigma)\bar{v}(\vec{k},\sigma) =
- \frac{\dsl{{k}} + m}{2m} \\
\end{eqnarray*}

In the second equation, we used $\gamma^0 = -\Case{\dsl{\hat{k}}}{m}$
and in the last equation we multiplied by $D(L)$ on the left and
$D^{-1}(L)$ on the right and used the covariance result. Here $L$ is a
boost which takes $\hat{k} \rightarrow \vec{k}$. Notice that multiplying
by $u(\vec{k},\sigma')$ and $v(\vec{k},\sigma')$ respectively, and using
the equations satisfied by the spinors, gives
$\boxed{\bar{u}(\vec{k},\sigma) u(\vec{k},\sigma') =
\delta_{\sigma,\sigma'} = - \bar{v}(\vec{k},\sigma)
v(\vec{k},\sigma')}$. These are consistent with the previously
chosen normalizations: $u^{\dagger}(\vec{k},\sigma)u(\vec{k},\sigma') =
2k^0 \delta_{\sigma,\sigma'} = v^{\dagger}(\vec{k},\sigma)

Thus, for all the field equations we have found unitary, irreducible
representations of the Poincare group. We have also discovered that the
field representations automatically also include `anti-particles'.
This is really a consequence of the requirement of manifest
Poincare covariance which necessitates the field representations.

Note: From the expressions above, it should be clear that
the ket vectors $|k,\sigma \rangle$ are in one-to-one correspondence
with the “Plane wave solutions” with positive and negative
frequencies, displayed above.

We need to study the behavior of these representations under the
space inversion, time reversal and the new possibility of `charge
conjugation' (particle-particle exchange).

§ PARITY, TIME REVERSAL AND CHARGE

For the field representations above, we focussed on the subgroup of the
Poincare group, connected to the identity. The remaining elements of the
Poincare group - improper and non-orthochronous are generated by the two
transformations of space inversion  and time reversal. Their actions on
the generators is given in eq.(<ref>). These
relations continue to hold in any representation, in particular
also on the field representations. Looking at the general form of the
Poincare action on the $\Psi^A(x)$'s, $\Psi^A_{\Lambda,a}(x) =
D^A_{~B}(\Lambda)\Psi^B(\Lambda^{-1}(x-a))$, we see that the action on
the space-time point is common to all $D(\Lambda)$ representations
and we may focus on the $D(\Lambda)$ part of it.  Recalling that the
time reversal operation is anti-linear and anti-unitary, we have to be
careful about taking complex conjugates.  Thus we define the actions as,

\begin{eqnarray}
\Psi^A_{\cal{P}}(x) & := & D^A_{~B}(\cal{P})\Psi^B(t, -\vec{x})
~ ~ , ~ ~ D(\cal{P}) =: \Pi \\
\Psi^A_{\cal{T}}(x) & := & D^A_{~B}(\cal{T})(\Psi^B(-t,
\vec{x}))^* ~ ~ , ~ ~ D(\cal{T}) =: \tau ,
\end{eqnarray}

It is customary to denote the anti-unitary, anti-linear operator
$\cal{T}$ as $\tau K$ where $\tau$ is a unitary operator and  $K$ takes
complex conjugate of numbers on its right. $\cal{T}^2 = \mathbb{1}$
implies $\tau\tau^* = \mathbb{1}$.

Consider the translation generators. These act as differential operators
on the $\Psi$'s: $P_{\mu}\Psi^A(x) = -i\partial_{\mu}\Psi^A$. Consider
the $\cal{P}$ and $\cal{T}$ actions on $P_{\mu}$ as given in eqn.
(<ref>). For instance,

\begin{eqnarray}
(\cal{T}^{-1}P_0\cal{T})\Psi^A(t,\vec{x}) & = &
\cal{T}^{-1}\{-i\partial_0 (\tau^A_{\ B}\Psi^B(-t, \vec{x})^*)\} ~ = ~
\cal{T}^{-1}\{+i\partial_{-t} (\tau^A_{\ B}\Psi^B(-t, \vec{x})^*)\}
\nonumber \\
& = & -i \partial_{t}(\tau\tau^*)^A_{\ B} \Psi^B(t,\vec{x}) ~ = ~
P_0\Psi^A(t,\vec{x})\ .
\end{eqnarray}

The time has been reversed twice and the complex conjugation has been
effected twice. For the spatial components, there is no `t' reversal and
the $\cal{T}$ introduces a sign. Space inversion is likewise straight
forward. We now focus on the Lorentz generators only.

The scalar case has no non-trivial matrices and the vector case is
simpler and is left as an exercise. The spinner case is non-trivial.  We
have the Lorentz generators:

\[
\Sigma^{\mu\nu} = \frac{i}{4}\left[\gamma^{\mu},
\gamma^{\nu}\right] ~ \leftrightarrow ~ J_i =
\frac{\epsilon_{ijk}}{2}\Sigma^{jk} ~ , ~ K^i = \Sigma^{i0} \ .
\]

The $\Pi$ and $\tau$ matrices are also $4\times 4$ matrices. We note a

Result: Any $4\times 4$ matrix can be written as a
linear combination of the 16 $\Gamma$ matrices, $\{\Gamma\} =
\{\mathbb{1}, \gamma^{\mu},\gamma^{[\mu}\gamma^{\nu]},
\gamma^{[\mu}\gamma^{\nu}\gamma^{\lambda]}, \gamma_5\}$, the
square brackets denoting anti-symmetrization (16 = 1 + 4 + 6 + 4 + 1).

