\sektion{14}{November 6}
\descriptionfourteen
% typeset by Yuan Sen TING
% edited by Bryan Gin-ge Chen
%\newcommand{\fourteensuminta}{\quad\;\int\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum}
%\newcommand{\fourteensumintb}{\;\,\,\int\!\!\!\!\!\!\!\!\!\sum}
Fourier transform (convention of Nov.~6):
$$ f(x) = \int \frac{d^4k}{(2\pi)^4} \tilde{f}(k) e^{ik\cdot x} $$
\noindent
This is a little unfortunate because
$$ e^{-iEt + i \vec{k}\cdot\vec{x}} \quad \quad (E>0)$$
\noindent
is generally called a positive frequency plane wave (because $i \frac{\partial}{\partial t}$ acting on it give $E$ and $\frac{1}{i} \frac{\partial}{\partial x}$ acting on it gives $\vec{k}$) and thus if $\tilde{f}(k)$ has support for positive $k^0$, $f(x)$ negative frequency. \\
A source with positive frequencies creates particles while a source with negative frequencies absorbs particles. This is summed up in the Feynman rule
$$\Diagram{ \bullet\momentum[bot]{f}{\rightarrow k}} \Longleftrightarrow i \tilde{\rho}(-k)$$
\noindent
(See Eqs.~(\ref{eq:14-frule1})-(\ref{eq:14-frule2}) if you don't know or remember how to get this Feynman rule)
\section*{Answer 1 (cont'd)}
We have found one meaning for our blob. We can use it to obtain another function, its Fourier transform.
Using the Fourier transform convention
$$ f(x) = \int \frac{d^4k}{(2\pi)^4} \tilde{f}(k) e^{ik\cdot x} $$
$$ \tilde{f}(k) = \int d^4 x f(x) e^{-ik\cdot x} $$
\noindent
(Which you'll notice has a different sign in the exponent from what we
used on Oct.~21) % *** check page number, pp10ff
\\
\noindent
(The power theorem is $\displaystyle \int d^4 x f(x) g(x) = \int \frac{d^4 k}{(2\pi)^4} \tilde{f}(k) \tilde{g}(-k) $)\\
We have
$$ G^{(n)} (x_1, \ldots, x_n) = \int \frac{d^4 k_1}{(2\pi)^4}
\cdots \frac{d^4 k_n}{(2\pi)^4} e^{ik_1\cdot x_1+\cdots + ik_n \cdot x_n} \tilde{G}^{(n)}(k_1, \ldots, k_n) $$
\noindent
where
$$ \begin{matrix} \begin{matrix}\displaystyle \tilde{G}^{(n)}(k_1, \ldots, k_n) =\\
\\ \\ \\ \\ \\ \\ \\ \\
\end{matrix}& \includegraphics[scale=0.35]{14-fig1.eps} \end{matrix}$$
\vspace{-1in}
This is the sum of all (let's make this definite) Feynman diagrams to
all orders (if the blob does not exist then the blob represents a
formal power series in $g$) and the blob includes factors for the
external propagators and the factor for overall energy momentum conservation $(2\pi)^4 \delta^{(4)}(k_1 + \cdots + k_n)$
\section*{Answer 2}
Consider modifying $\mathcal{H}$, say in model 3,
$$ \mathcal{H} \rightarrow \mathcal{H} - \rho(x) \phi(x) \quad \quad (\mathcal{L} \rightarrow \mathcal{L} + \rho(x) \phi(x))$$
\noindent
where $\rho(x)$ is a specified $c$ number source, not an operator. This adds a new vertex, a model 1 type vertex
$$\Diagram{ \bullet\momentum[bot]{f}{\rightarrow k}} \Longleftrightarrow i \tilde{\rho}(-k)$$
\noindent
The new Feynman rule was just quoted in class, let's see it arise in a simple example.\\
Let's suppose we have got the original Hamiltonian's vacuum
counterterm all calculated out to some high order in perturbation
theory so that there are no corrections to $\langle 0 | S | 0 \rangle
$ to this high order. The modification of the Hamiltonian (density)
$\mathcal{H} \rightarrow \mathcal{H} - \rho(x) \phi(x)$ spoils this.
There are now contributions to $\langle 0 | S | 0 \rangle $
proportional to $\rho^n$ at low orders in $g$. At order $\rho$ and
order $g$, we have $\Diagram{ \bullet f fs0 flSV flSuA fs0 } \quad
$ At order $\rho$ and $\mathcal{O}(g^3)$ we have
\includegraphics[width=1.5cm]{14-fig1b}%$\Diagram{\bullet f f
%fl fluV fA} $\\ % *** this diagram was wrong
\vspace{1cm}
Unfortunately, these are not interesting simple examples, because unless $\tilde{\rho}(0)$ is nonzero they vanish because of energy-momentum conservation.\\
At order $\rho^2$ and order $g^0$ we have $\Diagram{\bullet
\momentum[bot]{f}{} \bullet}$.
At order $\rho^2$ and order $g^2$ we have $\Diagram{ \bullet f fs0
flSA flSuV fs0 f \bullet }$, as well as $\begin{matrix}\Diagram{ \bullet f fs0
flSA flSuV fs0 } \\ \\\Diagram{ \bullet f fs0 flSA flSuV fs0
}\end{matrix}\,\,
$, $\quad\begin{matrix} \Diagram{ \bullet f \bullet} \\ \\ \Diagram{f fl fluA
fV} \end{matrix}\,\,$, and $\quad \quad \Diagram{flSA flSuV fs0 f fs0 flSA flSuV } \quad \quad \Diagram{\bullet f \bullet}$
\vspace{1cm}
This is a nice simple example, let's look at it. It comes from the term second order in $\rho$ and second order in $g$ in
$$ S = U_I(\infty, -\infty) = T e^{-i \int d^4 x ( g \psi^* \psi \phi - \rho \phi)} $$
\noindent
i.e.
