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\documentclass{article}
\usepackage{amsmath,amssymb}
\usepackage{hyperref}
\usepackage{graphicx}
\usepackage{todonotes}

\title{\bf{Laboratory Project Three: Laser Doppler Velocimetry}}
\author{Nicholas Malaya \\ Department of Mechanical Engineering \\
University of Texas at Austin} \date{}

\begin{document}
\maketitle
\date{}
\newpage
\section{Objectives}

\textbf{A short paragraph listing the specific objectives of this laboratory.}

I believe the purpose of this laboratory was to learn about
several of the considerations necessary to properly operate an
LDV. Unlike the previous labs, the LDV had quite a bit of operational
details required to even begin gathering data. Our group took over an
hour to first start up the laser, play around with the oscilloscope,
etc.

Beyond a familiarity with the LDV apparatus, this is also the first lab
where we have attempted to gather spatially resolved measurements. In
particular, we have characterized the free-stream turbulence (which we
expect to be statistically spatially independent) as well as the flow in
the wake of a cylinder.

Common with the previous efforts, this work was also focused on
estimating the uncertainties (and ultimately, plausibility ) of our
measurements, both through uncertainty estimates as well as comparing
with available results in the literature.

It was also my favorite lab. Felt very high tech.

\section{Background and Experimental Details}

\textbf{Describe the basic operation of an LDV.}

I see the LDV as a four step process.

\begin{itemize}
 \item Laser Crossing
 \item Particle movement through the fringes
 \item Light collected
 \item Signal postprocessing
\end{itemize}

Two laser beams cross, creating an interference pattern. Particles,
moving through this pattern reflect light in a similar oscillating
pattern, (some of) which is reflected back and collected. After signal
processing, the frequency of the signal of reflected light can be used
to calculate the quantity of interest, the particle's velocity.

\textbf{Explain what frequency
shifting is, and why it is used.}

We use frequency shifting so that we can measure flow
reversals. The frequency of light cannot be negative.
If we did not shift the frequency of light scattered off
particles, we would have instances where particles moving backward
appeared to have the same frequency as particles moving
forward. Instead, by shifting the frequency forward, we can measure
frequencies that are lower than the effective frequency shift, which
permits us to determine the sign of the velocity.

\textbf{Calculate the size of the Probe Volume}

The size of the probe volume is given by Stavros on page 267,
\begin{equation}
 V_p = \frac{\pi d_{fe}^3}{6\text{ Cos}(\theta/2)\text{
  Sin}(\theta/2)}.
\end{equation}

$d_{fe}$ is the focused beam diameter, which is expressed on the same
page as,
\begin{equation}
 d_{fe} \approx \frac{4f_T \lambda}{\pi d_e}
\end{equation}
with $f_T$ the lens focal length, $\lambda$ the laser wavelength, $d_e$
the beam diameter. The laser beam diameter was given as 1.3mm. The
wavelength of channel one was 514.5 nm, and channel two was 488 nm. The
lens focal length is 363mm for both beams. This provides an estimated
beam diameter of 0.1829 mm and 0.1735 mm.

However, we also need $\theta/2$, the half intersection angle, here. We
can calculate this using,
\begin{equation}
 \theta/2 = \text{tan}^{-1}\left(\frac{d/2}{f_T}\right)
\end{equation}
where d is the distance between beams and f is the focal length. For
both our beams, the focal length is 363 mm and the beam separation is 50
mm. This gives 0.687 radians or 3.94 degrees for the half angle. That
sounds approximately correct, the angle certainly appeared to be quite
shallow.

Our probe volume then is expected to be, $0.001397 m^3$ for probe volume
one, and $0.001325 m^3$ for probe volume two.

 \textbf{Calculate the expected frequency response of the glass sphere seed particles.}

Stavros give the characteristic time of the particles as,
\begin{equation}
 \tau_p = \frac{\rho_P d_p^2}{18 \mu_F}
\end{equation}
where $\rho_p $ and $d_p$ are the density and diameters of the
particles, respectively. $\mu_F$ is the dynamic viscosity of the fluid.
This provides a characteristic frequency, as
\begin{equation}
 \omega_c = 1/\tau_p.
\end{equation}
This is important because this represents the highest frequency of
turbulent fluctuation that the particles could accurately track in a
flow. With diameters of 12 $\mu m$ (this will give the most restrictive
frequency) and density of $1.1$ g/cc, we find a frequency response of 47
KHz.

This now naturally bears questioning: are we actually able to resolve the
velocities we are measuring? The doppler frequency is measured as,
\begin{equation}
 f_d = u_x / d_f
\end{equation}
Where $u_x = 0.42 m/s$ and $d_f$ is given as $3.75 \mu m$. Our doppler
frequency is therefore 112 kHz, which is higher than our mean. However,
the mean does not change, so the particles naturally equilibrate. A more
meaningful measure is the RMS, which is on the order of the velocity
fluctuations. This is in general much smaller than 0.42, and in the
freestream, was $\approx 0.01$, giving a doppler frequency of just 2
kHz. We can easily measure this. We expect to be able to resolve roughly
any fluctuations less than 0.2 m/s, which is more than
any fluctuations observed in our runs.

% the doppler frequency of the particle is the normal velocity over the
% fringe spacing, which was: X-direction: 3.744um; y-direction: 3.551um
% (taken from Flowsizer)

\section{Experiment One: Freestream}

\textbf{Indicate the frequency shift and bandpass filter settings used,
and why.}

The Bragg cell shifts by 40 MHz. Our downmix frequency was set to 39.85
for channel one, and 39.95 for channel two. The different settings were
due to the fact that the velocity components have different mean
velocities ($\approx 0.41$ and 0) and so produce a different shift. As a
result of this downmix, our effective shifting frequency was 150kHz (and
50 kHz for channel 2). As mentioned in the previous section, the mean
velocity shifts the doppler from the particles by as much as 112 kHz,
meaning that for both our cases, our total
frequency response was about 40-50kHz, which fits right in our
bandpass region of 10-100kHz.

We visually verified that the histograms were well centered in our
bandpass region, and that the tails were not being cut off.

% add histogram showing it is not cut off?
   % \begin{figure}[!htb]
   %  \begin{center}
   %   \includegraphics[width = 12 cm]{figs/Q_dP_fits.jpg}
   %   \caption{}
   %   \label{orif-zoom}
   %  \end{center}
   % \end{figure}

\textbf{Indicate the laser power and PMT voltage settings used, and why
these values were selected.}

The PMT for channel one was 600 V, and 500 for channel two. The laser
power was set at 1 W. We did not adjust the laser power much at all, as
we started with a reasonable count number. We did adjust the PMT down
for channel two because the number of events was much higher than the
freestream, and we were trying to get slightly more even numbers of
counts.

\textbf{Indicate the ``Burst Threshold'' and ``SNR'' settings used, and why
these values were selected.}

The burst threshold was set to 30 mV and 60 mV for channel's one and
two, respectively. We used a higher burst threshold for channel two
because the mean was typically zero here, and we were getting lower data
rates.

We set the SNR settings to ``very low'' because our burst efficiencies
were always quite high (often above 80\%). The lab guidelines mentioned
that anything over 50\% is good, so we felt there was no need to further
filter the data beyond the minimum settings.

We would often look at the oscilloscope to examine the bursts incoming.
After adjustment, we had very smooth and continuous (with reasonable
amplitude) bursts appearing. We thought that this was an indication that
the laser and signal processing was working quite well. Our filters were
much higher than the ambient noise, but seldom higher than obviously
incoming signals, so we were also happy with the settings in that
context.

These settings were used both both the freestream as well as the wake
measurements, so that we could make direct comparisons between data
gathered at each point.

%
% add chart of oscilloscope?
%
%

\textbf{Present the mean and RMS velocities measured, including
uncertainties for these measurements.}

The mean and RMS for the freestream velocity were measured as,
-0.4141 m/s and 0.00328 m/s, respectively. These were calculated by
taking the mean of 5 samples from the LDV. However, each of these
samples was taken from 2000 samples aggregated together as a
histogram. Thus, the uncertainty of these calculations
can be gleaned from our estimate of the of the variance of the mean, or
the ``sample error'', $2\sigma/\sqrt{n}$. This provides uncertainty
estimates of, .0016 m/s for the mean, e.g. $0.4141 \pm
0.0016$ which is 0.3\%. Likewise, our estimate for the uncertainty in the
streamwise RMS is 0.0002072 m/s, giving, $0.00328 m/s \pm 0.0002072$ or
about 6\%. We estimate the average and precision uncertainty for the v
mean and RMS as, $0.00344 m/s \pm 4.8e-5$ (1.4\%) and $0.00168 m/s \pm
6.7e-4$ (3.9\%).

These numbers seem reasonable to me. The fluctuating turbulence
provides an inherent variability, but we were throwing a fair number of
samples at the problem (2000) and this was resulting in very smooth
histograms. Furthermore, unlike the next results section, these
histrograms were not particularly skewed, but were evenly distributed
around the mean. The PDFs were not quite gaussian, with some visible
kurtosis, so they did exhibit some tendencies towards extreme events,
which we likely did have trouble catching with our sample
size.

Finally, as discussed in class, this data is also polluted by bias.
We can estimate the bias uncertainty as,
\begin{equation}
  \frac{U_\text{measured}}{U_\text{actual}} = 1 + \left(u_{\text{rms}}/U\right)^2
\end{equation}
At our peak turbulence intensities behind the cylinder, we find this can be as large at 1.03, or a 3\% bias error.
However, for the freestream velocities, with lower turbulence intensities, this is less than one percent.

\textbf{Comment on the turbulence intensities,
$\frac{u_{\text{rms}}}{U}$ and $\frac{v_{\text{rms}}}{U}$, do they make sense?}

$\frac{v_{\text{rms}}}{U}$ certainly does not as measured on the
laboratory equipment. I suspect that this was actually
$\frac{v_{\text{rms}}}{V}$, which was dividing essentially by a small number,
because the turbulence intensities were often shown to exceed 40\%.
$\frac{u_{\text{rms}}}{U}$ (the freestream component) appeared to be
reasonable. We observed turbulence intensities slightly below one
percent.

Calculating the turbulence intensities from our raw data, we found that
the freestream intensity was $0.00793 \pm 0.000510$, or less than one
percent in magnitude, and slightly less than six (relative to the
average ) percent uncertainty on that number. Likewise, the spanwise
velocity measurements were $0.004065 \pm 0.00015$ and an uncertainty of
closer to 4\%. I'll admit this result surprised me.
I expect the flow is roughly isotropic, since we are not near any walls.
Thus, I expected the turbulence intensities to be about the same as
well. However, they are patently not, and they are farther apart than
our uncertainty bounds, so I expect this is a real effect. Otherwise,
the turbulence intensity levels are roughly about where we would expect
them to be for free-stream turbulence.

% As for the bias stuff, I'm just looking at the equation he gave us in
% Lecture 13 where U_meas/U_actual=1+(u_rms/U)^2.  Calculating the right
% half gives values of ~1.03 for the higher turbulence intensities and
% ~1-1.01 for the lower, which seems to match up with the ranges he told
% us.

\section{Experiment Two: Wake Measurements}

\textbf{Present results from your wake measurements, including
uncertainty of the measurements. Compare your results with data in the
literature and comment on similarities and differences. }

I am comparing the wake measurements to an interesting paper by Ong and
Wallace, entitled, ``The velocity field of the turbulent very near wake
of a circular cylinder''. I like this paper because it provides
measurements from circular cylinders at Reynolds numbers approximately
near our conditions (Re $\approx 2000$ versus 3900, versus many papers
at substantially larger Re) and at values that are nominally identical
to our location, e.g. $x/d < 10$. A deficiency of this paper for our
purposes is that the measurements were conducted with a hot wire, and
not the present LDV system. Notice also that this paper was conducted in the air,
versus the water for our experiment. Of course, owing to similitude, this is not
important, aside from the difference in Re. It is worth noting to the reader, at least.

We measured the diameter of the cylinder with a set of calipers.
for a $d=0.374$ inches we desire to measure 5d downstream, or $1.87$
inches. We then moved another 1.87 inches and repeated out measurements
(10d). However, we had to convert from inches to mm for the arm
readings. This I wrote down as, $d=0.374$ inches or $9.5$ mm, so $5d =
47.5 mm$. The arm moves 1/10mm, so 475 was the reading placed into the
box to move the laser. We then checked both times the distance with the
micrometer, and it appeared to be correct.

