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\lhead{Physics 253a}
\chead{October 21}
\rhead{Sidney Coleman's Notes}
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\section*{\sc Diagrammatic Perturbation Theory}
Dyson's formula applied to $S=U_I(\infty,-\infty)$ is
\[ U_I(\infty,-\infty)=Te^{-i\int dtH_I(t)} \]
(Without the use of the time ordering notation, this formula for
$U_I(t,t')$ was written down by Dirac 15 years before Dyson wrote it
this way, and Dyson says he should not have credit for little more
than a change in notation.)
From this and Wick's theorem, which for those of you who really love
combinatorics can be written
\[T(\phi_1\cdots\phi_n)=:e^{\frac{1}{2}\sum_{i,j=1}^n\wick{1}
{<1\phi_i>1\phi_j}\frac{\partial}{\partial\phi_i}\frac{\partial}
{\partial\phi_j}}\phi_1\cdots\phi_n: \]
We have enough work done to write down diagrammatic perturbation
theory for $S=U_I(\infty,-\infty)$. The easiest way to see this is to
look at a specific model and a contribution to $U_I(\infty,-\infty)$
at a specific order in $g$.
In model 3, $\mathcal{H}_I=gf(t)\psi^*\psi\phi$, $H_I=\int
d^3x\mathcal{H}_I$,
\[ U_I(\infty,-\infty)=Te^{-i\int d^4x\mathcal{H}_I}=Te^{-i\int
d^4xgf(t)\psi^*\psi\phi} \]
\noindent the contribution at second order in $g$ is
\[\frac{(-ig)^2}{2!}\int d^4x_1d^4x_2f(t_1)f(t_2)T(\psi^*\psi\phi
(x_1)\psi^*\psi\phi(x_2)) \]
One of the terms in the expansion of the time ordered product into
normal ordered products by Wick's theorem is
\[\frac{(-ig)^2}{2!}\int d^4x_1d^4x_2f(t_1)f(t_2):\psi^*\psi\wick{1}
{<1\phi(x_1)\psi^*\psi>1\phi(x_2)}: \]
This term can contribute to a variety of physical processes. The
$\psi$ field contains operators that annihilate a ``nucleon'' and
operators that create an anti-``nucleon''. The $\psi^*$ field
contains operators that annihilate an anti-nucleon and create a
nucleon. The operator
\[:\psi^*\psi\wick{1}{<1\phi(x_1)\psi^*\psi>1\phi(x_2)}:=:\psi^*\psi
(x_1)\psi^*\psi(x_2):\wick{1}{<1\phi(x_1)>1\phi(x_2)} \]
\noindent can contribute to $N+N\rightarrow N+N$. That is to say
\[ \langle\substack{\text{final 2}\\ \text{nucleon state}}|
:\psi^*\psi(x_1)\psi^*\psi(x_2):|\substack{\text{initial 2}\\
\text{nucleon state}}\rangle\]
\noindent is nonzero because there are terms in the two $\psi$ fields
that can annihilate the two nucleons in the initial state and terms in
the 2 $\psi^*$ fields that can then create two nucleons, to give a
nonzero matrix element. It can also contribute to
$\bar{N}+\bar{N}\rightarrow\bar{N}+\bar{N}$ and $N+\bar{N}\rightarrow
N+\bar{N}$. You can see that there is no combination of creation and
annihilation operators in this operator that can contribute to
$N+N\rightarrow\bar{N}+\bar{N}$. The $\psi$ fields would have to
annihilate the nucleons and the $\psi^*$ fields cannot create
antinucleons. This is good because this process does not conserve the
$U(1)$ symmetry charge. However it looks like our operator can
contribute to vacuum$\rightarrow N+N+\bar{N}+\bar{N}$, which would be
a disaster. The coefficient of that term after integrating over $x_1$
and $x_2$ had better tun out to be zero.