$$ \frac{(-ig)^2}{2!} \frac{(i)^2}{2!} \int d^4x_1 d^4x_2 d^4x_3 d^4x_4 \rho (x_3) \rho (x_4) T ( \psi^* \psi \phi(x_1) \psi^* \psi \phi(x_2) \phi(x_3) \phi(x_4)) $$
\noindent
The process is vacuum $\rightarrow$ vacuum, so we are looking for the completely contracted terms in the Wick expansion of the time ordered product. They are
\begin{equation*}
\begin{split}
\wick{1}{<1 \psi^* >1 \psi} \wick{21}{<2 \phi (x_1) <1 \psi^* >1 \psi
>2\phi(x_2)} \wick{1}{<1 \phi(x_3) >1\phi(x_4)} \quad
&\longleftrightarrow \quad \quad \quad \Diagram{flSA flSuV fs0 f fs0 flSA
flSuV } \quad\quad \quad \Diagram{ \bullet f\bullet}\\
\\
\left.\begin{split}
\wick{1}{<1 \psi^* >1 \psi} \wick{213}{<2\phi (x_1) <1\psi^* >1\psi <3\phi(x_2) >3\phi(x_3) >2\phi(x_4)} \\
\wick{1}{<1 \psi^* >1 \psi} \wick{213}{<2\phi (x_1) <1\psi^* >1\psi <3\phi(x_2) >2\phi(x_3) >3\phi(x_4)}
\end{split}
\right\} \quad &\longleftrightarrow \quad \quad \Diagram{ \bullet
f fs0 flSA flSuV fs0 }\quad \quad \Diagram{ \bullet f fs0 flSA flSuV fs0 }
\\
\\
\wick{123}{<2 \psi^* <1 \psi <3\phi (x_1) >1\psi^* >2\psi >3\phi(x_2)} \wick{1}{<1\phi(x_3) >1\phi(x_4)} \quad &\longleftrightarrow \quad \quad \Diagram{ \bullet f \bullet} \quad \quad \quad \Diagram{f fl fluA fV} \\
\\
\left.\begin{split}
\wick{1234}{<2 \psi^* <1 \psi <3\phi (x_1) >1\psi^* >2\psi <4\phi(x_2) >4\phi(x_3) >3\phi(x_4)} \\
\wick{1234}{<2 \psi^* <1 \psi <3\phi (x_1) >1\psi^* >2\psi <4\phi(x_2) >3\phi(x_3) >4\phi(x_4)}
\end{split}\right\}\quad &\longleftrightarrow \quad \quad \Diagram{ \bullet f \bullet} \quad \quad \Diagram{ \bullet f fs0 flSA flSuV fs0 f \bullet}
\end{split}
\end{equation*}
\vspace{1cm}
The last two are the ones I want to look at in detail. They differ by an exchange of $x_1 \leftrightarrow x_2$ only, and since these are dummy variables of integration they together make a contribution to $\langle 0 | S | 0 \rangle$ of
$$ (-ig)^2 \frac{(i)^2}{2!} \int d^4x_1 d^4x_2 d^4 x_3 d^4 x_4 \rho(x_3) \rho(x_4) \underbrace{\wick{1}{<1\psi^*(x_1) >1\psi(x_2)}}_{(1)} \underbrace{\wick{1}{<1\psi^*(x_1) >1\psi(x_2)}}_{(2)} \underbrace{\wick{1}{<1\phi(x_1) >1\phi(x_3)}}_{(3)} \underbrace{\wick{1}{<1\phi(x_2) >1\phi(x_4)}}_{(4)}$$
$$ (1): \int \frac{d^4 p}{(2\pi)^4} e^{ip\cdot(x_1 - x_2)} \frac{i}{p^2 -m^2 + i\epsilon} $$
$$ (2): \int \frac{d^4 q}{(2\pi)^4} e^{iq\cdot(x_1 - x_2)} \frac{i}{q^2 -m^2 + i\epsilon} $$
$$ (3): \int \frac{d^4 k}{(2\pi)^4} e^{ik\cdot(x_1 - x_3)} \frac{i}{k^2 -m^2 + i\epsilon} $$
$$ (4): \int \frac{d^4 l}{(2\pi)^4} e^{il\cdot(x_2 - x_4)} \frac{i}{l^2 -m^2 + i\epsilon} $$
We have
\begin{multline*}
\int \frac{d^4 p}{(2\pi)^4}\frac{d^4 q}{(2\pi)^4}\frac{d^4
k}{(2\pi)^4}\frac{d^4 l}{(2\pi)^4} \frac{i}{p^2 -m^2 +
i\epsilon}\frac{i}{q^2 -m^2 + i\epsilon}\frac{i}{k^2 -m^2 +
i\epsilon}\frac{i}{l^2 -m^2 + i\epsilon} \\ \times (-ig)^2 (i)^2 \int
d^4 x_1 d^4 x_2 d^4 x_3 d^4 x_4 \rho(x_3) \rho(x_4)
e^{ix_1\cdot(p+q+k)}e^{ix_2\cdot (-p-q+l)}e^{ix_3\cdot k}e^{ix_4\cdot l}
\end{multline*}
Now if you go back to Eq.~%(1)
(\ref{eq:11-botpage2})
to Eq.~%(4)
(\ref{eq:11-page4})
and especially Eq.~%(1)
(\ref{eq:11-botpage2})
in the lecture of Oct.~28, % *** check equations
you'll see that when we have a
factor $e^{ix_1\cdot (p+q+k)}$ that corresponds to a picture with $p$, $q$ and $k$ flowing out of $x_1$. Our picture for the integral at hand is
$$ \Diagram{ \bullet \momentum[bot]{f}{\leftarrow k} fs0 \momentum[bot]{flSV}{\rightarrow q} \momentum[bot]{flSuA}{\rightarrow p} fs0 \momentum[bot]{f}{\rightarrow l} \bullet} $$
\vspace{1cm}
The left vertex, corresponding to space-time point $x_1$ (but I hate
to label it as such because it is just a dummy integration variable
which is going to be integrated over, and in our combinatoric
arguments for Feynman diagrams, we have kept the vertices unlabelled)
is creating a nucleon with momentum $p$, creating an antinucleon with momentum $q$, and creating a meson with momentum $k$.\\
The $x$ integrals are easy to perform, we have
\begin{multline*}
\int \frac{d^4 p}{(2\pi)^4}\frac{d^4 q}{(2\pi)^4}\frac{d^4 k}{(2\pi)^4}\frac{d^4 l}{(2\pi)^4} \frac{i}{p^2 -m^2 + i\epsilon}\frac{i}{q^2 -m^2 + i\epsilon}\frac{i}{k^2 -m^2 + i\epsilon}\frac{i}{l^2 -m^2 + i\epsilon} \\ \times (-ig)^2 (i)^2 \tilde{\rho}(k) \tilde{\rho}(l) (2\pi)^4 \delta^{(4)}(p+q+k) (2\pi)^4 \delta^{(4)}(-p-q+l)
\end{multline*}
This is just what you would have directly written down using our old Feynman rules supplanted by
$$\Diagram{ \bullet\momentum[bot]{f}{\leftarrow k}} \Longleftrightarrow i \tilde{\rho}(k)$$
\noindent
except for that $\cfrac{1}{2!}$ out front, which I'll explain in a moment.\\
This was a moderately interesting graph to show how the Feynman rule
comes out. If you know how the Feynman rule comes out, but you want to
check whether it's $\tilde{\rho}(k)$ or $\tilde{\rho}(-k)$ just look at the lowest order (in $\rho$, zeroth order in $g$) contribution to
\begin{align}\label{eq:14-frule1}
\langle 0 | S | k \rangle &= \langle 0 | U_I(\infty, -\infty) |k
\rangle \nonumber\\
&= \langle 0 | \left[ I + i \int d^4 x \rho(x) \phi(x) \right] |k
\rangle + \cdots \nonumber\\
&= 0 + \Diagram{ \bullet\momentum[bot]{f}{\leftarrow k}} + \cdots
\end{align}
\noindent
($|k\rangle$ is relativistically normalized and I'll use the
relativistically normalized creation and annihilation operators in the expansion for $\phi(x)$.)\\
\noindent
But,
\begin{equation} \langle 0 | \phi(x) | k \rangle = \int \frac{d^3 k^\prime}{(2\pi)^3
2 \omega_k^\prime} e^{-ik\cdot x}\!\!\!\!\! \underbrace{\langle 0 | a(k')
|k\rangle}_{(2\pi)^3 2 \omega_k \delta^{(3)}(\vec{k} -
\vec{k'})}\!\!\!\!\! =
e^{-ik\cdot x} \end{equation}
\noindent
So
\begin{equation} \label{eq:14-frule2}
\Diagram{ \bullet\momentum[bot]{f}{\leftarrow k}} = i \int d^4 x
\rho(x) e^{-ik\cdot x} =i \tilde{\rho}(k)\end{equation}
Now for that $\frac{1}{2!}$ out front (see also the short argument
after Eq.~(\ref{eq:14-shortarg})).