The result follows by showing that the 16 matrices are linearly

We now deduce the $\Pi$ and $\tau$ matrices from the commutation
relations (see eq.(<ref>)):

\[
\Sigma_{jk}\Pi = \Pi\Sigma_{jk} ~,~ \Sigma_{0i}\Pi = - \Pi\Sigma_{0i} ~
~ , ~ ~ \Sigma_{jk}\tau = - \tau(\Sigma_{jk})^* ~,~ \Sigma_{0i}\tau =
\tau(\Sigma_{0i})^* \ .
\]

The determination of $\Pi$ is straight forward. Commutation with
$\Sigma_{ij}$ implies that $\Pi$ is a linear combination of $\mathbb{1},
\gamma_5, \gamma_0$. Anti-commutation with $\Sigma_{0i} =
\Case{i}{2}\gamma_0\gamma_i$ implies that only $\gamma_0$ survives and
$\Pi = c\gamma^0$. Now $\Pi^2 = \mathbb{1}$ implies $c = \pm 1$. Hence,
$\boxed{(\Psi_{\cal{P}})^A(x) = \pm (\gamma^0)^A_{~B}\Psi^B(t,
-\vec{x})}$. It follows immediately that left(right) handed Weyl
spinor becomes right(left) handed Weyl spinor under space

The determination of $\tau$ involves complex conjugation. From the
unitarity of the $\gamma$ matrices and the defining Clifford relations,
the hermiticity properties are determined: $\gamma_0^{\dagger} =
\gamma_0, \ \gamma_i^{\dagger} = - \gamma_i$. The transpose/complex
conjugation properties however depend on the explicit choice of the
$\gamma$'s. The $\tau$ matrix thus depends on the explicit
representation of the $\gamma$'s. There are three commonly
employed representations:

\begin{eqnarray} \label{GammaRepresentations}
\mbox{Dirac-Pauli} & : & \gamma^0 := \left(\begin{array}{cc}
\mathbb{1} & \mathbb{0} \\ \mathbb{0} & -\mathbb{1}
\end{array}\right)\ ,\
\gamma^i := \left(\begin{array}{cc} \mathbb{0} & \sigma^i \\
-\sigma^i & \mathbb{0} \end{array}\right)\ ,\
\gamma_5 := \left(\begin{array}{cc} \mathbb{0} & \mathbb{1} \\
\mathbb{1} & \mathbb{0} \end{array}\right) \ ; \\
\mbox{Weyl} & : & \gamma^0 := \left(\begin{array}{cc} \mathbb{0}
& \mathbb{1} \\ \mathbb{1} & \mathbb{0} \end{array}\right)\ ,\
\gamma^i := \left(\begin{array}{cc} \mathbb{0} & \sigma^i \\
-\sigma^i & \mathbb{0} \end{array}\right)\ ,\
\gamma_5 := \left(\begin{array}{cc} -\mathbb{1} & \mathbb{0} \\
\mathbb{0} & \mathbb{1} \end{array}\right) \ ; \\
\mbox{Majorana} & : & \gamma^0 := \left(\begin{array}{cc}
\mathbb{0} & \sigma_2 \\ \sigma_2 & \mathbb{0}
\end{array}\right)\ ,\
\gamma^1 := \left(\begin{array}{cc} i\sigma_3 & \mathbb{0} \\
\mathbb{0} & i\sigma_3 \end{array}\right)\ ,\
\gamma^2 := \left(\begin{array}{cc} \mathbb{0} & -\sigma_2 \\
\sigma_2 & \mathbb{0} \end{array}\right) \ ; \\
& : & \gamma^3 := \left(\begin{array}{cc} -i\sigma_1 &
\mathbb{0} \\ \mathbb{0} & -i\sigma_1 \end{array}\right)\ ,\
\gamma_5 := \left(\begin{array}{cc} \sigma_2 & \mathbb{0} \\
\mathbb{0} & -\sigma_2 \end{array}\right)\ ,\
\mbox{All purely imaginary.} ~ ~
\end{eqnarray}

In the Dirac-Pauli and Weyl representations, only $\gamma_2$ is
imaginary. The $\tau$ matrix thus satisfies:

\begin{eqnarray*}
& & \Sigma_{12}\tau = - \tau\Sigma_{12} ~,~ \Sigma_{23}\tau = -
\tau\Sigma_{23} ~,~ \Sigma_{31}\tau =  \tau\Sigma_{31} ~, \\
& & \Sigma_{01}\tau = - \tau\Sigma_{01} ~,~ \Sigma_{02}\tau =
\tau\Sigma_{02} ~,~ \Sigma_{03}\tau = - \tau\Sigma_{03} ~,
\end{eqnarray*}

The last relation in the first line suggests $\tau = \lambda \gamma^0
\gamma^2$. It also checks with all other relations. The $\lambda$ is
determined as follows.

