\section{Quantum Field Theory}
{\bf Reminder:} To find extrema functional $x(t)$,
solve $\delta \int_{t_1}^{t_2} L(x, \dot{x}) dt= 0$ by solving
${\frac {\partial L} {\partial x}}- {\frac {d} {dt}}
({\frac {\partial L} {\partial {\dot x}}})= 0$.
$L=T-V$, $H= \sum_i p_i {\dot q}_i -L$, $p_i= {\frac {\partial L}{\partial {\dot{q}_i}}}$.
Canonical quantization replaces
scalars with operators (e.g. - $p$, etc).
\\
\\
{\bf Infinite degrees of freedom:}
Quantum Field Theory describes quantum systems with an infinite number of degrees of freedom.  Here is an example:
Consider $n$ particles of mass $m$ connected by identical springs with spring
constant $k$.  Let $y_i$ be the displacement of the
particles from their equilibrium position.
$L
= {\frac 1 2} \sum_{i=1}^n [m \dot{y_i}-k(y_{i+1}-y_i)^2]
= {\frac 1 2} \sum_{i=1}^n a [{\frac m a} \dot{y_i}-k {\frac {(y_{i+1}-y_i)^2} a}]
= a \sum_{i=1}^n L_i $.  Let
$n \rightarrow \infty$ then ${\frac m a} \rightarrow \mu$, the linear density,
${\frac {y_{i+1}-y_i} a}= {\frac {\partial y} {\partial x}}$ and $ka= Y$.
$\delta \int_{t_1}^{t_2} dt \int_{x_0}^{x_1} dx ({\frac 1 2}
[\mu {\dot{y}}^2 - Y ({\frac {\partial y}{\partial x}})^2])= 0$ gives
$
{\frac {\partial} {\partial x}} {\frac {\partial {\cal L}} {\partial {\frac {\partial y} {\partial x}}}} +
{\frac {\partial} {\partial t}} {\frac {\partial {\cal L}} {\partial {\frac {\partial y} {\partial t}}}} -
{\frac {\partial {\cal L}} {\partial y}}$; where the term in the second integral
is called the Lagrangian density denoted ${\cal L}$.  The corresponding Hamiltonian density is
${\cal H}= \dot{y} {\frac {\partial {\cal L}} {\partial {\dot{y}}}}- {\cal L}
$.
\\
\\
{\bf Maxwell's equations as four vectors:}
$b_{\mu}= ( {\vec b}, i b_0)$; Lorentz transform is $x_{\mu}'= a_{\mu, \nu} x_{\nu}$ and
$\partial_{\mu}' F = \partial_{\mu}' (x_{\nu}) \partial_{\nu} (F)$.
Let $j_{\mu}$ be the four-vector
$j_{\mu}= ({\vec j}, ic \rho)$ and
$$F_{\mu, \nu}=
\left(
\begin{array}{cccc}
0 & B_3 & -B_2 & -iE_1 \\
-B_3 & 0 & B_1 & -iE_2 \\
B_2 & -B_1 & 0 & -iE_3 \\
iE_1 & iE_2 & iE_3 & 0 \\
\end{array}
\right).$$
Maxwell's equations become ${\frac {\partial F_{\mu \nu}} {\partial x_{\nu}}} = {\frac {j_{\mu}} c}$.
Note the matrix is anti-symmetric.
This is also written as $\partial_{\nu} F_{\mu \nu}= {\frac {j_{\mu}} c}$.
$F_{\mu \nu} F_{\mu \nu} = 2(|B|^2-|E|^2)$ is a scalar.  $F_{uv}= \partial_u A_v- \partial_v A_u$.
Klein Gordon in four vector format:
${\cal L}= {\frac 1 2} \eta^{uv} \partial_u \partial_v \phi - {\frac 1 2} m \phi^2$ where
$\eta_{uv}=
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -1 \\
\end{array}
\right)
$; this is Minkowski's metric.
A tensor, $v_i$ is \emph{covariant} if it transforms as $(v_i)'= {\frac {\partial x_j} {\partial x_i}} v_j$.
A tensor, $v^i$ is \emph{contravariant} if it transforms as $(v^i)'= {\frac {\partial x^i} {\partial x^j}} v^j$.
Differentials are contravariant, $\nabla$ is covariant.
\\
\\
Lorentz rotation by $\theta$ followed by boost of $v$ along $x$ is
$$
\left(
\begin{array}{cccc}
\gamma & -\gamma v & 0 & 0 \\
-\gamma v & \gamma & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & cos(\theta) &  -sin(\theta) & 0 \\
0 &  sin(\theta) &  cos(\theta) & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right), \gamma= {\frac 1 {\sqrt {1-v^2/c^2)}}}.$$
\\
\\
{\bf Yukawa potential:}  Neutral scalar field, ${\cal L}= - ({\frac 1 2} ({\frac {\partial \phi} {\partial x_{\mu}}})^2
+ \mu^2 \phi^2)$.  $E^2- |p|^2 c^2= m^2 c^4$.
$E \rightarrow i \hbar {\frac {\partial} {\partial t}}$ and
$p \rightarrow - i \hbar {\frac {\partial} {\partial x_{\mu}}}$, $\mu \leftrightarrow {\frac {mc} {\hbar}}$.
Use the usual Fourier correspondance
$\tilde{\phi}(k)= {\frac 1 {(2 \pi)^{3/2}}} \int d^3 x (e^{- k \cdot x} \phi(x))$ and
${\phi}(x)= {\frac 1 {(2 \pi)^{3/2}}} \int d^3 k (e^{k \cdot x} \tilde{\phi}(k))$.
Now consider a scalar potential field, $\phi$, with a point source, $G$, satisfying the Klein-Gordon equation
$\Box^2 \phi = G \delta^{(3)}(x)$.  Multiply both sides by
${\frac 1 {(2 \pi)^{3/2}}} (e^{- i k \cdot x})$
and integrate remembering that $\phi$ and its derivatives
vanish at the limits of integration.  We get $(-|k|^2-\mu^2) \tilde{\phi}(k)= G {\frac 1 {(2 \pi)^{3/2}}}$,
so $\phi(x)= {\frac {e^{- \mu r}} r}$ and ${\cal H}_{int}= - {\cal L}_{int}$.
\\
\\
{\bf Transformations:}
$
\left(
\begin{array}{c}
\phi_1' \\
\phi_2' \\
\end{array}
\right)
=
\left(
\begin{array}{cc}
cos(\lambda) & - sin(\lambda)\\
sin(\lambda) &  cos(\lambda)\\
\end{array}
\right)
\left(
\begin{array}{c}
\phi_1 \\
\phi_2 \\
\end{array}
\right)
$ for free fields of identical mass.
Consider $n$ scalar fields $\phi_a$ with the same mass and ${\cal L}$.
${\cal L}= {\frac 1 2} \sum_{i=1}^n \partial_{\mu} \phi_a \partial_{\mu} \phi_a - {\frac 1 2} \sum_{i=1}^n
m^2 \phi_a^2 - g (\sum_{i=1}^n \phi_a^2)^2$,  ${\cal L}$ is invariant under $G= SO_n$.  Non-abelian symmetries
are global symmetries.
Translation invariance gives \emph{conservation of momentum}.
Time invariance gives \emph{conservation of energy}.
\emph{First quantization} promotes values (position, momentum, energy) to operators
($E \rightarrow i \hbar {\frac {\partial}{\partial t}}$,
$p \rightarrow -i \hbar \nabla$) and impose commutator realtionships.
\emph{Second quantization} position, momentum, energy etc. remain scalar but
fields like potential and conjugate momentum become operators; commutator relations imposed on field operators.