Another term in the expansion of the time ordered product into normal
ordered products is
\[\frac{(-ig)^2}{2!}\int d^4x_1d^4x_2f(t_1)f(t_2):\psi^*\wick{1}
{<1\psi\phi(x_1)>1\psi^*}\psi\phi(x_2): \]
This term can contribute to the following 2$\rightarrow$2 scattering
processes: $N+\phi\rightarrow N+\phi$, $\bar{N}+\phi\rightarrow
\bar{N}+\phi$, $N+\bar{N}\rightarrow 2\phi$, $2\phi\rightarrow
N+\bar{N}$.
A single term is capable of contributing to a variety processes
because a single field is capable of creating or destroying a
particle.
The terms in the Wick expansion can be written down in a diagrammatic
shorthand according to the following rules. At $N$th order in
perturbation theory, you start by writing down $N$ interaction
vertices and numbering them 1 to $N$. For model 3 at second order in
perturbation theory you write down
% insert fig oct 21 p.3/14 top, 109 in 1-214
{\large
\begin{align*}
&\substack{\text{\scriptsize This line is for the $\phi$ in}\\
\text{\scriptsize $\psi^*\psi\phi(x_1)$. It creates}\\
\text{\scriptsize or destroys a meson}}
\;\;\;\Diagram{ & & fuA \\
f \vertexlabel^1 \\
& &fdV\\
}&
\Diagram{ & & fuA \\
f\vertexlabel^{2} \\
& &fdV\\
}\;\;\;\substack{\text{\scriptsize This outgoing line is for the
$\psi^*$ in} \\ \text{\scriptsize $\psi^*\psi\phi(x_2)$. It can be
thought of as creating}\\
\text{\scriptsize a nucleon or annihilating an antinucleon}\\
\\
\\
\text{\scriptsize This incoming line is for the $\psi$ in
$\psi^*\psi\phi(x_2)$.} \\
\text{\scriptsize It can be thought of as annihilating}\\
\text{\scriptsize a nucleon or creating an antinucleon}\\ }
\end{align*}}
The vertex represents the factor of $f\psi^*\psi\phi$. From the fact
that there are two in this diagram you know to include
$\frac{(-ig)^2}{2!}\int d^4x_1d^4x_2$
Contractions are represented by connecting the lines. Any time there
is a contraction, join the lines of the contracted fields. The arrows
will always line up, because the contractions for which they don't are
zero. An unarrowed line will never be connected to an arrowed line
because that contraction is also zero.
Our first term in the expansion of the time ordered product corresponds
to the diagram
% insert fig oct 21 p 3/14 bot, 109 in 1-214
{\large \begin{align*}
\Diagram{ fdV & & & fuA \\
& \vertexlabel^{1}f & f\vertexlabel^{2} \\
fuA & & & fdV }
\end{align*}}
The second term corresponds to
% insert fig oct 21 p 4/14 #1, 110 in 1-214
{\large \begin{align*}
\Diagram{\;\;\,fd & \quad\quad fu \\
fV\vertexlabel^{1} fV & fV\vertexlabel^{2} fV }
\end{align*}}
The term in the Wick expansion
\[\frac{(-ig)^2}{2!}\int d^4x_1d^4x_2 f(t_1)f(t_2):\wick{1}{<1\psi^*
\psi\phi(x_1)>1\psi^*}\psi\phi(x_2): \]
\noindent is zero because $\wick{1}{<1\psi^*>1\psi^*}=0$, so we never
write down
% inserted figure oct 21 p 4/14 #2, 110 in 1-214
{\large \begin{align*}
\Diagram{\;\;\,fd & \quad\quad fu \\
fA\vertexlabel^{1} fA & fV\vertexlabel^{2} fV }
\end{align*}}
Because the arrows always line up, we can shorten
% inserted figure oct 21 p 4/14 #3, 110 in 1-214
{\large \begin{align*}
\Diagram{\;\;\;fd & \quad\quad fu \\
fV\vertexlabel^{1} fV & fV\vertexlabel^{2} fV }\;\;\;\text{\normalsize
to }\;\;\; \Diagram{\;\;\,fd \quad\;\;\; fu \\
fV\vertexlabel^{1} fV\vertexlabel^{2} fV }
\end{align*}}
These diagrams are in one-to-one correspondence with terms in the Wick
expansion of Dyson's formula.
