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\lhead{Physics 253a}
\chead{November 13}
\rhead{Sidney Coleman's Notes}
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What we have done so far has been rigorous, at least it can be made rigorous, with only a little effort. For the creation of multiparticle in and out states we are going to have to do some vigorous handwaving.\\
Consider two normalizable wave packets described by $F_1(\vec{k})$ and
$F_2(\vec{k})$ which have no common support in momentum space. This
excludes scattering at threshold. \\
What is $\displaystyle \lim_{t \rightarrow +(-) \infty} \langle \psi |
\phi'^{f_2} (t) | f_1 \rangle$ ?\\
The handwaving requires we picture this in position space. Because
$F_1$ and $F_2$ describe packets headed in different directions if you
wait long enough (go back far enough) the two wave packets will be
widely separated in position. Then the application of $\phi'^{f_2} (t)
$ to $|f_1 \rangle$ is for all purposes to an observer near the $f_2$ packet, like an application of $\phi'^{f_2}$ to the vacuum.\\
You may be bothered that a position space picture doesn't really
exist, there is no $\vec{x}$ operator. However there is still some
concept of localization up to a few Compton wavelengths. If a teensy
exponential tail with $\frac{1}{e}$ distance $\frac{1}{m}$ is too big
for you for some purpose, wait another thousand years.\\
This physical argument says
$$ \lim_{t \rightarrow +(-) \infty} \langle \psi | \phi'^{f_2} (t) |f_1 \rangle = \langle \psi | f_1, f_2 \rangle^{\text{out (in)}} $$
By definition, the $S$ matrix, is what tells you the probability
amplitude that looks like % added "like"
a given state in the far past will look like another given state in the far future.
\begin{align*}
\langle f_3, f_4 | S | f_1, f_2 \rangle &\equiv \ ^\text{out} \langle f_3, f_4 | S | f_1, f_2 \rangle^{\text{in}} \\
&= \lim_{t_4 \rightarrow \infty} \lim_{t_3 \rightarrow \infty}
\lim_{t_2 \rightarrow - \infty} \lim_{t_1 \rightarrow - \infty}
\langle 0 | \phi'^{f_3 \dagger}(t_3) \phi'^{f_4 \dagger}(t_4) \phi'^{f_2}(t_2) \phi'^{f_1}(t_1) | 0 \rangle
\end{align*}
Note that in the limits, this is time ordered and thus we have
succeeded in writing $S$ matrix elements in terms of the renormalized
Green's functions, So we could quit now, but we are going to massage
this expression. In doing so, we'll extend the idea of an $S$ matrix
element. Physically there is no way to create plane wave states. Thus
there is no way to measure of define $S$ matrix elements of plane wave
states. However, after we get done massaging the RHS of the above
equation, we will get an expression that you can put plane wave states
into without getting nonsense. We'll make this the definition of $S$
matrix elements of plane wave states, $\langle k_3, k_4 | (S-1) |k_1,
k_2 \rangle$. The utility of this object is that you can integrate
it, smear it a little, to recover physically measurable $S$ matrix elements. Now I'll tell you the answer, that is, what we will soon show is a sensible definition for
\begin{multline*} \langle k_3, k_4 | (S -1 )| k_1, k_2 \rangle = \int
d^4 x_1 \cdots d^4 x_4 e^{i k_3 \cdot x_3 + i k_4 \cdot x_4 - i k_1
\cdot x_1 - i k_2 \cdot x_2}\times\\
(i)^4 \prod_r (\Box_r + \mu^2) \langle 0 | T (\phi'(x_1) \cdots \phi'(x_4) ) |0 \rangle
\end{multline*}
That looks unfamiliar and messy, but it actually isn't. Recall that
\begin{align*}
\langle 0 | T (\phi'(x_1) \cdots \phi'(x_4)) |0 \rangle &\equiv G'(x_1, \ldots, x_4) \\