\\
\\
{\bf Invariance and conserved quantities:}
If $i \hbar {\frac {d O(t)}{dt}}= [O(t),H]=0$, $O$ is a constant of the motion.   Now let
$|\Phi\rangle \rightarrow |\Phi'\rangle= U |\Phi\rangle$ so
$O= U^{\dagger} O U$; often $U= e^{i \alpha T}$ where $\alpha$ is a continuous parameter.  For an
infinitesimal translation, $U= I+i \delta \alpha T$ and $\delta O= i \delta \alpha [T, O]$.
Invariance of the Lagrangian under a transformation,
leads to
${\frac {\partial f^{\alpha}}{\partial x^{\alpha}}}=0$ and
$F^{\alpha}(t)= \int d^3 x \thinspace {\frac {\partial f^{\alpha}}{\partial x^{\alpha}}}$
is a conserved quantity.
For example, suppose $\phi_r(x) \rightarrow \phi_r'(x)= \phi_r(x)+ \delta \phi_r(x)$.  The Lagrangian transforms
as
$\delta({\cal L})= {\frac {\partial {\cal L}} {\partial \phi_r}} \delta \phi_r +
{\frac {\partial {\cal L}} {\partial \phi_{r, \alpha}}} \delta \phi_{r, \alpha}$ which applying the
minimum condition becomes
$\delta({\cal L})= {\frac {\partial } {\partial x^{\alpha}}}
{\frac {\partial {\cal L}} {\partial \phi_{r, \alpha}}} \delta \phi$; here,
$f^{\alpha}= {\frac {\partial {\cal L}} {\partial \phi_{r, \alpha}}} \delta \phi$.
This gives $Q= {\frac {-i q}{\hbar}} \int d^3 x [\pi_r(x)\phi_r(x)-\pi^{\dagger} \phi^{\dagger}]$ as a conserved
quantity.  $Q$ is charge.  If instead we consider translation invariance we get conservation of energy/momentum.
Rotational invariance gives conservation of angular momentum.
\\
\\
{\bf Standard script for classical $\rightarrow$ quantum field theory:}
Promote conjugate variables to operators $\langle \phi_r, \pi_s \rangle$ and \emph{quantize} as
$[\phi_r(j,t), \pi_s(j',t)]= i \hbar \delta_{rs} \delta_{jj'}$,
$[\phi_r(j,t), \phi_s(j',t)]= [\pi_r(j,t), \pi_s(j',t)]= 0$, and
$[\phi_r(x,t), \pi_s(x',t)]= i \hbar \delta_{rs} \delta(x-x')$.
\\
\\
{\bf The complex potential:}  Consider two real scalar fields
$\phi_1$,
$\phi_2$ and put
$\phi = {\frac 1 {\sqrt 2}} (\phi_1 + i \phi_2)$.  If
$\phi$ is the solution to the Klein Gordon
equation in the presence of a potential $A_{\mu}$ with charge $e$, then
$\phi^*$ is the solution to the Klein Gordon
equation in the presence of a potential $A_{\mu}$ with charge $-e$.
The \emph{charge current density}, $s_{\mu}= i(
{\frac {\partial \phi^*} {\partial {x_{\mu}}}} \phi -
\phi^*{\frac {\partial \phi} {\partial {x_{\mu}}}})$
satisfies ${\frac {\partial s_{\mu}} {\partial x_{\mu}}}= 0$.
\\
\\
{\bf Creation and annihilation operators:}  We look at photons.
Let $\mu_{k, \alpha}(x)= \epsilon^{(\alpha)}(k) e^{i k \cdot x}$ where $\alpha$ is the linear polarization
selected so that $ \epsilon^{(1)}, \epsilon^{(2)}, k $ form an oriented orthogonal triad.  Consider the
potential bounded in space by a cube of side length $L$, $V= L^3$.
$A(x, t)= {\frac 1 {V^{1/3}}} \sum_k \sum_{\alpha=1,2} (
c_{k,\alpha} \mu_{k, \alpha}(x)+ c_{k,\alpha}^* \mu_{k, \alpha}^*(x))$.
We have $H= {\frac 1 2} \int (|B|^2+|E|^2) d^3x$, $H= \sum_k \sum_{\alpha}
2({\frac {\omega} c})^2 c_{k,\alpha}^* c_{k,\alpha} $.  Put
$Q_{k, \alpha}= {\frac 1 c}
(c_{k, \alpha}+
c_{k, \alpha}^*)$,
and
$P_{k, \alpha}= -{\frac {\omega} c} (c_{k, \alpha}- c_{k, \alpha}^*)$.
${\frac {\partial H} {\partial Q_{k, \alpha}}}= - \dot{P}_{k, \alpha}$ and
${\frac {\partial H} {\partial P_{k, \alpha}}}= \dot{Q}_{k, \alpha}$,
We get
$ {\frac 1 {V^{1/3}}} \int d^3 x
(\mu_{k, \alpha} \cdot
\mu_{k', \alpha'}^*) = \delta_{k, k'} \delta_{\alpha. \alpha'}$, $k_x, k_y, k_z= {\frac {2 \pi n} L}$ and
$\omega= |k|c$.  Make
$P_{k, \alpha}$ and
$Q_{k, \alpha}$
operators as usual giving
$[Q_{k, \alpha}, P_{k', \alpha'}]= \pm i \hbar \delta_{k, k'} \delta_{\alpha, \alpha'}$,
$[Q_{k, \alpha}, Q_{k', \alpha'}]= 0$,
$[P_{k, \alpha}, P_{k', \alpha'}]= 0$.
The \emph{annihilation operator} is
$a_{k, \alpha}= {\frac 1 {\sqrt {2 \hbar \omega}}} ( \omega Q_{k, \alpha} + i P_{k, \alpha})$.
The \emph{creation operator} is
$a_{k, \alpha}^{\dagger}= {\frac 1 {\sqrt {2 \hbar \omega}}} ( \omega Q_{k, \alpha} - i P_{k, \alpha})$.
$N_{k, \alpha}= a_{k, \alpha}^{\dagger} a_{k, \alpha}$.
$N |n \rangle= n |n \rangle$,
$N a^{\dagger} |n \rangle= (n+1) |n \rangle$,
$N a |n \rangle= (n-1) |n \rangle$.
\\
\\
{\bf Aharonov Bohm:}
For double slit setup with puddle of $B \neq 0$ completely inside the strip of the two slits.
$\phi=
\phi_1^{(0)} exp[{\frac {ie} {\hbar c}} \int_{\small path 1} A(x') \cdot dx'] +
\phi_2^{(0)} exp[{\frac {ie} {\hbar c}} \int_{\small path 2} A(x') \cdot dx'] $.
$
\int_{\small closed \; path} A(x') \cdot dx'=
\int_{\small surface} B \cdot n \; dS= {\frac {e \Phi} {\hbar c}}$.  For superconducting ring, with Cooper pair
quasi-particle,
$\phi= \phi^{(0)} exp[{\frac {2ie} {\hbar c}} \int_{\small closed \; path} A(x') \cdot dx']$.  Since $\phi$ is
the same whether or not the path encloses the flux;
${\frac {2e}{\hbar c}}
\int_{\small closed \; path} A(x') \cdot dx'
= 2n \pi$, so $\Phi= {\frac {n \pi \hbar c}{e}}$.
