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\lhead{Physics 253a}
\chead{October 14}
\rhead{Sidney Coleman's Notes}
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\section*{Lorentz transformation properties of conserved quantities}
We've worked with three currents, $J^\mu$, the current of meson
number, $T^{\mu\nu}$, the current of the $\nu$th component of
momentum, and $M^{\mu\nu\lambda}$, the current of the $[\nu\lambda]$
component of angular momentum. We integrated the zeroth component of
each of these currents to obtain $Q$, $P^\nu$ and $J^{\nu\lambda}$
respectively. These look like tensors with one less index, but do
they have the right Lorentz transformation properties?
We'll prove that $P^\nu=\int d^3xT^{0\nu}$ is indeed a Lorentz vector
given that $T^{\mu\nu}$ is a two index tensor and that $T^{\mu\nu}$ is
conserved, $\partial_\mu T^{\mu\nu}=0$. The generalization to
currents with more indices or 1 index will be clear. We could do this
proof in the classical theory or the quantum theory. We'll choose the
latter because we need the practice. The assumption that $T^{\mu\nu}$
is an operator in quantum field theory that transforms as a two-index
tensor is phrased mathematically as follows:
Given $U(\Lambda)$ the unitary operator that effects Lorentz
transformations in the theory
\[ U(\Lambda)^\dagger T^{\mu\nu}(x)U(\Lambda)=\Lambda^\mu_\sigma
\Lambda^\nu_\tau T^{\sigma\tau}(\Lambda^{-1}x) \]
Now we'll try to show that
\[ P^\nu=\int d^3x T^{0\nu}(\vec{x},0) \]
\noindent is a Lorentz vector. Introduce $n=(1,0,0,0)$, a unit vector
pointing in the time direction. Then we can write $P^\nu$ in a way
that makes its Lorentz transformation properties clearer.
\[ P^\nu=\int d^4x n_\mu T^{\mu\nu}\delta(n\cdot x) \]
Perform a Lorentz transformation on $P^\nu$
\begin{align*}
U(\Lambda)^\dagger P^\nu U(\Lambda)&=\int d^4x n_\mu
U(\Lambda)^\dagger T^{\mu\nu}(x)U(\Lambda)\delta(n\cdot x) \\
&=\int d^4x n_\mu \Lambda^\mu_\sigma\Lambda^\nu_\tau
T^{\sigma\tau}(\Lambda^{-1}x)\delta(n\cdot x)
\end{align*}
Change integration variables $x'=\Lambda^{-1}x$ and define
$n'=\Lambda^{-1}n$
\begin{align*}
U(\Lambda)^\dagger P^\nu U(\Lambda)&=\int d^4x' \Lambda_{\mu\rho}
n'^\rho \Lambda^\mu_\sigma \Lambda^\nu_\tau T^{\sigma\tau}(x')
\delta(n'\cdot x') \\
&=\int d^4x' n'^\rho \Lambda^\nu_\tau T^{\rho\tau}(x')\delta(n'\cdot
x')\;\;\;\substack{\text{Using $\Lambda_{\mu\rho}\Lambda^\mu_\sigma=
g_{\rho\sigma}$}} \\
&=\Lambda^\nu_\tau \int d^4x n'_\rho T^{\rho\tau}(x)\delta(n'\cdot x)
\end{align*}
In the last expression we've dropped the prime on the variable of
integration. The only difference between what we have and what we
would like to get
\[ U(\Lambda)^\dagger P^\nu U(\Lambda)=\Lambda^\nu_\tau P^\tau
=\Lambda^\nu_\tau \int d^4x n_\rho T^{\rho\tau}\delta(n\cdot x) \]
\noindent is that $n$ has been redefined. The surface of integration
is now $t'=0$, and we take the component $n'_\mu T^{\mu\nu}$ in the
$t'$ direction. Our active transformation has had the exact same
effect as if we had made a passive transformation changing coordinates
to $x'=\Lambda^{-1}x$. It's the same old story:
$\underbrace{\text{alias}}_{\text{passive}}$ (another name)
versus $\underbrace{\text{alibi}}_{\text{active}}$ (another place).
