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\begin{center}
{\bf Lecture 5}
\end{center}
\bigskip
Bound states at smaller length scales:
\begin{figure}
\includegraphics[width=14cm]{baryons.png}
\end{figure}
\begin{figure}
\includegraphics[width=14cm]{mesons.png}
\end{figure}
\newpage
Often in physics problems one needs to evaluate a definate integral. If it is
not possible to find the "anti-derivative" of the integrand then numerical
methods may be the only way to solve the problem. For our next assignment we
will calculate an integral that will give us an approximation to elastic
scattering. The particular problem we will do is to solve for $K^+$
particles scattering off of nuclei. First I will go over a simple way to
do numerical integration for one variable. Then, we will talk about the physics involved in
scattering experiments, and an approximate method for calculating
cross sections: the Born Approximation. \\
\bigskip
\centerline{\bf Numerical Integration}
\bigskip
\noindent{\bf Rectangle and Trapazoid Rules}
\bigskip
Consider the definate integral, $\int_a^b f(x) \; dx$. The integral is equal
to the area under the curve $f(x)$ from $a$ to $b$. Our task is then to estimate
the area under this curve. The simplest way is to divide the total area under the
curve into small rectangles and add up the areas of each small rectangle. Let the
base of each rectangle be on the x-axis and the height be the value $f(x)$. We
divide the x-axis into small segments of equal length $h$. Then the sum of the
areas of the rectangles is given by
\begin{eqnarray*}
\int_a^b f(x) \; dx & \approx & hf(a) + hf(a+h) + hf(a+2h) + \cdots + hf(b-h) \\
& \approx & h(f(a) + f(a+h) + f(a+2h) + \cdots + f(b-h) )\\
\end{eqnarray*}
\noindent The smaller that $h$ is, the better the approximation. We can write this
expression in summation notation as follows:
\begin{equation}
\int_a^b \approx \sum_{n=0}^{N-1} h \; f(a+nh)
\end{equation}
\noindent where $n$ is an integer and $h=(b-a)/N$. The larger that $N$ is, the more accurate is the sum.\\
\newpage
One can do slightly better by using trapazoids instead of rectangles. The area of
a trapazoid with a base $h$, a left height of $f(a)$ and a right height of $f(a+h)$
is $h(f(a)+f(a+h))/2$. Using trapazoids instead of rectangles yields
\begin{eqnarray*}
\int_a^b f(x) \; dx & \approx & h{{f(a)+f(a+h)} \over 2} + h{{f(a+h)+f(a+2h)} \over 2} + \cdots \\
& \approx & h({{f(a)} \over 2} + f(a+h) + f(a+2h) + \cdots + f(a+(N-1)h) + {{f(b)} \over 2})\\
\int_a^b f(x) \; dx &\approx & h({{f(a)} \over 2} + \sum_{n=1}^{N-1} h \; f(a+nh) + {{f(b)} \over 2}) \\
\end{eqnarray*}
\noindent where $n$ is an integer and $h=(b-a)/h$. As before, the approximation gets better as $N \rightarrow \infty$.
Also note that this expression is exact if $f(x)$ is a line.\\
There are other algorithms that are used to carry out numerical integation. At the end of
these notes I have summarize another simple one, Simpson's rule. If time allows, I'll
summarize the method of Gaussian Quadrature. Now, lets go over the integral in your
assignment and discuss the physics behind the integral.
\newpage
\bigskip
\begin{figure}
\includegraphics[width=14cm]{rectint.png}
\end{figure}
\newpage
\centerline{\bf Scattering Experiments}
\bigskip
Scattering experiments are often carried out to understand the interaction between
elementary particles as well as information on the excited states of nuclei and
particles. Basically, one uses a beam of particles, all with the same momentum,
that "hits" a target. One can also have colliding beams. One measures the probability
that a particle in the beam will be scattered off at a particular angle. Let's define some terms:\\
\noindent {\bf Incident Flux}: The number of particles per area per time.\\
\begin{figure}
\includegraphics[width=14cm]{lect5p1.png}
\end{figure}
\noindent {\bf Solid Angle}: If the detector has an area $A$ and is a distance $r$ from
the target, then the solid angle is $\Delta \Omega \equiv A/r^2$. The full solid angle, a spherical shell,
is $4\pi r^2/r^2 = 4 \pi$.\\
\noindent {\bf Differential Cross section}: The number of particles detected per sec per incident
flux per solid angle. It is given the symbol $\sigma (\theta , \phi)$, or $(d \sigma )/ (d\Omega )$.