&= \int \frac{d^4 l_1 }{(2\pi)^4} \cdots \frac{d^4 l_4 }{(2\pi)^4}
e^{i l_1 \cdot x_1 + \cdots + i l_4 \cdot x_4} \tilde{G}'(l_1, \ldots, l_4)
\end{align*}
If you substitute this in the expression for $\langle k_3, k_4 | (S -1 )| k_1, k_2 \rangle$, it collapses to
$$ \langle k_3, k_4 | (S -1 )| k_1, k_2 \rangle = \prod_r \frac{k_r^2 - \mu^2}{i} \tilde{G}'(k_1, k_2, -k_3, -k_4) $$
This says that an $S-1$ matrix element is equal to a Green's function
with the external propagators removed. This is almost exactly the
result that came out of our low budget scattering theory, with the
only difference being that the Green's function is of renormalized
fields. The result we will first obtain won't be an expression for
$\langle k_3, k_4 | (S -1 )| k_1, k_2 \rangle $. We get that by
abstracting the expression for $\langle f_3, f_4 | (S -1 )| f_1, f_2
\rangle$ which looks just like the expression for $\langle k_3, k_4
| (S -1 )| k_1, k_2 \rangle $ stated above except $e^{-k_1\cdot x_1}$
is replaced by $f_1(x_1)$, $e^{-ik_2\cdot x_2}$ by $f_2(x_2)$,
$e^{-ik_3\cdot x_3}$ by $f_3(x_3)$ and $e^{-ik_4\cdot x_4}$ by
$f_4(x_4)$. That is, what we will show is
\begin{multline*}
\langle f_3, f_4 | (S -1 )| f_1, f_2 \rangle = \int d^4 x_1 \cdots
d^4 x_4 f_3^*(x_3) f_4^*(x_4) f_1(x_1) f_2(x_2)\times \\
(i)^4 \prod_r (\Box_r + \mu^2) \langle 0 | T (\phi'(x_1) \cdots \phi'(x_4)) |0 \rangle
\end{multline*}
Let's get on with the proof, beginning with a lemma. \\
Given any function, $f(x)$, satisfying $(\Box + \mu^2) f(x) =0 $, and $f \rightarrow 0$ as $|x| \rightarrow \infty $ and a general field $A$, then
\begin{align*}
i \int d^4 x f(\Box + \mu^2) A &= i \int d^4 x \Big[ f \partial^2_0 A + A(-\nabla^2 + \mu^2) f \Big] \\
&= i \int d^4 x ( f \partial_0^2 A - A \partial_0^2 f ) \\
&= \int dt \; \partial_0 \underbrace{\int d^3 x \; i (f \partial_0 A -
A \partial_0 f )}_{\substack{\text{this is something that
appears}\\\text{often enough that it is worth}\\\text{giving it a
name. It is a function}\\\text{of time only, call it } -A^f(t)}}\\
&= - \int dt \; \partial_0 A^f(t) \\
&= \left(\lim_{t\rightarrow - \infty} - \lim_{t\rightarrow
\infty}\right) A^f(t)
\end{align*}
Also, if $A$ is hermitian,
$$ i \int d^x f^*(x) (\Box + \mu^2) A = \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\underbrace{\left(\lim_{t\rightarrow \infty} - \lim_{t\rightarrow
-\infty}\right)}_{\substack{\text{note difference in sign from conjugating}\\\text{the } i \text{ in the def'n of } A^f(t)}} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\! A^{f\dagger}(t)$$
Now we'll apply this equation to the RHS of the equation we want to
prove. First do % replaced "to" with "do"
the $x_1$ integration, you get
\begin{multline*}
\left(\lim_{t_1\rightarrow - \infty} - \lim_{t_1\rightarrow
\infty}\right) \int d^4 x_2 \, d^4 x_3 \, d^4 x_4 f_3^*(x_3)
f_4^*(x_4) f_2(x_2) \times\\
(i)^3 \prod_{r=2,3,4} (\Box_r + \mu^2) \langle 0 | T (\phi'^{f_1}(t_1) \phi'(x_2)\phi'(x_3) \phi'(x_4)) |0 \rangle
\end{multline*}
\noindent
We push a time derivative through a time-ordered product in these steps. It is OK in the limit we have. Then the $x_2$ integration
\begin{multline*}
\left(\lim_{t_1\rightarrow - \infty} - \lim_{t_1\rightarrow
\infty}\right)\left(\lim_{t_2\rightarrow - \infty} -
\lim_{t_2\rightarrow \infty}\right) \int d^4 x_2 \, d^4 x_3 \, d^4
x_4 f_3^*(x_3) f_4^*(x_4) \times\\
(i)^2 (\Box_3 + \mu^2) (\Box_4 + \mu^2)\langle 0 | T (\phi'^{f_1}(t_1) \phi'^{f_2}(t_2) \phi'(x_3) \phi'(x_4)) |0 \rangle
\end{multline*}
\noindent
Etcetera.