We'll call them Wick diagrams. They stand for operators and the
vertices are numbered. we are most of the way to Feynman diagrams
which stand for matrix elements, but these aren't them yet. The
vertices are numbered in Wick diagrams and
% inserted figure oct 21 p 4/14 #4, 110 in 1-214
{\large \begin{align*}
\Diagram{\;\;\,fd \quad\;\;\; fu \\
fV\vertexlabel^{2} fV\vertexlabel^{1} fV }
\;\;\;\text{\normalsize
is distinct from }\;\;\; \Diagram{\;\;\,fd \quad\;\;\; fu \\
fV\vertexlabel^{1} fV\vertexlabel^{2} fV }
\end{align*}}
\noindent (there are two distinct terms in the Wick expansion) even
though after integrating over $x_1$ and $x_2$ these are identical
operators. In Feynman diagrams the lines will be labelled by
momenta.
On the other hand
% insert fig oct 21 p 5/14 #1, 111 in 1-214
{\large \begin{align*}
\text{\normalsize 1}\feyn{f flV fluA f }\text{\normalsize
2}\;\;\;\text{\normalsize is identical to}\;\;\;
\text{\normalsize 2}\feyn{f flV fluA f }\text{\normalsize
1}
\end{align*}}
(there is only one way of contracting all three fields at one vertex
with all three at the other). Although these two have been written
down to look different they aren't. Rotate the right one by
180$^\circ$ and you see they are the same.
% insert fig oct 21 p 5/14 #2, 111 in 1-214
{\large \begin{align*}
\text{\normalsize \begin{sideways}\begin{sideways}1\end{sideways}
\end{sideways}}\feyn{f flV fluA f }\text{\normalsize
\begin{sideways}\begin{sideways}2\end{sideways}\end{sideways}}
\end{align*}}
The contraction this diagram corresponds to is
\[\frac{(-ig)^2}{2!}\int d^4x_1d^4x_2f(t_1)f(t_2):\wick{213}{<1\psi^*
<2\psi<3\phi(x_1)>2\psi^*>1\psi>3\phi(x_2)}: \]
In model 1 $\mathcal{H}_I=g\rho(x)\phi(x)$ (we don't have to insert a
turning on and off function because the interaction goes to zero as
$x\rightarrow\infty$ in any direction and in particular in the time
direction in the far past / future). $\rho(x)$ is a prescribed
$c$-number source, so strongly made we don't have to worry about the
back reaction of the field $\phi$ on the source. The vertex in this
model is
% insert fig oct 21 p 5/14 #3, 111 in 1-214
{\large \begin{align*}
\bullet\feyn{f }
\end{align*}}
That represents $\rho\phi(x)$. At $O(g)$ in $U_I$ we have $(-ig)\int
d^4x_1\rho\phi(x_1)$ which is represented by
$\overset{1}{\bullet}\feyn{f}$.
At $O(g^2)$ in $U_I$ we have $\overset{1}{\bullet}\!\feyn{f}\,
\overset{2}{\bullet}\negmedspace\feyn{f}$ and
$\overset{1}{\bullet}\feyn{f}\!\overset{2}{\bullet}$.
At $O(g^3)$ in $U_I$ we have
$\overset{1}{\bullet}\feyn{f}\,\overset{2}{\bullet}\!\feyn{f}\,
\overset{3}{\bullet}\!\feyn{f}$, $\overset{1}{\bullet}\feyn{f}\!