% inserted figure oct 14 p2/14, p78 in 1-214
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\includegraphics[width=8cm]{07-fig1.eps}
\end{center}
You can think of this as Lorentz transforming the field or Lorentz
transforming the surface.
This doesn't produce a vector $P^\nu$ for any tensor $T^{\mu\nu}$.
For an arbitrary tensor, $P^\nu$ is not even independent of what time
you compute at, let alone changing the tilt of the surface. We need
to use that $T^{\mu\nu}$ is conserved. More or less, that the current
that flows through the surface $t'=0$ is the same as the current that
flows through the surface $t=0$.
Note that $n_\mu\delta(n\cdot x)=\partial_\mu\theta(n\cdot x)$, so
that if I call the Lorentz transform of $P^\nu$, $P'^\nu$, what we are
trying to show is
\begin{align*}
0&=P^\nu-\Lambda^{-1\nu}_\sigma P'^\nu\\
&=\int d^4x[\partial_\mu\theta(n\cdot x)-\partial_\mu\theta(n'\cdot
x)] T^{\mu\nu}(x) \\
&=\int d^4x\partial_\mu [\theta(n\cdot x)-\theta(n'\cdot x)]
T^{\mu\nu}(x)
\end{align*}
This integral over all spacetime is a total divergence. In the far
future or the far past $\theta(n\cdot x)=\theta(n'\cdot x)$ so there
are no surface terms there, and as usual we won't worry about surface
terms at spatial infinity.
\section*{{\sc Discrete Symmetries}}
A discrete symmetry is a transformation $q^a(t)\rightarrow q'^a(t)$
that leaves the Lagrangian, $L$, unchanged $L\rightarrow L$. This
could be parity, this could be rotation by $\pi$ about the $z$ axis.
(It does not include time reversal, wait\footnote{See Dirac
\uline{Principles of QM}, pp.~103ff}.)
Since \uline{all} the properties of the theory are derived from the
Lagrangian, (the canonical commutation relations, the inner product,
the Hamiltonian all come from $L$) and since the Lagrangian is
unchanged, we expect that there is a unitary operator effecting the
transformation
\begin{align*}
U^\dagger q^a(t) U&=q'^a(t) & U^\dagger HU=H
\end{align*}
The discrete symmetry could be an element of a continuous symmetry
group. It could be rotation by 20$^\circ$. There is no conserved
quantity associated with a discrete symmetry however. What is
special about the continuous symmetry is that there is a parameter,
and that you have a unitary operator for each value of the parameter,
$\theta$, satisfying
\[U(\theta)^\dagger HU(\theta)=H\]
You can differentiate this with respect to the parameter to find that
\begin{align*}
[I,H]&=0 & I&\equiv-i\left.\frac{dU}{d\theta}\right|_{\theta=0}\equiv
-iDU
\end{align*}
There is nothing analogous for discrete symmetries.
\section*{Examples of internal symmetries (there is not a general
theory, but we'll do prototypical examples)}
\subsubsection*{Example (1). The transformation is
$\phi(x)\rightarrow-\phi(x)$ at every space-time point.}
This is a symmetry for any Lagrangian with only even powers of $\phi$,
in particular,
\[\mathcal{L}=\frac{1}{2}(\partial_\mu\phi)^2-\frac{1}{2}\mu^2\phi^2
-\lambda\phi^4\rightarrow\mathcal{L} \]
There should be a unitary operator effecting the transformation
\[ \phi\rightarrow U^\dagger \phi U=-\phi \]
For $\lambda=0$, we will actually be able to construct $U$, that is
give its action on the creation and annihilation operators, and thus
its action on the basis states.