\begin{align*}
\left(\lim_{t_1\rightarrow - \infty} - \lim_{t_1\rightarrow
\infty}\right)
\underbrace{\left(\lim_{t_2\rightarrow - \infty} - \lim_{t_2\rightarrow
\infty}\right) \left(\lim_{t_3\rightarrow \infty} - \lim_{t_3\rightarrow
-\infty}\right)}_{\text{note the difference in sign}}&
\left(\lim_{t_4\rightarrow - \infty} - \lim_{t_4\rightarrow
\infty}\right) \\
& \langle 0 | T (\phi'^{f_1}(t_1) \phi'^{f_2}(t_2)
\phi'^{f_3\dagger}(t_3) \phi'^{f_4\dagger}(t_4)) |0 \rangle
\end{align*}
If we had reduced the integrals in some other order we would have a
different order of limits here. All $4!$ orderings lead to the same
result however, and we'll just do the order we have arrived at. \\
When $t_4 \rightarrow - \infty$, it is the earliest time and thus the
time ordering puts it on the right. However $\phi'^{f_4\dagger}(t_4)$ with the vacuum on the right and any other state on the left vanishes in the limit. When $t_4 \rightarrow + \infty$, it is the latest time and the time ordering puts it on the left, and we get the matrix element of $\langle f_4 |$ with the rest of the mess. We have
$$ \left(\lim_{t_1\rightarrow - \infty} - \lim_{t_1\rightarrow
\infty}\right) \underbrace{\left(\lim_{t_2\rightarrow - \infty} -
\lim_{t_2\rightarrow \infty}\right) \left(\lim_{t_3\rightarrow
\infty} - \lim_{t_3\rightarrow -\infty}\right)}_{\text{note the
difference in sign}} \langle f_4 | T (\phi'^{f_1}(t_1)
\phi'^{f_2}(t_2) \phi'^{f_3\dagger}(t_3)) |0 \rangle$$
The exact same considerations apply to the $t_3$ limits except we get the matrix elements of $^{\text{out}}\langle f_3,f_4 |$ with the remaining mess `out' because both fields are applied to the vacuum in the far future. Doing the $t_2$ limits does not result in such a simplification. We get
$$ \left(\lim_{t_1\rightarrow - \infty} - \lim_{t_1\rightarrow
\infty}\right) \Big( \ ^{\text{out}}\langle f_3,f_4 | \phi'^{f_1}(t_1) |f_2 \rangle - \lim_{t_2 \rightarrow \infty} \ ^{\text{out}}\langle f_3,f_4 | \phi'^{f_2}(t_2) \phi'^{f_1}(t_1) |0 \rangle \Big) $$
The first term was expected. The second term looks real bad. Let's compartmentalize our ignorance by just giving a name to this state we have created
$$ \langle \psi | \equiv \lim_{t_2 \rightarrow \infty} \ ^{\text{out}}\langle f_3,f_4 | \phi'^{f_2}(t_2) $$
On to the evaluation of the $t_1$ limit. We get
$$ ^{\text{out}} \langle f_3, f_4 | f_1, f_2 \rangle^{\text{in}} - \
^{\text{out}} \langle f_3, f_4 | f_1, f_2 \rangle^{\text{out}} -
\cancel{\langle \psi | f_1 \rangle} + \cancel{\langle \psi | f_1
\rangle} $$
%$$ = ^{\text{out}} \langle f_3, f_4 | f_1, f_2 \rangle^{\text{in}} - \ ^{\text{out}} \langle f_3, f_4 | f_1, f_2 \rangle^{\text{out}}$$
The last two terms cancel, because there is no difference between a
matrix element of $\lim_{t_1 \rightarrow + \infty} \phi'^{f_2}(t_2) |
0 \rangle$ and $\lim_{t_1 \rightarrow - \infty} \phi'^{f_2}(t_2) | 0
\rangle$ $\ddot{\smile}$ . The two terms remaining are exactly what we wanted to get. We have obtained