\\
\\
{\bf Pauli matricies and the Dirac equation:}
$\sigma_1=
\left(
\begin{array}{cc}
0 & -i\\
i & 0\\
\end{array}
\right)$,
$\sigma_2=
\left(
\begin{array}{cc}
1 & 0\\
0 & -1\\
\end{array}
\right)$,
$\sigma_3=
\left(
\begin{array}{cc}
0 & 1\\
1 & 0\\
\end{array}
\right)$.
\begin{center}
\begin{tabular} {|c|c||c|c|}
\hline
Particle & Mass & Particle & Mass\\
\hline
1 & 2 & $\ldots$ & n\\
\hline
Neutrino & $10^{-2} eV$ & Proton/Neutron & $1 GeV$\\
Electron & $.5 MeV$ & $\tau$ & $ 2 GeV$ \\
Muon & $100 MeV$ & $W, Z$ Boson & $80-90 GeV$ \\
Pion & $140 MeV$ & $W, Z$ Higgs & $120-200 GeV$ \\
\hline
\end{tabular}
\end{center}
{\bf Particles from the vacuum:}
$ [H, a_p^{\dagger}] = \omega_p a_p^{\dagger}$,
$ [H, a_p] = - \omega_p a_p$, $|p \rangle = a_p^{\dagger} |0 \rangle$ and
$ H |p \rangle =  \omega_p |p \rangle$, $\omega_p^2= p^2 + m^2$.  $P|p \rangle = p | p \rangle$.
\\
\\
{\bf The Dirac equation:}
$(i \gamma^u \partial_u- {\frac {mc}{\hbar}}) \phi = 0$.
Time reversal transformation is $T: x^0 \rightarrow -x^0$; $x^i \rightarrow x^i$.
Parity reversal transformation is $P: x^0 \rightarrow x^0$; $x^i \rightarrow -x^i$.
$\gamma_k=
\left(
\begin{array}{cc}
0 & -i\sigma_k\\
i \sigma_k  & 0\\
\end{array}
\right)$,
$\gamma_0=
\left(
\begin{array}{cc}
I & 0\\
0  & -I\\
\end{array}
\right)$.
Dirac wanted to find a field equation with linear operators:
$E= \alpha \cdot p + \beta m$, $i {\frac {\partial \phi} {\partial t}} (-i \alpha \cdot \nabla + \beta m) \phi$
and require $E^2 = m^2 + |p|^2$.
$\alpha=
\left(
\begin{array}{cc}
0 & \sigma\\
\sigma  & 0\\
\end{array}
\right)$,
$\beta=
\left(
\begin{array}{cc}
1 & 0\\
0  & -1\\
\end{array}
\right)$.
Under spacetime transformation, spinor transforms as $\delta \phi = \epsilon^u \partial_u \phi$.
$T^{\mu \nu} \partial^{\mu} \phi \partial^{\nu} \phi - \eta^{\mu, \nu} {\cal L}$.
\\
\\
{\bf Angular momentum:}
$L= r \times p$.  If $[H, L]= 0$, angular momentum is conserved.
$[H, L] = [\alpha \cdot p, r \times p]= i \alpha \times p$.
$\Sigma= \left(
\begin{array}{cc}
\sigma & 0\\
0  & \sigma\\
\end{array}
\right)$, $[H, \Sigma] = [\alpha \cdot p, -i\alpha_1 \alpha_2 \alpha_3 \alpha]= 2i \alpha \times p$.
$J= L + {\frac 1 2} \Sigma$ is conserved, $[H,J]= 0$.
Lagrangian for free Dirac field is ${\cal L}= {\overline {\phi}} (i \gamma^u \partial_u -m) \phi=
\phi_i(i[\gamma^u]_{ij} \partial_u - m \delta_{ij}) \phi_j$.
\\
\\
{\bf Solving the Dirac equation for a free scalar field:}
$(i \gamma^0 \partial_t-m ) \phi=0$, gives
$i \gamma^0 \partial_t \phi= i
\left(
\begin{array}{cc}
I & 0\\
0  & -I\\
\end{array}
\right) $ or
$ \left(
\begin{array}{cc}
{\frac {\partial u} {\partial t}} \\
{\frac {\partial u} {\partial t}}\\
\end{array}
\right) =
m
\left(
\begin{array}{cc}
u \\
v \\
\end{array}
\right)
$, where
$u=
\left(
\begin{array}{cc}
\phi_1 \\
\phi_2 \\
\end{array}
\right)
$ and
$u=
\left(
\begin{array}{cc}
\phi_3 \\
\phi_4 \\
\end{array}
\right)
$.  Note $u$ and $v$ correspond to spin states;
$ \left(
\begin{array}{cc}
1\\
0\\
\end{array}
\right) $ is spin up and
$ \left(
\begin{array}{cc}
0\\
1\\
\end{array}
\right) $ is spin down.
Thus $i {\dot{u}}= mu$ and
$-i {\dot{v}}= mv$, so
$u(t)= u(0)e^{-imt}$ and
$v(t)= v(0)e^{imt}$.
\\
\\
{\bf Free field solution of Klein-Gordon:}
Here $\hbar=1$.  $\varphi(x, t) = A e^{Et-p \cdot x} = A e^{\omega_k x^0- k \cdot x}$ leading to
$\varphi(x) = \int {\frac {d^3k}{(2 \pi)^{3/2} {\sqrt {2 \omega_k}}}}
(\tilde{\varphi}(k) e^{\omega_k x^0- k \cdot x} +
\tilde{\varphi}(k)^* e^{\omega_k x^0- k \cdot x} )
$.  Promote
$\tilde{\varphi}(k) \rightarrow \hat{a}$ and
$\tilde{\varphi}(k)^* \rightarrow \hat{a}^{\dagger}$.
${\cal L}= {\frac 1 2} \partial_{\mu} \partial^{\mu} \varphi - {\frac 1 2} m^2 \varphi^2$ and
$\pi(x)= {\frac {\partial {\cal L}} {\partial ({\partial_0 \varphi)}}}$.
$\hat{\pi}(x) = -i \int {\frac {d^3k}{(2 \pi)^{3/2}}} {\sqrt {\frac {\omega_k}{2}}}
[(\tilde{a}(k) e^{-(\omega_k x^0- k \cdot x)} -
\tilde{a}^{\dagger}(k)^* e^{\omega_k x^0- k \cdot x} ) $.
$[x_i, p_j]= i \delta_{ij}$,
$[x_i, x_j]= 0$,
$[p_i, p_j]= 0$.
$|k_1, k_2\rangle=
\tilde{a}^{\dagger}(k_1)
\tilde{a}^{\dagger}(k_2) |00\rangle $.
Each $\hat{a}^{\dagger}(k_i)$ creates a single particle of
momentum $\hbar k_i$ and energy $\hbar \omega_k$.
$\varphi^+(x)=
\int {\frac {d^3k}{(2 \pi)^{3/2}}} {\sqrt {\frac {\omega_k}{2}}} \hat{a}^{\dagger}(k) e^{-(\omega_k x^0 - k \cdot x)}$
and positive frequency corresponds to annihilation.
$\varphi^-(x)=
\int {\frac {d^3k}{(2 \pi)^{3/2}}} {\sqrt {\frac {\omega_k}{2}}} \hat{a}^{\dagger}(k) e^{(\omega_k x^0 - k \cdot x)}$
and negative frequency corresponds to creation.
\\
\\
{\bf Normalization:}
$\langle 0| 0 \rangle= 1$,
$\langle k| k' \rangle= \delta(k-k')$ for bosons.
$\langle 0| \hat{H}|0 \rangle= \langle0| \int d^3k (N(k)+ {\frac 1 2})
|0\rangle = {\frac {\omega_k} 2} \int d^3k$.  \emph{Renormalized:}
$\hat{H}_R= \hat{H} -\int d^3k= \int d^3k (\omega_k \hat{a}^{\dagger}(k) \hat{a}(k)$ and
$\langle k | \hat{H}_R | k\rangle= \omega_k$.
\\
\\
{\bf Propagators:}
$\hat{P}= \int d^3k (k[\hat{a}^{\dagger}(k) \hat{a}(k) + \hat{b}^{\dagger}(k) \hat{b}(k)])$.
$\hat{Q}= \int d^3k [\hat{a}^{\dagger}(k) \hat{a}(k) + \hat{b}^{\dagger}(k) \hat{b}(k)]$.
$\hat{H}= \int d^3k (\omega_k [\hat{a}^{\dagger}(k) \hat{a}(k) + \hat{b}^{\dagger}(k) \hat{b}(k)])$.