The operator $D(\cal{T})$ is anti-unitary. Hence,

\[
(D(\cal{T})D(\cal{T})\Psi_1, D(\cal{T})D(\cal{T})\Psi_2) ~ = ~
(D(\cal{T})\Psi_2, D(\cal{T})\Psi_1) ~ ~ = ~ ~ (\Psi_1, \Psi_2) ~ ~
\Rightarrow ~ ~ D^2(\cal{T}) ~ \mbox{is unitary}.
\]
Next, $D^2(\cal{T}) = (\tau K)(\tau K) = \tau\tau^* =
|\lambda|^2(\gamma^0\gamma^2\gamma^0(-\gamma^2)) = -
|\lambda|^2\mathbb{1}$. The unitarity then implies that $|\lambda| = 1~$
i.e. $\lambda$ is a phase and $D^2(\cal{T}) = -\mathbb{1}$ in the spinor

In the Majorana representation, all $\gamma$'s are imaginary. Hence,
$\Sigma_{\mu\nu}^* = - \Sigma_{\mu\nu}.$ Hence $\Sigma_{jk}$ commute
with $\tau$ and anti-commute with $\Sigma_{0i}$. This is exactly as for
the space inversion and we deduce that $D(\cal{T}) = \lambda\gamma^0 K$.
Once again $D^2(\cal{T})$ is unitary and equal to $-\mathbb{1}$ and
$\lambda$ is a phase.

For vector representation, the `$A$' index will be a tensorial index and
it is left as an exercise to work out the $D$ matrices for space
inversion and time reversal.

The existence of the anti-particle representations suggest one more
discrete transformation of order 2, called Charge Conjugation.

Recall that the anti-particle representation is the subspace of negative
frequency solutions. These subspaces are spanned by plane wave solutions
and involve the operation of complex conjugation which takes $e^{ik\cdot
x} \to e^{-ik\cdot x}$. For the non-trivial $D(\Lambda)$
representations, complex conjugation also takes the complex conjugate of
the $D(\Lambda)$ matrices. However, a complex conjugation of a solution
need not be a solution again (especially for spinor as we will see) and
hence the charge conjugation must also involve additional
transformations over and above complex conjugation. For the
Klein-Gordon, Proca and Maxwell equations, complex conjugate of a
solution is also a solution as the differential operators are real. For
the Dirac equation we need to do further work.

For instance, let $\Psi$ be a positive frequency solution of the Dirac
equation, $(-i\gamma^{\mu}\partial_{\mu} + m)\Psi = 0$. Taking complex
conjugate of the equation, we get $(+i(\gamma^{\mu})^*\partial_{\mu} +
m)\Psi^* = 0$. If we could find an invertible matrix $B$ such that
$(\gamma^{\mu})^* = -B\gamma^{\mu}B^{-1}$, then $\Psi^c :=
(B^{-1}\Psi^*)$ satisfies the same Dirac equation and of course is a
negative frequency solution. $\Psi^c$ is called the charge
conjugate[The terminology comes when coupling to external
electromagnetic field is considered by the minimal substitution
$\partial_{\mu} \to \partial_{\mu} -ieA_{\mu}$. Under complex
conjugation, $e \to -e$. So if $\Psi$ is thought of as charge `e'
solution then $\Psi^c$ is a charge `-e' solution.] of the $\Psi$.
Apparently, $B$ depends on the choice of explicit $\gamma$ matrices.

As a preparation for subsequent development, we have a subsection
on properties of $\gamma$ matrices.

§.§ Representations of the Clifford algebra and relations among

Quite generally, for any group, given a (matrix) representation $R(G)$,
we have three other representations, namely, $R^*(G), (R^T)^{-1}(G)$ and
$(R^{\dagger})^{-1}(G)$. From the basic relation $R(g_1)R(g_2) =
R(g_1.g_2)$, taking complex conjugate, transpose inverse and adjoint
inverse immediately verifies the assertion. If in addition, the
representation $R(G)$ is unitary (always true for finite groups
and compact Lie groups), then $(R^T)^{-1}(G) =  R^*(G)$ and
$(R^{\dagger})^{-1}(G) = R(G)$. $R(G)$ and $R^*(G)$ are then the only
independent representations. These are either equivalent, $R^*(g) =
SR(g)S^{-1}$ or inequivalent. It turns out that for unitary, irreducible
representations, if $R$ and $R^*$ are equivalent, then $S$ is either
symmetric or anti-symmetric. The following terminology ensues.

For unitary, irreducible representations:

\begin{eqnarray*}
R(G) \ \mbox{is complex \hspace{0.55cm}} & \mbox{if} & R^*(G)
\nsim R(G) \\
R(G) \ \mbox{is pseudo-real \hspace{0.0cm}} & \mbox{if} & R^*(G)
= S R^*(G)
S^{-1} ~ ~,~ ~ S^T = - S \\
R(G) \ \mbox{is real \hspace{1.4cm}} & \mbox{if} & R^*(G) = S
R^*(G) S^{-1} ~ ~,~ ~ S^T =  S
\end{eqnarray*}

This is relevant for representations of internal symmetry groups as well
as the Clifford group of the 32 elements. The $D(\Lambda)$
representations are finite dimensional but not unitary since Lorentz
group does not have unitary finite dimensional representations.

For the Clifford group in 4 dimensions (and in even number of
dimensions) there is only one (up-to unitary equivalence) non-trivial
representation which is 4 dimensional (or $2^{N/2}$ dimensional for
$N$-even number of dimensions). Since $\pm\gamma^*, \pm\gamma^T$ and
$\pm\gamma^{\dagger}$ all satisfy the same Clifford algebra and the
representation is unique, $\exists$ matrices $B, C, D$ such that,

\[
(\gamma^{\mu})^* = - B\gamma^{\mu}B^{-1} ~,~ (\gamma^{\mu})^T =
- C\gamma^{\mu}C^{-1} ~,~ (\gamma^{\mu})^{\dagger} = +
D\gamma^{\mu}D^{-1} ~.
\]

The choice of signs above is conventional. Notice that replacing any of
the $B, C, D$ matrices by multiplying by $\gamma_5$ on the right,
reverses the signs.