$$ \langle f_3, f_4 | (S-1) | f_1, f_2 \rangle $$
\noindent
\textbf{REMARKS: }
\begin{enumerate}
\item The mathematical expression we started with makes sense even for $f_1$, $f_2$, $f_3$, $f_4$ plane waves. We'll make that expression the definition of an $S-1$ matrix element of plane waves. Of course you only get something physically measurable when you integrate, smear, the expression. The situation is very analogous to $V(\vec{x} - \vec{y})$ in the expression $U = \int d^3x d^3y V(\vec{x} - \vec{y}) \rho(\vec{x}) \rho(\vec{y})$. No one can build a point charge, and thus no one can make a charge distribution that directly measures $V(\vec{x} - \vec{y})$, that is, one for which the interaction energy is $V(\vec{x} - \vec{y})$. All you can do is measure $U$ for various charge distributions. Then you can abstract to the notion of $V(\vec{x} - \vec{y})$. ``the potential energy of between two point charges." You only recover something physically measurable when you integrate, smear, the expression for $S-1$ matrix elements of plane waves. The formula analogous to $U = \int d^3 x d^3 y \; V(\vec{x} - \vec{y}) \rho(x) \rho(y) $ is
\begin{multline*}
\langle f_3, f_4 | (S-1) | f_1,f_2 \rangle = \int \frac{d^3
k_1}{(2\pi)^3 2 \omega_{\vec{k_1}}} \cdots \frac{d^3 k_4}{(2\pi)^3 2
\omega_{\vec{k_4}}} F^*_3(\vec{k_3}) F^*_4(\vec{k_4}) F_1(\vec{k_1})
F_2(\vec{k_2}) \times\\
\langle k_3,k_4 | (S-1) |k_1,k_2 \rangle
\end{multline*}
\item The proof only required that the field you begin with have a
nonzero vacuum to one particle matrix element. Then you shift that field by some constant, and multiply it by another constant to get the renormalized field, whose Green's functions are what actually entered the proof. There is thus a many to one correspondence between fields and particle. From the point of view of the reduction formula, $\tilde{\phi} = \phi + \frac{1}{2} g \phi^2$ is just as good as field (at least except for one exceptional value of $g$ that makes the vacuum to one particle matrix element of $\tilde{\phi}$ vanish). You do not have to begin with one of the fields that seemed to be fundamental in the Lagrangian.
\item There is no problem in principle of obtaining scattering matrix elements of composite particles and bound states. In the QCD theory of the strong interaction, the mesons are bound states of a quark and an antiquark. If $q(x)$ is a quark field, you would expect $\bar{q}q(x)$ to have a nonvanishing vacuum to one meson matrix elements. ``All" you need to calculate $2\rightarrow 2$ meson scattering then would be
$$ G^{\prime(4)} (x_1,x_2,x_3,x_4) \equiv \langle 0 | T ( \bar{q} q (x_1) \bar{q}q(x_2) \bar{q}q (x_3) \bar{q} q (x_4) ) | 0 \rangle $$
where $\bar{q}q(x)$ is the renormalized field. Of course no one has gotten $G'^{(4)}$.
\item If we have some exact knowledge of the position space properties
of a field, it may be possible to use these properties in the LSZ
formula to get some exact knowledge about $S-1$ matrix elements.