The number of particles is $\hat{N}_a= \int d^3k [\hat{a}^{\dagger}(k) \hat{a}(k)]$;
the number of anti-particles is $\hat{N}_b= \int d^3k [\hat{b}^{\dagger}(k) \hat{b}(k)]$.
$[\hat{\psi}(x), \hat{\psi}(h)]= i \Delta(x-y)$ is a \emph{propagator}.  Feynman propagator is
$\Delta_F(x-y)= \langle 0 | T \psi(x) \psi(y) | 0 \rangle = D(x-y), x^0>y^0$, $T$ is the time ordering operation.
Note that an events are \emph{causal} if $[O_1(x), O_2(y)]=0$ when $(x-y)^2<0$.
\\
\\
{\bf Momentum space:}
$\langle x' | \alpha \rangle= \psi_{\alpha}(x')$, $\langle \beta | \alpha \rangle= \int dx' \langle \beta | x' \rangle
\langle x' | \alpha \rangle$.
$| \alpha \rangle = \sum_{a'}  |a'\rangle \langle a' | \alpha\rangle$,
$\langle x'| \alpha \rangle = \sum_{a'}  \langle x'|a'\rangle \langle a' | \alpha\rangle$.
$p | \alpha \rangle = \int dx' | x' \rangle (\langle x' | \alpha \rangle -
\Delta x' {\frac {\partial} {\partial x'}} \langle x' | \alpha \rangle$.
$p | \alpha \rangle = \int dp | p' \rangle (\langle p' | \alpha \rangle$,
$\langle p' | \alpha \rangle = \int dp \langle p | p' \rangle (\langle p' | \alpha \rangle$, $\langle p' | \alpha \rangle=
\psi_{\alpha}(p)$.  $p' \langle x' |p \rangle \langle p' | \alpha \rangle$ and thus
$p' \langle x' | p \rangle = - i \hbar {\frac {\partial} {\partial x'}} \langle x' | p' \rangle$
and $\langle x' | p' \rangle = N e^{i (p' \cdot x')/ \hbar}$.
$\psi_{a}(x')= {\frac 1 {\sqrt {2 \pi \hbar}}} \int dp' exp({\frac {i (p' \cdot x')} {\hbar}}) \phi_{a}(p')$;
$\phi_{a}(p') \rangle)= {\frac 1 {\sqrt {2 \pi \hbar}}} \int dx' exp({\frac {-i (p' \cdot x')} {\hbar}}) \psi_{a}(x')$.
\\
\\
{\bf Derivation of Hamiltonian from creation and annihilation operators:}
$\hat{a} \hat{a}^{\dagger}= ( {\frac {m \omega}{2 \hbar}}) [ \hat{x}^2 + {\frac {\hat{p}^2} {(m \omega)^2}}
- {\frac {i \hat{x} \cdot \hat{p}} {(m \omega)}} + {\frac {i \hat{x} \cdot \hat{p}} {(m \omega)}} ] =
({\frac {m \omega}{2 \hbar}}) [ \hat{x}^2 + {\frac {\hat{p}^2} {(m \omega)^2}}
- {\frac {i} {(m \omega)}} + {\frac {i \hat{x} \cdot \hat{p}} {(m \omega)}} ] =
({\frac {m \omega}{2 \hbar}}) [ \hat{x}^2 + {\frac {\hat{p}^2} {(m \omega)^2}} + {\frac {\hbar} {(m \omega)}} ]$.  Similarly,
$ \hat{a}^{\dagger} \hat{a} = ({\frac {m \omega}{2 \hbar}}) [ \hat{x}^2 + {\frac {\hat{p}^2} {(m \omega)^2}}
- {\frac {\hbar} {(m \omega)}} ] $.  Combining, $\hat{H}={\frac 1 2} \hbar \omega [ \hat{a}^{\dagger} \hat{a} + \hat{a} \hat{a}^{\dagger} ] =
\hbar \omega ( \hat{a}^{\dagger} \hat{a} + {\frac 1 2})$.  Note, $[\hat{a}, \hat{a}^{\dagger}]= 1$.
Now, put $N= \hat{a}^{\dagger} \hat{a}$ then $[\hat{a}^{\dagger}, N]= -\hat{a}^{\dagger}$ and
$[\hat{a}, N]= \hat{a}$.  Thus, $N \hat{a} = \hat{a}N - \hat{a}= \hat{a} (N-1)$ and
$N \hat{a}{\dagger} = \hat{a}{\dagger} (N+1)$.  Now, $\langle \psi | N | \psi \rangle \geq 0$ and is $0$ iff
$\hat{a} | \psi \rangle= 0$.  Now suppose $\lambda$ is an eignevalue of $N$:
$N |\psi \rangle = \lambda | \psi \rangle$.  By the above,
$N \hat{a}^{\dagger} | \psi \rangle = \hat{a}^{\dagger} (\lambda +1) | \psi \rangle$ and
$N \hat{a} | \psi \rangle = (\lambda -1) \hat{a} | \psi \rangle$,
so $\lambda+1$ and $\lambda-1$ are also an eigenvalues of $N$ corresponding with eigenvectors
$\hat{a}{\dagger}| \psi \rangle$ and $\hat{a} | \psi \rangle$ respectively.  Repeated application yields
$\hat{a}^n | \psi \rangle$ is an eigenvector of $N$ with eigenvalue $(\lambda -1) \ldots (\lambda -n)$
If $\lambda \notin {\mathbb Z}^{+}$, $\lambda -n$ is eventually negative and is never $0$  but then,
$\langle \psi | N | \psi \rangle \leq 0$ eventually which is impossible, so $\lambda \in {\mathbb Z}^{+}$.
Putting $| n \rangle= c_n \hat{a}^{\dagger} | 0 \rangle$ with $\langle n | n \rangle = 1$.  We get
$c_n = {\frac {c_{n-1}} {\sqrt n}}$.
$0= \langle x | \hat{a} | 0 \rangle =
({\frac {(m \omega)}{2 \hbar}})^{1/2} (\hat{x} + {\frac {\hbar}{m \omega}} {\frac {\partial} {\partial x}}) \langle x | 0 \rangle$.
$\psi_0(x) = \langle x | 0 \rangle$ satisfies $({\frac {\partial} {\partial x}} +{\frac {m \omega} {\hbar}}) \psi_0(x) =0$ and
$\psi_0(x)= N e^{{\frac {m \omega} {2 \hbar}} x^2}$ while
$\psi_1(x)= \langle x | 1 \rangle= ({\frac {2m \omega} {\hbar}})^{1/2} x \psi_0(x)$.
\\
\\
{\bf Multiple particles:}  For a two particle SHO,
$H= H_1 + H_2$ with $H_i= {\frac  {\hat{p}_i^2}{2m}} + {\frac 1 2} m \omega^2 \hat{x}_i^2$ with
$H_i |n \rangle_i = \hbar \omega (n + {\frac 1 2}) |n\rangle_i$ and
$ |n_1, n_2\rangle= |n_1\rangle  \otimes |n_2\rangle $ and $H |n_1, n_2 \rangle= \hbar \omega (n_1 + n_2 + 1) |n_1 , n_2 \rangle$.
\\
\\
{\bf Identical particles:}
$U x_i U^{-1}= x_{\sigma(i)}$, $U p_i U^{-1}= p_{\sigma(i)}$ and $U H U^{-1}= H$ with $H= \sum_i H_i$ and
$H_i | \psi_r \rangle = E_r | \psi_r \rangle$.  A basis for ${\cal H}_1 \otimes {\cal H}_2$ is
$ |\psi_r\rangle_1 |\psi_s\rangle_2; {\frac 1 {\sqrt 2}} ( |\psi_r\rangle_1 |\psi_s\rangle_2 + |\psi_s\rangle_1 |\psi_r\rangle_2), r \neq s $.