\overset{2}{\bullet}\;\overset{3}{\bullet}\!\feyn{f}$, $\overset{1}
{\bullet}\feyn{f}\;\overset{2}{\bullet}\!\feyn{f}\overset{3}
{\bullet}$, and $\overset{1}{\bullet}\!\feyn{f}\!\overset{3}{\bullet}
\;\overset{2}{\bullet}\!\feyn{f}$.
A diagram at $O(g^4)$ is $\overset{1}{\bullet}\feyn{f}\!\!
\overset{2}{\bullet}\,\overset{3}{\bullet}\!\feyn{f}\,\overset{4}
{\bullet}\negmedspace\feyn{f}$.
We have been putting the normal ordering inside the integrand. Of
course we could put it around the whole integral in which case we see
that this $O(g^4)$ diagram corresponds to
\begin{multline*}
\frac{(-ig)^4}{4!}:\int d^4x_1d^4x_2d^4x_3d^4x_4\rho(x_1)\rho(x_2)
\rho(x_3)\rho(x_4)\wick{1}{<1\phi(x_1)>1\phi(x_2)}\phi(x_3)\phi(x_4):
=\\
\frac{(-ig)^4}{4!}:\int d^4x_1d^4x_2\wick{1}{<1\phi(x_1)>1\phi(x_2)}
\rho(x_1)\rho(x_2) \int d^4x_3\phi(x_3)\rho(x_3)\int d^4x_4\phi(x_4)
\rho(x_4):
\end{multline*}
That is, the integrands factor into products of terms corresponding to
each connected\footnote{``Connected'' means (in any theory) that the
diagram is in one connected piece. It doesn't mean fully contracted.
$\overset{1}{\bullet}\!\!\!-\!\!\overset{2}{\bullet}\,\overset{3}
{\bullet}\!\!-\!\overset{4}{\bullet}$ is a fully contracted
diagram that is not connected. $\overset{1}{\bullet}\!\!-$ is not
contracted, but is connected.} part of the diagram. This suggests we
can sum the series and then normal order in this simple theory,
because at all orders in $g$ the diagrams only contain
$\bullet\!\feyn{f}\!\bullet$ and $\bullet\!\feyn{f}$ various numbers of
times. We could do the sum in this theory, but instead we will prove
a general theorem.
\[\sum\text{all Wick diagrams}=:e^{\sum\text{connected Wick
diagrams}}: \]
In a theory with only two connected diagrams this theorem is powerful
enough to solve the theory exactly in a couple of lines. It will
help a lot in model 3, but since there are still an infinite number of
connected diagrams in model 3, we won't solve it. This formula is also
useful in condensed matter physics where you develop a perturbation
theory for $\text{Tr }e^{-\beta H}$. The free energy which is the
logarithm of the partition function is what is actually of interest.
This theorem's analogue tells you that you don't have to calculate a
huge series for $\text{Tr }e^{-\beta H}$ and then try to take its
logarithm. The free energy is just the sum of the connected diagrams.
Let $D$ be a general diagram with $n(D)$ vertices. Associated with
this diagram is an operator
\[ \frac{:O(D):}{n(D)!} \]
We have explicitly displayed the $n(D)!$ and we have pulled the normal
ordering outside. For example for
\begin{align*}
D&= \Diagram{ fdV & & & fuA \\
& \vertexlabel^{1}f & fs\vertexlabel^{2} \\
fuA & & & fdV } & O(D)&=(-ig)^2\int d^4x_1d^4x_2
f(t_1)f(t_2)\wick{1}{<1\phi(x_1)>1\phi(x_2)}\psi^*\psi(x_1)\psi^*\psi
(x_2)
\end{align*}
I will define two diagrams to be of the same ``pattern'' if they
differ just by permuting the labels at the vertices, 1, 2, \dots,
$n(D)$.