\item Using methods of the same type as those used in the derivation of the LSZ formula, other formulas can be derived. For example, one can ``stop half way" in the reduction formula and obtain
\begin{multline*}
\langle k_3, k_4 | (S-1) |k_1 k_2 \rangle = \int d^4 x_3 \, d^4 x_4
e^{ik_3\cdot x_3} e^{ik_4\cdot x_4}\times \\
(i)^2 ( \Box_3 + \mu^2 ) ( \Box_4 + \mu^2 ) \langle 0 | T (\phi'(x_3) \phi'(x_4) )|k_1, k_2 \rangle^{\text{in}}
\end{multline*}
This is used to derive theorems about the production of ``soft" (low
energy) photons.\\
We can also use LSZ methods to derive expressions for the matrix elements of fields between in and out states. For example, I can show
\begin{multline*}
^\text{out}\langle k_1, \ldots, k_n | A(x) |0 \rangle = \int d^4 x_1
\cdots d^4x_n e^{ik_1\cdot x_1 + \cdots + i k_n\cdot x_n}\times\\
(i)^n \prod_r (\Box_r + \mu^2) \langle 0 | T (\phi'(x_1) \cdots \phi'(x_n) A(x) |0 \rangle
\end{multline*}
\noindent where $\phi'(x)$ is a correctly normalized field that can create the outgoing mesons and $A(x)$ is an arbitrary field. Of course this is really an abstraction of
\begin{align*}
^\text{out}\langle f_1, \ldots, f_n | A(x) |0 \rangle = \int d^4 x_1 \cdots d^4x_n & f^*(x_1) \cdots f^*(x_n) \\
& (i)^n \prod_r (\Box_r + \mu^2) \langle 0 | T (\phi'(x_1) \cdots \phi'(x_n) A(x) |0 \rangle
\end{align*}
Applying the methods used above, after the proof of the lemma, we get
$$ \left(\lim_{t_1 \rightarrow \infty} -\lim_{t_1 \rightarrow -
\infty}\right) \cdots \left(\lim_{t_n \rightarrow \infty} -\lim_{t_n
\rightarrow - \infty}\right)\langle 0 | T (\phi'^{f_1\dagger}(t_1) \cdots \phi'^{f_n \dagger}(t_n) A(x) |0 \rangle $$
Just as easily as we evaluated the $t_3$ and $t_4$ limits, these limits can be evaluated to get
$$ ^{\text{out}}\langle f_1, \ldots, f_n | A(x) |0 \rangle $$
% there's a remark about a homework problem in the notes here
\end{enumerate}
\newpage
\textbf{A second look at model 3 and its renormalization: }
$$ \mathcal{L} = \frac{1}{2} (\partial_\mu \phi)^2 - \frac{\mu_0^2}{2} \phi^2 + \partial_\mu \psi^* \partial^\mu \psi -m_0^2 \psi^* \psi - g_0 \psi^* \psi \phi $$
The upshot of what we have done so far is that some $0$ subscripts
have been added to the Lagrangian. The coefficient of $-\frac{1}{2}
\phi^2$ in $\mathcal{L}$ may not be the meson mass squared, $m_0^2$
may not be the charged muon (nucleon) mass squared. Furthermore $g_0$
may not be what we want to call the coupling constant. In real
electrodynamics there is a parameter $e$ defined by some experiment.
It would be lucky, extremely lucky, if that were the coefficient of
some term in the electrodynamics Lagrangian. In general it isn't.
We'll subscript the coupling constant, compute the conventionally
defined coupling constant from it, and then invert the equation to
eliminate $g_0$, which is not directly measured, from our expressions
for all other quantities of interest. Also, when calculating our
scattering matrix elements, we need Green's functions. What we have a
perturbative expansion for if we treat $-g_0 \psi^* \psi \phi$ as our interaction Lagrangian, is the Green's functions of $\phi$. Those Green's functions aren't exactly what we are interested in. We want the Green's function of $\phi'$, the field satisfying
$$ \langle 0 | \phi' |0 \rangle = 0 $$
$$ \langle p | \phi'(0) |0 \rangle = 1 $$
$$ \phi' = Z_3^{-\frac{1}{2}} ( \phi - \langle \phi \rangle_0 )$$
So along the way in calculating quantities of interest, we'll have to
calculate the Green's function of $\phi'$ from the Green's functions
of $\phi$. This determination of $\phi'$, $m$, $\mu$, and $g$ from
$m_0$, $\mu_0$, $g_0$ and the above conditions, and then the plugging
ins of the inverse of these equations into other quantities of interest sounds like a mess. It can be avoided.\\
We rewrite $\mathcal{L}$ with six new parameters, $A,\ldots, F$.