In order to investigate the histeresis and repeatability of our
measurements, we started at zero, moved down two diameter widths, then
back up through to positive two diameters, and then back down to
zero. We ``snaked'' through. In this way, we made a clean sweep through
two full sets of data, while simultaneously approaching from a different
direction for each set.

\begin{figure}[!htb]
 \begin{center}
  \includegraphics[width = 12 cm]{figs/defect}
  \caption{The streamwise velocity normalize by freestream velocity. We
  show both sets for a discussion of repeatability.}
  \label{repeat}
 \end{center}
\end{figure}

The streamwise velocity is shown in figure \ref{repeat}. Both repeated
experiments are shown. The good news is that the data is relatively
repeatable, with each point somewhat close to the other set.
Immediately one sees a major issue with the data,
namely, that several anomalous data points exist. For instance, in the
first data set, the point slightly less than zero at $y/d=0.25$ is
actually lower than the point directly behind the cylinder. Was this
just a very bad data point, or did some other more systematic error
appear? It is not clear, but it gives the appearance of skewness to the
entire output. Another issue is that we clearly did not take enough
points past $y/d=2$. From our literature search we had believed that
this was far enough to go to be back into the freestream, and actually
the data appears to support this, but we should have taken more points
so as to ensure that this is indeed truly the case, and that the points
beyond $y/d=2$ do continue to flatten out.

So far, this data compares reasonably well with the literature. The
bottom of our peak (at the centerline) corresponds to a value of
0.75. The paper by Ong et al. has a minimum at 5d of 0.73, which is at
least close, if not completely identical. Some of this difference might
be attributable the different reynolds numbers between the cases.

\begin{figure}[!htb]
 \begin{center}
  \includegraphics[width = 12 cm]{figs/backflow}
  \caption{The spanwise velocity at the centerline is
  significantly bimodal. }
  \label{back}
 \end{center}
\end{figure}

Notice that near the centerline we found significantly bimodal
distributions of velocity in the spanwise component of velocity. This is
shown in figure \ref{back}. This is as expected, because the spanwise
velocity is either going above or below the cylinder, and that tends to
favorably impart a spanwise velocity as it does. While at the centerline
these components were competing, above and below the centerline this
introduced considerable skewness in the observed histograms of spanwise
velocity.

\begin{figure}[!htb]
 \begin{center}
  \includegraphics[width = 12 cm]{figs/defect10}
  \caption{A comparison between the five and ten cylinder diameters
  streamwise velocity profiles. Notice that both series of the ten diameter runs are plotted, to
indicate the level of repeatibility. }
  \label{back}
 \end{center}
\end{figure}

An comparison between the mean velocity for 5 and 10 cylinder diameters
is shown in figure \ref{back}. It is very difficult to tell if this
is in agreement with the literature. What appears clear is that the data
for the streamwise velocities are close to each other even at a spacing
of 5 diameters. Our data appears consistent with the literature, in that
the wake region is broader at ten diameters. I looked at Pope to try to
quantitatively estimate how much it should grow, as if it was a
freestream jet, and the radius is the jet half-width. However, I
couldn't munge this into anything particularly interesting. In short,
while the fact these profiles are roughly equivalent surprised me, I
cannot find any evidence they should be exhibiting greater differences.

\begin{figure}[!htb]
 \begin{center}
  \includegraphics[width = 12 cm]{figs/urms}
  \caption{The five and ten diameter distances $U_{rms}$ profiles. }
  \label{urms}
 \end{center}
\end{figure}

In figure \ref{urms} we plot the five and ten diameter $U_{rms}$
profiles. First, the good news: the magnitudes of our measurements were
largely consistent with the values in the literature, with a sharp peak
existing below 0.6. Our five diameter measurements show, very modestly,
the characteristic double peak off center of the cylinder, which is also
consistent with our expectations. However, our data is noisy and did not
well resolve the peaks. We also did not take enough points around $|y/d|
=1$ to really resolve this peak. Furthermore, our colleagues paper
clearly shows a similar (although smaller) double peak at the
centerline, which we appear to have not captured at all. At this point I
am concerned that we did not measure the $y/d$ distance correctly, but
we really were very careful and did have an independent check (Nick did
it and it was checked by Owen).

\begin{figure}[!htb]
 \begin{center}
  \includegraphics[width = 12 cm]{figs/vrms}
  \caption{The five and ten diameter distances $V_{rms}$ profiles. Notice that both sets of data are plotted, to give an impression of the repeatability. }
  \label{vrms}
 \end{center}
\end{figure}

In figure \ref{vrms} we plot the five and ten diameter $V_{rms}$ values.
This figure actually is in very good agreement with the experimental paper!
The peak is clustered around 0, and trails off afterward. The 5d
peak is significantly higher than the 10d value, and the peaks largely
correspond to the values we would expect: in the paper they observed a
peak of 0.075 for 10d and 18.5 for 10d. Our 10d values closely match, with our
peak at roughly 0.8, but our 10d has a slightly larget discrepancy, rising above 0.2.
Nevertheless, the shape and structure of these data are largely consistent with the
literature search we performed. However, as plotted in figure \ref{vrms},
there is some non-trivial scatter between each series of experiments. While the experiments
are broadly repeatible, the scatter in the five diameter series was certainly appreciable, and is
larger than one would expect given the turbulence intensities.

%
% time to add mean V vel
%

\begin{figure}[!htb]
 \begin{center}
  \includegraphics[width = 12 cm]{figs/vmean}
  \caption{The five and ten diameter distances $\bar V$ profiles. }
  \label{vmean}
 \end{center}
\end{figure}

In figure \ref{vmean} we plot the five and ten diameter $\bar V$ values. This is clearly a disaster.
Essentially the only two points in favor of this plot are that the sign of the mean velocity is in the
correct direction, and that the magnitude is not patently absurd. We expect a positive value for
negative $y/d$ values, followed by an anti-symmetric reflection across the centerline. The peak values in the paper
are at a fraction of 0.1, but beyond that is is hard to tell and the raw data is not available. I do not
believe our values are off by a factor of 2, but that is as far as I am prepared to go on this plot. Furthermore, the data
obviously is noisy, with large fluctuations between points that are not expected to be physical. My guess here is that
becaues the low values of the velocity here, out data is contaminated by noise. It is also possible this occurs in the
paper, but the scale is so small that they have cleverly hidden the fluctuations we are seeing in our results. Either way,
these results are bad enough I consider them unreliable. Perhaps the only other arguement in favor of this plot is tha
the $y/d$ value at 5 diameters is higher than the data from 10, as in the paper we have for reference.

The uncertainty was bounded in estimation using either the repeated
measurements at the center ($y=0$) where the turbulence was at a peak,
versus the freestream uncertainty estimates provided previously. We
expect that the freestream is more accurate away from the cylinder, but
an underestimate of the variability as you approach the centerline. Vice
versa, the centerline error would be more accurate near the center, but
overestimating the uncertainty near the freestream. The estimate for the
uncetainty at the freestream was presented previously, for our velocities at the centerline we
use the same methodology and arrive at the following values.
The streamwise velocity mean is $-0.3109 \pm 0.002259$ (0.8\%). The streamwise RMS is:
$0.0536 \pm 0.000889$ (1.9\%). The Spanwise velocity ($\bar V$) is $0.001333 \pm 0.00036$,
or an astounding (27\%). This is the highly qustionable data. Finally, the V RMS is $0.1239 \pm 0.00388$ or 3.1\%.
Most of these numbers are quite sane, aside from the average of V, which remains anomalously highly.

What is also interesting is that while the values of the uncertainty are larger than in the freestream, they are seldom
so substantially larger than they are over-riding any conclusions that one could make. While the error bars are
larger in this region, they are actually smaller than I expected for a strongly turbulent region behind a cylinder.

\section{Conclusions}

\textbf{Conclude by giving your opinion about whether your measurements
are more or less reliable than measurements in the literature.}

I absolutely do not believe our measurements are more accurate than the
rival paper. I'll admit I am somewhat disappointed in the results
presented here, but in particular given the shortcomings presented in
the previous section with respect to several anomalous points, I would
want to perform additional measurements.

In theory, however, I do believe that an LDV could be accurate. Some of
the major disadvantages of LDV are the cost and the problems associated
with near wall features. However, neither of those impact the
measurements here. While we needed to perform more measurements and
possibly more laser and sensor adjustments, the LDV I believe in
principle can be more accurate than it appears here. Another
advantage of the LDV is that it can measure backflow which as mentioned
previously appeared in the spanwise component of velocity.

%
%
%
\end{document}

% LocalWords:  reynolds H20 piecewise RDG yar


% !TeX program = lualatex
\documentclass[]{article}
\usepackage{caption,subcaption,graphicx,float,url,amsmath,amssymb,amsthm,tocloft,cancel,thmtools,gensymb,braket,tikz-feynman}
\usepackage[toc,nonumberlist]{glossaries}
\usepackage{glossaries-extra}
\newcommand\numberthis{\addtocounter{equation}{1}\tag{\theequation}}

\newtheorem{thm}{Theorem}
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%opening
\title{Theoretical Minimum\\Particle Physics 1\\Basic Concepts}
\author{Simon Crase (compiler)\\simon@greenweaves.nz}

\begin{document}

\maketitle

\begin{abstract}
These are my notes for the \emph{New Revolutions in Particle Physics 1} lectures\cite{susskind2009particles} from Leonard Susskind's \emph{Theoretical Minimum} series\cite{susskind2007theoretical}.

\begin{quotation}
	Revolutionary new concepts about elementary particles, space and time, and the structure of matter began to emerge in the mid-1970s. Theory got far ahead of experiment with radical new ideas such as grand unification and supersymmetry, but the concepts have never been experimentally tested. Now all that is about to change. The Large Hadron Collider, or LHC, has finally been built and is about to confront theory with experiment. This course is devoted to these theoretical ideas and how they will be tested.
\end{quotation}

Disclaimer: I have created these notes as an aide-m\'emoire for my own use; if you find them useful, you are welcome, but I'd appreciate hearing from you. They are not intended
as a substitute for listening to the lectures. The intellectual property for all material derived from the lectures belongs, of course, to Professor Susskind; any mistakes, however, are my own.

The notes were created using TexStudio\cite{TexStudio}, which I recommend for compiling them to a PDF, and the bibliography was created using JabRef\cite{Jabref}.
\end{abstract}

\tableofcontents
\listoffigures
\listoftables
\listoftheorems

\section{Particles and Light}

This lecture contains  facts from other courses, which will be used in later lectures.

Particle physics is about the question: is matter discrete? If so we will call the smallest things "particles". If matter forms a continuum, we have fields.

Which is correct? Both and neither.

The first evidence for atoms came from chemistry. John Dalton found that the mass of one mole of each substance was an integer multiple of the mass of a mole of hydrogen; this suggested that there are building blocks. We now know that mass of molecule is mass of protons, electrons (very small), and neutrons, which have nearly the same mass as protons.

Figure \ref{fig:em:wave} illustrates an electromagnetic wave. We need the concepts of wavelength, $\lambda$, and period, $T$.

We have
\begin{align*}
\frac{\lambda}{T}=&c\\
f=&\frac{1}{T} \text{, so}\\
\lambda f =& c\text{. Physicists tend to measure frequency in radians per second:}\\
\omega =& 2 \pi f \text{, so} \\
\omega =& \frac{2 \pi c}{\lambda} \numberthis \label{eq:omega:lambda}
\end{align*}

\begin{figure}[H]
	\caption{An electromagnetic wave}\label{fig:em:wave}
	\includegraphics[width=0.9\textwidth]{Wavelength}
\end{figure}

Wave/particle duality.

A photon has energy:
\begin{align*}
E=&\hslash \omega \text{. Energy of single photon.} \numberthis\label{eq:E_omega}\\
E_{ray}=&n \hslash \omega \text{. Energy of ray.}
\end{align*}

In modern usage, mass is what used to be called rest mass. $E = m c^2$ only for stationary object.

In particle physics we choose units such that $c=1$ and $\hslash=1$. In this course, $c$ and $\hslash$ will appear explicitly only when Prof. Susskind wants to show the magnitude of some quantity.

Energy depends on the motion of the object as seen by the observer: it isn't a universal thing that everyone will agree on.

Light has energy and momentum (not mass):

\begin{align*}
\left|P\right|=&\frac{E}{c} \\
=&\frac{\hslash \omega}{c} \text{, from (\ref{eq:E_omega})}\\
=& \frac{2 \pi \hslash}{\lambda} \text{, from (\ref{eq:omega:lambda}). c.f. Harmonic Oscillator.}\numberthis\label{eq:P:lambda}
\end{align*}

This is why we need larger and larger instruments to observe smaller particles!
To see really small things, we need particles with high energy and momentum.

\section{Quantum field theory}

\subsection{Mathematical Preliminaries}

Consider a classical wave $A \sin(kx)$. Energy is proportional to $A^2$, but also to the number of photons, $n$: $E \propto n \hslash \omega$, $A \propto \sqrt(n)$.

Generally, equations containing $c$ apply for photons only. For non-relativistic particles:
\begin{align*}
E=&\frac{p^2}{2m}\\
f =& \frac{h}{2 m \lambda^2} \text{. c.f. Schr\"odinger equation!}
\end{align*}

\begin{itemize}
	\item Phase Velocity: velocity of wave packet;
	\item Group velocity: velocity of troughs and peaks--identified with particle velocity.
\end{itemize}

For light waves, phase and group velocities the same, but this isn't true for most particles. Phase velocity can exceed $c$, but it can't transmit information--Section \ref{section:phase:group:velocities}.