It follows immediately that,

\[
-(\Sigma^{\mu\nu})^* ~ = ~ B\Sigma^{\mu\nu}B^{-1} ~ ~,~ ~
-(\Sigma^{\mu\nu})^T ~ = ~ C\Sigma^{\mu\nu}C^{-1} ~ ~,~ ~
(\Sigma^{\mu\nu})^* ~ = ~ D\Sigma^{\mu\nu}D^{-1}
\]

Given a representation $D(\Lambda)$ we have the
$(D^{\dagger})^{-1}(\Lambda)\ , (D^T)^{-1}(\Lambda)$ and $D^*(\Lambda)$
representations. The infinitesimal forms are $(\mathbb{1} +
\frac{i}{2}\omega\cdot\Sigma), (\mathbb{1} -
\frac{i}{2}\omega\cdot\Sigma)^{\dagger}, (\mathbb{1} -
\frac{i}{2}\omega\cdot\Sigma^T), (\mathbb{1} -
\frac{i}{2}\omega\cdot\Sigma^*)$ which imply that the corresponding
generators are $\Sigma^{\mu\nu}, (\Sigma^{\mu\nu})^{\dagger},
-(\Sigma^{\mu\nu})^T, -(\Sigma^{\mu\nu})^*$. The $B, C, D$ matrices
precisely relate these generators.

These relations can be exponentiated to get corresponding relations
among the $D(\Lambda), D^{\dagger}(\Lambda), D^T(\Lambda),
D^*(\Lambda)$, namely,

\[
D^{\dagger}(\Lambda) = DD^{-1}(\Lambda)D^{-1} ~ ~,~ ~
D^{T}(\Lambda) = CD^{-1}(\Lambda)C^{-1} ~ ~,~ ~ D^{*}(\Lambda) =
BD(\Lambda)B^{-1} ~ .
\]

These can be checked easily by using the series form of the exponentials
derived from $D(\Lambda) = \sum_{k=0}^{\infty} \frac{(i \omega\cdot
\Sigma)^k}{k!}$.

§.§ Dirac-Majorana-Charge Conjugates

This allows us to construct Lorentz invariant bi-linears  from the
spinors. For instance, define the Majorana conjugate $\tilde{\Psi}
:= \Psi^T C$. Then,

\[
(\tilde{\Psi})_{\Lambda} := (\Psi_{\Lambda}^T) C ~ = ~
\Psi^T(D^T(\Lambda)) C ~ = ~ \Psi^T C D^{-1}(\Lambda) ~ = ~
\tilde{\Psi}D^{-1}(\Lambda) \ .
\]

Thus the Majorana conjugate, like the Dirac conjugate $\bar{\psi} :=
\Psi^{\dagger}D$, transforms by $D^{-1}(\Lambda)$. Recall that the
charge conjugate, $\Psi^{c}$ transforms by $D(\Lambda)$. Consequently, $\bar{\Psi}\Psi, \bar{\Psi}\Psi^c, \tilde{\Psi}
\Psi, \tilde{\Psi} \Psi^c$ are all Lorentz scalars.

Recalling that $D^{-1}(\Lambda)\gamma^{\mu}D(\Lambda) =
\Lambda^{\mu}_{~\nu}\gamma^{\nu}$, it is easy to see that,

\begin{eqnarray*}
\bar{\Psi}\Psi & \mbox{is a} & \mbox{Scalar} \\
\bar{\Psi}\gamma^{\mu}\Psi & \mbox{is a} & \mbox{Vector} \\
\bar{\Psi}\gamma^{\mu}\gamma^{\nu}\Psi & \mbox{is a} &
\mbox{Tensor of rank 2} \\
\bar{\Psi}\gamma^{\mu}\gamma_5\Psi & \mbox{is a} & \mbox{Axial
(or pseudo) vector} \\
\bar{\Psi}\gamma_5\Psi & \mbox{is a} & \mbox{Pseudo-scalar}
\end{eqnarray*}

Combine these with the definitions,

\begin{eqnarray*}
\mbox{Dirac Spinor} & : & \Psi = \Psi ~ ~,~ ~ (\Psi)_{\Lambda} =
D(\Lambda)\Psi ; \\
\mbox{Dirac Conjugate} & : & \bar{\Psi} = \Psi^{\dagger}D ~ ~,~
~ (\bar{\Psi})_{\Lambda} = \bar{\Psi}D^{-1}(\Lambda) ; \\
\mbox{Charge Conjugate} & : & \Psi^c = B^{-1}\Psi^* ~ ~,~ ~
(\Psi^c)_{\Lambda} = D(\Lambda)\Psi^c ; \\
\mbox{Majorana Conjugate} & : & \tilde{\Psi} = \Psi^TC ~ ~,~ ~
(\tilde{\Psi})_{\Lambda} = \tilde{\Psi}D^{-1}(\Lambda) .
\end{eqnarray*}

Thus, the same Lorentz transformation properties hold with $\Psi \to
\Psi^c$ and/or $\bar{\Psi} \to \tilde{\Psi}$. The last two need a little

