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\lhead{Physics 253a}
\chead{September 30}
\rhead{Sidney Coleman's Notes}
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In ordinary QM, \underline{any} hermitian operator is observable.
This can't be true in relativistic quantum mechanics. Imagine two
experiments that are at space-like separation. If $x_1\in R_1$ and
$x_2\in R_2$ then $(x_1-x_2)^2<0$.
% inserted figure from sep 30, 1/10 p.26 in 1-214
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\includegraphics[width=.8\textwidth]{03-fig1.eps}
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Suppose observer 2 has an electron in his lab and she measures
$\sigma_k$. If observer 1 can measure $\sigma_y$ of that electron it
will foul up observer 2's experiment. Half the time when she
remeasures $\sigma_x$ it will have been flipped. This tells her that
observer 1 has made a measurement. This is faster than light
communication, an impossibility. It is a little hard to
mathematically state the obvious experimental fact that I can't
measure the spin of an electron in the Andromeda galaxy. We don't
have any way of localizing particles, any position operator, yet. We
can make a mathematical statement in terms of observables:
If $O_1$ is an observable that can be measured in $R_1$ and $O_2$ is
an observable that can be measured in $R_2$ and $R_1$ and $R_2$ are
space-like separated then they commute:
\[ [O_2,O_2]=0\]
Observables are attached to space time points. A given observer
cannot measure all observables, only the ones associated with his or
her region of space-time.
It is not possible, even in principle for everyone to measure
everything. Out of the hordes of observables, only a restricted set
can be measured at a space-time region. Localization of measurements
is going to substitute for localization of particles.
The attachment of observables to space-time points has no analog in
NRQM, nor in the classical theory of a single particle, relativistic
or nonrelativistic, but it does have an analog in classical field
theory. In electromagnetism, there are six observables at each point:
$E_x(x)=E_x(\vec{x},t)$, $E_y(x)$, $E_z(x)$, $B_x(x)$, $B_y(x)$,
$B_z(x)$. We can't design an apparatus here that measures the $E_x$
field now in the Andromeda galaxy.
In classical field theory, these observables are numbers. In quantum
mechanics, observables are given by operators. The fields will become
quantum fields, an operator for each spacetime point. We can see in
another way that the electric field is going to have to become a
quantum field: How would you measure an electric field? You mount a
charged ball, pith ball, to some springs and see how much the springs
stretch. The location of the ball is given by an equation like:
\[q\ddot{x} = E_x(x)\longleftarrow\begin{smallmatrix}
\text{might be $\int d^4xf(x)E_x(x)$ where $f(x)$}\\
\text{gives some suitable average over the pith ball.}
\end{smallmatrix}\]
The amount the ball moves is related to the $\vec{E}$ field, and if
the world is quantum mechanical, $x$ must be an operator, and so $E_x$
must be an operator.
We don't have a proof, but what is strongly suggested is that QM and
relativistic causality force us to introduce quantum fields. In fact,
relativistic QM is practically synonymous with quantum field theory.
We will try to build our observables from a complete commuting set of
quantum fields.
\[\underset{a=1,\dots,N}{\phi^a(x)}\;\;\;\text{ operator valued
functions of space-time}.\]
Observables in a region $R$ will be built out of $\phi^a(x)$ with
$x\in R$. Observables in space-like separated regions will be
guaranteed to commute if
\begin{enumerate}
\item $[\phi^a(x),\phi^b(y)]=0$ whenever $(x-y)^2<0$. \\
\null\\
We are going to construct our fields out of the creation and
annihilation ops. These five conditions will determine them:
\item $\phi^a(x)=\phi^{a\dagger}(x)$ hermitian, observable\\
\null\\
and that they have proper translation and Lorentz transformation
properties
\item $e^{-iP\cdot a}\phi^a(x)e^{iP\cdot a}=\phi^a(x-a)$
\item $U(\Lambda)^\dagger\phi^a(x)U(\Lambda)=\underset{\substack{
\text{If this were not a scalar}\\ \text{field, there would be extra}
\\ \text{factors here reflecting a} \\ \text{change of basis; as well
as} \\ \text{a change of argument}}}{\underbrace{}\phi^a(\Lambda^{-1}x)}$
\item $\phi^a(x)=\int
d^3k[F^a_k(x)a_{\vec{k}}+G^a_k(x)a_{\vec{k}}^\dagger]$
\end{enumerate}
We can think of our unitary transformations in two ways; as
transformations on the states $|\psi\rangle\rightarrow U|\psi\rangle$,
or as transformations on the operators $A\rightarrow U^\dagger AU$.