Space is infinite. If physics is the art of getting problems into a shape that computers can solve, we need to remove infinities.

Let's look at 1 dimensional waves. We can restrict to fixed length $L$: reflecting boundary condition at ends. But this violates conservation of momentum. Instead use a topological "circle" circumference $L$--periodic boundary conditions. As a consequence momentum is quantized:

\begin{align*}
\lambda =& \frac{L}{N}\\
P =& \frac{h}{\lambda}\\
=& \frac{h N}{L}
\end{align*}

Now make into a real circle, radius $R$, and redefine $L$ to be \emph{angular momentum}.

\begin{align*}
P =& \frac{h N}{2 \pi R}\\
L =& N \frac{h}{2 \pi}\\
=& N \hslash \text{ independent of $R$.}
\end{align*}

\subsection{Quantum field theory}

Start with harmonic oscillator: a wave is a collection of harmonic oscillators. Energy is quantized: 0, $\hslash \omega$, $2\hslash \omega$...

Introduce operators:

\begin{align*}
a^+ \ket{n} =& \sqrt{n+1} \ket{n+1} \text{, creation operator} \numberthis \label{eq:creation}\\
a^- \ket{n} =& \sqrt{n} \ket{n-1} \text{, annihilation operator} \numberthis \label{eq:annihilation} \\
a^+ a^- \ket{n} =& n \ket{n} \text{, or}\\
a^+ a^-  =& n \text{. Similarly}\\
a^- a^+  =& n+1\\
[a^-, a^+] =& 1
\end{align*}

Return to world on a circle--$\omega_N$--equivalent to oscillator. State is $\ket{n_1,n_2, n_3,...}$--occupation numbers.
\begin{defn}[quantum field]
	A quantum field is a collection of harmonic oscillators, together with annihilation operators and creation operators.
\end{defn}

\section{Quantum fields and particles}

\begin{align*}
\text{Wave} =& e^{i k x} \text{, $k$ is wave number}\\
P=&\hslash k\\
n(k) =& \text{ occupation number.}
\end{align*}

We represent the state of a 1D circular system by occupation numbers, $\ket{...n(-1), n(0), n(1), n(2)...}$, and introduce operators $a^+(k)$, $a^-(k)$.
\begin{align*}
a^+(1)\ket{...n(-1), n(0), \underbrace{n(1)}_\text{target of $a^+(1)$}, n(2)...}=& \sqrt{n(1)+1}\ket{...n(-1), n(0), n(1)+1, n(2)...}
\end{align*}
$a^+(k)$, $a^-(k)$ are quantum mechanical versions of the Fourier coefficients of a field $\Psi$.

Here is a classical wave:
\begin{align*}
\Psi(x) =& \sum_k \alpha(k) e^{ikx}\\
\Psi^*(x) =& \sum_k \alpha^*(k) e^{-ikx}
\end{align*}

We quantize as shown in (\ref{eq:q:1}) and (\ref{eq:q:2}) to produce a quantum field:
\begin{align*}
\alpha(k) \rightarrow& a^-(k) \numberthis \label{eq:q:1}\\
\alpha^*(k) \rightarrow& a^+(k) \numberthis \label{eq:q:2}\\
\Psi(x) =& \sum_k a^-(k) e^{ikx} \numberthis \label{eq:Psi}\\
\Psi^\dagger(x) =& \sum_k a^+(k) e^{-ikx} \numberthis \label{eq:Psi:dagger}
\end{align*}

We need to justify (\ref{eq:q:1}) and (\ref{eq:q:2}) in some appropriate limit.

How can we describe scattering? Imagine scattering so a particle with momentum $k_7$ becomes a particle with momentum $k_9$: $a^+(k_9)a^-(k_7)\ket{0,0,..1,0,0}$.

What if laws of physics allows number of particles to change? $a^+(k_{16})a^+(k_9)a^-(k_7)\ket{0,0,..1,0,0}$. We can't do this in regular quantum mechanics.

What does $\Psi^\dagger$ do? Start with the vacuum $\ket{0}$.

\begin{align*}
\Psi^\dagger(x)\ket{0} =& \sum_k a^+(k) e^{-ikx} \ket{0} \text{ from (\ref{eq:Psi:dagger})}\\
=& \sum_k e^{-ikx} \underbrace{a^+(k) \ket{0}}_\text{One particle state with momentum $k$} \\
=& \sum_k e^{-ikx} \ket{k} \text{. Superposition gives particle at definite position.}
\end{align*}

If we make the circle larger and larger, $\sum \rightarrow \int$.

What does $\Psi$ do? It annihilates a particle at $x$.

Locality: when something happens, it happens at one spot. Figure \ref{fig:ex:locality} is an example of one particle being replaced by two.

\begin{figure}[H]
	\caption[Example of locality: one particle is replaced by two]{Example of locality: one particle is replaced by two. $\Psi^\dagger(x)\Psi^\dagger(x)\Psi(x)\ket{...}$}\label{fig:ex:locality}
	\includegraphics[width=0.8\textwidth]{split-particle}
\end{figure}

Start with state $\ket{0,...\underbrace{1}_\text{$k_i$},0,000}$.

We assume that $k_i$ denotes the index of the only momentum with a non-zero index.
\begin{align*}
\Psi^\dagger(x)\Psi^\dagger(x)\Psi(x)\ket{0,...1,0,000}=&\sum_m a^+(m) e^{-imx} \sum_l a^+(l) e^{-ilx} \sum_k a^-(k) e^{ikx} \ket{...}\\
=&\sum_m a^+(m) e^{-imx} \sum_l a^+(l) e^{-ilx} a^-(k_i) e^{ik_ix} \ket{...}\\
=&\sum_m a^+(m) e^{-imx} \sum_l a^+(l) e^{-ilx}  e^{ik_ix} \ket{0}\\
=&\sum_{l,m}  e^{-imx}  e^{-ilx}  e^{ik_ix} \underbrace{\ket{001001...}}_\text{$l$ and $m$ are indices of $1s$.}\\
\triangleq &\sum_{l,m}   e^{i(k-l-m)x} \ket{l,m}
\end{align*}

This isn't true if $l=m$! We have two application of $a^+(m)$, so vacuum becomes $\sqrt{2}\ket{...2...}$. Probability of two particles coming out in same state is twice probability of them being in different states.

For the time being we ignore conservation laws!

Consider generating photon at a position where there is a pre-existing photon with momentum $l$--Figure \ref{fig:simple:decay}.
\begin{figure}[H]
	\caption[Generating photon  where there is a pre-existing photon]{Generating photon at a position where there is a pre-existing photon with momentum $l$.}\label{fig:simple:decay}
	\includegraphics[width=0.8\textwidth]{decay}
\end{figure}

\begin{align*}
\Psi(x)^\dagger\ket{l} =& \sum_k a^+(k) e^{-ikx} \ket{l}\\
=& \sum_{k\ne l} e^{-ikx} \ket{k,l} + \sqrt{2} e^{-ilx} \ket{l,l}
\end{align*}

Stimulated emission: higher probability of generating particles that are already there (c.f. spontaneous emission). This is true for Bosons, but not Fermions.

 Laws for Bosons
 \begin{align*}
 [a^+(k),a^+(l)] =& 0\\
 [a^-(k),a^-(l)] =& 0\\
 [a^-(k),a^+(l)] =& \delta_{k,l}\\
 \Psi(x) =& \sum_k a^-(k) e^{ikx} \\
 \Psi^\dagger(x) =& \sum_k a^+(k) e^{-ikx}
 \end{align*}

 \begin{align*}
 \frac{1}{L}\int_\frac{-L}{2}^\frac{L}{2} dx \Psi^\dagger(x) \Psi(x) =& \frac{1}{L} \int_\frac{-L}{2}^\frac{L}{2} dx \sum_k a^+(k) e^{-ikx} \sum_l a^-(l) e^{ilx}\\
 =& \sum_k a^+(k) \sum_l a^-(l) \frac{1}{L} \int_\frac{-L}{2}^\frac{L}{2} e^{-ikx} e^{ilx} dx\\
 =& \sum_k a^+(k) \sum_l a^-(l) \frac{1}{L} \int_\frac{-L}{2}^\frac{L}{2} e^{-i(k-l)x} dx\\
  =& \sum_k a^+(k) \sum_l a^-(l) \delta_{k,l}\\
 =&  \sum_k \underbrace{a^+(k) a^-(k)}_\text{Occupation number}\\
 =& N \text{, total number of particles.}\\
 \Psi^\dagger(x) \Psi(x) =& \text{ density of particles at $x$}.
 \end{align*}

\section{More quantum field theory}

\subsection{Mathematical Preliminaries}

\subsubsection{Dirac Delta function}

\begin{align*}
\text{For } k =& \frac{2 \pi n}{L}\\
\int_{-\frac{L}{2}}^{\frac{L}{2}} e^{ikx} dx =& L \text{, if $k =0$}\\
=& 0 \text{, otherwise}\\
\rightarrow& 2 \pi \delta(k) \text{ as $k \rightarrow \infty$}
\end{align*}

We will use \emph{ket} vectors for initial states, \emph{bra} for final. This will help with bookkeeping.

\subsubsection{Creation and annihilation operators operating on bra vectors}

\begin{align*}
\braket{n|m} =& \delta_{n,m}\numberthis \label{eq:orthogonal} \text{. We want the following}\\
\braket{n|(a^+|m)} =& \braket{(n|a^+)|m}  \numberthis \label{eq:consistency}\\
=& \sqrt{m+1} \braket{n|m+1} \text{, from (\ref{eq:creation})}\\
=& \sqrt{m+1} \delta_{n,m+1} \text{, from (\ref{eq:orthogonal})} \text{, so we define}\\
\bra{n} a^+ \triangleq & \sqrt{n} \bra{n-1} \text{ to satisfy (\ref{eq:consistency}), and} \numberthis \label{eq:bra:create}\\
\bra{n} a^= \triangleq & \sqrt{n+1} \bra{n-1} \numberthis \label{eq:bra:annihilate}
\end{align*}
Example: calculate an expression two different ways.
\begin{align*}
\braket{n|(a^+a^-|n)} =& \sqrt{n}\braket{n|(a^+|n-1)}\\
=& (\sqrt{n})^2 \braket{n|n} \text{ from (\ref{eq:bra:create})}\\
=& n\\
\braket{(n|a^+a^-)|n} =& \sqrt{n} \braket{(n-1|a^-)|n}\\
=& n \braket{n|n} \text{ from (\ref{eq:bra:annihilate})}
\end{align*}

\subsection{Quantum Fields}

The only way we can study particles is to collide them; quantum field theory is the mathematical tool.

We will set $\hslash=1$, so $P=k$, and we will take $X$ and $P$ to be 3 dimensional.
\begin{align*}
\Psi^\dagger(X)(X,t) \triangleq& \sum_k a^+(k) e^{-ikX} e^{i \omega(k) t} \\
\Psi(X)(X,t) \triangleq& \sum_k a^-(k) e^{ikX} e^{- i \omega(k) t}
\end{align*}

\begin{align*}
\omega =& E \text{ and}\\
E =& \frac{P^2}{2m} \text{, whence}\\
\omega =& \frac{k^2}{2m} \numberthis \label{eq:omega:k}
\end{align*}
We  determine the differential equation satisfied by $\Psi^\dagger(X)$.

\begin{align*}
\dot{\Psi^\dagger(X)} =&  \sum_k i \omega(k) a^+(k) e^{-ikX} e^{i \omega(k) t}\\
\frac{\partial^2 \Psi^\dagger(X)}{\partial X^2} =& \sum_k -k^2 a^+(k) e^{-ikX} e^{i \omega(k) t} \numberthis \label{eq:omega:dot}\\
 =& - \sum_k 2m \omega a^+(k) e^{-ikX} e^{i \omega(k) t} \text{ from (\ref{eq:omega:k})}\\
 \frac{1}{2m} \frac{\partial^2 \Psi^\dagger(X)}{\partial X^2} =& - \sum_k \omega a^+(k) e^{-ikX} e^{i \omega(k) t}\\
 =& -\frac{1}{i} \frac{\Psi^\dagger(X)}{\partial t}\\
  i \frac{\Psi^\dagger(X)}{\partial t} =&  \frac{1}{2m} \frac{\partial^2 \Psi^\dagger(X)}{\partial X^2} \text{ From (\ref{eq:omega:dot}). This is Schr\"odinger's equation for field!}
\end{align*}

Consider particle scattering from heavy target; we show it conserves energy, but not momentum--Figure \ref{fig:particle:heavy:target}. Assume scattering probability independent of time.

\begin{figure}[H]
	\caption{Particle being absorbed and then re-emitted}\label{fig:particle:heavy:target}
	\includegraphics[width=0.8\textwidth]{particle-heavy-target}
\end{figure}

We want the probability of emitting one particle in final state (Introduce coupling constant $g$)--$\big| \bra{k_f} g \int dt \Psi^\dagger(0,t)\Psi(0,t)\ket{k_i} \big|^2$

\begin{align*}
&\bra{k_f} g \int dt \sum_l \underbrace{a^+(l) e^{i \omega_l t}}_\text{contributes iff $l==k_f$} \sum_k \underbrace{a^-(k) e^{-i \omega_k t} \ket{k_i}}_\text{contributes iff $k==k_i$}\\
=&\bra{k_f} g \int dt a^+(k_f) e^{i \omega_{k_f} t}  a^-(k_i) e^{-i \omega_{k_i} t} \ket{k_i}\\
=& g \int dt e^{i(\omega_{k_f}-\omega_{k_i})}\cancel{<0|0>}\\
=& 2 \pi g \delta(\omega_{k_f}-\omega_{k_i}) \text{--conservation of energy.} \numberthis \label{eq:conservation:energy}
\end{align*}

Conservation of energy works only because of time symmetry!