Since after integration over $x_1$, \dots, $x_{n(D)}$ two different
diagrams of the same pattern give identical contributions to $U_I$ and
since there are $n(D)!$ permutations of the numbers 1, \dots, $n(D)$,
you might expect the sum over all diagrams of a given pattern to
exactly cancel the $n(D)!$. This is not quite right however. For
some diagrams there are permutations of the vertices that have no
effect, for example
{\large\begin{align*}
\Diagram{ fd & & & fu \\
& \vertexlabel^{1}fA \vertexlabel^{2} \\
&fvA & fvV\\
& \vertexlabel_{3}fV \vertexlabel_{4} \\
fu & & & fd }
\;\;\;\text{\normalsize is not distinct from }\;\;\;
\Diagram{ fd & & & fu \\
& \vertexlabel^{4}fA \vertexlabel^{1} \\
&fvA & fvV\\
& \vertexlabel_{3}fV \vertexlabel_{2} \\
fu & & & fd }
\end{align*}}
\noindent (and there are two more cyclic permutations) but it is
distinct from the diagrams with noncyclic permutations. This is in
exact correspondence with the question of whether or not there is a
new term in the Wick expansion from permuting $x_1$, \dots,
$x_{n(D)}$.
For any pattern, there will be some symmetry number, $S(D)$, which is
the number of permutations that have no effect on the diagram $D$
(and of course there is the analogous statement, that there are $S(D)$
permutations of $x_1$, \dots, $x_{n(D)}$ that do not give additional
contributions in the Wick expansion). Summing over all distinct
diagrams of the same pattern as $D$ yields
\[ \frac{:O(D):}{S(D)}\]
Let $D_1$, $D_2$, \dots, $D_r$, \dots be a complete set of connected
diagrams, with one diagram of each pattern. A general diagram, $D$,
has $n_r$ components of pattern $D_r$. Because of the factorization
of the integrands and because we have explicitly pulled out the
$n(D)!$
\[:O(D):=:\prod_{r=1}^\infty[O(D_r)]^{n_r}: \]
Summing over all diagrams with the same pattern as $D$ gives
$\frac{:O(D):}{S(D)}$. What is $S(D)$? $S(D)$ certainly
contains $\prod_r[S(D_r)]^{n_r}$. If I have 2 identical factors, I can
take all the indices on one of them and exchange them with the other.
If I have $n$ identical factors, there are $n!$ whole exchanges, So
$S(D)$ contains $\prod_r n_r!$. The sum over all diagrams with the
same pattern as $D$ gives
\[ \frac{:O(D):}{S(D)}=\frac{:\prod_{r=1}^\infty[O
(D_r)]^{n_r}: }{\prod_{r=1}^{\infty}[S(D_r)^{n_r}n_r!]} \]
Now that we have done the sum over all diagrams of a given pattern, we
have to sum over all patterns. Notice that there is a 1-1
correspondence between patterns and sets $\{n_r\}$. Thus summing over
all patterns is the same as summing over all sets $\{n_r\}$.
So,
\begin{align*}
\sum\text{all Wick diagrams}&=\sum_{n_1=0}^\infty\sum_{n_2=0}^\infty
\cdots\frac{:\prod_{r=1}^\infty[O(D_r)]^{n_r}: }
{\prod_{r=1}^{\infty}[S(D_r)^{n_r}n_r!]} \\
&=:\sum_{n_1=0}^\infty\sum_{n_2=0}^\infty \cdots\prod_{r=1}^{\infty}
\frac{[O(D_r)]^{n_r}}{S(D_r)^{n_r}n_r!}: \\
&=:\prod_{r=1}^{\infty}\left(\sum_{n_r=0}^\infty\frac{\left[\frac{
O(D_r)}{S(D_r)}\right]^{n_r}}{n_r!}\right): \\
&=:\prod_{r=1}^{\infty}e^{\frac{O(D_r)}{S(D_r)}}: \\
&=:e^{\sum_{r=1}^\infty\frac{O(D_r)}{S(D_r)}}: \\
&=:e^{\sum\text{connected Wick diagrams}}:
\end{align*}
This is a neat theorem because it expresses a fact about diagrams,
pictures, algebraically.