For unpolarized targets and beams, the differential cross section will only depend on
$\theta $, the scattering angle, and is labeled as $\sigma (\theta )$\\
\noindent {\bf Total Cross section}: The total number of particles scattered. The total
cross section $\sigma$ equals the differential cross section integrated over all solid
angle: $\sigma_{total} = \int \sigma(\theta ,\phi ) d\Omega$.\\
A graph of the differential cross section for $K^+$ particles scattering off $^{12}C$
is given on the next page. The units for the differential cross section are
Area/(solid angle), and the units for the total cross section is Area. The
area represented by the total cross section can be thought of as follows:
It is the effective area per target such that if the particle "hits"
this area it is scattered.
The cross sectional area of a nucleon is around $\pi r^2 \approx \pi \; fm^2$, since
the nucleon radius is around one Fermi. The nucleon's cross sectional area is
thus around $10^{-30} \; fm^2$ or $10^{-26} \; cm^2$. The unit for area in
nuclear and particle physics is the barn, which is defined as
$1 \; barn \equiv 10^{-24} \; cm^2 = 100 \; fm^2$. A barn is around 100 times the area of
a nucleon, and is quite large on the nuclear scale. Note that the units on the
graph for the differential cross section, $\sigma (\theta )$ are milli-barns per
ster-radian (mb/sr).
\newpage
\begin{figure}
\includegraphics[width=14cm]{kc12.png}
\end{figure}
\newpage
\begin{figure}
\includegraphics[width=14cm]{kca40.png}
\end{figure}
\newpage
\centerline{\bf Calculating the Cross Section}
\bigskip
If the interaction between the incident particle and the target can be treated
non-relativistically and be represented by a potential energy function
$V(r)$, then the scattering cross section can be calculated using the
Schroedinger equation. For our assignment we will use an approximation
to the differential cross section that only requires the computation of an
integral. The approximation is called "the Born approximation". If you
are interested in how it is derived, you can look at my previous notes
or an undergraduate quantum text. At the heart of the approximation is
the evaluation of the following integral:
\begin{equation}
\int_0^\infty V(r) r sin(qr) dr
\end{equation}
\noindent where we will take $V(r) = V_0/(1+e^{(r-R)/a})$. The
quantity $q=p/\hbar$ and represents the momentum/$\hbar$ that is transfered
to the target. In terms of the kaon's momentum $p$ and the scattering angle,
$\theta$, $q=2pc/(\hbar c)sin(\theta /2)$. As before, we will take the size
of the potential to be the size of the nucleus, so
$R=1.28A^{1/3} \; fm$. We will take $a=0.6 \; fm$. The cross section is
most easily expressed in terms of the scattring amplitude,
$f(\theta )$. The Born approximation gives:
\begin{equation}
f(\theta ) = -{1 \over {\hbar c}} {{mc^2} \over {pc}} {1 \over {sin(\theta /2)}}
\int_0^\infty V(r) r sin(qr) dr
\end{equation}
\noindent for the scattering amplitude. In terms of the scattering amplitude,
the differential cross section is
\begin{equation}
{{d \sigma} \over {d \Omega}} = |f(\theta )|^2
\end{equation}
Putting everything together, the differential cross section is
\begin{equation}
\sigma (\theta ) = |-{1 \over {\hbar c}} {{mc^2} \over {pc}} {1 \over {sin(\theta /2)}}
\int_0^\infty V(r) r sin(qr) dr|^2
\end{equation}
\newpage
\begin{figure}
\includegraphics[width=14cm]{lect5p2.png}
\end{figure}
\newpage
\noindent {\bf The Born Approximation}
\bigskip
The Born approximation can be derived from time dependent perturbation theory.
The probability for a transition to occur from an initial state
$\psi_i \propto e^{ik_i \cdot r}$ to a final state $\psi_f \propto e^{ik_f \cdot r}$
(where $k = p/\hbar$) is proportional to the matrix element
\begin{equation}
<\psi_f |V|\psi_i> \; \propto \int_0^\infty V(r) e^{i(\vec{k}_f-\vec{k}_i)\cdot \vec{r}} d^3 \vec{r}
\end{equation}
\noindent This integral, which is the Fourier transform of V(r) with respect to the
momentum transfer, produces a scattering diffraction pattern. The kinematics
enters the problem only with the momentum transfer, which is most conveniently defined
as $\vec{q} = \vec{k}_f - \vec{k}_i = (\vec{p}_f - \vec{p}_i)/\hbar$.