\\
\\
{\bf Spinless Bosons:}
For $N$ spinless bosons, $H= \sum_i H_i$.  The symmetric basis is
${\frac 1 {\sqrt {N!}}}  \sum_{\sigma} ( | \psi_{\sigma(1)} \rangle_1 | \psi_{\sigma(2)} \rangle_2 \ldots | \psi_{\sigma(N)} \rangle_N)$.
Note that spinless bosons are fully characterized by $x, p$.
\\
\\
{\bf Spin ${\frac 1 2}$ Fermions:}
State includes spin $| s \rangle$, the full state is $|x\rangle | s \rangle$ with
$\psi(x,s)= | x, s \rangle$ and $\psi_s(x)= \langle x, s | \psi \rangle$.  The basis for two particles is
$\chi_A(s_1 , s_2)= {\frac 1 {\sqrt {2}}} ( \chi_{1/2}(s_1) \chi_{-1/2}(s_2) - \chi_{-1/2}(s_1) \chi_{1/2}(s_2))$.
\\
\\
{\bf Bell's argument restated:}
Consider two spin-${\frac 1 2}$ electrons.
$|\uparrow\rangle$ and $|\downarrow\rangle$ are eigenvectors of $S_3= {\frac 1 2} \hbar \sigma_3$.
$\chi_{\uparrow}=
\left(
\begin{array}{c}
1\\
0\\
\end{array}
\right)$,
$\chi_{\downarrow}=
\left(
\begin{array}{c}
0\\
1\\
\end{array}
\right)$.
$\sigma \cdot n=
\left(
\begin{array}{cc}
cos(\theta) & sin(\theta)\\
sin(\theta) & -cos(\theta)\\
\end{array}\right)$ and $n= ( sin(\theta), 0, cos(\theta))$.
$ \chi_{\uparrow;n}= cos({\frac {\theta} 2}) \chi_{\uparrow}+
sin({\frac {\theta} 2}) \chi_{\downarrow} $ and
$ \chi_{\downarrow;n}= -sin({\frac {\theta} 2}) \chi_{\uparrow}+
cos({\frac {\theta} 2}) \chi_{\downarrow}$,
set $| \Phi \rangle= {\frac 1 {\sqrt 2}} [
| \uparrow \rangle_1 | \downarrow \rangle_2 + | \downarrow \rangle_1 | \uparrow \rangle_2 ]$.  If there are
``hidden variables,'' we'd expect a distribution $0 \leq
p(S_z^{(1)}, S_n^{(1)}, S_m^{(1)}, S_z^{(2)}, S_n^{(2)}, S_m^{(2)}) \leq 1$.  Define
$p_{bc}(b,c)= \sum_a p(a,b,c)$, $p_{ac}(a,c)= \sum_b p(a,b,c)$, $p_{ab}(a,b)= \sum_c p(a,b,c)$.
$ p_{bc}(1, -1) \leq p_{ab}(1, 1) + p_{ac}(-1, -1) $.  Applying this to two electrons,
$
P( S_n^{(1)}=1, S_m^{(2)}=1) \leq
P( S_z^{(1)}=1, S_n^{(2)}=-1)+
P( S_z^{(1)}=-1, S_m^{(2)}=1)
$.
$P( S_n^{(1)}=1, S_n^{(2)}=-1) = cos^2({\frac {\theta} 2})$ and
$ P( S_z^{(1)}=1, S_m^{(2)}=-1)= cos^2({\frac {\phi+\theta} 2}) $ which is not generally true.
\\
\\
{\bf Definition:}
The \emph{Compton wavelength} is  $\Delta x \geq {\frac {\hbar} {mc}}$.
\\
\\
{\bf Lie Groups:}
$R(\theta, \hat{z})=
\left(
\begin{array}{ccc}
cos(\theta) & sin(\theta) & 0\\
-sin(\theta) & cos(\theta) & 0\\
0 & 0 & 1\\
\end{array}
\right).$
$v'= R(\theta, \hat{z})$.  $L_z= {\frac 1 i} {\frac {\partial R(\theta, \hat{z})} {\partial \theta}}_{|\theta= 0}$.
$R(\theta, \hat{z})= 1 + \delta \theta L_z$.  $SO(3)$ is generated by
$\langle L_x , L_y , L_z \rangle$, $R(\theta, \hat{z})=
lim_{N \rightarrow \infty} R({\frac {\theta} {N}}, \hat{z})$.  $[L_i, L_j]= i \epsilon_{ijk} L_k$.
$J^2 | j, m\rangle= j(j+1) |j, m \rangle$, $J_z |j,m\rangle= m |j, m \rangle$.  $SU(2)$ acts on spinors.
$A= A(\theta, \hat{n})= exp(i {\frac {\theta} 2} \sigma \cdot n)= 1+ i \theta J \cdot n$.  $J \cdot n=
{\frac 1 i} {\frac {\partial A(\theta, \hat{n})} {\partial \theta}}_{| \theta=0}= {\frac {\sigma} 2} \cdot n$.
$\langle {\frac {\sigma_x} 2}, {\frac {\sigma_y} 2}, {\frac {\sigma_z} 2}, \rangle$ is a basis for $SU(2)$.
\\
\\
{\bf Representations of symmetry groups:}
Let $G$ be the symmetry group of a physical system with Hamiltonian $H$, if $g \in G$ then
$[g,H]=0$ (Note: it is sufficient if this holds on generators.).
$\Phi: G \rightarrow V_n(F)$ is a representation of the system.  Given a Hermitian representation,
$\Phi(g)$ of $G$, $\Phi'(g)= exp(i \Phi(g))$ is a unitary representation.  These are called $D$ functions.
$\langle j,m' | U(\phi, \theta, \chi ) | j, m \rangle= D^{(j)}_{m', m} (\phi, \theta, \chi)=
e^{im \phi} d^{(j)}_{m', m}(\theta) e^{-i m \chi}$.  $J^2$ acts on a $2j+1$ dimensional Hilbert space.
$J_{\pm}= J_x \pm i J_y$ and $J_{\pm} | j, m \rangle=
{\sqrt {j(j+1)-m(m \pm 1)}} | j, m \pm 1 \rangle$.  Spin
is a representation of $SU(2)$.  There is a homomorphism of $SU(2) \rightarrow SO(3)$ with kernel $\{ \pm 1\}$
given by $A(\theta, \hat{n}) \mapsto R(\theta, \hat{n})$ or
$
\left(
\begin{array}{cc}
e^{-i \alpha/2} & 0\\
0 & e^{i \alpha/2}\\
\end{array}
\right)
\mapsto
\left(
\begin{array}{ccc}
cos(\alpha) & sin(\alpha) & 0\\
-sin(\alpha) & cos(\alpha) & 0\\
0 & 0 & 1\\
\end{array}
\right)$.
The Klein Gordon equation is invariant under the Lorentz group.  The \emph{Poincare group} is the Lorentz group plus
translations $T(x)= \Lambda x+a$.  $|\Lambda_0^0| \ge 1$.
\begin{center}
\begin{tabular} {|c||c|c|}
\hline
Subgroup & $det$ & $\Lambda_0^0$\\
\hline
$L_+^{\uparrow}$ & $1$ & $\geq 1$\\
$L_+^{\downarrow}$ & $1$ & $\leq -1$\\
$L_-^{\uparrow}$ & $-1$ & $\geq 1$\\
$L_-^{\downarrow}$ & $-1$ & $\leq -1$\\
\hline
\end{tabular}
\end{center}
Two fundamental transformations are ``rotations'' and ``boosts''(time change).
Rotations, $(J_1, J_2, J_3)$: $\Lambda_R(\theta, \hat{z})= exp(i \theta)$.
Boosts, $(K_1, K_2, K_3)$: $\Lambda_B(\theta, \hat{z})= exp(\theta)$.