Since $\phi$ is a linear function of $a_{\vec{k}}$ and
$a_{\vec{k}}^\dagger$ it must be that
\[ U^\dagger a_{\vec{k}}U=-a_{\vec{k}} \]
\noindent and the h.c.~equation
\[ U^\dagger a_{\vec{k}}^\dagger U=-a_{\vec{k}}^\dagger \]
We'll also make a choice in the phase of $U$ by specifying
\[ U|0\rangle=|0\rangle \]
We can determine the action of $U$ on the basis states
\begin{align*}
U|k_1,\dots,k_n\rangle &= Ua_{\vec{k}_1}^\dagger a_{\vec{k}_2}^\dagger
\cdots a_{\vec{k}_n}^\dagger|0\rangle \\
&= Ua_{\vec{k}_1}^\dagger U^\dagger Ua_{\vec{k}_2}^\dagger U^\dagger
\cdots U a_{\vec{k}_n}^\dagger U^\dagger U|0\rangle \\
&= (-1)^na_{\vec{k}_1}^\dagger a_{\vec{k}_2}^\dagger
\cdots a_{\vec{k}_n}^\dagger|0\rangle =(-1)^n |k_1,\dots,k_n\rangle
\end{align*}
As an operator statement we see that
\[ U=(-1)^N(=e^{i\pi N}\text{ if you prefer
})\;\;\;\substack{(N=\text{meson number}\\ =\int d^3k
a_{\vec{k}}^\dagger a_{\vec{k}} )}\]
Suppose we could construct this operator when $\lambda\neq0$. The
existence of this unitary operator tells you that you'll never see 2
mesons scatter into 43 mesons or any odd number of mesons. Mesons are
always produced in pairs. More formally, it must be that $0=\langle
n|S|m\rangle$ where $n\rangle$ is a state with $n$ mesons, $|m\rangle$
is a state with $m$ mesons and $S$ is the scattering matrix made from
the Hamiltonian, whenever $n+m$ is odd. For if $n+m$ is odd
\begin{align*}
\langle n|S|m\rangle &=\langle n|U^\dagger USU^\dagger U|m\rangle =
\langle n|U^\dagger\!\!\!\!\!\!\!\!\!\!\underbrace{(USU^\dagger)}_{S\text{ because
}UHU^\dagger=H}\!\!\!\!\!\!\!\!\!U|m\rangle \\
&=(-1)^{n+m}\langle n|S|m\rangle =-\langle n|S|m\rangle
\end{align*}
\subsubsection*{Our second example of a discrete internal symmetry
will turn out to be \sc Charge Conjugation.}
Recall that
\[\mathcal{L}=\sum_{a=1}^2\left(\frac{1}{2}(\partial_\mu\phi^a)^2
-\frac{1}{2}\mu^2(\phi^a)^2\right)-\lambda[(\phi^1)^2+(\phi^2)^2]^2 \]
\noindent had an $SO(2)$ symmetry. In fact it has an $O(2)$ symmetry,
rotations and rotations with reflections in the $\phi^1$, $\phi^2$
plane. It is invariant under proper and improper rotations. We'll
take one standard improper rotation
\begin{align*}
\phi^1&\rightarrow \phi^1 & \phi^2&\rightarrow-\phi^2
\end{align*}
Any other one can be obtained by composing this one with an element of
the internal symmetry group $SO(2)$.
It is easy to write down the unitary operator in the case $\lambda=0$.
Then we have two independent free scalar field theories
\[ U_C=(-1)^{N_2} \]
As mathematicians are fond of saying, we have reduced it to the
previous case.
The action of $U$ is especially nice if we put it in terms of the
fields $\psi$ and $\psi^\dagger$ already introduced
\begin{align*}
\psi=\frac{\phi^1+i\phi^2}{\sqrt{2}}&\rightarrow U_C^\dagger \psi U_C
= \frac{\phi^1-i\phi^2}{\sqrt{2}} =\psi^\dagger \\
\psi^\dagger&\rightarrow\psi
\end{align*}
For this reason, this is sometimes called a conjugation symmetry.