Earlier (after Eq.~%(5)
(\ref{eq:11-page5})
of lecture on Oct.~28),
we sung and danced about how there were no symmetry factors in model 3. That argument still goes through, and it still only applies to diagrams where \underline{each connected part has at least one external line}. What we have here is a vacuum to vacuum diagram, no external lines, and we now have to worry about symmetry factors.\\
Suppose we have a graph in the Wick expansion with $n$ powers of $\rho \phi$ and $m$ powers of $g \psi^* \psi \phi$, which comes with a $\frac{1}{n!} \frac{1}{m!}$ from the exponential.
\begin{figure}[H]
\centering
\includegraphics[scale=0.35]{14-fig2.eps}
\end{figure}
\noindent
(In this example $n=9$ and $m=5$)\\
Each of the $m!$ permutations of the model 3 vertices, keeping the
source vertices fixed, is a new term in the Wick expansion. They are
all uniquely identified by the way they are attached to the source
vertices. There are no model 3 vertices that are not somehow attached
to a source vertex. That would be a disconnected bubble, which is a
contribution to $\langle 0 | S | 0 \rangle$ with the source off, and
by assumption, the vacuum energy counterterm has been adjusted so that
there are no corrections to $\langle 0 | S | 0 \rangle$ with the source
off.\\
Now what about the $n!$ permutations of source vertices. Some of them make no new contributions to the Wick expansion. For example $5\leftrightarrow 6$ and $8 \leftrightarrow 9$, but also $2 \leftrightarrow 3$, because that has already been counted as $1^\prime \leftrightarrow 2^\prime$.\\
Any of the $n!$ permutations that do make a new contribution to the Wick expansion, that is, that have not already been counted in the permutations of the $m$ model 3 vertices, are accounted for in another way.\\
\noindent
For example, $7 \leftrightarrow 8$ or $6 \leftrightarrow 8$ gives a new term in the Wick expansion.\\
To see the accounting work, look at the messy example in momentum space. It is
$$\begin{matrix}\begin{matrix} \displaystyle\frac{(i)^9}{9!} \int \frac{d^4
k_1}{(2\pi)^4} \cdots \frac{d^4 k_9}{(2\pi)^4} \tilde{\rho}(-k_1)
\cdots \tilde{\rho}(-k_9) \times\\ \\ \\ \\ \\ \\ \\\end{matrix}
&\!\!\!\!\!\!\!\!\!\!\underbrace{\includegraphics[scale=0.35]{14-fig3.eps}}_{\substack{\text{Feynman
diagram with}\\ \text{external lines off the mass shell}}}\end{matrix}$$
\vspace{-.5in}
\noindent
Instead of using the permutation $7 \leftrightarrow 8$ to partially cancel off the $\frac{1}{9!}$ out front consider it as the same mess $m$ with a new diagram in the integrand.
\begin{figure}[h]
\centering
\includegraphics[scale=0.35]{14-fig4.eps}
\end{figure}
\noindent
Of course when the momenta are integrated over, this is identical to the integral above.\\
Now both of these would be counted in $\tilde{G} (k_1, \cdots, k_q)$ and in fact every permutation of the 9 source vertices that leads to a new term in the Wick expansion corresponds to a diagram in $\tilde{G} (k_1, k_2, \cdots, k_q)$.\\
Of course there are diagrams with $n=9$ and $m=5$ that are not of the
same pattern as the one drawn (differ by more than a permutation of vertices). For example
\begin{figure}[h]
\centering
\includegraphics[scale=0.35]{14-fig5.eps}
\end{figure}
\noindent
There is a Feynman diagram in $\tilde{G}(k_1, \ldots, k_q)$ for this too
\begin{figure}[H]
\centering
\includegraphics[scale=0.35]{14-fig6.eps}
\end{figure}
Let
\begin{figure}[H]
\centering
\includegraphics[scale=0.35]{14-fig7.eps}
\end{figure}
\noindent
denote the sum of all diagrams to all orders in $g$ (i.e.~all $m$) and at $n$th order in $\rho$ that contribute to $\langle 0 | S | 0 \rangle$. \\
What these combinatoric arguments say is
\begin{align} \label{eq:14-shortarg}
\includegraphics[scale=0.2]{14-fig7.eps} &\begin{matrix}\displaystyle =
\frac{(i)^n}{n!} \int \frac{d^4 k_1}{(2\pi)^4} \cdots \frac{d^4
k_n}{(2\pi)^4} \tilde{\rho}(-k_1) \cdots \tilde{\rho}(-k_n)
\times \\ \\ \\ \\ \\ \\\end{matrix}
\includegraphics[scale=0.2]{14-fig1.eps} \nonumber\\
\rule{0pt}{-1cm}&= \frac{(i)^n}{n!} \int \frac{d^4 k_1}{(2\pi)^4} \cdots \frac{d^4 k_n}{(2\pi)^4} \tilde{\rho}(-k_1) \cdots \tilde{\rho}(-k_n) \tilde{G}(k_1,\ldots,k_n)
\end{align} % *** vertical alignment?