Probability $\propto g^2$.

Prof. Susskind has shown:
\begin{itemize}
	\item Definition of coupling constant;
	\item Integration over time is the thing that ensures energy conservation;
	\item Scattering amplitude.
\end{itemize}

NB.\begin{itemize}
	\item  If X not zero (say $X_s$), then amplitude gets a phase of $e^{i(k_f-k_i)S_s}$, which doesn't affect probability.
	\item if $g$ is a function of time, it generally means that we are ignoring something.
\end{itemize}

To reiterate: $\Psi$ and $\Psi^\dagger$ are the tools that let us discuss particle interactions.

What if we are dealing with an electron? Each particle has its own $\Psi$. What if one electron goes in but two come out-- $\Psi^\dagger(0,t)\Psi^\dagger(0,t)\Psi(0,t)$? Or two in, two out--$\Psi^\dagger(0,t)\Psi^\dagger(0,t)\Psi(0,t)\Psi(0,t)$? Which are allowed? One rule is that there must be the same number of $\Psi^\dagger$s and $\Psi$s. Imagine multiplying by a phase:

\begin{align*}
\Psi \rightarrow & e^{i\alpha} \Psi \numberthis \label{eq:conservation:charge1}\\
\Psi^\dagger \rightarrow & e^{-i\alpha} \Psi^\dagger \numberthis \label{eq:conservation:charge2}
\end{align*}

Allowable combinations are invariant under multiplication by $e^{i\alpha}$--charge is conserved. Table \ref{table:conserve} shows some pairs of symmetries and conservation laws.

\begin{table}[H]
	\caption{Invariance and Conservation Laws}\label{table:conserve}
	\begin{center}
			\begin{tabular}{|l|l|l|}  \hline
				Invariance&Conservation&Ref\\ \hline
				Time& Energy&(\ref{eq:conservation:energy})\\  \hline
				Space&Charge&(\ref{eq:conservation:charge1}),  (\ref{eq:conservation:charge2})\\ \hline
				Phase&Momentum&Section \ref{sec:momentum:conservation}\\ \hline
			\end{tabular}
	\end{center}
\end{table}

NB, this isn't a real electron, as $\Psi$ is for Bosons. A negative pion would be OK.

\section{Energy conservation and waves}\label{sec:energy:conservation}

\subsection{Phase Velocity and Group Velocity}\label{section:phase:group:velocities}

\subsubsection{Phase Velocity and Group Velocity of non-relativistic particles}
Generally phase velocity doesn't have anything to do with measurable things in quantum mechanics. Group velocity is what counts.

Consider $\sin(kx-\omega t)$. The wave moves so that $kx-\omega t$ remains constant.
\begin{align*}
x =& \underbrace{\frac{\omega}{k}}_\text{phase velocity} t\\
\omega =& \frac{2 \pi}{T}\\
k =& \frac{2 \pi}{\lambda}
\end{align*}

This is connected with Schr\"odinger equation, since $\omega$ is energy, and $k$ momentum, so:
\begin{align*}
\omega=&\frac{k^2}{2m} \text{, whence}\\
\frac{\omega}{k}=&\frac{k}{2m} \text{, phase velocity.}\\
=& \frac{v}{2}  \text{, where $v$ is velocity of classical particle. Let's add a constant energy:}\\
\omega=&\frac{k^2}{2m}+c \text{, changes phase velocity--but changing $E$ by an additive constant has no physical effect!} \numberthis \label{fig:phase:c}
\end{align*}
Now we write:
\begin{align*}
	\Psi =& \sum_k a^-(k) e^{ikx}e^{- i \omega(k) t} e^{-i c t}
\end{align*}
Given solution to Schr\"odinger equation, adding constant to phase just multiplies $\Psi$ by $e^{-i c t}$. Now all physical quantities depend on $\Psi\Psi^\dagger=\cancel{e^{-i c t}}\Psi \cancel{e^{+i c t}}\Psi^\dagger$: the phase velocity isn't physical.

We will try to follow where the constructive interference takes place between two waves.

\begin{align*}
	\sin(kx-\omega(k)t) +& \sin(k^\prime x-\omega(k^\prime)t)\text{ will reinforece each other when}\\
	kx-\omega(k)t=&k^\prime x-\omega(k^\prime)t\\
	x =& \frac{\omega-\omega^\prime}{k-k^\prime}t\\
	\rightarrow & \big[\underbrace{\frac{d \omega}{dk}}_\text{group velocity}\big] t \text{, as $k-k^\prime \rightarrow 0$}\\
	\frac{d \omega}{dk}=&\frac{k}{m} \text{, from (\ref{fig:phase:c})}
\end{align*}

\begin{itemize}
	\item Group velocity does not depend on $c$
	\item Group velocity matches velocity of relativistic particle.
\end{itemize}

\subsubsection{Phase Velocity and Group Velocity of relativistic particles}

\begin{align*}
E =& \sqrt{P^2 + m^2}\\
\omega =& \sqrt{k^2 + m^2}\\
\omega =& \left| k \right| \text{ at speed of light(m=0), so}\\
V_p =& \frac{\omega}{k} \text{ phase velocity of photon waves}\\
 =&1\\
 V_g =& \frac{d\omega}{dk} \text{ group velocity of photon waves}\\
 =&1 \text{. Now lets us make $m>0$}\\
 V_p =& \frac{\omega}{k}\\
 =&\sqrt{\frac{k^2+m^2}{k^2}}\\
 =&\sqrt{1+\frac{m^2}{k^2}}\\
 >&1 \\
 V_g =& \frac{d\omega}{dk} \\
 =& \frac{\cancel{2}k}{\cancel{2}\sqrt{k^2+m^2}}\\
 =& \frac{1}{\sqrt{1+\frac{m^2}{k^2}}}\\
 <&1\\
 V_p V_g =& 1
\end{align*}
Difference between group and phase velocities is connected to dispersion.

\subsection{Momentum Conservation}\label{sec:momentum:conservation}

Consider a bunch of operators action on an initial state to produce a final state. If Figure \ref{fig:particle:heavy:target} we derived conservation of energy. Figure \ref{fig:split:particle} show conservation of momentum for splitting a particle.

\begin{figure}[H]
	\caption{Split Particle at any position $x$}\label{fig:split:particle}
	\includegraphics[width=0.8\textwidth]{split-particle1}
\end{figure}

\begin{align*}
\bra{k_1^\prime,k_2^\prime}&\int \Psi^\dagger(x)\Psi^\dagger(x)\Psi(x) dx\ket{k}\\
=&\bra{k_1^\prime,k_2^\prime}\int a^+(k_1^\prime) a^+(k_2^\prime) a^-(k) \underbrace{e^{ikx} e^{-i(k_1^\prime+k_2^\prime)x}}_\text{integrates to Dirac-$\delta$--(\ref{eq:conservation:energy})} dx\ket{k}
\end{align*}

So invariance under space translation yields conservation of momentum.

Another possible symmetry: $\Psi \rightarrow e^{i \lambda} \Psi$: no change to probabilities if number of $\Psi$s is same as number of $\Psi^\dagger$s. This is connected with conservation of charge.

\subsection{Fermions}

\subsubsection{Introduction to Fermions}

\begin{itemize}
	\item Can have any number of Bosons in same state (same place);
	\item They have a tendency to be in same state. E.g. particles decays and emits photon. If lots of other photons already on some state, decay will preferentially favour that state.
\end{itemize}

Fermions cannot be in same state (Pauli Exclusion Principle).

\begin{align*}
\Psi(x) =& \sum_k a^-(k) e^{ikx}\\
State:& \ket{001...01010...} \text{, ones and zeroes only!}\\
\Psi^\dagger(x) =& \sum_k a^+(k) e^{-ikx}
\end{align*}

Focus on one state, either full $(\ket{1})$ or empty $(\ket{0})$: those are only two possibilities. What are the algebraic rules for creation and annihilation operators?

\begin{align*}
c^+\ket{0} =& \ket{1} \text{ for Fermions use $c$ rather than $a$}\\
c^+\ket{1} =& 0 \text{, i.e. $\psi(x) \equiv 0$}\\
c^-\ket{1} =& \ket{0}\\
c^-\ket{0} =& 0\\
\end{align*}

We deduce:
\begin{align*}
c^+c^-\ket{0} =& 0\\
c^-c^+\ket{0} =& \ket{0}\\
(c^+c^-+c^-c^+)\ket{0} =& \ket{0}\\
c^+c^-\ket{1} =& \ket{1} \\
c^-c^+\ket{1} =&0 \\
(c^+c^-+c^-c^+)\ket{1} =& \ket{1}\\
\{c^+,c^-\} =& 1 \text{, where anticommutator is defined}\\
\{c^+,c^-\} \triangleq(c^+c^-+c^-c^+)& \text{. We can also show}\\
\{c^+,c^+\} =&0 \text{ and}\\
\{c^-,c^-\} =&0
\end{align*}

So Fermions have parallel mathematics: $[,] \leftrightarrow \{,\}$. Cannot put two particles into same state, as $c^+c^+=0$.

What about different momenta? We can show that:
\begin{align*}
\{c^+_k,c^-_{k^\prime} \} =& \delta_{k,k^\prime} \\
\{c^+_k,c^+_{k^\prime} \} =&0 \\
\{c^-_k,c^-_{k^\prime} \} =&0
\end{align*}

Can we have two Fermions at the same position? Let's try to create them at the origin.

\begin{align*}
\Psi^\dagger(0) \Psi^\dagger(0) \ket{0} =& \sum_{k,k^\prime} c^+_k c^+_{k^\prime} \ket{0}\\
=& 0
\end{align*}

So the answer is no: we can't have two Fermions at the same position. So we can't have an electron laser.

\subsubsection{Ground State}
We now compare the ground state of a system of Bosons with the ground state of a system of Fermions. In Figure \ref{fig:electons:confined:momentum} we have some particles confined in momentum space.

\begin{figure}[H]
	\caption{Particles confined in momentum space (3D)}\label{fig:electons:confined:momentum}
	\includegraphics[width=0.8\textwidth]{electons-confined-momentum}
\end{figure}

Allowable values form a lattice.
\begin{itemize}
	\item We can put lots and lots of Bosons in the lowest energy (zero momentum), giving a Bose condensate. Complete uncertainty of position.
	\item Have to assign different momentum to electrons--Figure \ref{fig:adding:more:fermions}. Many Fermion configuration has much more energy that corresponding Boson configuration.
\end{itemize}

\begin{defn}[Bose condensate]
	Many Bosons in the lowest energy  state.
\end{defn}

\begin{figure}[H]
	\caption{Adding more Fermions}\label{fig:adding:more:fermions}
	\begin{subfigure}{0.32\textwidth}
		\caption{Two electrons, lowest possible energy}\label{fig:electons:lowest:energy}
		\includegraphics[width=0.9\textwidth]{electons-lowest-energy}
	\end{subfigure}
	\begin{subfigure}{0.32\textwidth}
		\caption{Filled lowest two energy levels.}
		\includegraphics[width=0.9\textwidth]{5-electrons}
	\end{subfigure}
	\begin{subfigure}{0.32\textwidth}
		\caption{$E=\frac{k^2}{2m}$, so there are more positions in each energy level as energy increases.}
		\includegraphics[width=0.9\textwidth]{more-fermions}
	\end{subfigure}
\end{figure}

\begin{defn}[Fermi Sphere]
	Electrons fill a sphere in momentum space, whose radius depends on number of electrons. This is called the Fermi Sphere.
\end{defn}

We can move Bosons and Fermions out of the ground state, as shown in Figures \ref{fig:bosons:move:ground:state} and \ref{fig:fermions:move:ground:state}.
\begin{figure}[H]
	\caption{Moving Bosons and Fermions from lowest state}
	\begin{subfigure}{0.5\textwidth}
		\caption{Bosons in Ground State}\label{fig:bosons:move:ground:state}
		\includegraphics[width=0.9\textwidth]{bosons-ground-state}
	\end{subfigure}
	\begin{subfigure}{0.5\textwidth}
		\caption{Moving Fermions. Take from surface of Fermi Sphere to minimize energy required. Leaves hole behind.}\label{fig:fermions:move:ground:state}
		\includegraphics[width=0.9\textwidth]{moving-outside-fermi}
	\end{subfigure}
\end{figure}

In Figure \ref{fig:fermions:move:ground:state} a charged hole is left behind.
If an electron drops into hole, a photon is emitted.