Because $U_C^\dagger QU_C=-Q$ it is also called charge\footnote{the
charge from the $SO(2)$ symmetry} conjugation or
particle-anti-particle conjugation. From the action on $\psi$ and
$\psi^\dagger$ we see
\[ U_C^\dagger b_{\vec{k}} U_C=c_{\vec{k}}\;\;\;\text{ and }\;\;\;
U_C^\dagger c_{\vec{k}}U_C=b_{\vec{k}} \]
\noindent and the equations obtained from these by hermitian
conjugation. $U_C$ is unitary and hermitian:
\[U_C^2=1=U_CU_C^\dagger\Rightarrow U_C=U_C^\dagger\]
We could continue the discussion of discrete internal symmetries, but
it is boring. You could write down a Lagrangian with four fields that
is invariant under rotation in four-dimensional space and under
permutations of any of the four fields. You could write down a theory
with the icosahedral group.
\subsubsection*{\sc Parity Transformations}
Any transformation that takes
\[ \phi^a(\vec{x},t)\rightarrow\sum_b M^a_b\phi^b(-\vec{x},t) \]
\noindent we'll call a parity transformation. (We have used the fact
that we live in an odd number of dimensions (3) and thus that
$\vec{x}\rightarrow-\vec{x}$ is an improper rotation, in our
definition. If we lived in two space dimensions we could do a similar
thing with only $x^2\rightarrow-x^2$.) A parity transformation
transforms each fundamental observable at the point $\vec{x}$
into some linear combination of fundamental observables at the point
$-\vec{x}$.
Usually parity takes $\mathcal{L}\overset{\substack{\text{The usual}\\
\text{confusing notation}}}{\overbrace{(\vec{x},t)}\rightarrow
\mathcal{L}(-\vec{x},t)}$, when it is a symmetry, but all we really
demand is that parity takes
\[L(t)\rightarrow L(t)\;\;\;\text{ as usual.}\]
\newcounter{07count1}
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\begin{list}{\sc Example (\arabic{07count1}) }
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\item Parity is a symmetry of
\begin{align*}
\mathcal{L}&=\frac{1}{2}(\partial_\mu\phi)^2-\frac{\mu^2}{2}\phi^2 &
\begin{split}
P:\phi(\vec{x},t) &\rightarrow\phi(-\vec{x},t)\\
L&\rightarrow L
\end{split}\;\;\;(M=1)
\end{align*}
From
\begin{align*}
\phi(\vec{x},t)&=\int \frac{d^3k}{(2\pi)^{3/2}
\sqrt{2\omega_{\vec{k}}}} [a_{\vec{k}}^ae^{-i\vec{k}\cdot \vec{x}}
e^{-i\omega_{\vec{k}}t}+ a_{\vec{k}}^{a\dagger}e^{i\vec{k}\cdot
\vec{x}}e^{i\omega_{\vec{k}}t}]\\
\text{and }\;\;\;U_P^\dagger\phi(\vec{x},t)U_P&=\phi(-\vec{x},t)
\end{align*}
\noindent we can see that the action on the creation and annihilation
operators must be
\[ U_P^\dagger\begin{Bmatrix}
a_{\vec{k}}\\
a_{\vec{k}}^\dagger
\end{Bmatrix} U_p=\begin{Bmatrix}
a_{-\vec{k}}\\
a_{-\vec{k}}^\dagger
\end{Bmatrix} \]
\noindent and on the basis states
\[ U_P|\vec{k}_1,\dots,\vec{k}_n\rangle=|-\vec{k}_1,\dots,-\vec{k}_n
\rangle \]
But there is a second possibility for parity
\begin{align*}
P':\phi(\vec{x},t)&\rightarrow-\phi(-\vec{x},t) &(M=-1)\\
L&\rightarrow L
\end{align*}
\uline{Whenever} there is an internal symmetry in a theory I can
multiply one definition of parity by an element of that symmetry group
(discrete or continuous) and get another definition of parity. In the
case at hand the unitary operator $U_{P'}$ is given by
\[ U_{P'}=(-1)^NU_P\;\;\;\text{ or }\;\;\;
U_{P'}|\vec{k}_1,\dots,\vec{k}_n\rangle=(-1)^n|-\vec{k}_1,\dots,
-\vec{k}_n\rangle \]
Sometimes people distinguish between a theory with invariance under
$P:\phi(\vec{x},t)$ $\rightarrow\phi(-\vec{x},t)$ and a theory with
invariance under $P':\phi(\vec{x},t)\rightarrow-\phi(-\vec{x},t)$, by
calling the first the theory of a scalar meson and the second a theory
of a pseudo scalar meson. Our theory is invariant under both; it's the
old plate of cookies problem again. The theory has a set of
invariances. As long as you are in agreement about the total set of
invariances of the theory, you shouldn't waste time arguing about what
you'll call each one. This is why the conventions on the parity of
some particles is arbitrary (relative parity). If $-\lambda\phi^4$ is
added to $\mathcal{L}$ both $P$ and $P'$ are still invariances of $L$.