\begin{equation}
<\psi_f |V|\psi_i> \; \propto \int_0^\infty V(r) e^{i \vec{q}\cdot \vec{r}} d^3 \vec{r}
\end{equation}
At first glance, the integral looks difficult. However, if we choose the coordinate
system for the integral such that the z-axis is directed along $\vec{q}$, then
$\vec{q} \cdot \vec{r} = |q|\; |r|\; cos(\alpha)=qr\;cos(\alpha)$, where
$\alpha$ is the angle between $\vec{q}$ and $\vec{r}$. With this
choice of coordinate system, the integral becomes:
\begin{equation}
\int_0^\infty V(r) e^{i \vec{q}\cdot \vec{r}} d^3 \vec{r} =
\int_0^\infty V(r)\int_0^\pi e^{iqr cos(\alpha )} 2 \pi sin(\alpha ) \; d\alpha r^2 dr
\end{equation}
\noindent The angular part of the integral can be evaluated since $sin(\alpha)$ is the derivative of
$cos(\alpha)$:
\begin{eqnarray*}
\int_0^\infty V(r) e^{iqr cos(\alpha )} 2 \pi sin(\alpha )r^2 \;d\alpha dr & = &
\int_0^\infty V(r)2\pi ({{-e^{iqrcos(\alpha)}} \over {iqr}})|_0^\pi r^2 dr \\
& = & \int V(r)2\pi \; {{e^{iqr}-e^{-iqr}} \over {iqr}} \; r^2 dr \\
& = & {{4 \pi} \over q} \int_0^\infty V(r) r \; sin(qr) dr
\end{eqnarray*}
\noindent which is the integral you need to solve numerically.
\newpage
Note that the angle enters through $q$, since $q=(2p/\hbar)sin(\theta /2)$. When integrating
over $r$, sin(qr) can change sign and the integral can be zero. That is, there can be
angles where this integral will be zero. The cancelation in the integral is in essence
destructive interference from the particle scattering off different parts of the
target. One sees the same effect in double slit interference. As with the
double slit interference pattern, the interference occurs at smaller angles
for larger targets. This effect is seen in the comparison of the $K^+$
scattering data off $^{12}C$ versus $^{40}Ca$.
The diffraction interference occurs at around $28^\circ$ for $^{12}C$ and
around $17^\circ$ for $^{40}Ca$ for kaons with the same momentum.
If $V(r)$ is proportional to the density of the target, this integral is often refered to
as the {\bf form factor} of the interaction.
\newpage
\centerline{\bf Simpson's Rule}
\bigskip
The trapaziod rule used two values of $f(x)$ for each interval, and is exact if $f(x)$ is a line.
One can do a little bit better using equal spacing and three values of $f(x)$. We will derive
in class the following formula for the area under a parabola using values of the function
evaluated at $x-h$, $x$, and $x+h$:
\begin{equation}
AREA = h({{f(x-h)} \over 3} + {{4 f(x)} \over 3} + {{f(x+h)} \over 3} )
\end{equation}
We can do the same treatment with parabola fits that we did with trapazoids. We
divide the interval from $a$ to $b$ into $N$ equal segments. For each consequitive
triplet of segments we can use the parabola formula above. The sum over all
the segments is:
\begin{eqnarray*}
\int_a^b f(x) \; dx & \approx & h(({{f(a)} \over 3} + {{4 f(a+h)} \over 3} + {{f(a+2h)} \over 3}) + \\
& & ({{f(a+2h)} \over 3} + {{4 f(a+3h)} \over 3} + {{f(a+4h)} \over 3}) + \\
& & ({{f(a+4h)} \over 3} + {{4 f(a+5h)} \over 3} + {{f(a+6h)} \over 3}) + \cdots ) \\
\end{eqnarray*}
\noindent Notice that the even increments of $h$ are counted twice resulting in a $2/3$ factor on
even multiples of $h$:
\begin{eqnarray*}
\int_a^b f(x) \; dx & \approx & h({{f(a)} \over 3} + {{4 f(a+h)} \over 3} + {{2f(a+2h)} \over 3} + \\
& & {{4f(a+3h)} \over 3} + {{2 f(a+4h)} \over 3} + {{4f(a+5h)} \over 3} + \\
& & \cdots +{{4 f(a+(N-1)h)} \over 3} + {{2f(b)} \over 3}) \\
\end{eqnarray*}
\noindent where $h=(b-a)/N$ and $N$ must be an even number. $N$ must be even so that the function
is evaluated at an odd number of points. This is to insure that the triplet of evenly spaced
segments fits properly into the interval between $a$ and $b$.
The parabolic formula for numerical integration is called Simpson's Rule. It is exact if $f(x)$
is a parabola.
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