Now we'll apply this to model 1.
\section*{\uline{Model 1 Solved}}
\begin{align*}
D_1&=\overset{1}{\bullet}\!\feyn{f},\;\;\;
D_2=\overset{1}{\bullet}\!\feyn{f}\!\overset{2}{\bullet},\;\;\;
S(D_2)=2\\
U_I(\infty,-\infty)&=:e^{O_1+\frac{O_2}{2}}:=
:e^{\overset{1}{\bullet}\!\feyn{f}\;+\frac{\overset{1}{\bullet}\!
\feyn{f}\!\overset{2}{\bullet}}{2}} \\
O_1&=-ig\int d^4x_1\rho(x_1)\phi(x_1)\\
O_2&=(-ig)^2\int d^4x_1d^4x_2\wick{1}{<1\phi(x_1)>1
\phi(x_2)}\rho(x_1)\rho(x_2)=\substack{\text{some}\\
\text{number}}=\alpha+i\beta
\end{align*}
You will compute $\alpha$ in the homework. We'll get it here by a
consistency argument, demanding that $U_I$ be unitary.
Let's rewrite $O_1$, using the expansion for $\phi(x)$.
\begin{align*}
O_1&=-ig\int \frac{d^3k}{(2\pi)^{3/2} \sqrt{2\omega_{\vec{k}}}}\int
d^4x\rho(x)\left(e^{-ik\cdot x}a_{\vec{k}} + e^{ik\cdot
x}a_{\vec{k}}^\dagger\right)\\
&=-ig\int \frac{d^3k}{(2\pi)^{3/2} \sqrt{2\omega_{\vec{k}}}}
\Bigl(\underbrace{\tilde{\rho}(-k)}_{\tilde{\rho}(k)^*}a_{\vec{k}}
+\tilde{\rho}(k)a_{\vec{k}}^\dagger\Bigr)
\end{align*}
$\biggl($Using the Fourier transform convention (this convention will
\uline{not} be adhered to, see Nov.~6)
\begin{align*}
\tilde{f}(k)&=\int d^4x e^{ik\cdot x}f(x) &
f(x)&=\int\frac{d^4k}{(2\pi)^4} e^{-ik\cdot x}\tilde{f}(k)
\end{align*}
\noindent also in three space dimensions
\begin{align*}
\tilde{f}(\vec{k})&=\int d^3x e^{i\vec{k}\cdot \vec{x}}f(\vec{x}) &
f(\vec{x})&=\int\frac{d^3\vec{k}}{(2\pi)^3} e^{-i\vec{k}\cdot \vec{x}}
\tilde{f}(\vec{k})\biggr)
\end{align*}
So as not to carry around so many factors, define
\[ f(\vec{k})\equiv\frac{-ig}{(2\pi)^{3/2} \sqrt{2\omega_{\vec{k}}}}
\tilde{\rho}(\vec{k},\omega_{\vec{k}})\]
then
\[ U_I(\infty,-\infty)=e^{\frac{1}{2}(\alpha+i\beta)}e^{\int
d^3kf(\vec{k})a_{\vec{k}}^\dagger}e^{-\int d^3kf(\vec{k})^*
a_{\vec{k}}}\]
Now that we have solved the model we can answer the usual questions
you ask about when a field is driven by an external source.
Given that you start with nothing in the far past, $|0\rangle$, what
is the probability of finding $n$ mesons in the far future?