\noindent
Having gone all the way back to Wick expansion arguments to show the combinatorics are right for this, I'll try to make a shorter argument.\\
The source creates $n$ mesons, which are distinguishable by virtue of the fact that they all carry different momenta $k_1, \ldots, k_n$. They interact in all possible ways. That gives us \\
$(i)^n \tilde{\rho}(-k_1)\cdots \tilde{\rho}(-k_n) \tilde{G}(k_1,\dots, k_n)$. Now we integrate over \underline{all} momenta $k_1, \dots, k_n$, and in doing so we make an overcounting by $n!$ .\\
\noindent
\textbf{BEST ARGUMENT}\\
One last way of arguing this. Instead of considering this as an $n$th
order calculation in $\rho$, temporarily think of it as a first order
calculation in each of $n$ different sources $\rho_1(x), \dots,
\rho_n(x)$. Then the diagram where source 1 creates a particle with
momentum $k_1$ and source 2 creates a particle with momentum $k_2$
really is distinguishable from a diagram where source 1 creates $k_2$
and source $2$ creates $k_1$. There is no overcounting when you integrate over all momenta. That contribution to $\langle 0 | S | 0 \rangle $ would be
$$ (i)^n \int \frac{d^4 k_1}{(2\pi)^4} \cdots \frac{d^4 k_n}{(2\pi)^4} \tilde{\rho}(-k_1) \cdots \tilde{\rho}(-k_n) \tilde{G}(k_1,\ldots,k_n) $$
How does this imagined calculation differ from ours? Well, in the exponential $\tilde{\rho_1}(x_1) \cdots \tilde{\rho_1}(x_n)$ comes with coefficient $1$, while $\rho(x_1) \cdots \rho(x_n)$ comes with coefficient $\frac{1}{n!}$.\\
To all orders
\begin{align*}
\langle 0 | S | 0 \rangle &= 1 + \sum_{n=1}^\infty \frac{(i)^n}{n!} \int \frac{d^4 k_1}{(2\pi)^4} \cdots \frac{d^4 k_n}{(2\pi)^4} \tilde{\rho}(-k_1) \cdots \tilde{\rho}(-k_n) \tilde{G}^{(n)}(k_1,\ldots,k_n) \\
&= 1 + \sum_{n=1}^\infty \frac{(i)^n}{n!} \int d^4 x_1\, d^4 x_2\, d^4 x_3\, d^4 x_4\, \rho(x_1) \cdots \rho(x_n) G^{(n)}(x_1,\ldots,x_n)
\end{align*}
This is the second answer to our question. The Fourier transform of
the sum of Feynman diagrams with $n$ external lines off the mass shell
is a Green's function (that's what $G$ stands for). In a theory with
linear response only $G^{(1)} \neq 0$. From conservation of
probability alone, you can see that the response of a quantum
mechanical vertex
can't be linear. Green introduced the first Green's function in the
early 19th century. From a prescribed charge distribution, $\rho(\vec{x})$, his Green's function gave you the electrostatic potential, $\phi (\vec{x})$.
$$ \phi(\vec{x}) = \int d^3 x' G(\vec{x},\vec{x}') \rho(\vec{x}')$$
$$ G(\vec{x}, \vec{x}') = \frac{1}{|\vec{x} - \vec{x}'|}$$
\noindent
satisfies
$$ \nabla^2 \phi = - \vec{\nabla } \cdot \vec{E} = - 4 \pi \rho \quad \quad (\vec{E} = - \vec{\nabla} \phi) $$
Let's explicitly note that the vacuum to vacuum transition amplitude
depends on $\rho$ by writing $\langle 0 | S | 0 \rangle_\rho $ (don't
confuse the subscript $\rho$ with a $p$).
$\langle 0 | S | 0
\rangle_\rho $ is a function\underline{al} of $\rho$. You give me a
\underline{function} on
spacetime, $\rho(x)$, and I give you back a number, $\langle 0 | S | 0
\rangle_\rho$. Actually, it is just a function of an infinite number
of variables, the value of the source at each spacetime point, and the
nomenclature ``functional" is redundant, we could just say
``function". Mathematicians don't call a vector in an infinite
dimensional space ``vector\underline{al}". $\langle 0 | S | 0 \rangle_\rho $ comes up often enough, it gets a name, $Z [\rho]$
$$ Z \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\underbrace{[\rho]}_{\substack{\text{The square brackets}\\
\text{remind you that}\\ \text{this is a function}\\ \text{of a
function } \rho}} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \equiv \langle 0 | S | 0 \rangle_\rho $$
$Z [\rho]$ is called a generating functional for the Green's functions because in the infinite dimensional generalization of a Taylor series, we have
$$ \frac{\delta^n Z [\rho]}{\delta \rho(x_1) \cdots \delta \rho(x_n)} \Bigg|_{\rho=0} = (i)^n G^{(n)} (x_1, \ldots, x_n)$$
\noindent
The $\delta$ instead of a $\partial$ reminds you that you are taking a
partial derivative of $z$ with respect to $\rho(x)$, holding a $(4-d)$
continuum of other variables fixed. These are called functional
derivatives. \\
\underline{Ex nihil omnes}: All physical information (all Green's
functions, and hence all $S$ matrix elements) about the system is coded in the vacuum persistence amplitude in the presence of an external source $\rho$. \\
$Z[\rho]$ is called a generating functional in analogy with the
functions of two variables which when you Taylor expand in one
variable, the coefficients are a set of functions like the Legendre
polynomials in the other. Sometimes it is very useful to put a whole set of functions in one neat package like that. An example will be the Ward identities, which are a statement about Green's functions resulting from a symmetry. It can be put very compactly in term of $Z$.\\
Because of our great theorem
$$ \sum \text{all Wick diagrams} = \\\ :e ^{\sum \text{connected diagrams}} : $$
\noindent
The sums are sums of normal ordered terms.
Apply this to a model which has had a source added. Take the vacuum expectation value of both sides. The LHS is just $\langle 0 | S | 0 \rangle_\rho $, i.e.~$Z[\rho]$.
We have
$$ Z[\rho] = \langle 0 | :e^{\sum \text{connected diagrams} } : | 0 \rangle= e^{\langle 0 | \sum \text{connected diagrams } |0 \rangle} $$
\noindent
This is true because the terms in the sum in the exponential are themselves normal ordered, convince yourself. Taking the natural logarithm,
\begin{align*}
\text{ln } Z [\rho] &= \langle 0 | \sum \text{connected diagrams } | 0 \rangle \\
&= \sum_{n=1}^\infty \frac{(i)^n}{n!} \int \frac{d^4 k_1}{(2\pi)^4} \cdots \frac{d^4 k_n}{(2\pi)^4} \tilde{\rho}(-k_1) \cdots \tilde{\rho}(-k_n) \tilde{G}_c(k_1,\ldots,k_n)
\end{align*}
\noindent
$\tilde{G}_c$ is the sum of all connected Feynman diagrams, with $k_1, \ldots, k_n$ possibly off shell, including the overall energy momentum conserving $\delta$ function, and the external propagator, which blow up on mass shell.