NOT BOTH!
What's embodied in assumption (3):
Given $U(\vec{a})$ the unitary operator of space translation by
$\vec{a}$ ($U(\vec{a})=e^{-i\vec{P}\cdot\vec{a}}$) the translation of
a state $|\psi\rangle$ is a state $|\psi'\rangle =
U(\vec{a})|\psi\rangle$. Suppose the value of some observable, like
charge density is
\[ f(\vec{x})=\langle\psi|\rho(\vec{x})|\psi\rangle\]
\noindent then it should be that
\[ \langle\psi'|\rho(\vec{x})|\psi'\rangle = f(\vec{x}-\vec{a}).\]
% inserted figure sep 30 p 4/10, p.29 in 1-214
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\includegraphics[width=7cm]{03-fig2.eps}
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Rewrite the first equation with $\vec{x}\rightarrow \vec{x}-\vec{a}$
\[f(\vec{x}-\vec{a})=\langle \psi|\rho(\vec{x}-\vec{a})|\psi\rangle \]
\begin{align*}
\text{Equate }\langle\psi|\rho(\vec{x}-\vec{a})|\psi\rangle &=\langle
\psi'|\rho(\vec{x})|\psi'\rangle \\
&=\langle\psi|e^{i\vec{P}\cdot\vec{a}}\rho(\vec{x})e^{-i\vec{P}\cdot
\vec{a}}|\psi\rangle
\end{align*}
A hermitian operator is determined by its expectation values
\[ e^{i\vec{P}\cdot\vec{a}}\rho(\vec{x})e^{-i\vec{P}\cdot\vec{a}}=
\rho(\vec{x}-\vec{a}) \]
(3) is just the full relativistic form of this equation. The equation
with $a=(t,\vec{0})$ is just the time evolution for Heisenberg fields.
For the exact same reason as $x-a$ appears in the RHS of (3),
$\Lambda^{-1}x$ appears in the RHS of (4). (4) gives the Lorentz
transformation properties of a \uwave{scalar} field. This is not much
of an assumption. We can get fields transforming as vectors or
tensors by taking derivatives of the $\phi^a$. Out of vector or
tensor fields, we could make scalars.
In order to apply condition (4), it is nice to have the discussion
phrased in terms of relativistically normalized creation and
annihilation operators.
Recall the relativistically normalized one particle states
\[ |k\rangle = (2\pi)^{3/2}\sqrt{2\omega_{\vec{k}}}|\vec{k}\rangle \]
Introduce $\alpha^\dagger(k)=(2\pi)^{3/2}\sqrt{2\omega_{\vec{k}}}
a_{\vec{k}}$, $\alpha^\dagger(k)|0\rangle=|k\rangle$.
Multiparticle states are made by
\[ \alpha^\dagger(k_1)\cdots\alpha^\dagger(k_n)|0\rangle=|k_1,\dots,
k_n\rangle \]
The Lorentz transformation properties of the states are
\begin{align*}
U(\Lambda)|0\rangle &=0 \\
U(\Lambda)|k_1,\dots,k_n\rangle &=|\Lambda k_1,\dots,\Lambda
k_n\rangle
\end{align*}
\noindent and $U(a)=e^{iP\cdot a}$ is found from
\begin{align*}
P^\mu|0\rangle&=0\\
P^\mu|k_1,\dots,k_n\rangle &=(k_1+\cdots+k_n)^\mu|k_1,\dots,k_n\rangle
\end{align*}
In Eqs.~(9)-(12) % (\ref{eq:01-lor1})-(\ref{eq:01-lor4})
and Eqs.~(1)-(3) % (\ref{eq:01-trans1})-(\ref{eq:01-trans3})
of the Sept.~23 lecture we set up the criteria
that $U(\Lambda)$ and $U(a)$ must satisfy. At the time our Hilbert
space consisted only of the one particle part of the whole Fock space
we have now. You should check that the criteria are satisfied in Fock
space.