The state in the far future is
\begin{align}
U_I(\infty,-\infty)|0\rangle&=e^{\frac{1}{2}(\alpha+i\beta)}e^{\int
d^3kf(\vec{k})a_{\vec{k}}^\dagger}\nonumber\\
&=e^{\frac{1}{2}(\alpha+i\beta)}\sum_{n=0}^\infty\frac{1}{n!}\int
d^3k_1\cdots d^3k_n f(\vec{k}_1)\cdots f(\vec{k}_n)|\vec{k}_1,\dots,
\vec{k}_n\rangle\label{eq:09-UI}
\end{align}
The probability, $P_n$, of finding $n$ mesons is thus
\begin{align*}
P_n&=\left|e^{\frac{1}{2}(\alpha+i\beta)}\frac{1}{n!}\int d^3k_1\cdots
d^3k_n f(\vec{k}_1)\cdots f(\vec{k}_n)|\vec{k}_1,\dots, \vec{k}_n
\rangle\right|^2\\
&=e^\alpha\frac{1}{(n!)^2}\int d^3k_1\cdots d^3k_n |f(\vec{k}_1)|^2
\cdots|f(\vec{k}_n)|^2n!\\
&=e^\alpha\frac{1}{n!}\left(\int d^3k_1|f(\vec{k}_1)|^2\right)^n
\end{align*}
Now is where we demand unitarity of $U_I$ to get $\alpha$.
\begin{align*}
1&\overset{!}{=}\sum_nP_n=e^\alpha\sum_n\frac{1}{n!}\left(\int d^3
k_1|f(\vec{k}_1)|^2\right)^n=e^\alpha e^{\int d^3 k_1|f(\vec{k}_1)|^2}
\\
\alpha&=-\int d^3k|f(\vec{k})|^2\\
\text{So }P_n&=e^{-|\alpha|}\frac{|\alpha|^n}{n!}\;\;\;\text{Poisson
distribution.}
\end{align*}
This state, created by a classical source, is called a coherent state.
Coherent states of the harmonic oscillator are
\[|\lambda\rangle \equiv e^{\lambda a^\dagger}|0\rangle \]
They are special because they diagonalize $a$
\[ a|\lambda\rangle=ae^{\lambda a^\dagger}|0\rangle = [a,e^{\lambda
a^\dagger}]|0\rangle=\lambda e^{\lambda a^\dagger}|0\rangle
=\lambda|\lambda\rangle \]
$\langle\lambda|x(t)|\lambda\rangle$ and $\langle\lambda|p(t)|\lambda
\rangle$ oscillate sinusoidally like the classical variables.
The coherent states we have constructed are eigenvectors of
$\phi^{(+)}(x)$ with eigenvalue
\[ \int \frac{d^3k}{(2\pi)^{3/2} \sqrt{2\omega_{\vec{k}}}} e^{-ik\cdot
x} f(\vec{k}) \]
Except for the $\frac{1}{n!}$ this state's $n$ particle part is just
the product of $n$ 1 particle states. It is about as uncorrelated as
a state of mesons can be. If you remove a particle with
$\phi^{(+)}(x)$, you get the same state back. Expectations of normal
ordered products factorize.
What is the average number of mesons created?
\begin{equation}\label{eq:09-N}
\langle N\rangle=\sum_{n=0}^\infty nP_n=\sum_{n=1}^\infty
\frac{e^{-|\alpha|}|\alpha|^n}{(n-1)!}=|\alpha|\;\;\;\substack{
\text{(pull out an $|\alpha|$ and}\\ \text{reindex the
sum)}}\end{equation}
What is the average energy of the final state, i.e.~the total energy
of all the mesons created?
\begin{align*}
\langle H\rangle &=\sum_{n=0}^\infty\frac{e^{-|\alpha|}}{(n!)^2}
\int d^3k_1\cdots d^3k_n |f(\vec{k}_1)|^2\cdots|f(\vec{k}_n)|^2
\underbrace{(\omega_{\vec{k}_1}+\cdots+\omega_{\vec{k}_n})}_{n
\omega_{\vec{k}_1}}n!\\
&=\sum_{n=1}^\infty \frac{e^{-|\alpha|}}{(n-1)!}|\alpha|^{n-1}\int
d^3k|f(\vec{k})|^2\omega_{\vec{k}} = \int
d^3k|f(\vec{k})|^2\omega_{\vec{k}}
\end{align*}
Average momentum?
\[\langle \vec{P}\rangle = \int d^3k|f(\vec{k})|^2\vec{k} \]
\section*{\uline{Model 2 solved} (beginning)}
Combinatorically, model 2 is identical to model 1, but physically the
content is different. The interaction doesn't actually turn off in
the far past / future. We put that in by hand.