\section*{Answer 3}
One more way of interpreting $G^{(n)}(x_1,\ldots, x_n)$. By a cunning
trick we will show that $G^{(n)}(x_1,\ldots, x_n)$ is a VEV (``Vacuum
Expectation Value") of a time ordered string of Heisenberg fields. As
we did in obtaining answer 2, let
$$ \mathcal{H} \rightarrow \mathcal{H} - \rho \phi(x) $$
$$ \mathcal{H}_0 + \mathcal{H}' \rightarrow \mathcal{H}_0 + \mathcal{H}' - \rho \phi(x) $$
\noindent
As far as Dyson's formula is concerned, you can break the Hamiltonian up into a ``free" and interacting part in any way you please. Let's take the ``free" part to be $\mathcal{H}_0 + \mathcal{H}'$ and the interaction to be $-\rho \phi(x)$. I put quotes around ``free", because in this new interaction picture, the fields evolve according to
$$ \phi(\vec{x},t) = e^{iHt} \phi(\vec{x}, 0), e^{-iHt} $$
$$ H = \int d^3 x \mathcal{H} \quad \quad \mathcal{H} = \mathcal{H}_0 + \mathcal{H}' $$
These fields are not free. They do not obey the free field equations
of motion. You can't define a contraction for these fields, and thus
you can't do Wick's theorem. These fields are what we would have called Heisenberg fields if there was no source. For this reason we'll subscript them with an $H$.\\
Let's see what this tells us about $Z[\rho]$.
\begin{align*}
Z [\rho] &= \langle 0 | S | 0 \rangle_\rho = \langle 0 | T e^{+i \int d^4 x \rho(x) \phi_H (x)} | 0 \rangle \\
\text{\small just expand }\quad &= 1 + \sum_{n=1}^\infty
\frac{(i)^n}{n!} \int d^4 x_1\, d^4 x_2\, d^4 x_3\, d^4 x_4\,
\rho(x_1) \cdots \rho(x_n) \langle 0 | T(\phi_H(x_1) \cdots
\phi_H(x_n) )|0 \rangle
\end{align*}
\noindent
and we read off
\begin{equation}\label{eq:14-1}
G^{(n)}(x_1,\ldots,x_n) = \langle 0 | T(\phi_H(x_1) \cdots
\phi_H(x_n)) |0 \rangle \end{equation}\\
To summarize, we have found three meanings for the (sum of all) Feynman diagrams with ($n$) external lines off the mass shell.
\begin{enumerate}
\item It is a handy blob we can plaster into the interior of a larger diagram.
\item Its Fourier transform (times $\frac{(i)^n}{n!}$) is the coefficient of the $n$th order term in $\rho$ in the expansion of the vacuum to vacuum persistence amplitude in the presence of a source, $\rho$.
\item Its Fourier transform is the VEV of a time ordered string of Heisenberg fiels.
\end{enumerate}
\noindent
This can all be taken as motivation, because we are going to start from scratch and do a
\begin{center}
REFORMULATION OF SCATTERING THEORY \\
No more turning on and off function
\end{center}
Imagine you have a well-defined theory, with a time independent
Hamiltonian, $H=\int d^3x\mathcal{H}$ (the turning on and off function
is gone for good), whose spectrum is bounded below, whose lowest lying
state, is not part of a continuum, and the Hamiltonian has actually
been adjusted so that this state, $\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\underbrace{|0\rangle_p}_{\text{don't confuse } p \text{ with } \rho}
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!$, the physical vacuum, satisfies
$$ H |0\rangle_p = 0$$
\noindent
The vacuum is translationally invariant and normalized to one
$$ \vec{p} |0 \rangle_p = 0 \quad \text{and} \quad _p\langle 0 | 0 \rangle_p = 1 $$
Now let $\mathcal{H} \rightarrow \mathcal{H} - \rho(x) \phi(x) $ and define
\begin{align*}
Z [\rho] &\equiv \ _p\langle 0 | S | 0 \rangle_p \, \Big|_{\text{in the presence of the source }\rho} \\
&=\ _p\langle 0 | \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \underbrace{U(-\infty,
\infty)}_{\substack{\text{Schr\"odinger picture evolution
operator}\\\text{for the Hamiltonian} \int d^3x (\mathcal{H} - \rho
\phi)}} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! |0 \rangle_p
\end{align*}
\noindent
and define
$$ G^{(n)} (x_1, \ldots, x_n) = \frac{1}{i^n} \frac{\delta^{n} Z
[\rho]}{\delta \rho(x_1) \cdots \delta\rho(x_n)} $$
\textbf{Two Questions:}
\begin{enumerate}
\item Is $G^{(n)}$ defined this way (the F.T.) of the sum of all
Feynman graphs? Let's call the $G^{(n)}$ defined as the sum of all
Feynman graphs $G_F^{(n)}$ and the $Z$ which generated those $Z_F$.
The question is: Is $G^{(n)} = G^{(n)}_F$ ? or equivalently, is $Z =
Z_F$?
\begin{center} Answer will be ``yes."\end{center}
\item Are $S-1$ matrix elements obtained from Green's functions in the same way as before? For example, is
$$ \langle k'_1, k'_2 | (S-1) |k_1, k_2 \rangle = \prod_{a=1,2,1',2'} \frac{k_a^2 - \mu^2}{i} \tilde{G}(-k_1', k_2', k_1, k_2) \quad \text{?} $$
\begin{center}Answer will be ``almost."\end{center}
\end{enumerate}
\textbf{Answer to question 1:} Is $G^{(n)} = G_F{(n)}$?
Using Dyson's formula, in the exact same way as we did in Eq.~(\ref{eq:14-1}),
gives
\begin{equation}\label{eq:14-2} G^{(n)} (x_1,\ldots, x_n) =\ _p
\langle 0 | T (\phi_H(x_1) \cdots \phi_H(x_n)) |0 \rangle_p
\end{equation}
\noindent
Does $Z_F[\rho]$, the generating functional you get by blindly summing graphs generate the same Green's functions? \\
$\mathcal{H}$ splits up into $\mathcal{H}_0 + \mathcal{H}'$. Let
$\mathcal{H}_I = \mathcal{H}'(\phi_I)$. The thing which after Wick's
theorem and a combinatoric argument or two had a graphical expansion is
$$ Z_F [\rho] = \lim_{t_\pm \rightarrow \pm \infty} \!\!\!\!\!\!\!\!\! \underbrace{\langle 0 |}_{\substack{\text{Base eigenstate}\\\text{of } H_0\\ H_0 | 0 \rangle = 0}} \!\!\!\!\!\!\!\!\!\! T e^{-i \int^{t_+}_{t_-} d^4 x [ \mathcal{H}_I - \rho \phi_I ]} |0 \rangle $$
We used to adjust the constant part of $\mathcal{H}_I$ to eliminate
vacuum bubbles in our old scattering theory when $\rho = 0$. That is,
we adjusted the vacuum energy counterterm, so that the vacuum to
vacuum graphs (with no source vertices) summed to zero. There is an
equivalent way of throwing away the vacuum bubbles. You divide out of
$Z_F[\rho]$ the sum of all vacuum to vacuum graphs with no source
vertices explicitly, and then you don't have to worry about a vacuum
energy c.t., i.e.~you divide by the same thing with $\rho = 0$.