With a $\phi^3$ interaction $P$ is a symmetry, $P'$ isn't. All the
physicists in the world would agree that this is a scalar meson. One
of the cookies has been poisoned.
\item This awful Lagrangian has been cooked up to illustrate a point.
After this lecture you won't see anything this bad again.
\[\mathcal{L}=\sum_{a=1}^4\left[\frac{1}{2}(\partial_\mu\phi^a)^2-
\frac{\mu_a^2}{2}(\phi^a)^2\right]-g\epsilon^{\mu\nu\lambda\sigma}
\partial_\mu\phi^1\partial_\nu\phi^2\partial_\lambda\phi^3
\partial_\sigma\phi^4\]
All four meson masses are different. The new interaction involving
the totally antisymmetric tensor in four indices is invariant under
L.T.~for the same reason that $\vec{a}\cdot(\vec{b}\times\vec{c})$ is
invariant under proper rotations. ($\vec{a}\cdot(\vec{b}\times
\vec{c})$ is multiplied by $\det R$ under a rotation). Because
$\epsilon^{\mu\nu\lambda\sigma}$ is nonzero only if one of its indices
is timelike, the other three spacelike, three of the derivatives are
on space coordinates. We get three minus signs under parity. An odd
number of mesons are going to have to be pseudoscalar to get a net
even number of minus signs. Because any $\phi^a\rightarrow-\phi^a$ is
an internal symmetry of the free Lagrangian, it doesn't matter which
one you choose or which three you choose to be pseudoscalar.
\item Take the perverse Lagrangian of the last example by and make it
worse by adding
\[\sum_{a=1}^4(\phi^a)^3 \]
Now there is no definition of parity that gives a symmetry. This
theory violates parity.
\item Sometimes people say that because the product of two reflections
is 1, the square of parity is 1. This example is cooked up to show
that a theory with parity can have indeed $U_P^2\neq 1$. Indeed $U_P$
cannot be chosen to satisfy $U_P^2=1$.
\begin{multline*}
\mathcal{L}=\sum_{a=1}^4\left[\frac{1}{2}(\partial_\mu\phi^a)^2-
\frac{\mu_a^2}{2}(\phi^a)^2\right]+\partial_\mu\psi^*\partial^\mu\psi
-m^2\psi^*\psi\\
-h\sum_{a=1}^4(\phi^a)^3-g\epsilon_{\mu\nu\lambda\sigma}
\partial^\mu\phi^1\partial^\nu\phi^2\partial^\lambda\phi^3
\partial^\sigma\phi^4[\psi^2+\psi^{*2}]
\end{multline*}
The transformation of the $\phi^a$'s must be $\phi^a(\vec{x},t)
\rightarrow+\phi^a(\vec{x},t)$. The only way to make the last term
parity invariant is for $\psi$ ot transform as
\begin{align*}
\psi&\rightarrow\pm i\psi & \psi^*&\rightarrow\mp i\psi^*
\end{align*}
In either case $U_P^2\neq1$, $U_P^\dagger U_P^\dagger \psi
U_PU_P=-\psi$. Fortunately, nothing like this occurs in nature (as
far as we know). If it did, and if parity were a symmetry (or an
approximate symmetry) of the world we would have a name for fields
transforming like $\psi$, a ``semi-pseudo-scalar''.