Consider an atom, with all electrons in their ground state. Low energy electron is struck by photon, which takes electron to excited state.

\begin{defn}[Fermi Sea]
	Electrons in Fermi Sphere are said to be in Fermi Sea.
\end{defn}

Electrons that are deep in Fermi Sea can't be excited by low energy photons. Only shallow electrons are accessible.

\subsubsection{Dirac and antiparticles}

We consider the 1-dimensional case. Imagine a wave propagating to the tight with speed $c$.
\begin{align*}
\omega =& k\\
\frac{\partial \psi}{\partial t} =& - \frac{\partial \psi}{\partial x}  \numberthis \label{eq:rh:dirac}\\
\psi =& e^{i(kx-\omega t)} \numberthis \label{eq:rh:dirac:wave}
\end{align*}
What if $k$ negative? Then we have negative energy. But if we had arbitrarily negative energies, particle could keep emitting photons all the way down. So (\ref{eq:rh:dirac}) would not make sense for Bosons.

Dirac said to fill all negative energy states: so lowest energy state has all negative energies filled. Now we can take an electron out, leaving hole. Now we have two positive energy objects, electron and positron.

\section{Dirac equation and Higgs particles}

\subsection{Dirac equation}

The right-handed Dirac equation (\ref{eq:rh:dirac}) has the following solution:
\begin{align*}
\psi_R =& e^{i(kx-\omega t)} \text{, from (\ref{eq:rh:dirac:wave}), whence:}\\
V_g =& \frac{\omega}{k} \text{, from which}\\
\frac{\partial \psi_R}{\partial t} + \frac{\partial \psi_R}{\partial x} =&0
\end{align*}

If $k<0$, $\omega<0$, i.e. negative energy, which is not a good thing. So we fill up the Dirac Sea. Then we can remove one electron, creating a hole in the Dirac Sea; particle and hole both have positive momentum and positive energy.

We now introduce left moving electrons.

\begin{align*}
\frac{\partial \psi_L}{\partial t} =& \frac{\partial \psi_L}{\partial x} \numberthis \label{eq:lh:dirac:wave}\\
\frac{\omega}{k} =& -1
\end{align*}
Fill sea with left moving negative energy.

Define $\psi=\begin{pmatrix}
	\psi_R\\
	\psi_L
\end{pmatrix}$, and write (\ref{eq:rh:dirac:wave}) and (\ref{eq:lh:dirac:wave}):

\begin{align*}
	\frac{\partial \psi}{\partial t} =& - \alpha \frac{\partial \psi}{\partial x} \text{, where}\\
	\alpha =& \begin{pmatrix}
		1&0\\
		0&-1
	\end{pmatrix} \text{. We will find the following useful:} \numberthis\label{eq:alpha}\\
	\omega =& \alpha k \numberthis \label{eq:dirac:1}
\end{align*}

Now we give the particles mass $m$.

\begin{align*}
	\omega^2 =& k^2 + m^2 \text{. Now we generalize (\ref{eq:dirac:1})}\\
	\omega =& \alpha k + m \beta \text{. This is the 1D Dirac equation.} \numberthis \label{eq:Dirac:1D}\\
	\omega^2 =& \alpha^2 k^2 + m^2 \beta^2 + m k (\alpha \beta + \beta \alpha) \text{, so we want:}\\
	\alpha^2 =& 1 \text{, satsfied by (\ref{eq:alpha})}\\
	\beta^2 =& 1 \numberthis \label{eq:beta2}\\
	\{\alpha,\beta\} =& 0 \numberthis \label{eq:alpha:beta}\\
	\beta =& \begin{pmatrix}
	0 & 1\\
	1 & 0
	\end{pmatrix} \text{ satisfies (\ref{eq:beta2}) and (\ref{eq:alpha:beta}).}
\end{align*}

The mass term couples equations for $\Psi_L$ and $\Psi_R$. Dirac equation is Lorentz covariant.

Let us make particle stand still, i.e. $k=0 \implies \omega = \beta m$.

\begin{align*}
	i \frac{\partial \psi}{\partial t} =& \beta m \Psi\\
	i \dot{\psi_R} =& m \psi_L\\
	i \dot{\psi_L} =&  m \psi_R \\
	i \dot{\psi_+} =&  m \psi_+ \text{, where $\psi_+\triangleq\psi_R + \psi_L$--corresponds to $\omega=m$}\\
	i \dot{\psi_-} =&  m \psi_- \text{, where $\psi_-\triangleq\psi_R - \psi_L$--corresponds to $\omega=-m$}
\end{align*}

So particle at rest still has positive and negative energies: throw latter in Dirac Sea.

\subsection{Higgs particles}

In the presence of the Higgs field, $\phi(x)$, the Dirac equation becomes:
\begin{align*}
i \dot{\psi_R} =& -i \partial_x \psi_R + g \phi \psi_L\\
i \dot{\psi_L} =& i \partial_x \psi_L + g \phi \psi_R \text{, which gives massless electron if $\phi=0$}
\end{align*}
What if the lowest energy of Higgs favours a non-zero value, $\phi(x)=\phi$? The $m=g\phi$. This could come about as the result of non-lienar interactions between Boson and Fermions.

The Higgs field is a Bosonic field, which has an energy from a potential, $V(\phi)$. There is a symmetry so $V(\phi)=V(-\phi)$.

\begin{figure}[H]
	\caption[Higgs potential]{Higgs potential: reason for shape not fully understood at time of lecture.}
	\begin{subfigure}{0.5\textwidth}
		\caption{Higgs potential, showing two minima. Either one gives rise to mass of $g\phi$}\label{fig:higgs:potential}
		\includegraphics[width=0.8\textwidth]{higgs-potential}
	\end{subfigure}
	\begin{subfigure}{0.5\textwidth}
		\caption{Higgs field vibrating around vacuum. Frequency is related to mass of Higgs particle: Higgs corresponds to quanta of vibration.}
		\includegraphics[width=0.8\textwidth]{higgs-vibrating}
	\end{subfigure}
\end{figure}

If Higgs field is coupled in an interesting way to Fermions, Higgs will respond to collisions of Fermions and create Higgs particles. In the Standard Model there is only one Higgs, Supersymmetry has two.

If vacuum were symmetric--Figure \ref{fig:higgs:potential}-- $\phi$ would be zero: it is the breaking of the symmetry which gives electron its mass. All particles except photon get their mass through the Higgs Mechanism. Except for Higgs particle: it gets its mass from a different mechanism.

\subsection{3+1 Dimensional Dirac Equation}

Dirac started with generalization.

\begin{align*}
\omega =& \underbrace{\alpha . k}_{\alpha_1 k_1 + \alpha_2 k_2 + \alpha_3 k_3} +\beta m\\
\omega =& \alpha_1 k_1 + \alpha_2 k_2 + \alpha_3 k_3 +\beta m\\
\omega^2 =& k_1^2 + k_2^2 +k_3^2 +m^2\\
\omega^2 =& (\alpha_1 k_1 + \alpha_2 k_2 + \alpha_3 k_3 +\beta m)(\alpha_1 k_1 + \alpha_2 k_2 + \alpha_3 k_3 +\beta m) \text{, whence}\\
\alpha_i^2 =& 1  \text{ and} \numberthis \label{eq:dirac:ac1}\\
\beta^2 =& 1 \text{ and} \numberthis \label{eq:dirac:ac2}\\
\{\alpha_i,\alpha_j\} =& 2 \delta_{i,j} \numberthis \label{eq:dirac:ac3}\\
\{\alpha_i,\beta\} =& 0 \numberthis \label{eq:dirac:ac4}
\end{align*}

Theorem \ref{thm:dirac:anticommutation} shows 2 or 3 dimensional matrices cannot satisfy these equations; $4 \times 4$ is the first case.

Here is a set of Dirac Matrices (not unique, but equivalent--Theorem \ref{thm:dirac:anticommutation} presents a different set).

\begin{align*}
	\beta =& \begin{pmatrix}
		1&0&0&0\\
		0&1&0&0\\
		0&0&-1&0\\
		0&0&0&-1
	\end{pmatrix} \numberthis \label{eq:beta:1}\\
	\alpha_1 =& \begin{pmatrix}
		0&0&0&1\\
		0&0&1&0\\
		0&1&0&0\\
		1&0&0&0
	\end{pmatrix}\numberthis \label{eq:alpha1:1}\\
	\alpha_2 =& \begin{pmatrix}
		0&0&0&-i\\
		0&0&i&0\\
		0&-i&0&0\\
		i&0&0&0
	\end{pmatrix} \numberthis \label{eq:alpha2:1} \\
	\alpha_3 =& \begin{pmatrix}
		0&0&1&0\\
		0&0&0&-1\\
		1&0&0&0\\
		0&-1&0&0
	\end{pmatrix} \numberthis \label{eq:alpha3:1}
\end{align*}

Partitioning, and using the Pauli matrices (\ref{eq:pauli:x}), (\ref{eq:pauli:y}), and (\ref{eq:pauli:z}):
\begin{align*}
	\beta = \begin{pmatrix}
		I&0\\
		0&I
	\end{pmatrix}\\
	\alpha = \begin{pmatrix}
		0&\sigma\\
		-\sigma&0
	\end{pmatrix}
\end{align*}

\begin{align*}
	\psi =& \begin{pmatrix}
		\psi_1\\
		\psi_2\\
		\psi_3\\
		\psi_4
	\end{pmatrix} \text{ spinor--left/right and spin up/down}\\
	i \frac{\partial \psi}{\partial t} =& -i \alpha_i \frac{\partial \psi}{\partial x^i} + \beta m \psi \text{, the famous Dirac equation} \numberthis \label{eq:famous:dirac}
\end{align*}

\section{Angular momentum}

Angular momentum is a vector, which points along the direction of rotation and follows right hand rule: fingers follow rotation, thumb points along vector.

There are two kinds:
\begin{itemize}
	\item orbital angular momentum--motion of centre of mass;
	\item spin angular momentum--angular momentum in frame where momentum is zero.
\end{itemize}

Is a rotating nucleus the same object as a non-rotating nucleus? Note that angular momentum is quantized, so it takes a discrete amount of energy to start rotation. Now an atom with too much rotation will fly apart, which is definitely different.

What about electron? Is it too small for us to rotate? The smaller the object, the more energy it needs to rotate with given amount of angular momentum.Maybe electron too small.

Angular momentum for small object like electron \emph{characterizes} object.

Let us invent a particle with definite angular momentum, a "spintron". Can it point in any direction? Yes, since laws of physics are rotationally invariant. Angular momentum is quantized.

Theory is mathematical: it seems abstract, but corresponds to experiment.

Classically:
\begin{align*}
I\triangleq&mr^2 \text{, moment of inertia}\\
E=&\frac{L^2}{2I}\\
=&\frac{L^2}{2mr^2}\\
\rightarrow &\infty \text{ as } r \rightarrow 0 \text{ for discrete $L$}
\end{align*}

Figure \ref{fig:two:particles:rotating} depicts two particles rotating: total momentum is zero, but there is angular momentum.
\begin{figure}[H]
	\caption{Two particles rotating.}\label{fig:two:particles:rotating}
	\includegraphics[width=0.6\textwidth]{two-particles-rotating}
\end{figure}

Orbital angular momentum of a point particle moving about origin.
\begin{align*}
\vec{L} =& \vec{R} \times \vec{p} \text{ accordng to convention -- c.f. right hand rule.}\\
L_x =& y P_z - z P_y\\
L_y =& z P_x - x P_z\\
L_z =& x P_y - y P_x
\end{align*}

We will determine commutation relations of angular momentum operators.