We can determine Lorentz transformation and translation properties of
the $\alpha^\dagger(k)$. Consider,
\begin{align*}
U(\Lambda)\alpha^\dagger(k)U(\Lambda)^\dagger|k_1,\dots,k_n\rangle &=
U(\Lambda)\alpha^\dagger(k)|\Lambda^{-1}k_1,\dots,\Lambda^{-1}k_n
\rangle \\
&=U(\Lambda)|k,\Lambda^{-1}k_1,\dots,\Lambda^{-1}k_n \rangle\\
& =|\Lambda k,k_1\dots,k_n\rangle
\end{align*}
That is $U(\Lambda)\alpha^\dagger(k)U(\Lambda)^\dagger|k_1,\dots,
k_n\rangle =\alpha^\dagger(\Lambda k)|k_1,\dots,k_n\rangle$
$|k_1,\dots,k_n\rangle$ is an arbitrary state in our complete basis so
we have determined its action completely
\[ U(\Lambda)\alpha^\dagger(k)U(\Lambda)^\dagger=\alpha^\dagger
(\Lambda k)\]
Similarly, or by taking the adjoint of this equation
\[ U(\Lambda)\alpha(k)U(\Lambda)^\dagger=\alpha(\Lambda k)\]
An analogous derivation shows that
\begin{align*}
e^{iP\cdot x}\alpha^\dagger(k)e^{-iP\cdot x}&= e^{ik\cdot x}
\alpha^\dagger(k)\\
e^{iP\cdot x}\alpha(k)e^{-iP\cdot x}&= e^{ik\cdot x} \alpha(k)
\end{align*}
Now to construct the field $\phi$ (if there is more than one we'll
label them when we've found them) satisfying all 5 conditions. First
we'll satisfy condition (5) except we'll write the linear combination
of $a_{\vec{k}}$ and $a_{\vec{k}}^\dagger$ in terms of our new
$\alpha(\vec{k})$ and $\alpha^\dagger(\vec{k})$
\[ \phi(x)=\int\!\!\!\!\!\!\overbrace{\frac{d^3k}{(2\pi)^32\omega_{\vec{k}}}}^{
\substack{\text{It would be stupid}\\ \text{not to use the} \\
\text{L.I.~measure}}}\!\!\!\!\!\![f_k(x)\alpha(k) +g_k(x)\alpha^\dagger(k)] \]
By (3), $\phi(x)=e^{iP\cdot x}\phi(0)e^{-iP\cdot x}$, that is,
\begin{align*}
\phi(x)&=\int \frac{d^3k}{(2\pi)^32\omega_{\vec{k}}}[f_k(0)e^{iP\cdot
x}\alpha(k)e^{-iP\cdot x} +g_k(0)e^{iP\cdot x}\alpha^\dagger(k)
e^{-iP\cdot x}]\\
&=\int \frac{d^3k}{(2\pi)^32\omega_{\vec{k}}}[f_k(0)e^{-ik\cdot x}
\alpha(k) +g_k(0)e^{ik\cdot x}\alpha^\dagger(k) ]
\end{align*}
We have found the $x$ dependence of $f_k(x)$ and $g_k(x)$ now we will
use (4) to get their $k$ dependence. A special case of (4) is
\begin{align*}
\phi(0)&=U(\Lambda)\phi(0)U(\Lambda)^\dagger \\
\int\frac{d^3k}{(2\pi)^32\omega_{\vec{k}}}[f_k(0)\alpha(k) +g_k(0)
\alpha^\dagger(k) ]&=\\
\int\frac{d^3k}{(2\pi)^32\omega_{\vec{k}}}&[f_k(0)
\underbrace{U(\Lambda)\alpha(k)U(\Lambda)^\dagger}_{\alpha(\Lambda k)}
+g_k(0)\underbrace{U(\Lambda)\alpha^\dagger(k)U(\Lambda)^\dagger}_{
\alpha^\dagger(\Lambda k)} ] \\
&\substack{\text{change variables}\\ \text{measure is unchanged}}
\;\;\; k\rightarrow \Lambda^{-1}k\\
&=\int\frac{d^3k}{(2\pi)^32\omega_{\vec{k}}}[f_{\Lambda^{-1}k}(0)
\alpha(k) +g_{\Lambda^{-1}k}(0)\alpha^\dagger(k) ]
\end{align*}
The coefficients of $\alpha(k)$ and $\alpha^\dagger(k)$ must be
unchanged $\Rightarrow f_k(0)=f_{\Lambda^{-1}k}(0)$ and $g_k(0)=
g_{\Lambda^{-1}k}(0)$.