$$ Z_F [\rho] = \lim_{t_\pm \rightarrow \pm \infty} \frac{\langle 0 | T e^{-i \int^{t_+}_{t_-} d^4 x [ \mathcal{H}_I - \rho \phi_I ]} |0 \rangle}{\langle 0 | T e^{-i \int^{t_+}_{t_-} d^4 x \mathcal{H}_I} |0 \rangle} $$
\noindent
To get $G_F^{(n)} (x_1, \ldots, x_n)$ we do $n$ functional
derivatives w.r.t.~$\rho$ and then set $\rho = 0$ (and divide by
$i^n$)
$$ G_F^{(n)} (x_1, \ldots, x_n) = \lim_{t_\pm \rightarrow \pm \infty} \frac{\langle 0 | T e^{-i \int^{t_+}_{t_-} d^4 x [ \phi_I(x_1) \cdots \phi_I(x_n)\mathcal{H}_I ]} |0 \rangle}{\langle 0 | T e^{-i \int^{t_+}_{t_-} d^4 x \, \mathcal{H}_I} |0 \rangle} $$
We have got a little work to do to show this is the same as $G^{(n)}$
in Eq.~(\ref{eq:14-2}). Fortunately, both these expressions are
manifestly symmetric under the $n!$ permutations of the $x_1, \ldots, x_n$, so it suffices to prove they are equal for one ordering which for convenience we choose so that
$$ x_1^0 > x_2^0 > \cdots > x_n^0 \quad \quad \text{or for short } t_1 > t_2 > \cdots > t_n $$
The time ordering in the expression for $G^{(n)}$ is just lexicographic ordering
\begin{align*}
G^{(n)}(x_1, \ldots, x_n) &= \ _p \langle 0 | T (\phi_H(x_1) \cdots \phi_H(x_n)) | 0 \rangle_p \\
&= \ _p \langle 0 | \phi_H(x_1) \cdots \phi_H(x_n) | 0 \rangle_p
\end{align*}
\noindent
Using the standard shorthand for $\displaystyle e^{-i \int^{t_b}_{t_a} d^4x \mathcal{H}_I} = U_I(t_b, t_a) $, the time ordering in the expression for $G_F^{(n)}$ is
$$ G^{(n)}_F(x_1,\ldots,x_n) = \lim_{t_\pm \rightarrow \pm \infty} \frac{ \langle 0 | U_I(t_+, t_1) \phi_I(x_1) U_I(t_1, t_2) \phi_I(x_2) \cdots \phi_I(x_n)U_I(t_n, t_-) |0 \rangle }{ \langle 0 | U(t_+, t_-) |0 \rangle } $$
\noindent
at least in the $\lim_{t_\pm \rightarrow \pm \infty}$ when $t_+ > t_1
> \cdots > t_n > t_- $. Convince yourself.\footnote{I usually put ``convince
yourself" when I haven't written enough to make something clear, but
if I wrote more it would take just as long to figure out what I was
saying as it would take to convince yourself.}\\
Everywhere $U_I(t_a,t_b)$ appears, rewrite it as $U_I(t_a,0) U_I(0,t_b)$ and then use
$$ \phi_H(x_i) = U_I(t_i, 0 )^\dagger \phi_I(x_i) U_I(t_i,0) = U_I(0, t_i ) \phi_I(x_i) U_I(t_i,0) $$
to get
$$ G^{(n)}_F(x_1,\ldots,x_n) = \lim_{t_\pm \rightarrow \pm \infty} \frac{ \langle 0 | U_I(t_+, 0) \phi_H(x_1) \phi_H(x_2) \cdots \phi_H(x_n)U_I(0, t_-) |0 \rangle }{ \langle 0 | U(t_+, 0) U(0, t_-)|0 \rangle } $$
Considering the whole mess sandwiched with $U_I(0,t_-) |0 \rangle$ in
the numerator or denominator as some fixed state $\langle \phi |$,
let's work on
\begin{align*}
\lim_{t_- \rightarrow -\infty} \langle \phi | U_I(0,t_-) |0 \rangle &=
\lim_{t_- \rightarrow -\infty} \langle \phi | U_I(0,t_-) \!\!\!\!
\underbrace{e^{i H_0 t_-}}_{\substack{\text{fancy way
of}\\\text{inserting 1}}} \!\!\!\! |0 \rangle\\
&\!\!\!\overbrace{=}^{\substack{*\text{see}\\\text{below}}}
\lim_{t_- \rightarrow -\infty} \langle \phi | \!\!\!\!\!\!\!
\underbrace{U(0,t_-)}_{\text{Schr\"odinger picture}} \!\!\!\!\!\!\!\!
|0 \rangle \quad \quad \substack{\text{Using an easily
derivable}\\\text{relationship between the}\\\text{evolution operator
in the}\\\text{various pictures, Oct.~16, Eq.~%(1)}} \\ % ***
(\ref{eq:08-interactionpicture})}}\\
\substack{\text{insert a}\\\text{complete}\\\text{set}} &= \lim_{t_-
\rightarrow -\infty} \langle \phi | U(0,t_-) \underbrace{\left[ |0
\rangle_p \, _p\langle 0 | + \fourteensumintb |n \rangle \langle n|
\right]}_{\substack{\text{all other eigenstates of}\\\text{the full
Hamiltonian, } H\\ H |0 \rangle_p = 0, \, H|n\rangle = E_n | n \rangle }} |0 \rangle \\
&= \langle \phi | 0 \rangle_p \ _p\langle 0 | 0 \rangle + \lim_{t_-
\rightarrow - \infty} \fourteensuminta_{\substack{\text{all
other}\\\text{eigenstates}}} e^{i E_n t_-} \langle \phi | n \rangle \langle n | 0 \rangle
\end{align*}
\noindent
*: You see it doesn't really matter from this point on that it is a
bare vacuum that $U(0,t_0)$ is acting on. What we are showing is
that for two arbitrary fixed states $\lim_{t_- \rightarrow - \infty}
\langle \phi | U(0,t_-) | \psi \rangle = \langle \phi | 0 \rangle_p
\ _p \langle 0| \psi \rangle $ .\\
Now every state but the vacuum is part of a continuum. As long as
$\langle \phi | n \rangle \langle n | 0 \rangle $ is a continuous
function, the limit vanishes. The integral is a continuous function
that oscillates more and more wildly as $t_- \rightarrow - \infty$. In
the limit it integrates to zero. A similar argument shows
$$\lim_{t_+ \rightarrow \infty} \langle 0 | U_I(t_+,0) |\psi \rangle = \langle 0 | 0 \rangle_p \ _p\langle 0 | \psi \rangle $$
Physically what this theorem about oscillation integrands (the
Riemann-Lebesgue lemma) says is that if you look at a state in some
fixed region (take its inner product with some fixed state $\langle
\phi |$) and you wait long enough, the only trace of it that will
remain is its (true) vacuum component. All the one and multi-particle
states will have run away.