$[J^i, J^j]= i \epsilon^{ijk} J^k$,
$[J^i, K^j]= i \epsilon^{ijk} K^k$,
$[K^i, K^j]= i \epsilon^{ijk} J^k$.
$M= exp({\frac i 2} \theta \cdot \sigma) exp(\pm {\frac 1 2} \phi \cdot \sigma)$,
$\hat{\phi}= \phi \cdot n$,
$\hat{\sigma}= \sigma \cdot n$.
Type I representation ($M$): $J= {\frac {\sigma} 2}$, $K= -i {\frac {\sigma} 2}$.
Type II representation ($\overline{M}$): $J= {\frac {\sigma} 2}$, $K= -i {\frac {\sigma} 2}$.
$
\left(
\begin{array}{c}
\xi\\
\eta\\
\end{array}
\right)
\rightarrow
\left(
\begin{array}{cc}
M(\Lambda) & 0\\
0 & {\overline M}(\Lambda)\\
\end{array}
\right)
\left(
\begin{array}{cc}
\xi\\
\eta\\
\end{array}
\right)
$, ${\overline M}(\Lambda)= \epsilon M^* \epsilon^{-1}, \epsilon= i \sigma^2$.
\\
\\
{\bf Symmetries and fields:}
$(t,x) \rightarrow (t, -x)$ is the \emph{parity} symmetry.
$\phi \rightarrow {\overline \phi}^T$ is the \emph{charge conservation} symmetry.
Spin $0$ scalar fields have Lagrangian ${\cal L}= {\frac 1 2} \partial_{\mu} \phi \partial_{\nu} \phi - {\frac 1 2} M^2 \phi^2$.
Spin ${\frac 1 2}$ Dirac fields have Lagrangian ${\cal L}= {\overline \phi}(i\partial_{\mu} \partial_{\nu} - M) \phi$.
Spin $1$ vector fields have Lagrangian
${\cal L}= -{\frac 1 4} F_{\mu, \nu} F^{\mu, \nu} + {\frac 1 2} m^2 V_{\mu} V^{\mu}=
{\frac {\lambda} 2} (\partial_{mu} V^{\mu})^2$.  The \emph{Poisson Bracket}
is $[A, B]_P= {\frac {dA} {dq}} {\frac {dB} {dp}} - {\frac {dA} {dp}} {\frac {dB} {dq}}$.
${\frac {dO}{dt}}= [O, H]_P$ and
${\frac {dO}{dq}}= [O, p]_P$.
\\
\\
{\bf Summary:}
\\
(1) Spin $0$ boson, Klein Gordon equation.
${\cal L}= {\frac 1 2} (\dot{\phi}_{\alpha} \dot{\phi}_{\alpha} - \mu^2 \phi^2)$,
$\pi(x)= {\frac {\partial {\cal L}} {\partial {\dot{\phi}}}}$.
$[\phi, \phi^+]= i \hbar c^2 \delta(x-x')$.
$\phi^+= \sum_k ({\frac {\hbar c^2}{2 V \omega_k}})^{1/2} a+(k) e^{-ikx/\hbar}$.
$H= \sum_k \hbar \omega_k (a^{\dagger} a(k) + {\frac 1 2})$.
$P= \sum_k \hbar k (a^{\dagger} a(k) + {\frac 1 2})$.
Propagator:
$[\phi^+, \phi]= i \hbar c \Delta(x-y)$
$\Delta(x)= {\frac {i c}{(2 \pi)^3}} \int d^4 k \thinspace \delta(k^2-\mu^2) \epsilon(k) e^{-ikx}$.
\\
(2)
Spin $1/2$ fermion, Dirac equation.
${\cal L}= c {\overline {\phi}}[c \alpha \cdot (i \hbar \nabla) + \beta m c^2]\phi$,
$\pi(x)= {\frac {\partial {\cal L}} {\partial {\dot{\phi}}}}$.
$\{ a_r, a_s^{\dagger} \}= \delta_{rs}$, $\{ a_r, a_s \}= 0$.
$H= c \int d^3x {\overline {\phi}}
[-i \hbar c \gamma^j {\frac{\partial}{\partial x^j}}+mc^2]
\phi$.
$P= -i \hbar \int  (\phi^{\dagger} \nabla \phi(k) $.
Propagator:
$S_F(x)= {\frac {- \hbar}{(2 \pi \hbar)^4}} \int d^4 p \;
e^{-ip \cdot x/ \hbar} {\frac {\gamma^{\mu} p_{\mu} + mc} {p^2-m^2c^2+1\epsilon}}$.
\\
(3)
Spin $1$ photon, Maxwell's equation.
${\cal L}= -{\frac 1 4} (F_{\mu \nu} F^{\mu \nu} - {\frac 1 c} s_{\mu}(x) A^{\mu}(x))$.
$[ A^{\mu}(x), A^{\mu}(x')]= i \hbar x D^{\mu \nu}(x-x')$,
$H= \sum_{k, r} d^3 x \thinspace \hbar \omega_k \eta_r a_r^{\dagger}(k) a_r(k)$.
\section{More on Dirac and QED}
{\bf The equations:}
Spin $0$ particles are characterized by Klein-Gordon.
Spin ${\frac 1 2}$ particles are characterized by Dirac.
Spin $1$ particles are characterized by Proca.
\\
\\
{\bf Dirac:} $(i \hbar \gamma^{\mu} {\partial}_{\mu} - mc= 0)$.  The time
independent solution is
$\psi(x) = a e^{- i k \cdot x} u(k)$ for particles and
$\psi(x) = a e^{i k \cdot x} u(k)$ for antiparticles.
$u u^{\dagger} = {\frac {2E} {c}}$.  $N= {\sqrt {\frac
{E+mc^2} {2}}}$.
$\gamma^0=
\left(
\begin{array}{cc}
1 & 0\\
0 & -1\\
\end{array}
\right)$,
$\gamma^i=
\left(
\begin{array}{cc}
0 & \sigma^i\\
-\sigma^i & 0\\
\end{array}
\right)$.
$\psi=
\left(
\begin{array}{c}
u^{(1)}\\
u^{(2)}\\
v^{(1)}\\
v^{(2)}\\
\end{array}
\right)$.
$u^{(1)}= N
\left(
\begin{array}{c}
1 \\
0 \\
{\frac {c p_z} {E+mc^2}} \\
{\frac {c (p_x + i p_y)} {E+mc^2}} \\
\end{array}
\right)$.
$u^{(2)}= N
\left(
\begin{array}{c}
0 \\
1 \\
{\frac {c (p_x + i p_y)} {E+mc^2}} \\
{\frac {c p_z} {E+mc^2}} \\
\end{array}
\right)$.
$v^{(1)}= N
\left(
\begin{array}{c}
{\frac {c (p_x - i p_y)} {E+mc^2}} \\
{\frac {-c p_z} {E+mc^2}} \\
0 \\
1 \\
\end{array}
\right)$.
$v^{(2)}= N
\left(
\begin{array}{c}
{\frac {-c p_z} {E+mc^2}} \\
{\frac {c (p_x - i p_y)} {E+mc^2}} \\
1 \\
0 \\
\end{array}
\right)$.
For photon, $A_{\mu}= ae^{i p \cdot x / \hbar} \epsilon^{\mu}(p)$, $\epsilon^{\mu}$ is the polarization vector.
$\epsilon^0 = 0$ so $\epsilon \cdot p = 0$,
$\epsilon^1 = (1, 0, 0)$,
$\epsilon^2 = (0, 1, 0)$.
\begin{center}
\begin{tabular} {|c|c|c|}
\hline
Spinor component & Particle & Spin\\
\hline
$u^{(1)}$ & $e^-$ & up\\
$u^{(2)}$ & $e^-$ & down\\
$v^{(1)}$ & $e^+$ & up\\
$v^{(2)}$ & $e^+$ & down\\
\hline
\end{tabular}
\end{center}
$S= {\frac {\hbar} 2}
\left(
\begin{array}{cc}
\sigma & 0 \\
0 & \sigma\\
\end{array}
\right)$.