$k$ ranges all over the mass hyperboloid ($k^0>0$ sheet), but a
Lorentz transformation can turn any of these $k$'s into any other. So
$f_k(0)$ and $g_k(0)$ are constants, independent of $k$.
\[\phi(x)=\int \frac{d^3k}{(2\pi)^32\omega_{\vec{k}}}[fe^{-ik\cdot x}
\alpha(k) +ge^{ik\cdot x}\alpha^\dagger(k) ]\]
We have two linearly independent solutions of conditions (3), (4) and
(5), the coefficients of the complex constants $f$ and $g$. We'll
name them. (Switching back to our old creation and annihilation ops.)
\begin{align*}
\phi^+(x)&=\int \frac{d^3k}{(2\pi)^{3/2}\sqrt{2\omega_{\vec{k}}}}
a_{\vec{k}}e^{-ik\cdot x} &
\phi^-(x)&=\int \frac{d^3k}{(2\pi)^{3/2}\sqrt{2\omega_{\vec{k}}}}
a^\dagger_{\vec{k}}e^{ik\cdot x}\\
\text{Note }\phi^-(x)&=\phi^+(x)^\dagger & &\substack{\text{$\pm$
convention is bananas, but it}\\ \text{was est'd by Heisenberg and Pauli
50 years ago.} }
\end{align*}
Now we'll apply hermiticity. Two independent combinations satisfying
(2) are
\[\phi(x)=\phi^+(x)+\phi^-(x)\;\;\text{ and }\;\;
\phi(x)=\frac{1}{i}[\phi^+(x)-\phi^-(x)]\]
These are two independent cases of the most general choice satisfying
(2):
\[\phi(x) = e^{i\theta}\phi^+(x)+e^{-i\theta}\phi^-(x)\]
Now to satisfy (1). There are three possible outcomes of trying to
satisfy (1).
% define "Lcount" as a counter
\newcounter{Lcount}
% set the "default" label to print counter as a Roman numeral
\begin{list}{Possibility \Alph{Lcount}: }
% inform the list command to use this counter
{\usecounter{Lcount}
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\item Both of the above combinations are OK. We have two fields
$\phi^1$ and $\phi^2$ commuting with themselves \uline{and each other}
at spacelike separation. In this possibility $\phi^+(x)$ and
$\phi^-(x)$ commute with each other at spacelike separation.
\item Only one combination is acceptable. It is of the form
$\phi=e^{i\theta}\phi^++e^{-i\theta}\phi^-$. While $\theta$ may be
arbitrary, only one $\theta$ is acceptable.
\item The program crashes, and we could weaken (5) or think harder.