\end{list}
\subsubsection*{\sc Time Reversal}
First a famous example from classical particle mechanics, a particle
moving in a potential
\begin{align*}
L&=\frac{1}{2}m\dot{q}^2-V(q) & T:q(t)&\rightarrow q(-t)
\end{align*}
$T$ is not a discrete symmetry the way we have defined it
\[ T:L(t)\rightarrow L(-t) \;\;\;\substack{\text{in the usual
confusing } \\ \text{notation where $L(t)$ refers} \\ \text{to the
time dependence} \\ \text{through the coordinates}}\]
Nevertheless $T$ does take one solution of the equations of motion
into another. You might still hope there is a unitary operator that
does the job in the quantum theory.
\[U_T^\dagger q(t)U_T=q(-t)\]
There are two paradoxes I'll give to show this can't happen.
\begin{description}
\item{\sc 1st Paradox } Differentiate $U_T^\dagger q(t)U_T=q(-t)$ with
respect to $t$. Because $p(t)\propto \dot{q}(t)$
\[ U_T^\dagger p(t)U_T=-p(t)\]
Consider $U_T^\dagger[p,q]U_T=-i$. From the two relations we have
juts obtained we also have
\[ U_T^\dagger[p(t),q(t)]U_T=-[p(-t),q(-t)]=-i\;\;\;\substack{
\text{Particularly poignant}\\ \text{at $t=0$}} \]
Looks like we would have to give up the canonical commutation
relations to implement time reversal. If that isn't enough to make
you abandon the idea of a unitary time reversal operator I'll continue
to the
\item{\sc 2nd Paradox } Roughly, $U_T$ should reverse time evolution,
i.e.~$U_T^\dagger e^{-iHt}U_T=e^{iHt}$. I can prove this. For any
operator $\mathcal{O}(t)$
\[ \mathcal{O}(t)=e^{iHt}\mathcal{O}e^{-iHt} \]
Apply $U_T^\dagger$ the unitary transformation to both sides to obtain
\begin{align*}
\mathcal{O}(-t)&=U_T^\dagger e^{+iHt}U_T\mathcal{O}U_T^\dagger
e^{-iHt}U_T & \mathcal{O}&=\mathcal{O}(0) \\
\text{but }\mathcal{O}(-t)&= e^{-iHt}U_T\mathcal{O}e^{iHt} &
\substack{\text{let }V\equiv U_T^\dagger e^{-iHt}U_T}
\end{align*}
I'd like to show $V=e^{iHt}$. What we have is $e^{-iHt}\mathcal{O}
e^{iHt}=V^{-1}\mathcal{O}V$ which implies $Ve^{-iHt}\mathcal{O}=
\mathcal{O}Ve^{-iHt}$. $Ve^{-iHt}$ commutes with any operator
$\mathcal{O}$. $Ve^{-iHt}=1$. Now that I've proved
\[ U_T^\dagger e^{-iHt}U_T=e^{iHt} \]
Take $\left.\frac{d}{dt}\right|_{t=0}$ of this relation,
\[U_T^\dagger(\cancel{-i}H)U_T=\cancel{i}H,\]
\noindent canceling the $i$'s, we see that $H$ is unitarily related
to $-H$. The spectrum of $H$ cannot be bounded below, because (the
spectrum of unitarily related operators is the same and) the spectrum
of $H$ is not bounded above. AUGGH!
\end{description}
A unitary time reversal operator is an object that makes no sense
whatsoever. The answer is that time reversal is implemented by an
antiunitary operator. Antiunitary operators are antilinear. Dirac
notation is designed to automate the handling of linear operators, so
for a while we'll use some more cumbersome notation that does not
automate the handling of linear operators.
$\begin{matrix}
\text{Let }& a,b &\text{denote states,}\\
& \alpha,\beta &\text{denote complex numbers and}\\
& A,B &\text{denote operators.}\\
&(a,b)&\text{is the inner product of two states.}
\end{matrix}$
A \uline{unitary operator} is an invertible operator, $U$, satisfying
\begin{align*}
(Ua,Ub)&=(a,b) \;\;\;\text{ for all }\;\;\;a,b & \substack{\text{
unitarily}}
\end{align*}
This is enough of an assumption to show $U$ is linear (see related
proof below). The simplest unitary operator is 1.