We have \cite{susskind2014quantum}:
\begin{align*}
	[X_i,X_j] =& 0\\
	[P_i,P_j]=& 0\\
	[X_i,P_j]=& i \hslash \delta_{i,j}`
\end{align*}

\begin{thm}[Commutators for angular momentum]
	\begin{align*}
	[L_x,L_y]=& i \hslash L_z \numberthis \label{eq:lx:ly}\\
	[L_y,L_x]=& i \hslash L_x \numberthis \label{eq:ly:lz}\\
	[L_z,L_x]=& i \hslash L_y \numberthis \label{eq:lz:lx}
	\end{align*}
\end{thm}
\begin{proof}
	\begin{align*}
	[L_x,L_y]=& [y P_z - z P_y,z P_x - x P_z]\\
	=&[y P_z, z P_x] -\cancel{[yP_z,xP_z]} -\bcancel{[zP_y,zP_x]} + [z P_y,x P_z] \\
	=& y P_x[P_z,z] + P_y x[z,P_z]\\
	=& -i \hslash y P_x + i \hslash P_y x\\
	=& i \hslash L_z \text{, which is (\ref{eq:lx:ly})}
	\end{align*}
	The other two equations, (\ref{eq:ly:lz}) and (\ref{eq:lz:lx}) follow by cyclically permuting indices.
\end{proof}

 These relationships:
\begin{itemize}
	\item remain true if we add angular momenta for many particles;
	\item are true for spin
	 \item are relationally invariant.
\end{itemize}

We focus on the $z$-axis ("quantization axis") and introduce two new quantities--raising and lowering operators. Choosing units where $\hslash-1$:

\begin{align*}
	L_\pm \triangleq& L_x \pm i L_y \numberthis \label{eq:L_pm} \\
	[L_+,L_z] =& [Lx+i L_y,L_z]\\
	=& -iL_y + i^2 L_x \text{, from (\ref{eq:lz:lx}) and (\ref{eq:lx:ly})}\\
	=& -L_+ \text{ from (\ref{eq:L_pm}). Similarly} \numberthis \label{eq:L+:Lz}\\
	[L_-,L_z] =& + L_- \numberthis \label{eq:L-:Lz}
\end{align*}

\begin{thm}[$L_+$ is a creation operator for $L_z$]\label{thm:angular:momentum:creation}
	\begin{align*}
		L_z \ket{m} =& m \ket{m} \numberthis \label{eq:L:eigen}\\
		\implies&\\
		L_z (L_+ \ket{m}) =& (m+1) (L_+ \ket{m})
	\end{align*}
\end{thm}

\begin{proof}
	\begin{align*}
		(\ref{eq:L+:Lz})\implies\\
		L_+L_z-L_zL_+ =& -L_+ \text{, whence:}\\
		L_zL_+ =& L_+L_z + L_+\ \text{ and}\\
		L_z L_+ \ket{m} =& (L_+L_z + L_+)\ket{m}\\
		=& L_+ (L_z + I)\ket{m}\\
		=& L_+ (m + 1) \ket{m} \text{ from (\ref{eq:L:eigen})}\\
		=&  (m + 1) (L_+ \ket{m})
	\end{align*}
\end{proof}

\begin{cor}[$L_-$ is an annihilation  operator for $L_z$]
	\begin{align*}
	L_z \ket{m} =& m \ket{m} \\
	\implies&\\
	L_z (L_- \ket{m}) =& (m-1) (L_- \ket{m})
	\end{align*}
\end{cor}

The angular momentum spectrum is spaced by integers. How far can we bump it up before the object changes? Imagine a classical object with fixed $L^2$. We can maximize $L_z$  by pointing it upward.

Theorem \ref{thm:angular:momentum:creation} ignores the possibility that $L^+m_{max}=0$ and $L^-m_{min}=0$. (c.f. harmonic oscillator). By relational invariance, $m_{min}=-m_{max}$: Spectrum must be symmetric about zero.

There are two possibilities for the spectrum:
\begin{itemize}
	\item $-m_{max},...,-1,0,1,...m_{max}$
	\item $-m_{max},...,-\frac{1}{2},\frac{1}{2},...m_{max}$
\end{itemize}

\begin{defn}[Spin]
	$m_max$ is known as the spin. It can be 0. There are no elementary particles known with spin 3, but there are atoms. Similarly no half-spin particle over $\frac{3}{2}$.
\end{defn}

What about $L^2$? If you know $L_z$ at maximum, there will be uncertainty in other two components, so $L^2$ is not quite $m_{max}^2$.

\begin{align*}
	L_- L_+ =& (L_x-iL_y)(L_x+iL_y)\\
	=& L_x^2 +i L_x L_y -i L_y L_x + L_y^2\\
	=& L_x^2 + L_y^2 + i[L_x,L_y]\\
	=& L^2 - L_z^2 + i[Lx,Ly] \numberthis \label{eq:L-:L+}\text{, hence}\\
	L^2 =& L_- L_+ + L_z^2 \text{, classically, but this is not true for quantum world. }\\
	(\ref{eq:L-:L+}) \implies& \\
	 L^2 =& L_- L_+ + L_z^2 -\underbrace{i[Lx,Ly]}_\text{$i^2L_z$ from (\ref{eq:lx:ly})}\text{, in the quantum world}\\
	 =& L_- L_+ + L_z^2 +L_z\\
	L^2 \ket{m_{max}}=& L_- L_+ \ket{m_{max}} + L_z^2 \ket{m_{max}} +L_z \ket{m_{max}}\\
	=& (0 +m_{max}^2 + m_{max})\ket{m_{max}} \\
	=& m_{max} (m_{max}+1)\ket{m_{max}}
\end{align*}

All states have same value of $\braket{L^2}$.

\begin{thm}[$L^2$ commutes with all its components.]
	 \begin{align*}
	 [L^2,L_i]=0  \;\forall i \in {x,y,z}
	 \end{align*}
\end{thm}
\begin{proof}
	We need the following lemma.
	\begin{lemma}[Commutator with square]\label{lemma:comm:sq}
		\begin{align*}
			[A^2,B] =& A[A,B] + [A,B]A
		\end{align*}
	\end{lemma}
	\begin{proof}[Proof of lemma]
		\begin{align*}
			[A^2,B] =& AAB - BAA\\
			=& AAB - ABA +ABA -BAA\\
			=& A[A,B] + [A,B]A
		\end{align*}
	\end{proof}
	\begin{align*}
	[L^2,L_x]=&[L_x^2,L_x] + [L_y^2,L_x] + [L_z^2,L_x] \numberthis \label{eq:L2lx}\\
	[L_y^2,L_x] =&L_y[L_y,L_x] + [L_y,L_x]L_y \text{, using Lemma \ref{lemma:comm:sq}}\\
	=& -i L_y L_z -i L_z L_y \text{ from (\ref{eq:ly:lz}). Similarly}\\
	[L_z^2,L_x] = &L_z [L_z,L_x] + [L_z,L_x] L_z\\
	=& i L_z L_y  + i L_y L_z \text{ from (\ref{eq:lz:lx}), whence (\ref{eq:L2lx}) yields}\\
	[L^2,L_x]=&\cancel{[L_x^2,L_x] }+\bcancel{ [L_y^2,L_x]} + \bcancel{[L_z^2,L_x]}
	=& 0
	\end{align*}
\end{proof}
Consider spin $\frac{1}{2}$--call them $\ket{+}$ and  $\ket{-}$.

\begin{align*}
e^{i\theta} \alpha \ket{+} + e^{i\theta}\beta \ket{-} \text{, has probability $\alpha^2$ of being in $\ket{+}$}
\end{align*}
So can ignore 1 degree of freedom. $\alpha$ and $\beta$ have 2 degrees of freedom each. Phase eliminates one, and $\alpha^2+\beta^2=1$ eliminates one other, leaving two. This is the same as number of degrees of freedom in direction.

Measured value is always $\pm \frac{1}{2}$; average may be zero; it behaves like a classical variable.

\begin{thm}[Spin statistics]
		All particles with half integral spin are Fermions;
		All particles with  integral spin are Bosons.
\end{thm}

\begin{proof}
	TBP
\end{proof}
Take two half spin particles: composite behaves like a Boson--e.g. Hydrogen atom; deuteron is Boson, deuterium is Fermion.

\section{Spin}

\subsection{Combining Fermions to make Bosons}

Paradox: if two Fermions make a Boson, can we have two of these composite Bosons in the same state?

Wave function of Boson is supposed to be symmetric:

\begin{align*}
	\psi(x,y) =&\psi(y,x) \text{ e.g. }\\
	\psi(x,y) =& \phi(x)\phi(y)\\
	\psi(x,y) =&-\psi(y,x) \text{ for Fermions}
\end{align*}

Now imagine a wave function, $\psi(x,y)$, that isn't symmetric:
\begin{align*}
\psi(x,y)+& \psi(y,x) \text{ symmetric}\\
\psi(x,y)-& \psi(y,x) \text{ antisymmetric}
\end{align*}

Now imagine 4 Fermions--2 electons(X) and 2 protons(Y).

\begin{align*}
\psi(x_1,x_2,y_1,y_2) = &-\psi(x_2,x_1,y_1,y_2)\\
=& \psi(x_1,x_2,y_2,y_1)
\end{align*}

Hydrogen atom at origin plus another hydrogen atom:$\psi(x_1,y_1) \phi(x_2,y_2)$.
If we antisymmetrize:
\begin{align*}
	\psi(x_1,y_1) \phi(x_2,y_2) -& \psi(x_2,y_1) \phi(x_1,y_2)
	-\psi(x_1,y_2) \phi(x_2,y_1) + \psi(x_2,y_2) \phi(x_1,y_1)\\
	x_1 \leftrightarrow& x_2 \text{ changes sign, and}\\
	y_1 \leftrightarrow& y_2 \text{ changes sign, but}\\
	1 \leftrightarrow& 2 \text { preserves sign.}
\end{align*}

What if both atoms have same wave function? Ignoring a factor of $2$:

\begin{align*}
	\psi(x_1,y_1) \psi(x_2,y_2) -& \psi(x_2,y_1) \psi(x_1,y_2)
\end{align*}

Both atoms can be in same place, but we can't find 2 electrons in same place.

\subsection{Mathematics of Spin}

Mathematics of Spin is needed for Colour, Isospin, etc.

\subsubsection{Half spin}

We have two values along the z axis: $S_z = m \hslash$. We will use $S$ for spin, $L$ for orbital angular momentum.

\begin{align*}
	[S_x,S_y] =& i S_z \numberthis \label{eq:sx:sy}\\
	[S_y,S_z] =& i S_x  \numberthis \label{eq:sy:sz}\\
	[S_z,S_x] =& i S_y  \numberthis \label{eq:sz:sx}\\
	S_\pm =& S_x \pm S_y
\end{align*}
Compare with (\ref{eq:lx:ly}),  (\ref{eq:ly:lz}), and  (\ref{eq:lz:lx}).

Can express spin as $\begin{pmatrix}
\alpha\\
\beta
\end{pmatrix}$ or $\alpha\ket{+} + \beta\ket{-}$

Components are 2D matrices satisfying algebra (\ref{eq:sx:sy}), (\ref{eq:sy:sz}), and (\ref{eq:sz:sx}). They are not unique, but equivalent. We will use the Pauli spin\footnote{Without $\frac{1}{2}$ they are known as Pauli matrices.}:
\begin{align*}
	S_x =& \frac{1}{2}\begin{pmatrix} \numberthis \label{eq:pauli:x}
		0&1\\
		1&0
	\end{pmatrix}\\
	S_y =& \frac{1}{2}\begin{pmatrix} \numberthis \label{eq:pauli:y}
		0&-i\\
		i&0
	\end{pmatrix}\\
	S_z =& \frac{1}{2}\begin{pmatrix} \numberthis \label{eq:pauli:z}
		1&0\\
		0&-1
	\end{pmatrix}
\end{align*}

\subsubsection{Spin 1}

We have three states $\{-1,0,1\}$, a Boson. Are there $3\times3$ matrices satisfying (\ref{eq:sx:sy}), (\ref{eq:sy:sz}), and (\ref{eq:sz:sx})?

\begin{align*}
	S_z = i \begin{pmatrix}
		0&1&0\\
		-1&0&0\\
		0&0&0
	\end{pmatrix}\\
	S_y = i \begin{pmatrix}
		0&0&1\\
		0&0&0\\
		-1&0&0
	\end{pmatrix}\\
	S_x = i \begin{pmatrix}
		0&0&0\\
		0&0&1\\
		0&-1&0
	\end{pmatrix}
\end{align*}

Find eigenstates of $S_z$:
\begin{align*}
	i \begin{pmatrix}
		0&1&0\\
		-1&0&0\\
		0&0&0
	\end{pmatrix}\begin{pmatrix}
		\alpha\\
		\beta\\
		\gamma
	\end{pmatrix}=&\begin{pmatrix}
		\alpha\\
		\beta\\
		\gamma
	\end{pmatrix}\\
	=&i\begin{pmatrix}
		\beta\\
		-\alpha\\
		0
	\end{pmatrix}
\end{align*}

Return to real electron that can move in space. We turn the 2D spinors into functions of position: $\begin{pmatrix}
\psi_u(x)\\
\psi_d(x)
\end{pmatrix}$.

For spin 1: $\begin{pmatrix}
\phi_1(x)\\
\phi_2(x)\\
\phi_3(x)
\end{pmatrix}$.