\begin{align*} G_F^{(n)}(x_1, \ldots, x_n) &= \frac{\cancel{\langle 0
| 0 \rangle_p} \ _p\langle 0 |\phi_H(x_1)\cdots \phi_H(x_n) |0
\rangle_p \cancel{\ _p\langle 0 | 0 \rangle}}{\cancel{\langle 0 | 0
\rangle_p} \ _p\langle 0 | 0 \rangle_p \cancel{\ _p\langle 0 | 0
\rangle}} \\
&= G^{(n)}(x_1, \ldots, x_n)
\end{align*}
\noindent
and there is no longer any reason to distinguish between them. \\
\textbf{Question 2}: Are $S-1$ matrix elements obtained from Green's
functions in the same way as before?\\
By introducing a turning on and off function, we were able to show that
$$ \langle l_1, \ldots, l_s | (S-1) | k_1, \ldots, k_r \rangle = \prod_{a=1,\ldots,s} \frac{l_a^2 - \mu^2}{i} \prod_{b=1,\ldots,r} \frac{k_b^2 - \mu^2}{i} \tilde{G}^{(r+s)}(-l_1, \ldots, -l_s, k_1, \ldots, k_r)$$
The real world does not have a turning on and off function. Is this formula right? \\
The answer is ``almost."\\
We will show how to obtain $S-1$ matrix elements from Green's function
without resorting to perturbation theory. We will make no reference to free Hamiltonia, bare vacua, interaction picture fields, etc. Accordingly, take
$$ \phi(x) \equiv \phi_H(x) $$
$$ | 0 \rangle \equiv | 0 \rangle_p $$
\noindent
$|0 \rangle$ is the ground state of the full Hamiltonian which as
usual we assume to be translationally invariant and not part of a
continuum, i.e.~normalizable.
$$ P^\mu | 0 \rangle = 0 $$
$$ \langle 0 | 0 \rangle = 0 = 1$$
We will assume there are physical one meson states, $|k \rangle$, relativistically normalized,
$$ H | k \rangle = \sqrt{\vec{k}^2 + \mu^2 } |k \rangle $$
$$ \vec{P} | k \rangle = \vec{k} | k \rangle $$
$$ \langle k' | k \rangle = (2 \pi)^3 2 \omega_{\vec{k}} \delta^{(3)}(\vec{k} - \vec{k}') $$
The reason that the answer to question 2 is ``almost" is because the
field, $\phi$, which enters the formula for $S-1$ matrix elements
through $G^{(n)}$ ($G^{(n)}(x_1,\ldots,x_n) = \langle 0 | T(\phi(x_1)
\cdots \phi(x_n)) |0 \rangle$), does not have quite the right
properties. First, it may have a VEV, and second, in general it is not
normalized so as to create a one particle state from the vacuum with a
standard amplitude. It is normalized to obey the canonical commutation
relations. We correct for these things by defining a renormalized field, $\phi^\prime$ in terms of $\phi$.\\
More precisely, $\langle 0 | \phi(x) | 0 \rangle $ may not be zero. However this VEV is independent of $x$ by translational invariance.
$$ \langle 0 | \phi(x) | 0 \rangle = \langle 0 | e^{iP\cdot x} \phi(0)
e^{-iP\cdot x} |0 \rangle = \langle 0 | \phi(0) | 0 \rangle $$
\noindent
We also have by translational invariance
$$ \langle k | \phi(x) |0 \rangle = \langle k | e^{iP\cdot x} \phi(0)
e^{-iP\cdot x} |0 \rangle = e^{ik\cdot x} \langle k | \phi(0) |0 \rangle $$
By Lorentz invariance, you can easily see that $\langle k | \phi(0) |0 \rangle$ is independent of $k$. It is some number, $Z_3^{\frac{1}{2}}$ (traditionally called the ``wave function renormalization"), in general $\neq 1$,
$$ Z_3^{\frac{1}{2}} \equiv \langle k | \phi(0) |0 \rangle$$
\noindent
which we hope is nonzero. \\
We define a new field $\phi'$ which has zero VEV and is normalized to have a standard amplitude to create one meson
$$ \phi'(x) = Z_3^{-\frac{1}{2}} (\phi(x) - \langle 0 | \phi(0) | 0
\rangle) $$
$$ \langle 0 | \phi'(x) | 0 \rangle = 0 $$
$$ \langle k | \phi'(x) | 0 \rangle = e^{ik\cdot x} $$
\section*{LSZ formula stated}
Define the renormalized Green's functions, $G'^{(n)}$,
$$ G'^{(n)} (x_1, \ldots, x_n) \equiv \langle 0 | T (\phi'(x_1) \cdots \phi(x_n)) |0 \rangle $$
\noindent
and $\tilde{G}'^{(n)}$, their Fourier transforms, then $S-1$ matrix elements are given by
$$ \langle l_1, \ldots, l_s | (S-1) | k_1, \ldots, k_r \rangle =
\prod_{a=1,\ldots,s} \frac{l_a^2 - \mu^2}{i} \prod_{b=1,\ldots,r}
\frac{k_b^2 - \mu^2}{i} \tilde{G}'^{(r+s)}(-l_1, \ldots, -l_s, k_1, \ldots, k_r)$$
This is the Lehmann-Symanzik-Zimmermann reduction formula.\\
The \underline{only} assumptions needed about the total scalar field
$\phi'$ is that it satisfy
$$ \langle 0 | \phi'(x) | 0\rangle \quad \text{and} \quad \langle k |
\phi'(x) |0 \rangle = e^{ik\cdot x} $$
In particular, its relationship to $\phi(x)$, the field that appears
in the Lagrangian with a standard kinetic term, is not
used. Any field that satisfies these properties, whose Green's
functions you have, gives you the $S$ matrix.\\
The proof of the LSZ reduction formula is as follows. First we'll find
a way of making one meson wave packets. The method will be inspired by
the way a limiting process gave us the physical vacuum when we started
with the bare vacuum. Once we know how to make one meson states we'll
wave our arms some and describe how to get many meson in and out
states. Then we'll be set to get $S$ matrix elements in terms of the
Green's function of the renormalized fields that were used to create the in and out states.
\subsection*{LSZ reduction formula proof}
A notation for normalizable wave packet states
$$ |f \rangle \equiv \int \frac{d^3 k}{(2\pi)^3 2 \omega_{\vec{k}}} F(\vec{k}) |k \rangle $$
\noindent
(You can recover $F(\vec{k})$ from $|f \rangle$: $F(\vec{k}) = \langle k | f \rangle $.)\\
\noindent
associated with each of these wave packets, we have a negative
frequency solution of the K.-G.~equation.
$$ f(x) \equiv \int \frac{d^3k}{(2\pi)^3 2\omega_{\vec{k}}} F(\vec{k})
e^{-ik\cdot x} $$
$$ (\Box + \mu^2) f(x) = 0 $$
\noindent
Note that as $|f\rangle \rightarrow | k \rangle $ (i.e.~$F(\vec{k'}) \rightarrow (2 \pi)^3 2 \omega_{\vec{k}} \delta^{(3)} (\vec{k} - \vec{k'})$)
$$ f(x) \rightarrow e^{-ik\cdot x} $$
Define an operator which is a function of time only out of any operator which is a function of $\vec{x}$ and $t$ (here taken to be $\phi'$):
$$ \phi'^f(t) = i \int d^3x (\phi' \partial_0 f - f \partial_0 \phi') $$
\noindent
Those of you who are familiar with the initial value theory of the
K.-G.~equation will not find this a strange combination. It will turn out to create a particle in state $|f \rangle$, in the limit $t \rightarrow \pm \infty $.