\begin{align*}
U(\alpha a+\beta b)&= \alpha Ua+\beta Ub &
\substack{\text{linearity}}
\end{align*}
The adjoint of a linear operator $A$ is denoted $A^\dagger$ and is the
operator defined by (this definition is not consistent if $A$ is not
linear)
\[(a,A^\dagger b)=(Aa,b)\;\;\;\text{ for all }\;\;\; a,b \]
I'll show that $U^\dagger=U^{-1}$ (which is sometimes given as the
definition of unitarity).
\[(a,U^{-1}b)\overbrace{=}^{\substack{\text{unitarity}\\ \text{of
$U$}}} (Ua,UU^{-1}b)=(Ua,b) \]
A transformation of the states, $a\rightarrow Ua$, can also be thought
of as a transformation of the operators in the theory
\[(a,Ab)\rightarrow(Ua,AUb)=(a,U^\dagger AUb)\;\;\;\text{ can
alternatively be thought of as $A\rightarrow U^\dagger AU$.}\]
An antiunitary operator is an invertible operator, $\Omega$, (this is
a notational gem, an upside down $U$) satisfying
\begin{align*}
(\Omega a,\Omega b)&=(b,a)\;\;\; \text{ for all }
\;\;\;a,b & \substack{\text{antiunitarity}}
\end{align*}
We can immediately make a little table showing the result of taking
products of unitary and antiunitary operators. (The product of a
unitary operator with an antiunitary operator is an antiunitary
operator, etc.) $\begin{matrix}
& \begin{matrix}U\hfill & \Omega\end{matrix}\\
\begin{matrix}
U\\
\Omega\end{matrix} &
\begin{array}{|c|c|}
\hline
U&\Omega \\
\hline
\Omega&U\\
\hline
\end{array}
\end{matrix}$
We can prove (this is the related proof referred to above) that any
operator (not necessarily invertible) satisfying the antiunitarity
condition is antilinear.
\begin{align*}
(\Omega a,\Omega b)&=(b,a)\Rightarrow \Omega(\alpha a+\beta
b)=a^*\Omega a+\beta^*\Omega b &\substack{\text{antilinearity}}
\end{align*}
Consider $(\Omega(\alpha a+\beta b)-\alpha^*\Omega a-\beta^*\Omega b,
\Omega(\alpha a+\beta b)-\alpha^*\Omega a-\beta^*\Omega b)$. (You ask
why?!) This is the inner product of $\Omega(\alpha a+\beta
b)-\alpha^*\Omega a-\beta^*\Omega b$ with itself. If this is zero,
the fact that the inner product is positive definite implies that
$\Omega(\alpha a+\beta b)-\alpha^*\Omega a-\beta^*\Omega b=0$.
The result we want$!$ Indeed, it is simply a matter of expanding this
inner product out into its 9 terms, applying the antiunitarity
condition to each term, and then expand the 5 terms containing $\alpha
a+\beta b$ some more to show this is zero. (The analogous proof for
operators satisfying the unitarity condition also only uses properties
of the inner product and is even easier.)
The simplest antiunitary operator is complex conjugation, $K$. For
the elements of some basis, $b_i$, $Kb_i=b_i$ and on any linear
combination
\[K(\sum_i\alpha_ib_i)=\sum_i\alpha_i^*b_i \]
For consistency $(b_i,b_j)$ must be real. This is the familiar
complex conjugation of nonrelativistic quantum mechanics of position
space wave functions. The basis is a complete set of real wave
functions.
A useful fact (especially conceptually) is that any antiunitary
operator, $\Omega$, is equal to $UK$ for some unitary $U$. Proof by
construction: take $U=\Omega K$.
In a more limited sense, the transformation of the states by an
antiunitary operator $\Omega$, $a\rightarrow \Omega a$, can also be
thought of as a transformation of the operators in the theory.