% inserted figure oct 21 p 13/14, p 119 in 1-214
\[ \mathcal{H}_I=g\phi(x)\rho(\vec{x})f(t)
\quad\quad\includegraphics[width=8 cm]{09-fig1.eps}\]
Assuming the theory has a ground state, the vacuum-to-vacuum
scattering matrix element ought to be easy to calculate. \uline{ex
nihil nihil}. You start out with nothing you end up with nothing.
If you calculate $\langle 0|S|0\rangle$ however, you will not get one.
\begin{align*}
\text{Let }\;\;\;|0\rangle_P&=\substack{\text{ground state of the
whole}\\ \text{Hamiltonian, with energy $E_0$}}\\
|0\rangle&= \substack{\text{ground state of $H_0$ as usual}}
\end{align*}
Let's look at the scattering process in the Schr\"odinger picture.
For $t<-\frac{T}{2}$ we have $|0\rangle$; At $t\approx-\frac{T}{2}$,
in time $\Delta$, the interaction turns on adiabatically. The
adiabatic hypothesis says that $|0\rangle$ turns into $|0\rangle_P$
with probability 1. It can pick up a phase, $e^{-i\gamma_-}$. From
$t=-\frac{T}{2}$ to $t=+\frac{T}{2}$, the state evolves with the full
Hamiltonian, it rotates as $e^{-iE_0t}$ and picks up a total phase
$e^{-iE_0T}$. At $t\approx \frac{T}{2}$, as the interaction turns off
adiabatically $|0\rangle_P$ turns back into $|0\rangle$, getting one
more phase $e^{-i\gamma_+}$. The state we have for $t>\frac{T}{2}$ is
$e^{-i(\gamma_-+\gamma_++E_0T}|0\rangle$.
We can transfer this to the interaction picture, to get
\[ \langle0|U_I(\infty,-\infty)|0\rangle=e^{-i(\gamma_-+\gamma_+
+E_0T)}\]
This is disgusting. A divergent phase. How will we get rid of it?
We'll change the theory. The problem is that there is a mismatch
between the ground state energy of the full Hamiltonian and the ground
state energy of the free Hamiltonian. Subtract the mismatch and we'll
eliminate the problem.
\begin{align*}
H_I&\rightarrow \left[g\int d^3x\phi(\vec{x},t)\rho(\vec{x})-a\right]
f(t) & a&=E_0
\end{align*}
It's obvious what will happen. The number, $a$, just exponentiates
while the interaction is on.
\begin{align*}
\langle0|S|0\rangle&=e^{-i\left[(\gamma_++\gamma_-+E_0T)-aT(1+O\left(
\frac{\Delta}{T}\right)\right]}\overset{!}{=}1\\
\text{take }\;\;\; a&=E_0+O\left(\frac{\Delta}{T}\right)
\end{align*}
This is the first example of what is called a counterterm. It
counters a problem we ran into in scattering theory. It doesn't
change the physics, but it fixes up the problem.
You might worry that there will be energy mismatches in the one or
many particle energy states even after we get the energy mismatch in
the ground states fixed up. There shouldn't be though. Because the
physical states get far away from the potential at large times, we
expect the energy difference between a state with one physical
particle and the physical vacuum, to be the same as the energy
difference between the bare particle and the bare vacuum. If the
vacuum energies are lined up, the one particle state energies should
be lined up. We don't expect this to be true in model 3. The
particles interact with themselves and they can never get away from
that as a particle can get away from an external potential.
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