We defined $\gamma_5 := i \gamma^0\gamma^1\gamma^2\gamma^3 =
\Case{i}{4!}\varepsilon_{\mu\nu\alpha\beta} \gamma^{\mu} \gamma^{\nu}
\gamma^{\alpha} \gamma^{\beta}$, where $\varepsilon$ is the completely
antisymmetric symbol, with $\varepsilon_{0123} := 1$. This symbol is
also used to define the determinant of a matrix, eg,
$\boxed{\varepsilon_{\mu'\nu'\alpha'\beta'} \Lambda^{\mu'}_{~\mu}
\Lambda^{\nu'}_{~\nu} \Lambda^{\alpha'}_{~\alpha}
\Lambda^{\beta'}_{~\beta} = det(\Lambda)
\varepsilon_{\mu\nu\alpha\beta}}$. It follows immediately that
$\boxed{D^{-`}(\Lambda)\gamma_5D(\Lambda) = det(\Lambda)\gamma_5}$.

Consider now an axial vector combination,

\[
[\bar{\Psi}\gamma^{\mu}\gamma_5\Psi]_{\Lambda} ~ = ~
\bar{\Psi}D^{-1}(\Lambda)\gamma^{\mu}\gamma_5D(\Lambda)\Psi ~ =
~ \Lambda^{\mu}_{~\nu} \bar{\Psi} \gamma^{\nu} D^{-1} \gamma_5
D(\Lambda)\Psi ~ = ~ det(\Lambda) \Lambda^{\mu}_{~\nu}
\bar{\Psi} \gamma^{\nu} \gamma_5 \Psi \ .
\]

Note that for proper Lorentz transformations, the axial vector and the
pseudo-scalar transform as vector and scalar respectively.

Quantities that transform as tensors but with an extra factor of
determinant of the transformation are  called pseudo-tensors.
There are also tensor looking quantities that transform with additional
factor of $|det|^w$ and these are  called tensor densities of
weight $w$.

Let us return to determine the matrices $B, C, D$.

Since the $\gamma$'s are unitary, we can always choose any of the $B, C,
D$ matrices to be unitary. Quite generally, if $R' = S R S^{-1}$ for two
irreducible, equivalent unitary representations, then $R'^{\dagger}R' =
\mathbb{1} \Rightarrow R^{\dagger}S^{\dagger}S = S^{\dagger}S R
\Rightarrow S^{\dagger} S = \alpha\mathbb{1}$. Positivity of
$S^{\dagger}S$ gives $\alpha > 0$. Hence $S' := S/\sqrt{\alpha}$ is a
unitary matrix. This proves the claim.

Next, $\gamma^{\mu} = (\gamma^{\mu*})^* = -B^*\gamma^{\mu*}(B^{-1})^* =
+ B^*B \gamma^{m} (B^*B)^{-1} \Rightarrow B^*B = \lambda\mathbb{1}$ with
$\lambda$ being real since $B^*B$ is.  The determinant of $B$ being a
phase (unitarity of $B$) fixes $\lambda := \epsilon_B := \pm 1$ and
$\boxed{B^*B = \epsilon_B\mathbb{1}}$. Similar manipulation with
$\gamma^{\mu} = (\gamma^{\mu T})^T$ leads to $C^{-1} C^T =
\lambda\mathbb{1}$. Unitarity gives $C^{-1}C^T = C^{\dagger}C^T =
(C^T)^*C^T$, hence $\lambda$ is real and determinant gives $\lambda^4 =
1$. Hence, $\boxed{C^T = \epsilon_C C,  \epsilon_C = \pm 1}$. For
$\gamma = (\gamma^{\dagger})^{\dagger}$ we get $D^{\dagger} = \lambda D$
for some phase $\lambda$. Since $D$ is determined to within a phase, we
can define $D' := e^{i\alpha}D$ and choose $\alpha$ so that $\lambda =
e^{-2i\alpha}$ to get $(D')^{\dagger} = D'$. Thus, without loss of
generality, we can always choose $D^{\dagger} = D$. Now, knowing the
hermiticity properties of the $\gamma$'s, we get $\boxed{D = \gamma^0}$,
independent of any choice of explicit $\gamma$ matrices.

The $\epsilon_B, \epsilon_C$ are correlated since $\gamma^T =
(\gamma^*)^{\dagger} = (\gamma^{\dagger})^*$.

$\gamma^T = (\gamma^*)^{\dagger}$ leads to,

\[
-C\gamma C^{-1} ~ = ~ -(B^{\dagger})^{-1}D\gamma
D^{-1}B^{\dagger} ~ ~ \Rightarrow ~ ~ C = \lambda B D .
\]

$\gamma^T = (\gamma^{\dagger})^*$ leads to,

\[
-C\gamma C^{-1} ~ = ~ -(D^{*})B\gamma
B^{-1}(D^{-1})^* ~ ~ \Rightarrow ~ ~ C = \lambda' D^* B .
\]

Eliminating $C$ gives $BD = \Case{\lambda'}{\lambda}D^*B$ or $D^* =
\Case{\lambda}{\lambda'}B D B^{-1}$. However, $D = \gamma^0$ can always
be achieved as shown above. Therefore, $\lambda' = - \lambda$ and
$C = \lambda B D = - \lambda D^* B$. Next,

\[
C^T = \lambda D^T B^T = \lambda(D^{\dagger})^*(B^{\dagger})^* =
\lambda D^* (B^*)^{-1} = \lambda D^* (\epsilon_B B) = -
\epsilon_B C ~ \Rightarrow ~ \boxed{\epsilon_C = -\epsilon_B .}
\]

Note that this is independent of explicit $\gamma$ matrices and also
independent of the phase $\lambda$! We still need to determine
$\epsilon_B$, say.