Consider the expectation value of a Hermitian operator (observable)
in the state $a$. It transforms as
\begin{align*}
(a,Aa)\rightarrow(\Omega a,A\Omega a)&=(A\Omega a,\Omega a) &
\substack{\text{(hermiticity)}} \\
&= (\Omega\Omega^{-1}A\Omega a,\Omega a)
&\substack{\text{invertibility}} \\
&= (a,\Omega^{-1}A\Omega a) & \substack{\text{antiunitarity}}
\end{align*}
This transformation can alternatively be thought of as
\[ A\rightarrow\Omega^{-1} A\Omega \]
We don't write $\Omega^\dagger A\Omega$ because adjoint is not even
defined for antilinear ops.
50 years ago [1931], Eugene Wigner proved a beautiful theorem
telling us why unitary and antiunitary operators are important in QM.
He showed that (up to phases) they are the only operators that
preserve probabilities. It is not necessary to preserve inner
products; they aren't measurable. It is the probabilities that are
measurable. Look in the appendix of his book on group theory.
Given that $F(a)$ ($F:\mathcal{H}\rightarrow\mathcal{H}$) is
continuous and for any $a$ and $b$
\begin{align*}
|(F(a),F(b))|^2&=|(a,b)|^2\\
\text{then }\;\;\;F(a)&=e^{i\phi(a)}x\left\{\substack{\text{a unitary
op}\\ \text{or an}\\ \text{antiunitary op}}\right. &
(\phi:\mathcal{H}\rightarrow\mathbb{R})
\end{align*}
Time reversal is not a unitary operator. Time reversal is
antiunitary.
It is easy to see now how the two paradoxes are avoided.
\[\Omega_T^{-1}i\Omega_T=-i\]
You can't cancel the $i$'s as we did in paradox 2.
\[\Omega_T^{-1}(-iH)\Omega_T=iH\Rightarrow \Omega_T^{-1}H\Omega_T=H \]
We can explicitly construct the time reversal operator in free field
theory. The simpler thing to look at in a relativistic theory is
actually $PT$. Let's find
\[\Omega_{PT}\;\;\;\text{ such that }\;\;\;\Omega_{PT}^{-1}\phi(x)
\Omega_{PT}=\phi(-x) \]
The simplest candidate is just complex conjugation, in the momentum
state basis. That is $\Omega_{PT}$ does nothing, \uline{absolutely
nothing} to $a_{\vec{k}}$ and $a_{\vec{k}}^\dagger$
\[\Omega_{PT}^{-1}a_{\vec{k}}\Omega_{PT} = a_{\vec{k}}\;\;\;\text{
and }\;\;\;\Omega_{PT}^{-1}a_{\vec{k}}^\dagger\Omega_{PT} =
a_{\vec{k}}^\dagger\]
Furthermore we'll take $\Omega_{PT}|0\rangle=|0\rangle$ and it follows
that
\[\Omega_{PT}|\vec{k}_1,\dots,\vec{k}_n\rangle=|\vec{k}_1,\dots,
\vec{k}_n\rangle \]
\noindent again nothing, they just lie there. What does this operator
do to $\phi(x)$?
\[\phi(x)=\int \frac{d^3k}{(2\pi)^{3/2} \sqrt{2\omega_{\vec{k}}}}
[a_{\vec{k}}^ae^{-ik\cdot x} + a_{\vec{k}}^{a\dagger}e^{ik\cdot x}] \]
Apply $\Omega_{PT}$ to $\phi(x)$. It does nothing to
$\frac{1}{(2\pi)^{3/2}\sqrt{2\omega_{\vec{k}}}}$, it does nothing to
$a_{\vec{k}}$, and nothing to $a_{\vec{k}}^\dagger$.
But what about that $i$ up in the exponential. It turns that into
$-i$!
\begin{align*}
\Omega_{PT}^{-1}\phi(x)\Omega_{PT}&=\int \frac{d^3k}{(2\pi)^{3/2}
\sqrt{2\omega_{\vec{k}}}} [a_{\vec{k}}^ae^{ik\cdot x} +
a_{\vec{k}}^{a\dagger}e^{-ik\cdot x}] \\
&=\phi(-x)\;\;\;\text{\large !}
\end{align*}
$PT$ does nothing to momentum states. That is expected. Parity turns
$\vec{k}\rightarrow-\vec{k}$ and time reversal changes it back again.
\end{document}