\\
\\
{\bf The vacuum:}  The \emph{Casimir effect} is the force between two plates seperated
by $d$ due to vacuum fluctuation of the EM field.   In QFT,
$P= \int {\frac {d^3p} {(2 \pi)^3}} p a_p^{\dagger} a_p$.
We can recover the position operator by
$X= \int d^3 x (x \phi^{\dagger}(x)\phi(x)$ giving $X |x\rangle= x |x\rangle$.
\\
\\
{\bf Summary:}
\begin{center}
\begin{tabular} {|c|c|c|c|}
\hline
Characteristic & Electrons & Positrons & Photons\\
\hline
Ket &
$ae^{-i p \cdot x / \hbar} u^{(s)}(p)$ & $ae^{i p \cdot x / \hbar} v^{(s)}(p)$ & $A_{\mu}= a e^{-i \epsilon \cdot x/ \hbar} \epsilon_{\mu}^{(s)}$\\
EOS &
$(\gamma^{\mu} p_{\mu} -mc)u=0$ & $(\gamma^{\mu} p_{\mu} +mc)v=0$ &  $\Box^2 A^{\mu}= 0$\\
Adjoint & ${\overline u}(\gamma^{\mu} p_{\mu} -mc)=0$ & ${\overline v}(\gamma^{\mu} p_{\mu} +mc)=0$ & - \\
Constraints &
${\overline u}^{(1)} \cdot u^{(2)} = 0$, ${\overline u} u = 2mc$ &
${\overline v}^{(1)} \cdot v^{(2)} = 0$ , ${\overline v} v = -2mc$ & $p^{\mu} \epsilon_{\mu}= 0$\\
Basis & $\sum_s {\overline u}^{(s)} \cdot u^{(s)} = (\gamma^{\mu} p_{\mu} +mc)$ &
$\sum_s {\overline v}^{(s)} \cdot v^{(s)} = (\gamma^{\mu} p_{\mu} -mc)$ &
$\sum_s \epsilon_i^{(s)} (\epsilon_j^{(s)})^*= \delta_{ij} - \hat{p}_i \hat{p}_j $\\
\hline
\end{tabular}
\end{center}
{\bf Feynman rules for QED:}  The rules for QED.
To calculate ${\cal M}$: (1) To eac external line, draw directed segment labeled by momentum,(2) for electron into (out of)
vertex use $u$ (${\overline u}$) switch for positron, (3) use vertex coupling $ig_e \gamma^{\mu}$, $g_e= e {\sqrt {\frac {4 \pi} {\hbar c}}}$,
(4) Use propagators ${\frac {i(\gamma^{\mu} q_{\mu} + mc)} {q^2-m^2 c^2}}$ for $e^+, e^-$ and ${\frac {ig_{\mu, \nu}} {q^2}}$ for
$\gamma$, (5) apply $\delta$ function $\delta^{(4)}(k_1+k_2+k_3)$ (inwards) at each vertex, (6) integrate over internal momentum
${\frac {d^4 q} {(2 \pi)^4}}$, (7) cancel the $\delta$'s replacing them wiht $i$, (8) do anti-symmetrization in diagrams interchanging only
direction.  Example: \emph{electron-electron} scattering ${\cal M}=
- {\frac {g_e^2}{(p_1-p_3)^2}} [{\overline u}(3) \gamma^{\mu}u(1)] [{\overline u}(3) \gamma_{\mu} u(2)]
+ {\frac {g_e^2}{(p_1-p_4)^2}} [{\overline u}(4) \gamma^{\mu}u(1)] [{\overline u}(3) \gamma_{\mu} u(2)] $.
\chapter{Quantum Computing}
\section{Basics}
${\cal G}= \langle CNOT, X,Y,Z,H,T \rangle$ can be efficiently simulated on a probabalistic
computer if there is little entanglement.  This is the theorem of \emph{Gottesman-Knill}.
\\
\\
$CNOT:
{\frac {(| 0 \rangle + |1 \rangle)} {\sqrt 2}}
{\frac {(| 0 \rangle - |1 \rangle)} {\sqrt 2}}
\mapsto
{\frac {(| 0 \rangle - |1 \rangle)} {\sqrt 2}}
{\frac {(| 0 \rangle - |1 \rangle)} {\sqrt 2}}
$.  Note that
${\frac {(| 0 \rangle + |1 \rangle)} {\sqrt 2}}$ and
${\frac {(| 0 \rangle - |1 \rangle)} {\sqrt 2}}$ are eigenvectors of $CNOT$.
$CNOT: |b \rangle
{\frac {(| 0 \rangle - |1 \rangle)} {\sqrt 2}} = (-1)^b |b \rangle
{\frac {(| 0 \rangle - |1 \rangle)} {\sqrt 2}}
$
and
$CNOT:
(\alpha_0 |0 \rangle +
(\alpha_1 |1 \rangle)
{\frac {(| 0 \rangle - |1 \rangle)} {\sqrt 2}} =
(\alpha_0 |)\rangle-\alpha_1 |1\rangle)
{\frac {(| 0 \rangle - |1 \rangle)} {\sqrt 2}}
$.
\\
\\
Let
$U_f: |x\rangle
|y\rangle \mapsto
|x\rangle |y \oplus f(x)\rangle $.
$U_f: |x\rangle {\frac {(| 0 \rangle - |1 \rangle)} {\sqrt 2}} =
|x\rangle=
{\frac {(| 0  \oplus f(x)\rangle - |1 \oplus f(x) \rangle)} {\sqrt 2}} =
(-1)^{f(x)}
|x\rangle
{\frac {(| 0 \rangle - |1 \rangle)} {\sqrt 2}} $ and
$U_f:
(\alpha_0 |0 \rangle+
(\alpha_1 |1 \rangle)
{\frac {(| 0 \rangle - |1 \rangle)} {\sqrt 2}} \mapsto
((-1)^{f(0)}\alpha_0 |0\rangle-(-1)^{f(1)}\alpha_1 |1\rangle)
{\frac {(| 0 \rangle - |1 \rangle)} {\sqrt 2}}
$. $H$ decodes information encoded in the phase.
\\
\\
{\bf Proof of no-cloning theorem:}  Suppose such a unitary transforamtion exists.
$U(|\psi\rangle |0\rangle)= |\psi\rangle |\psi \rangle$ and
$U(|\phi\rangle |0\rangle)= |\phi\rangle |\phi \rangle$.
On one hand,
$U( a | \phi \rangle + b | \psi \rangle )= a | \phi \rangle | \phi \rangle + b | \psi \rangle \psi \rangle )$.
On the other hand,
$U( a | \phi \rangle + b | \psi \rangle |0\rangle )= (a | \phi \rangle + b | \psi \rangle )
(a | \phi \rangle + b | \psi \rangle )$.  This is a contradiction.  There is no ``approximate cloning'' either.
\\
\\
{\bf Phase Estimation Problem:}
Given
$|\psi\rangle= {\frac 1 {{\sqrt {2^n}}}} \sum_{|y\rangle} e^{2 \pi i \omega y} |y\rangle$.
($\omega= \{0,1\}$.
$QFT=
(|0>+e^{2 \pi i 2^{n-1} \omega}|1\rangle) \otimes
(|0>+e^{2 \pi i 2^{n-2} \omega}|1\rangle) \otimes \ldots A.$
$R_n=
\left(
\begin{array}{cc}
1 & 0\\
0 & e^{2 \pi i/ 2^n}\\
\end{array}
\right)
$.