Claim: The $\epsilon_B$ does not depend on the choice of $\gamma$

This is easily proved. Let $\gamma'$ and $\gamma$ be two distinct
choices of explicit $\gamma$-matrices. Both being unitary are related by
some unitary matrix, $S$ as $\gamma = S\gamma' S^{-1}$. Then
substitution in $\gamma^* = -B\gamma B^{-1}$ gives, $(\gamma')^* = -
B'\gamma' (B')^{-1}$ with $B' := (S^*)^{-1} B S$. It follows that
$(B')^* B' = (S^{-1}B^*S^*)( (S^*)^{-1}B S ) = \epsilon_B\mathbb{1}$,
proving the claim. It therefore suffices to choose an explicit
representation and evaluate $\epsilon_B$.

Referring to say the Dirac-Pauli representation, see
(<ref>), only $\gamma^2$ is pure imaginary. Hence,
$\gamma^{0,1,3}B = -B\gamma^{0,1,3}$ and $\gamma^2B = +B\gamma^2$.
Furthermore, $\gamma_5^* = \gamma_5 \Rightarrow \gamma_5B = - B\gamma_5$
as well. Therefore $B$ must be made up of odd number of $\gamma$'s. By
inspection, $B = \alpha\gamma^2$ satisfies the conditions. Unitarity of
$B$ restricts $\alpha$ to be a phase and $B^*B =
|\alpha|^2(-\gamma^2)(\gamma^2) = \mathbb{1} \Rightarrow \epsilon_B =
+1$ and $\epsilon_C = -1$. Also $C = \lambda BD =
\lambda\alpha\gamma^2\gamma^0 = (-\lambda\alpha)\gamma^0\gamma^2$. The phases $\lambda,\ \alpha$ are arbitrary and convention dependent.
We choose $\lambda = 1$ and $\alpha = i$ so that $\boxed{B =
i\gamma^2 ~,~ C = i\gamma^2\gamma^0\ .}$

This also verifies for the Weyl representation. For Majorana
representation, all $\gamma^{\mu}$'s are imaginary and thus commute with $B$. Hence $B$ must be a phase multiple of $\mathbb{1}$.
This is also consistent with commutation with $\gamma_5$. Clearly
$\epsilon_B = +1$ and $C = (\lambda\alpha)\gamma^0$.

This completes the discussion of representations and relations among

It is customary to introduce a charge conjugation operator,
$\cal{C} := B^{-1} K$, where $K$ instructs to take the complex conjugate
of the numbers on its right. This operator is anti-linear and since $B$
is unitary, it is also anti-unitary operator. It follows,

\[
\cal{C}\Sigma^{\mu\nu} = B^{-1}(\Sigma^{\mu\nu})^*K =
B^{-1}(-B\Sigma^{\mu\nu}B^{-1})K = - \Sigma^{\mu\nu}\cal{C} ~
\Rightarrow ~ \cal{C}D(\Lambda) = D(\Lambda)\cal{C} \ .
\]

Thus charge conjugation acts invariantly on Lorentz representations.
Provided $\cal{C}^2 = \mathbb{1}$, we have $(\Case{\mathbb{1} \pm
\cal{C}}{2})^2 = \Case{\mathbb{1} \pm \cal{C}}{2}$ and the Lorentz
representation can be reduced further. The corresponding subspaces
satisfy $\cal{C} \psi = \pm \Psi$ and these spinors are  called
Majorana spinors. Thus, Majorana spinors are Dirac spinors which
satisfy $\cal{C} \Psi = \pm \Psi \leftrightarrow B^{-1}\psi^* = \pm
\Psi$. So is $\cal{C}^2 = \mathbb{1}$? We have $\cal{C}^2 =
B^{-1}KB^{-1}K = (B^*B)^{-1}K^2 = \epsilon_B\mathbb{1} = \mathbb{1}$.
Therefore, we do have Majorana spinors (in 4 dimensions).

Do we have Weyl-Majorana spinors? Well, $\cal{C}\gamma_5 =
B^{-1}\gamma_5^*K = - \gamma_5B^{-1}K = -\gamma_5 \cal{C}$ and we cannot have Weyl-Majorana spinors in 4 dimensions.

A similar analysis can be carried out for spinors in any dimensions and
with any metric signature. This may be seen for instance in

§ RELATIVISTIC ACTIONS: CLASSICAL FIELDS

So far we focused on the representation theory of the Poincare group.
The abstract, algebraic approach revealed the attributes of these
representations, namely mass and spin/helicity. In order to have a
framework which is manifestly Poincare covariant, we constructed and
analyzed representations on vector valued (complex in general) function,
$\Psi^A$ - henceforth these will be generically referred to as fields. The irreducibility condition emerged as “field equations”.
These equations are all homogeneous, linear and with at the most two
derivatives. They all admit plane waves and their linear combinations as
solutions, but no other `phenomenon' involving (say) different types of
waves modifying their propagation etc. Intuitively, there are no

We would like to have a framework which is not only Poincare covariant
but also involves “interaction” or non-trivial “dynamics”. The well
tested and successful strategy is to have an action formulation.
Let us quickly note several advantages of having an action formulation:

* The equations of motion are retrieved invoking a
variational principle as the Euler-Lagrange equations of

* “Interactions” can be understood naturally as leading to
non-linear and/or coupled equations which can be easily
introduced as more than quadratic order terms in the

* Covariance of equations of motion (or dynamics) can be
easily incorporated by requiring appropriate invariance
of the action;

* The Noether's theorem gives a recipe for obtaining
quantities conserved by equations of motion (and hence
during interactions as well);

* It leads to a canonical framework of symplectic structure
(“Poisson brackets”) and a Hamiltonian evolution. This
provides a systematic method of identifying “degrees of
freedom” (eg Dirac's theory of constrained systems).