\end{list}
Let's first calculate some commutators. Using their expansions in
terms of the $a_{\vec{k}}$ and $a_{\vec{k}}^\dagger$ and the
commutation relations for $a_{\vec{k}}$ and $a_{\vec{k}}^\dagger$ we
find
\begin{align*}
[\phi^+(x),\phi^-(y)]&=0=[\phi^-(x),\phi^+(y)]\\
\text{and }\;\;[\phi^+(x),\phi^-(y)]&=\int \frac{d^3k}{(2\pi)^{3}
2\omega_{\vec{k}}}e^{-ik\cdot(x-y)}\equiv \underbrace{\Delta_+(x-y,
\mu^2)}_{\text{or just }\Delta_+(x-y)}\\
\text{also }\;\;[\phi^-(x),\phi^+(y)]&=-\Delta_+(x-y,\mu^2)\\
\text{$\Delta_+$ is manifestly Lorentz invariant }\;\;\Delta_+(\Lambda
x)&=\Delta_+(x).
\end{align*}
Possibility A runs only if $\Delta_+(x-y)=0$ for $(x-y)^2<0$. We have
encountered a similar integral when we were looking at the evolution
of one particle position eigenstates. In fact,
\[i\partial_0\Delta_+(x-y)=\int\frac{d^3k}{(2\pi)^3}e^{-ik\cdot(x-y)}\]
\noindent is the very integral we studied, and we found that it did
\uwave{not} vanish when $(x-y)^2<0$.
Possibility A is DEAD.
On to possibility B. Take
$\phi(x)=e^{i\theta}\phi^+(x)+e^{-i\theta}\phi^-(x)$ and calculate
\[ [\phi(x),\phi(y)]=\Delta_+(x-y)-\Delta_+(y-x)\;\;\substack{
\text{$\theta$ dependence} \\ \text{drops out}}\]
Does this vanish when $(x-y)^2<0$? Yes, and we can see this without
any calculations. A space-like vector has the property that it can be
turned into minus itself by a (connected) Lorentz transformation.
This and the fact that $\Delta_+$ is Lorentz invariant, tells us that
$[\phi(x),\phi(y)]=0$ when $x-y$ is space-like. We can choose
$\theta$ arbitrarily, but we can't choose more than one $\theta$.
We'll choose $\theta=0$. Any phase could be absorbed into the
$a_{\vec{k}}$ and $a_{\vec{k}}^\dagger$.
Possibility B is ALIVE, and we don't have to go on to C.
We have our free scalar field of mass $\mu$
\[\phi(x)=\phi^+(x)+\phi^-(x)=\int \frac{d^3k}{(2\pi)^{3/2}\sqrt{
2\omega_{\vec{k}}}}[ a_{\vec{k}}e^{-ik\cdot x}+ a_{\vec{k}}^\dagger
e^{ik\cdot x} ] \]
Our field satisfies an equation (show using $k^2=\mu^2$)
\begin{align*}
(\square+\mu^2)\phi(x)&=0 & \square&=\partial^\mu\partial_\mu
\end{align*}
This is the Heisenberg equation of motion for the field. It is called
the Klein-Gordon equation. If we had quantized the electromagnetic
field it would have satisfied Maxwell's equations.
Actually, Schr\"odinger first wrote down the Klein-Gordon equation.
He got it at the same time as he got the Schr\"odinger equation:
\[i\partial_0\psi=-\frac{1}{2\mu}\nabla^2\psi \]
This equation is obtained by starting with $E=\frac{\vec{p}^2}{2m}$
noting that $E=\omega$ (when $\hbar=1$) and $p^i=k^i$ and for plane
waves $\omega=i\partial_0$ and $k^i=\frac{1}{i}\partial_i$.
Schr\"odinger was no dummy, he knew about relativity, so he also
obtained $(\square+\mu^2)\phi(x)=0$ from $p^2=\mu^2$.
He immediately saw that something was wrong with the equation though.
The equation has both positive and negative energy solutions. For a
free particle the energies of its possible states are unbounded
below$!$ This is a disgusting relativistic generalization of a single particle
wave equation, but with 50 years hindsight we see that this is no
problem for a field that can create and destroy particles.
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