\subsection{Dirac Equation: connection with Spin}

\begin{align*}
i \frac{\partial \psi}{\partial t} =& - i \alpha_i \frac{\partial \psi}{\partial x^i} + \beta m \psi\\
\omega =& \sum_i \alpha_i k_i + \beta m\\
\omega^2 =& k^2 + m^2
\end{align*}
Where the $\alpha_i$ and $\beta$ satisfy (\ref{eq:dirac:ac1}), (\ref{eq:dirac:ac2}), (\ref{eq:dirac:ac3}), and (\ref{eq:dirac:ac4}).

\begin{thm}[Dirac anticommutation relations.]\label{thm:dirac:anticommutation}
	Equations (\ref{eq:dirac:ac1}), (\ref{eq:dirac:ac2}), (\ref{eq:dirac:ac3}), and (\ref{eq:dirac:ac4}) can be satisfied by suitably chosen $2\times4$ matrices, and they cannot be satisfied by $n \times n$ for $n<4$.
\end{thm}

One set of solutions, which is not the same as (\ref{eq:beta:1}), (\ref{eq:alpha1:1}), (\ref{eq:alpha2:1}), and (\ref{eq:alpha3:1}):
\begin{align*}
	\alpha_i=&\begin{pmatrix}
		\sigma_i&0\\
		0&-\sigma_i
	\end{pmatrix}\\
	\beta =& \begin{pmatrix}
		0&I\\
		I&0
	\end{pmatrix}
\end{align*}

There are two kinds of $2\times2$: 2 blocks of $2\times2$ matrices. This corresponds to the physics - there are two energies, positive and negative, and two spins.
The 4 components are not a relativistic 4-vector! The easiest way to see this is to set momentum to zero, then the $\alpha_i$ drop out.

\begin{align*}
i \frac{\partial \psi}{\partial t} =&  \beta m \psi \text{ for particle at rest.}\\
i \frac{\partial}{\partial t}\begin{pmatrix}
\psi_+\\
\psi_-
\end{pmatrix}=&m \begin{pmatrix}
0&I\\
I&0
\end{pmatrix}\begin{pmatrix}
\psi_+\\
\psi_-
\end{pmatrix}\\
i \frac{\partial \psi_+}{\partial t} =&   m \psi_-\\
i \frac{\partial \psi_-}{\partial t} =&   m \psi_+ \\
i \frac{\partial (\psi_++\psi_-)}{\partial t} =&   m (\psi_++\psi_-) \text{ positive energy--$\omega=m$}\\
i \frac{\partial (\psi_+-\psi_-)}{\partial t} =&   -m (\psi_+-\psi_-) \text{ negative energy--$\omega=-m$}
\end{align*}

So we end up with two uncoupled equations! $2\times2$ is connected to positive and negative energies. $2\times2$ of $\psi_+$ arises from spin.

\section{Equations of motion of particles and fields}\label{sect:equation:motion}

The equations of motion of fields that describe  particles all come from a Lagrangian. In quantum field theory, Lagrangian describes everything.

\subsection{Review of classical fields}

\begin{align*}
\mathcal{L}=& \mathcal{L}(\phi,\partial_t \phi, \partial_x \phi)\\
=&\mathcal{L}(\phi,\partial_{\mu} \phi) \text{ in Minkowski space}\\
\frac{\partial}{\partial t} \frac{\partial \mathcal{L}}{\partial \big(\frac{\partial \phi}{\partial t}\big)}+\sum_i \frac{\partial}{\partial x_i} \frac{\partial \mathcal{L}}{\partial \big(\frac{\partial \phi}{\partial x_i}\big)}=&\frac{\partial \mathcal{L}}{\partial \phi} \text{, equation of motion} \numberthis \label{eq:lagrange}
\end{align*}

(If $\mathcal{L}$ depends on several fields, there will be a version of (\ref{eq:lagrange})  for each field.)

Example of $\mathcal{L}$.

\begin{align*}
	\mathcal{L} =& \frac{1}{2} (\frac{\partial \phi}{\partial t})^2 - \frac{1}{2} (\frac{\partial \phi}{\partial x})^2 - \frac{1}{2} (\frac{\partial \phi}{\partial y})^2 - \frac{1}{2} (\frac{\partial \phi}{\partial z})^2 - V(\phi)
\end{align*}

Substituting in (\ref{eq:lagrange}):
\begin{align*}
	\ddot{\phi} - \frac{\partial^2 \phi}{\partial x^2}  - \frac{\partial^2 \phi}{\partial y^2}  - \frac{\partial^2 \phi}{\partial z^2} =& - \frac{\partial V}{\partial \phi} \text{. The simplest interesting case is:}\\
	V(\phi) =& \frac{m^2 \phi^2}{2}\\
	\ddot{\phi} - \frac{\partial^2 \phi}{\partial x^2}  - \frac{\partial^2 \phi}{\partial y^2}  - \frac{\partial^2 \phi}{\partial z^2} =& - m^2 \phi \numberthis \label{eq:simplest:non:trivial}
\end{align*}

Differentiating a wave function $e^{-i \omega t} e^{-i k x}$ twice by $t$ gives a factor $-\omega^2$. Substituting in (\ref{eq:simplest:non:trivial}):
\begin{align*}
-\omega^2 \cancel{\phi} - (k_x^2+k_y^2+k_z^2)\cancel{\phi} =& -m^2 \cancel{\phi}\\
-\omega^2  - (k_x^2+k_y^2+k_z^2) =& -m^2 \\
\omega^2 + P^2 = m^2
\end{align*}

If $\mathcal{L}$ is a scalar, equations of momentum will be Lorentz Invariant.

What about a more complex $V(\phi)$?
\begin{align*}
	V(\phi) =& \frac{m^2 \phi^2}{2} + g \phi^3\\
	\ddot{\phi} - \frac{\partial^2 \phi}{\partial x^2}  =& - m^2 \phi - 3 g \phi^2
\end{align*}

Quadratic terms in Lagrangian lead to linear equations; higher powers lead to non-linearities, so cannot superpose. May scatter.

Consider two fields:
\begin{align*}
	\mathcal{L} =& \frac{1}{2}(\partial_{\mu} \phi)^2 + \frac{1}{2}(\partial_{\mu} \rho)^2 - \frac{m^2 \phi^2}{2} - \frac{M^2 \rho^2}{2} - \rho \phi^2\\
	\ddot{\phi} - \partial_x^2 \phi =& -m^2 \phi - \rho \phi
	\\
	\ddot{\rho} - \partial_x^2 \rho =& -m^2 \rho -phi^2
\end{align*}

These are non-linear and coupled, so the particles scatter each other.

Consider the Dirac equation.

\begin{align*}
\dot{\psi} =& i \alpha \frac{\partial \psi}{\partial x} + m \beta \psi\\
\mathcal{L} =& i \big( \psi^\dagger \partial_t \psi +  \psi^\dagger \alpha \partial_x \psi + \psi^\dagger \beta \psi m \big)
\end{align*}

Suppose there is another field: can add scalar field to Dirac field and get scattering, and generate equations of motion for both fields.

\begin{align*}
\mathcal{L} =& i \big( \psi^\dagger \partial_t \psi +  \psi^\dagger \alpha \partial_x \psi + \psi^\dagger \beta \psi m \big) - \psi^\dagger \beta \psi \phi
\end{align*}

Lagrangian is what specifies a field theory.

\subsection{Quantum Fields}

Imagine Lagrangian containing $\phi^3$, where $\phi = (a_+ + a_-)$.
Figure \ref{fig:phi:3:x} shows the kinds of interaction that may occur. Quadratic terms in Lagrangian give rise to motion, higher powers to interactions: 1 particle $\rightarrow$ 2, 2 particles $\rightarrow$ 1, 0 particles $\rightarrow$ 3, 3 particles $\rightarrow$ 0.

\begin{figure}[H]
	\caption{$\phi^3(x)$ gives rise to many possible interactions.}\label{fig:phi:3:x}
	\includegraphics[width=0.8\textwidth]{phi-3-x}
\end{figure}

How to change position of particle. We need a pair of operators $\phi(x) \phi(x^\prime)$ to absorb particle at $x$, and re-emit at $x^\prime$--i.e. move particle

\begin{align*}
(\frac{\partial \phi}{\partial x})^2 =& \frac{\big( \phi(x) - \phi(x^\prime)\big)^2}{...}\\
=& \frac{\phi(x)^2 + \phi(x^\prime)^2 - 2  \phi(x) \phi(x^\prime)}{...}
\end{align*}

Electromagnetic field and Dirac electron.

\begin{align*}
	\psi =& c_e^-(\text{+ve energy}) + c_e^-(\text{+ve energy}) \text{ Dirac field}\\
	\psi =& c_e^-(\text{+ve energy}) + c_p^+(\text{+ve energy}) \text { Dirac sea--$p$ for positron}\\
	\psi^\dagger =& c_e^+(\text{+ve energy}) + c_e^+(\text{+ve energy})\\
	=& c_e^+(\text{+ve energy}) + c_p^-(\text{+ve energy})
\end{align*}

Basic interaction of electrodynamics.
\begin{align*}
	e \big( \psi^\dagger \psi A_0 + 	\underbrace{\psi^\dagger \alpha_i  \psi}_\text{vector from $\psi$} A_i \big) \text{ polarization of photon $A$} \numberthis \label{eq:basic:interaction}
\end{align*}

\begin{figure}[H]
	\caption{Interactions arising from Lagrangian (\ref{eq:basic:interaction})}
	\begin{subfigure}{0.48\textwidth}
		\caption{Annihilation + 2 creation: photon emitted}
		\includegraphics[width=0.7\textwidth]{em-dirac1}
	\end{subfigure}
	\begin{subfigure}{0.48\textwidth}
		\caption{2 Annihilation +  creation: photon absorbed}
		\includegraphics[width=0.7\textwidth]{em-dirac2}
	\end{subfigure}
	\begin{subfigure}{0.48\textwidth}
		\caption{2 Annihilation +  creation: Electron + positron produces photon. NB, by convetion, positron is drawn with downward arrow.}
		\includegraphics[width=0.7\textwidth]{em-dirac3}
	\end{subfigure}
	\begin{subfigure}{0.48\textwidth}
		\caption{Photon absorbed, electron and positron created.}
		\includegraphics[width=0.7\textwidth]{em-dirac4}
	\end{subfigure}
	\begin{subfigure}{0.48\textwidth}
		\caption{Photon, electron and positron absorbed (energy and momentum not conserved!).}\label{fig:em:dirac5}
		\includegraphics[width=0.7\textwidth]{em-dirac5}
	\end{subfigure}
\end{figure}

$e$ is strength of interaction, probability of interaction $e^2$.

Remember that energy conservation arose from integrating over all time, and momentum conservation from integrating over space. If we integrate (\ref{eq:basic:interaction}) over all space and all time,  energy and momentum will both be conserved, which rules out the process depicted in Figure \ref{fig:em:dirac5}.

Protons, Neutrons, Pi mesons. Pions are scalar particles, and they come in 3 types, $\pi_+$, $\pi_-$, $\pi_0$.

\begin{table}[H]
	\begin{center}
		\caption{Protons, Neutrons, Pi mesons}
		\begin{tabular}{|l|l|} \hline
			Operators &Description\\ \hline
			$\psi_p$, $\psi^\dagger_p$& create and annihilate proton\\ \hline
		    $\psi_n$, $\psi^\dagger_n$& create and annihilate neutron\\ \hline
			$\pi_+$& annihilate pion\\ \hline
			$\pi_-$&annihilate pion\\ \hline
			$\pi_0$&annihilate pion\\ \hline
			$\psi^\dagger_p \psi_n \pi_+$&Absorb pion and neutron, emit proton\\ \hline
			$\psi^\dagger_n \psi_p \pi_-$&Absorb pion and proton, emit neutron\\ \hline
			$\psi^\dagger_p \psi_p \pi_0$&Absorb pion and proton, emit proton\\ \hline
			$\psi^\dagger_n \psi_n \pi_0$&Absorb pion and neutron, emit neutron \\ \hline
		\end{tabular}
	\end{center}
\end{table}
The fundamental process of course involve quarks; if energy is high enough to break up protons, cannot use proton fields..

\begin{figure}[H]
	\caption{Two processes from the same interaction.}
	\begin{subfigure}{0.45\textwidth}
		\caption{Proton $\rightarrow$ neutron + pion}
		\includegraphics[width=0.8\textwidth]{pion1}
	\end{subfigure}
	\begin{subfigure}{0.45\textwidth}
		\caption{Proton + antineutron $\rightarrow$ pion}
		\includegraphics[width=0.8\textwidth]{pion2}
	\end{subfigure}
\end{figure}

The Lagrangian describes things that happen at a point, or a pair of neighbouring points. Figure \ref{fig:proton:electron:photon} shows an interaction arising from a product of terms. There is no action at a distance.
\begin{figure}[H]
	\caption[Proton photon electron interaction]{Proton photon electron interaction--product of two separate terms in Lagrangian}\label{fig:proton:electron:photon}
	\includegraphics[width=0.8\textwidth]{proton-electron-photon}
\end{figure}

NB: creation operators are functions of P or X, never both (uncertainty principle). They are related through Fourier transforms, which are the essence of the uncertainty principle.