A canonical structure is already inherent in the quantum
framework: the imaginary part of the inner product
provides the symplectic structure while the Schrodinger
equation gives a Hamiltonian evolution;

* Path integral quantization - very well suited for gauge
theories especially the non-abelian ones - has action as
the central quantity.

Without further ado, let us proceed with an action formulation. Our
first aim is to obtain the Klein-Gordon, the Dirac and the Proca/Maxwell
equations. Because the equations are local, partial differential
equations, the action must be an integral over the Minkowski space-time,
of a Lagrangian density built out of the Poincare covariant fields and
their derivatives i.e. it must of the form $S = \int
d^4x\sqrt{|det(\eta)|} \cal{L} $ with $\cal{L}$ is a Lorentz scalar
built out of the fields. It is  called the Lagrangian density. Here,
$\eta$ denotes the Minkowski metric and is necessary to absorb the
Jacobian of Lorentz transforms of the space-time coordinates. The
absolute value of the determinant happens to be 1 and is suppressed
throughout. A variational principle has the form,

\[
\delta S[\Phi] := S[\Phi + \delta\Phi] - S[\Phi] ~ = ~ \int d^4x
\delta \cal{L} ~ := ~ \int d^4x ~ \delta\Phi ~ `` ~\frac{\delta
\cal{L}}{\delta\Phi} ~ " \ .
\]

Extremization of the action under arbitrary variation $\delta\Phi$ leads
to the Euler-Lagrange equations: $\frac{\delta \cal{L}}{\delta\Phi} =
0$. Since the equation we want to derive are linear, it suffices to have
$\cal{L}$ to be quadratic in the fields. Furthermore, the equations have
no more than two derivatives which can be obtained by restricting to
first derivatives in the Lagrangian.

Let us begin with the scalar field,

\begin{eqnarray*}
\phi_{\Lambda,a}(x) & = & \phi(\Lambda^{-1}(x - a)) ~ ~ , ~ ~
\Lambda_{\mu}^{~\alpha}\partial_{\alpha}\phi(\Lambda^{-1}(x -
a)) ~ ~ \Rightarrow ~ ~ \\
\left[\partial^{\mu}\phi\partial_{\mu}\phi\right]_{\Lambda,a} &
= & \eta^{\mu\nu} \left[\partial_{\mu} \phi\partial_{\nu}
\phi\right]_{\Lambda,a} ~ = ~ \eta^{\mu\nu}
\Lambda_{\mu}^{~\alpha} \Lambda_{\nu}^{~\beta}
\left[\partial_{\alpha} \phi\partial_{\beta}\phi\right]
(\Lambda^{-1}(x-a)) \\
& = &
\eta^{\alpha\beta}\partial_{\alpha}\phi\partial_{\beta}\phi
(\Lambda^{-1}(x-a)) ~ ~ = ~ ~ \left[\partial^{\alpha}\phi
\partial_{\alpha}\phi\right] (\Lambda^{-1}(x-a)) \ .
\end{eqnarray*}

When integrated over the space-time, we can change the dummy variable $x
\to \Lambda x + a$ which restores the argument of the $\phi$'s without
changing the integration measure. Hence $\int d^4x \partial^{\mu}\phi
\partial_{\mu}\phi$ is clearly Poincare invariant. Since all fields have
the same shift of the space-time argument, we will suppress the
translation part and focus on the Lorentz. This works as long as there
are no externally prescribed fields/functions, which break translation
invariance. Thus, let

\begin{eqnarray}
S[\phi] & := & \int_{M}d^4x\left[ -\frac{1}{2}\partial^{\mu}
\phi\partial_{\mu}\phi - \frac{1}{2}m^2\phi^2\right] ~ ~ =: ~ ~ \int_M
d^4x \cal{L}(\phi, \partial\phi) \\
\delta S\left[\phi\right] & = & \int_M d^4x \left[ - \partial_{\mu}\phi
\partial^{\mu}\delta\phi - m^2\phi \delta\phi\right] \nonumber \\
& = & \int_M d^4x \left[ - \partial_{\mu}(\partial^{\mu}\phi \delta\phi)
+ (\Box\phi - m^2\phi)\delta\phi \right] \nonumber \\
& = & \int_M d^4x \ \delta\phi\left[(\Box - m^2)\phi\right] -
\int_{\partial M}d^3x \ n_{\mu}(\partial^{\mu}\phi \delta\phi)
\end{eqnarray}

The integration is over all of Minkowski space-time which has no
boundary and hence the second term should be zero. In practice, such
space-times extending to infinite coordinates  are handled/defined putting
the system in a large box and taking a limit. The fields must satisfy
suitable boundary conditions so that the boundary contribution again
vanishes. Alternatively requiring the field to vanish fast enough in the
asymptotic regions, makes the boundary term vanish. We can always choose
the variational principle to set $\delta\phi = 0$ at the
boundary/asymptotic regions. While the boundary contribution are
important in some context, we will restrict ourselves to boundary
contribution being zero.