$$ \mathcal{L} = \frac{1}{2} (\partial_{\mu} \phi')^2 - \frac{\mu^2}{2} \phi^{\prime 2} + \partial_{\mu} \psi^{*\prime} \partial^\mu \psi' - m^2 \psi^{*\prime} \psi' - g \psi^{*\prime} \psi' \phi' + \mathcal{L}_{CT} $$
$$ \mathcal{L}_{CT} = A \phi' + \frac{B}{2}(\partial_\mu \phi')^2 - \frac{c}{2}\phi^{\prime 2} + D \partial_\mu \psi^{\prime *} \partial^\mu \psi' - E \psi^{*\prime}\psi' - F \psi^{*\prime} \psi' \phi' + \text{const} $$
The six new parameters are going to be determined order by order in
perturbation theory by six renormalization conditions.
\begin{enumerate}
\item $\langle 0 | \phi' | 0 \rangle = 0 $
\item $\underbrace{\langle q |}_{\text{one meson}} \!\!\!\!\!\! \phi'(0) | 0 \rangle = 1 $
\item $\underbrace{\langle p |}_{\text{one anti-nucleon}} \!\!\!\!\!\!\!\!\!\!\! \psi'(0) | 0 \rangle = 1$
\item The meson mass is $\mu$
\item The nucleon mass is $m$
\item $g$ agrees with the conventionally defined $g$.
\end{enumerate}
Six unknowns, six conditions.\\
Of course, if you actually wanted to know the relationship of $\phi'$
to $\phi$, the field whose kinetic term has coefficient 1 in the
Lagrangian (and thus obeys the canonical commutation
relation)\footnote{This parenthetical remark should be emphasized more. It is $\phi$ that satisfies $[\phi, \dot{\phi}] = i \delta^{(3)}$ with coefficient 1, not $\phi'$.}, and has no linear term, you can read it off. You can also read off the bare meson mass, the bare nucleon mass, $g_0$, and the relationship of $\psi'$ to $\psi$.
$$\frac{1}{2}(1+B)(\partial_\mu \phi')^2 + \frac{1}{2}(\mu^2 +C)\phi^{\prime 2} + A \phi' + \text{const} = \frac{1}{2} (\partial_\mu \phi)^2 - \frac{\mu_0^2}{2} \phi^2 $$
$$ Z_3 = 1+B$$
$$ \mu_0^2 = \frac{1}{Z_3}(\mu^2 +C) \text{, etc.} $$
\noindent
(See % *** check page 13-15
Oct.~23 for these same ideas expressed before we knew / worried about $Z_3$ and $\langle 0 | \phi | 0 \rangle$)\\
What you have now is a perturbation theory for the quantities you are really interested in in terms of the conditions on $\phi'$ and $\psi'$, and experimentally input parameters. \\
The differences in the two kinds of perturbation theory is what you
call the interaction Lagrangian. We'll be taking $-g
\psi^{*\prime}\psi'\phi' + \mathcal{L}_{CT}$ as the interaction. This
is called renormalized perturbation theory.\\
Instead of computing scattering matrix elements $\mu^2$, $m^2$ and $g$
from $\mu_0$, $m_0$ and $g_0$, the wrong parameters to hold fixed, and
then inverting to get $S$ matrix elements in terms of $\mu^2$, $m^2$
and $g$, we compute everything in terms of the right quantities, the
experimentally input parameters, $\mu^2$, $m^2$ and $g$.\\
This procedure has a bonus, as long as you stick to observable quantities expressed in terms of physical parameters, you avoid the infinities which plague quantum field theory. \\
There are three technical obstacles we will have to overcome to implement this program.
\begin{enumerate}
\item There are derivative interactions in $\mathcal{L}_{CT}$.
\item Renormalization conditions (4), (5), (6) are not expressed in terms of Green's functions, the things we usually compute.
\item We have to make contact with the committee definition of $g$.
Then we may still have to worry about defining it in terms of Green's
functions [(2)].
\end{enumerate}
\end{document}