\\
\\
{\bf Hidden subgroup problem:}  Let $f: G \rightarrow X$,
$\exists S<G$ with $f(x)=f(y)$ iff $x+S=y+S$.
\\
\\
\begin{quote}
\emph{Deutsch:}
$G= {\mathbb Z}_2$, $X= \{ 0, 1 \}$.
$S= \{ 0 \}$ if $f$ is balenced.
$S= \{ 0,1 \}$ if $f$ is constant.
\\
\\
\emph{Order Finding:}
$G= {\mathbb Z}$, $X= H<G$, $r=|a|, a \in H$.
$S= r{\mathbb Z}$ so $S$ gets $r$.
\\
\\
\emph{Discrete Log:}
$G= {\mathbb Z}_r \times {\mathbb Z}_r$, $X= H<G$, $a:a^r=1$,
$b=a^k$, $f(x_1, x_2) =a^{x_1} b^{x_2}$.
$f(x_1, x_2)= f(y_1 , y_2)$ iff
$a^{x_1-y_1}b^{x_2-y_2}=1$ iff
$\langle z_1-y_1, x_2-y_2\rangle= (t,-tk), t= 0,1, \ldots, r-1$.
$S=\langle 1, -k \rangle$, $k= log(b)$.
\\
\\
\emph{Hidden Linear function:}
$G= {\mathbb Z} \times {\mathbb Z}$, $g \in S_n$.
$h:
{\mathbb Z} \times {\mathbb Z}
\rightarrow
{\mathbb Z}_n
$ by $h(x,y)= x+ya \jmod{n}$.  $f= g \circ h$,
$G= {\mathbb Z}$, $X= \{ 0, 1 \}$.
$S= \langle -a, r \rangle$.
\\
\\
\emph{Abelian Stabilizer:}
$G$ acts on $X$, $f_x: g \rightarrow X$ by $f_x(g)=x^g$.  $S= G_x$.
\\
\\
\emph{Graph Isomorphism:}
$G= S_n$ and ${\cal G}_n$ is a graph on $n$ vertices.  For $\sigma \in G$,
$f_{\cal G}(\sigma({\cal G}))$.  The hidden subgroup is the automorphisms of ${\cal G}_n$.
\end{quote}
${\cal G}= \langle CNOT, X,Y,Z,H,T \rangle$ can be efficiently simulated on a probabalistic
computer if there is little entanglement.  This is the theorem of \emph{Gottesman-Knill}.
\\
\\
$CNOT:
{\frac {(| 0 \rangle + |1 \rangle)} {\sqrt 2}}
{\frac {(| 0 \rangle - |1 \rangle)} {\sqrt 2}}
\mapsto
{\frac {(| 0 \rangle - |1 \rangle)} {\sqrt 2}}
{\frac {(| 0 \rangle - |1 \rangle)} {\sqrt 2}}
$.  Note that
${\frac {(| 0 \rangle + |1 \rangle)} {\sqrt 2}}$ and
${\frac {(| 0 \rangle - |1 \rangle)} {\sqrt 2}}$ are eigenvectors of $CNOT$.
$CNOT: |b \rangle
{\frac {(| 0 \rangle - |1 \rangle)} {\sqrt 2}} = (-1)^b |b \rangle
{\frac {(| 0 \rangle - |1 \rangle)} {\sqrt 2}}
$
and
$CNOT:
(\alpha_0 |0 \rangle +
(\alpha_1 |1 \rangle)
{\frac {(| 0 \rangle - |1 \rangle)} {\sqrt 2}} =
(\alpha_0 |)\rangle-\alpha_1 |1\rangle)
{\frac {(| 0 \rangle - |1 \rangle)} {\sqrt 2}}
$.
\\
\\
Let
$U_f: |x\rangle
|y\rangle \mapsto
|x\rangle |y \oplus f(x)\rangle $.
$U_f: |x\rangle {\frac {(| 0 \rangle - |1 \rangle)} {\sqrt 2}} =
|x\rangle=
{\frac {(| 0  \oplus f(x)\rangle - |1 \oplus f(x) \rangle)} {\sqrt 2}} =
(-1)^{f(x)}
|x\rangle
{\frac {(| 0 \rangle - |1 \rangle)} {\sqrt 2}} $ and
$U_f:
(\alpha_0 |0 \rangle+
(\alpha_1 |1 \rangle)
{\frac {(| 0 \rangle - |1 \rangle)} {\sqrt 2}} \mapsto
((-1)^{f(0)}\alpha_0 |0\rangle-(-1)^{f(1)}\alpha_1 |1\rangle)
{\frac {(| 0 \rangle - |1 \rangle)} {\sqrt 2}}
$. $H$ decodes information encoded in the phase.


%!TEX root = ../../report.tex

\subsubsection{Fractals} % (fold)
\label{ssub:fractals}

A fractal is defined in \cite{Ebert2002} as ``a geometrically complex object, the complexity of which arises through the repetition of a given form over a range of scales''.
This concept is observed in some forms that exist in nature. Trees, mountains, coastlines and the network of neurons on a human cortex can be seen as examples of fractals. Natural shapes tend to be irregular and fragmented and exhibit a complexity incomparable to regular geometry \cite{mandelbrot1984fractal}.
Fractals were proposed to be seen as a new form of symmetry \cite{Ebert2002}, \emph{Dilation Symmetry}, which is when an object is invariant over a change of scale. This invariance might be only qualitatively and not exact. For instance, a river network exhibit dilation symmetry if \textit{zooming in} in some part looks the same as the whole image. As this example, many others show dilated symmetry such as clouds, tree branches and some vegetables as shown in Figure~\ref{fig:NFractals}.

\begin{figure}
        \centering
        %\begin{subfigure}[b]{0.4\textwidth}
                \includegraphics[width=0.45\textwidth]{img/Theory/Fractals/Leaf.png}
        %        \label{fig:Fleaf}
        %\end{subfigure}%
        ~ %add desired spacing between images, e. g. ~, \quad, \qquad, \hfill etc.
          %(or a blank line to force the subfigure onto a new line)
        %\begin{subfigure}[b]{0.4\textwidth}
                \includegraphics[width=0.45\textwidth]{img/Theory/Fractals/Fractal_Broccoli.jpg}
        %        \label{fig:Fbrocoli}
        %\end{subfigure}
        \caption{Fractals in Nature}
        \label{fig:NFractals}
\end{figure}

This idea was applied in maths and resulted in a new area in this science called fractal mathematics. The objective of this field is to describe very complex shapes with simple rules such as repeating a substitution pattern.

\begin{figure}[htbp]
	\centering
	\includegraphics[width=0.7\textwidth]{img/Theory/Fractals/Fractal1_1000.png}
	\caption{Geometric Fractals}
	\label{fig:GFractals}
\end{figure}

In Figure~\ref{fig:GFractals} there are four examples of Geometric Fractals, with the first five iterations of each one. All of them are built by the substitution of a part of the image by another one.

The example of the second row is known as the Koch snowflake. In this example, at each iteration, all the line segments are replaced by four segments with 1/3 of the size of the original one with the two in the middle being placed in a angle forming a equilateral triangle with the original line that is removed.

It is clear that the detail that is presented in each iteration increases as the scale changes. There is the concept of \emph{Fractal Dimension} that tries to measure this evolution, in which the detail in a pattern changes in comparison with the scale in which it is measured.

As stated before, the world is visually very complex, so when synthesizing worlds, ``\emph{complexity} equals \emph{work}''\cite{Ebert2002}. This work can be done by the programmer/artist or by a computer. Fractals as being defined as a simple mathematical function, it is relatively easy to implement a procedure that model one fractal.

This technique is used to model many natural forms that present fractal properties. Mountains, for instance, are usually modeled using of fractals. Other natural forms that present fractal properties are trees, river systems, lightning or vascular systems in living beings.

% subsection fractals (end)