Demanding that the action be stationary for arbitrary variations of the
field, we get the equation of motion $(\Box - m^2)\phi = 0$.
Incidentally, we will get the same equation of motion from another
Lagrangian $\cal{L}' = \cal{L} + \partial_{\mu}\Lambda^{\mu}$. Thus,
several Lagrangians can give the same equations of motion (although the
Hamiltonian formulation will vary with the Lagrangians.)

Consider now a vector field. For the massive case, the equations we want
are $(\Box - m^2)v^{\mu} = 0 = \partial_{\mu}v^{\mu}$.

Consider $\cal{L} = aF^{\mu\nu}F_{\mu\nu} + b m^2v^\mu v_\mu,$ where
$F_{\mu\nu} := \partial_{\mu}v_{\nu} - \partial_{\nu}v_{\mu}$ and $a, b$
are non-zero constants to be determined.

\begin{eqnarray}
\delta\cal{L} & = & 2a F^{\mu\nu}(\partial_{\mu}\delta v_{\nu} -
\partial_{\nu}v_{\mu}) + 2b m^2v^{\mu}\delta v_{\mu} ~ ~ = ~ ~
4aF^{\mu\nu}\partial_{\mu}\delta v_{\nu} + 2 b m^2 v^{\nu}\delta
v_{\nu} \nonumber \\
& = & \partial_{\mu}\left(4a F^{\mu\nu}\delta v_{\nu}\right) -
4a(\partial_{\mu}F^{\mu\nu})\delta v_{\nu} + 2b m^2
v^{\nu}\delta v_{\nu} ~ ~ \Rightarrow ~ ~
-4a\partial_{\mu}F^{\mu\nu} + 2b m^2 v^{\nu} ~ = ~ 0 .\nonumber
\\
\therefore 0 & = & \left(\Box v^{\nu} - \frac{b}{2a}m^2
v^{\nu}\right) - \partial^{\nu}\partial\cdot v ~ ~ ~ ~ \mbox{
taking divergence and choosing $b = 2a$ gives} \nonumber \\
0 & = & (\Box - m^2)v^{\nu} ~ \mbox{and} ~ \partial\cdot v = 0 .
\end{eqnarray}

We got $b = 2a$ and we choose $b = -\Case{1}{2}$ similar to the
scalar field case. The reason for this choice will be clear little

Note: For $m = 0$, we denote the vector field by $A^{\mu}$. Now we
cannot conclude $\partial\cdot A = 0$, and the equation we get is:
$\partial_{\mu}F^{\mu\nu} = 0$ which is just the Maxwell equation. It
has the well known gauge invariance: $A_{\mu} \to A_{\mu} +
\partial_{\mu}\Lambda$ which allows us to take one of the components of
$A_{\mu}$ to be zero.

Exercise: Another natural choice of the Lagrangian is: $\cal{L}' =
a \partial^{\mu}v^{\nu}\partial_{\mu}v_{\nu}  + b m^2 v_{\mu}v^{\mu} + c
(\partial_{\mu}v^{\mu})^2$. Repeat the steps and deduce the choices
for the constants so as to get the Proca equations. Check that $\cal{L},
\cal{L}'$ differ by a divergence.

Lastly, let us consider the action for getting the Dirac equation. We
have already noted the Lorentz transformations of $\Psi, \bar{\Psi},
\gamma^{\mu}, \partial_{\mu}$. From these it follows that we can have
the following Lorentz invariant terms, quadratic  in the fields and with
a single derivative: $\bar{\Psi}\gamma^{\alpha}\partial_{\alpha}\Psi,\
\partial^{\alpha}\bar{\Psi}\partial_{\alpha}\Psi,\ \bar{\Psi}\Psi$ and
the same terms with a $\gamma_5$ inserted between the spinors. The terms
with the $\gamma_5$ are all pseudo-scalars and may be dropped if parity
(space-inversion plus a rotation) is required to be a symmetry. We
assume so for the present. The candidate Lagrangian is then,

\begin{equation}
\cal{L} := a \bar{\Psi}\gamma^{\alpha}\partial_{\alpha}\Psi + b
\partial^{\alpha}\bar{\Psi}\partial_{\alpha}\Psi + c m
\bar{\Psi}\Psi ~\mbox{and}~ \delta_{\bar{\Psi}}\cal{L} =
\delta\bar{\Psi}[ a
\dsl{\partial}\Psi - b\Box\Psi  + c m\Psi] .
\end{equation}

For the choice $a = -i, b = 0$ and $c = 1$, we get the Dirac equation.
If we varied $\psi$, then for the same choice, we will get the conjugate
Dirac equation: $i\partial_{\alpha}\bar{\Psi}\gamma^{\alpha} + m
\bar{\Psi} = 0$.

In summary, we have $S = \int_M d^4x\sqrt{|det\eta|}\cal{L}$ with,

\begin{eqnarray} \label{LagrangiansEqn}
\cal{L}_{scalar} & = & -
\frac{1}{2}\partial^{\mu}\phi\partial_{\mu}\phi  -
\frac{1}{2}m^2 \phi^2 \ ; \\
\cal{L}_{vector} & = & - \frac{1}{4}F^{\mu\nu}F_{\mu\nu} -
\frac{1}{2}m^2 v^{\mu}v_{\mu}\ ; \\