From the properties of $\phi'(x)$ and $f(x)$
$$ \langle 0 | \phi'^f (t) | 0 \rangle = 0 \quad \quad \text{and} $$
\begin{align*}
\langle k | \phi'^f(t) | 0 \rangle &= i \int d^3 x \int \frac{d^3
k'}{(2 \pi )^3 2 \omega_{\vec{k'}}} F(\vec{k'}) \langle k |
\Big[ \phi' \partial_0 e^{-ik'\cdot x} - e^{-ik'\cdot x} \partial_0 \phi'(x,t) \Big] | 0 \rangle \\
&= i \int d^3 x \int \frac{d^3 k'}{(2 \pi )^3 2 \omega_{\vec{k'}}}
F(\vec{k'}) \langle k | \Big[-i \omega_{\vec{k'}} e^{-ik'\cdot x} -
e^{-ik'\cdot x} \partial_0 \Big] \underbrace{\langle k | \phi'(x,t) | 0
\rangle}_{e^{ik\cdot x}} \\
&= i \int d^3 x \int \frac{d^3 k'}{(2 \pi )^3 2 \omega_{\vec{k'}}}
F(\vec{k'}) ( -i \omega_{\vec{k'}} -i \omega_{\vec{k}} ) e^{-ik'\cdot x +
ik\cdot x} \\
&= F(\vec{k}) \quad \quad \text{independent of time}
\end{align*}
A similar derivation except for one crucial minus sign shows
$$ \langle 0 | \phi'^f(t) |k\rangle =0 $$
In these few matrix elements, $\phi'^f(t)$ is acting like a creation operator for a physical meson wave packet. We will now take the limit $t\rightarrow \pm \infty$, and in this limit we'll see that many more matrix elements of $\phi'^{f}(t)$ look like the matrix elements of a creation operator.\\
Consider any state with two or more particles satisfying $P^\mu |n\rangle = P^{\mu}_n |n\rangle $.
\begin{align*}
\langle n | \phi'^f(t) | 0 \rangle &= \langle n | i \int d^3 x \Big[ \phi'(\vec{x},t) \partial_0 f - f \partial_0 \phi'(\vec{x},t) \Big] | 0 \rangle \\
&= i \int d^3 x \Big[ \partial_0 f - f \partial_0 \Big]
\underbrace{\langle n | \phi'^(\vec{x},t) | 0 \rangle}_{e^{iP_n\cdot x} \langle n | \phi'^(0) | 0 \rangle} \\
&= i \int d^3 x \Big[ \partial_0 f - f i P_n^0 \Big] e^{iP_n\cdot x} \langle n | \phi'^(0) | 0 \rangle \\
&= i \int d^3 x \Big[ \partial_0 - i P_n^0 \Big] \int
\frac{d^3k}{(2\pi)^3 2\omega_{\vec{k}}} F(\vec{k}) e^{-ik\cdot x}
e^{iP_n\cdot x} \langle n | \phi'^(0) | 0 \rangle \\
&= i \int \frac{d^3k}{(2\pi)^3 2\omega_{\vec{k}}} F(\vec{k})
(-i\omega_{\vec{k}} - i P^0_n) \int \!\!\!\!\!\!\!\!\!\!
\underbrace{d^3 x e^{-ik\cdot x} e^{iP_n\cdot x}}_{(2\pi)^3 \delta^{(3)} (\vec{k} - \vec{P}_n) \cdot e^{-i \omega_{\vec{k}}t + i P^0_n t}} \!\!\!\! \langle n | \phi'^(0) | 0 \rangle \\
&= \frac{\omega_{\vec{P}_n} + P_n^0}{2 \omega_{\vec{P}_n}} F(\vec{P}_n) \langle n | \phi'(0) | 0 \rangle e^{-i(\omega_{\vec{P}_n} - P_n^0)t}
\end{align*}
The important thing to notice is that this matrix element contains $
e^{-i(\omega_{\vec{P}_n} - P_n^0)t}$ and that $\omega_{\vec{P}_n} <
P_n^0$ for any multiparticle state. A multiparticle state with momentum $\vec{P}_n$ always
has more energy than a single particle state with momentum
$\vec{P}_n$. A two particle state with $\vec{P} = 0$ can have any
energy from $2\mu$ to $\infty$. The one particle state with $\vec{P} =
0$ has energy $\omega_{\vec{P}} = \mu$. \\
Now consider $\langle \psi | \phi'^{f}(t) | 0 \rangle $ in the limit $t \rightarrow \pm \infty$ where $|\psi \rangle$ is a definite (not varying with $t$) normalizable state. Insert a complete set.
\begin{align*}
\lim_{t \rightarrow \pm \infty} \langle \psi | \phi'^f(t) |0 \rangle
&= \lim_{t \rightarrow \pm \infty} \langle \psi | \left( |0 \rangle
\langle 0 | +\int \frac{d^3 k}{(2\pi)^3 2 \omega_{\vec{k}}} |k \rangle
\langle k | + \fourteensuminta_{\substack{\text{multiparticle}\\\text{states} |n\rangle }} |n\rangle \langle n | \right) \phi'^f(t) |0 \rangle \\
&= 0 + \frac{d^3 k}{(2\pi)^3 2 \omega_{\vec{k}}} \langle \psi | k
\rangle F(\vec{k}) + \lim_{t \rightarrow \pm \infty} \fourteensumintb_{|n \rangle} \langle \psi | n \rangle \frac{F(\vec{P}_n) (\omega_{\vec{P}_n} + P_n^0)} {2 \omega_{\vec{P}_n}} \\
&= \langle \psi | f \rangle + 0
\end{align*}
The integrals over the various \underline{continua} of multiparticle states have
integrands which oscillate more and more wildly as $t\rightarrow \pm
\infty$. They integrate to zero in the limit by the Riemann-Lebesgue lemma.\\
The phases were arranged to cancel in the one particle state matrix elements only. \\
The analogous derivation showing $\lim_{t \rightarrow \pm \infty}
\langle 0 | \phi'^{f} (t) | \psi \rangle = 0$ goes through because the
phases add (and thus obviously never cancel) for every momentum eigenstate, one or multiparticle in the inserted complete set. \\
Physically what we have shown is this: We have created a state which
in part looks like a one meson wave packet, plus a little multiparticle
garbage. If we look at this state in some definite region in space
time (take its inner product with some definite state $|\psi \rangle$,
and send the time of reaction to $ - \infty$), all the multiparticle
states will run away. Of course the one particle state may run away
too. We prevent this by modifying the state we create in such a way
that the one meson packet always has the same relationship to the
meson that's the funny combination $\phi'^f(t)$. No multiparticle state has the right dispersion relation to keep the same relationship to the observer under this modification.