$\bra{0}\psi^\dagger\psi^\dagger\psi^\dagger...\mathcal{L}\mathcal{L}\mathcal{L}...\psi\psi\psi\ket{0}$

\begin{align*}
	\Pi_i& \big(1+\mathcal{L}_{interaction}(x_i)\big)\\
	=&1 + \sum_i \underbrace{ \mathcal{L}(x_i)}_\text{one vertex}  +\sum_{i,j}\underbrace{\mathcal{L}(x_i)\mathcal{L}(x_j)}_\text{two vertices} +...
\end{align*}

We get an infinite number of places where 1 thing, 2 things, etc, happen.

The art form is figuring out the Lagrangian; then QFT becomes a precise tool.

Often can ignore parts of Lagrangian: e.g. quarks don't have much to do with electrons. QFT may be a coarse grained description of something else.

\section{Field Lagrangians and Path Integrals}

The basic technique that Lecture \ref{sect:equation:motion} alluded to is called the \emph{path integral method of quantum mechanics}. It is the most direct route to the theory of particle interactions based on Feynman diagrams. The path integral is the quantum mechanical version of the Principle of Least Action\cite{susskind2013quantum}. If we think of a classical Newtonian particle, its motion from one point of space-time to another is determined by a Lagrangian $\mathcal{L}(\dot{X},X)$. We use the Lagrangian to construct the Action.

\begin{defn}[Classical Action of Path]
	For every trajectory, $P$ from the initial point to the final point, whether or not the trajectory is a solution to Newton's Laws of motion,  the Classical Action is:
	\begin{align*}
		\mathcal{A}=\int_P dt \mathcal{L}(\dot{X},X)
	\end{align*}
\end{defn}

The classical form of the Principle of Least Action is that the true trajectory minimizes the action. We can use the calculus of variations to derive differential equations which satisfy Newton's Laws. The same idea works in classical field theory, where $\mathcal{L}=\mathcal{L}(\phi,\partial_\mu\phi)$. The theory of relativity requires that the Lagrangian be a scalar\cite{susskind2017special}.

\begin{defn}[Action over Region]
	For a region of spaces time, $R$,  the Action is:
	\begin{align*}
		\mathcal{A}=\int_R dx^4 \mathcal{L}(\phi,\partial_\mu\phi)
	\end{align*}
\end{defn}

The analogue of the starting point for a particle is the initial value of the field. If we prescribe the starting and final values of the field, we can ask whether there is a solution that starts with starting values, and ask whether there is a solution that minimizes the action. The principle of least action states that the correct value of the field is the one that minimizes the action. So the original principal of least action gets extended to a space-time principal of least action, where the degrees of freedom are fields. NB, we are looking for a way to interpolate the field which minimizes the Action.
\url{https://youtu.be/q-QaaKCvDUs?t=555}
We now look at the quantum mechanical use of Action: \emph{we won't attempt to derive the classical version from the quantum mechanical version.}

According to quantum mechanics, a thing that you might want to calculate is the \emph{amplitude}, so we can calculate probabilities. Imagine that particle is injected at $(x,t)$: what is amplitude that particle will be found at $(x^\prime,t^\prime)$?

Feynman's rule is:

\begin{align*}
	Amplitude =& \sum_{Histories} e^{-\frac{i}{\hslash} \int \mathcal{L} dt}
\end{align*}

where the sum is taken over all classically \emph{possible} histories--Figure \ref{fig:sum:histories}--i.e. they can not go back in time for non-relativistic QM. Probability is the square of the magnitude of the Amplitude.

\begin{figure}[H]
	\caption{Sum over histories}\label{fig:sum:histories}
	\includegraphics[width=0.8\textwidth]{sum-histories}
\end{figure}

Imagine starting systems with field $\phi(x)$, and asking for probability of $\phi^\prime(x)$ at some later time--Figure \ref{fig:path-integral1}. Again this is determined by an amplitude:
\begin{align*}
	\sum e^{-\frac{i}{\hslash} \int dx dy dz dt\mathcal{L}(\phi,\partial \phi) }
\end{align*}
summed over all possible histories between $\phi(x)$ and $\phi^\prime(x)$.

\begin{figure}[H]
	\caption{Initial and final conditions}\label{fig:path-integral1}
	\includegraphics[width=0.8\textwidth]{path-integral1}
\end{figure}

(A question was asked about convergence: trajectories tend to cancel, and those with the least cancellation tend to be near the ones that have least action classically.)

There are two ways to think of quantum field theory, in terms of fields, and in terms of particles. What if we specify field quanta (particles)? We ask the probability of some specified particle content at a later time. The answer involves a path integral. Start by partitioning space into cells--Figure \ref{fig:path-integral2}--and think about the value of the field in each cell. We compute:
\begin{align*}
	\Pi_{i\in cells}(1-\frac{i}{\hslash} \mathcal{L_i}a^4) \text{, where $a^4$ is volume of a cell.}
\end{align*}

Now we have the following approximation for small $\epsilon$
\begin{align*}
	1 + \epsilon \approx &e^\epsilon\\
	\approx& 1 + \frac{\epsilon}{1!} + \frac{\epsilon^2}{2!}  + \frac{\epsilon^3}{3!} +...\\
\end{align*}

\begin{align*}
	\sum_{histories} e^{-\frac{i}{\hslash} \int dx dy dz dt\mathcal{L}(\phi,\partial \phi) } =& \sum_{histories} e^{-\frac{i}{\hslash} A_i } \text{, where $A_i$ is action in $i$th cell}\\
	=& \sum_{histories} \Pi_i e^{-\frac{i}{\hslash} A_i }
\end{align*}

Let's forget particles for a moment and think of fields.

\begin{figure}[H]
	\caption[Space partitioned into cells]{Space partitioned into cells. Initial conditions in bottom row, final in top. We consider all possible histories that match initial and final conditions. NB: functions aren't necessarily analytic or even continuous.}\label{fig:path-integral2}
	\includegraphics[width=0.8\textwidth]{path-integral2}
\end{figure}

Our question is now: if initial condition matches our initial configuration of particles, and final condition matches desired configuration, what is the amplitude? Figure \ref{fig:path-integral3} depicts a particle moving in one history.

\begin{figure}[H]
	\caption[Particle moving]{Creation operator in one cell, annihilation in neighbour amounts to movement}\label{fig:path-integral3}
	\includegraphics[width=0.8\textwidth]{path-integral3}
\end{figure}

There are terms in the Lagrangian associated with pairs of boxes--the derivatives:
\begin{align*}
	\frac{1}{2}(\partial_t \phi)^2 \rightarrow& \frac{1}{2}\frac{(\phi(t,x)-\phi(t^\prime,x))^2}{...}\\
	\frac{1}{2}(\partial_x \phi)^2 \rightarrow& \frac{1}{2}\frac{(\phi(t,x)-\phi(t,x^\prime))^2}{...} \numberthis \label{eq:phi:phi}
\end{align*}
(\ref{eq:phi:phi}) gives rise to terms such as:
\begin{alignat*}{2}
	e^{- i \sum_{neighbours} \phi(x) \phi(x^\prime)}=1& -&& i \sum_{neighbours} \underbrace{\phi(x) \phi(x^\prime)}_\text{Figure \ref{fig:path-integral5}}\\
&-&& \frac{1}{2!}\underbrace{\sum_{neighbours} \phi(x) \phi(x^\prime)\sum_{neighbours} \phi(x^{\prime\prime}) \phi(x^{\prime\prime\prime})}_\text{Figure \ref{fig:path-integral7}}\\
&+&&\underbrace{i\frac{1}{3!}\sum_{neighbours}\phi(..)\phi(..)\sum_{neighbours}\phi(..)\phi(..)\sum_{neighbours}\phi(..)\phi(..)}_\text{Figure \ref{fig:path-integral9}}\\
&+&& ... \numberthis \label{eq:path:integral:expanded}
\end{alignat*}

\begin{figure}[H]
	\caption{Rules for Path Integrals}
	\begin{subfigure}{0.45\textwidth}
		\caption{Create at one point, annihilate at neighbour}\label{fig:path-integral5}
		\includegraphics[width=1.0\textwidth]{path-integral5}
	\end{subfigure}
	\begin{subfigure}{0.45\textwidth}
		\caption{A dangling endpoint is illegal unless one endpoint creates, other annihilates: since we are only putting particles in from initial state and removing from final, this is illegal.}\label{fig:path-integral4}
		\includegraphics[width=1.0\textwidth]{path-integral4}
	\end{subfigure}
	\begin{subfigure}{0.45\textwidth}
		\caption{The only way Figure \ref{fig:path-integral5} can contribute is if space is only two layers thick.}\label{fig:path-integral6}
		\includegraphics[width=1.0\textwidth]{path-integral6}
	\end{subfigure}
	\begin{subfigure}{0.45\textwidth}
		\caption{What about thicker space? How to we get from \ref{fig:path-integral6} to here? We need the quadratic term in (\ref{eq:path:integral:expanded})}\label{fig:path-integral7}
		\includegraphics[width=1.0\textwidth]{path-integral7}
	\end{subfigure}
	\begin{subfigure}{0.45\textwidth}
		\caption{How to avoid dangling ends in \ref{fig:path-integral7}? Restrict to 3 layers. This allows us to find $x^\prime=x^{\prime\prime}$ in (\ref{eq:path:integral:expanded})}\label{fig:path-integral8}
		\includegraphics[width=1.0\textwidth]{path-integral8}
	\end{subfigure}
	\begin{subfigure}{0.45\textwidth}
		\caption{Including the cubic term in (\ref{eq:path:integral:expanded}) allows us to find two 3-step paths in (\ref{eq:path:integral:expanded}).}\label{fig:path-integral9}
		\includegraphics[width=1.0\textwidth]{path-integral9}
	\end{subfigure}
\end{figure}

So we can sum up contributions $\frac{i^n}{n !}$ in (\ref{eq:path:integral:expanded}) to determine complex amplitude. We include backward paths from special relativity--Figure \ref{fig:path:backward:time}.

\begin{figure}[H]
	\caption{Path going backwards in Time. We can think of it in two ways.}\label{fig:path:backward:time}
	\begin{subfigure}{0.45\textwidth}
		\caption{New rule allows particle to go back in time}
		\includegraphics[width=0.8\textwidth]{path-backward-time}
	\end{subfigure}
	\begin{subfigure}{0.45\textwidth}
		\caption{Particle anti-particle pair used instead.}
		\includegraphics[width=0.8\textwidth]{path-backward-time1}
	\end{subfigure}
\end{figure}

Two particles.

\begin{figure}[H]
	\caption{Two particles}\label{fig:path:integral:2particles:1}
	\begin{subfigure}{0.45\textwidth}
		\caption{Quadratic term in (\ref{eq:path:integral:expanded}) used to advance two particles by one box}
		\includegraphics[width=0.8\textwidth]{path-integral-2particles-1}
	\end{subfigure}
	\begin{subfigure}{0.45\textwidth}
		\caption{Weight particle by $m^2$ from $\frac{m}{2}\phi^2$}\label{fig:path:integral:mass}
		\includegraphics[width=0.8\textwidth]{path-integral-mass}
	\end{subfigure}
\end{figure}

What goes in Lagrangian?

\begin{align*}
\mathcal{L}=& \frac{1}{2} (\partial_t \phi)^2 - \frac{1}{2} (\partial_x \phi)^2 + \underbrace{\frac{m}{2}\phi^2}_\text{ absorb and emit at same point} +\underbrace{g \phi^3}_\text{Figure \ref{fig:path:integral:cubic}}
\end{align*}

This weights path by $m^2$--Figure \ref{fig:path:integral:mass}. If not used we have theory of massless particles.

\begin{figure}[H]
	\caption{Some possibilities from $\phi^3$}
	\begin{subfigure}{0.4\textwidth}
	\caption{Particle absorbed and two new particles created.}\label{fig:path:integral:cubic}
		\includegraphics[width=0.8\textwidth]{path-integral-cubic}
	\end{subfigure}
	\begin{subfigure}{0.4\textwidth}
		\caption{Particle emitted and later reabsorbed.}\label{fig:path:integral:cubic:split}
		\includegraphics[width=0.8\textwidth]{path-integral-cubic-split}
	\end{subfigure}
	\begin{subfigure}{0.4\textwidth}
		\caption{Electron emits and absorbs photon.}\label{fig:path:integral:cubic:split1}
		\includegraphics[width=0.5\textwidth]{path-integral-cubic-split1}
	\end{subfigure}
	\begin{subfigure}{0.4\textwidth}
		\caption{More complex version of Figure \ref{fig:path:integral:cubic:split1}.}\label{fig:path:integral:cubic:split2}
		\includegraphics[width=0.5\textwidth]{path-integral-cubic-split2}
	\end{subfigure}
\end{figure}

How come answer not infinite? Powers of coefficients from Lagrangian $(g)$ appear with different powers; hope they are small enough.

Feynman diagrams and Lagrangians are complementary.

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