EM from U

\documentclass[12pt]{report} \setcounter{tocdepth}{5} \setcounter{secnumdepth}{5} \usepackage{quantikz} \usepackage{mathtools} \usepackage{atbegshi} \usepackage{braket} \usepackage{geometry} \geometry{ a4paper, total={170mm,257mm}, left=20mm, top=20mm, } \usepackage{calligra} \DeclareMathAlphabet{\mathcalligra}{T1}{calligra}{m}{n} \DeclareFontShape{T1}{calligra}{m}{n}{←>s*[2.2]callig15}{} \newcommand{\scripty}[1]{\ensuremath{\mathcalligra{#1}}} \usepackage{bm} \newcommand{\uvec}[1]{\hat{#1}} \setlength{\footskip}{45pt} \renewcommand{\familydefault}{\sfdefault} \usepackage{float} \usepackage{sfmath} \usepackage{parskip} \usepackage{graphicx} \usepackage{amsmath, fourier} \everymath{\displaystyle} \begin{document} tableofcontents \chapter{Electrodynamics} \section{Charges} {\bfseries Electric Charge:}\par [electrostatic impulse] $\left[Q=q(r^{\prime })=\int \rho (r^{\prime })d^3{r}^{\prime }\right]$ {\bfseries Electrostatics:}\par [field equations] $\left[\vec{E}(r)=k\frac{q(r^{\prime })}{\text{scripty}{r}^2}\widehat{\text{scripty}r}\left[\frac{Q}{4\pi {ϵ}_0{r}^2}\hat{r}\right]\sim k\int \frac{\rho (r^{\prime })}{\text{scripty}{r}^2}\text{uvec}\text{scripty}r{d}^3{r}^{\prime }\right]:\left[\vec{\nabla }\times \vec{E}=0,\vec{\nabla }\cdot \vec{E}=\frac{\rho }{{ϵ}_0}\right]$ [derives as] $\left[{\phi }_E=\oint \vec{E}(r)\cdot d\vec{a}(r)=\sum \int k\frac{q_i(r^{\prime })}{4\pi {ϵ}_0{r}^2}\hat{r}\cdot (r^2\sin \theta d\theta d\varphi \hat{r})=\frac{\sum q_i(r^{\prime })}{{ϵ}_0}=\frac{Q_{\text{enc}}(r^{\prime })}{{ϵ}_0}\right]$

$\left[\vec{\nabla }\cdot \vec{E}(r)=k\int \vec{\nabla }\cdot \left(\frac{\text{uvec}\text{scripty}r}{\text{scripty}{r}^2}\right)\rho (r^{\prime })d^3{r}^{\prime }=k\int 4\pi {\delta }^3(\text{scripty}r)\rho (r^{\prime })d^3{r}^{\prime }=\frac{\rho (r)}{{ϵ}_0}\right]$

$\left[{\lambda }_E={\int }_a^b\vec{E}(r)\cdot d\vec{l}(r)={\int }_a^{b}k\frac{q(r^{\prime })}{4\pi {ϵ}_0{r}^2}\hat{r}\cdot (dr\hat{r}+rd\theta \hat{\theta }+r\sin \theta d\theta \hat{\varphi })=k\left(\frac{1}{r_a}-\frac{1}{r_b}\right)\right]$

$\left[\vec{\nabla }\times \vec{E}(r)=k\int \vec{\nabla }\times \left(\frac{\text{uvec}\text{scripty}r}{\text{scripty}{r}^2}\right)\rho (r^{\prime })d^3{r}^{\prime }=0\right]$

[which allows] $\left[\vec{E}(r)=-\vec{\nabla }V(r)\text{ so, }V(r)=-{\int }_O^r\vec{E}(r)\cdot \overrightarrow{dl}(r)\right]$ {\bfseries Physical Meaning Of V:} \par [work to take a unit +ve charge from the reference O to designated r which for, $O=\infty$,] $\left[V(r)=k\frac{q(r^{\prime })}{\text{scripty}r}\sim k\int \frac{\rho (r^{\prime })}{\text{scripty}r}{d}^3{r}^{\prime }={\sum }_{n=0}^{\infty }\frac{k}{r^{n+1}}\int {r^{\prime }}^n{P}_n(\cos {\theta }_{r-r^{\prime }}^{\prime })\rho (r^{\prime })d^3{r}^{\prime }\right]$[individuals] $\left[V_{\text{mon}}=k\frac{Q}{r}:{\vec{E}}_{\text{mon}}=k\frac{Q}{r^2},V_{\text{dip}}=k\frac{\vec{p}\cdot \text{uvec}r}{r^2}:{\vec{E}}_{\text{dip}}=k\frac{3(\vec{p}\cdot \text{uvec}r)\text{uvec}r-\vec{p}}{r^3}\left[\vec{p}=\int {\vec{r}}^{\prime }\rho (r^{\prime })d^3{r}^{\prime }=q\vec{d}\right]\right]$ \par [formation of electric system] $\left[w=\frac{1}{2}q(r)V(r)\sim \frac{1}{2}\int \rho (r)V(r)d^{3}r\sim \frac{{ϵ}_0}{2}\left[-\int \vec{E}\cdot (\nabla V)d^{3}r+\oint V\vec{E}\cdot d\vec{a}\right]\sim \frac{{ϵ}_0}{2}\int \vec{E}(r{)}^2{d}^{3}r\right]$ [other systems] $\left[V_{\text{spherical}}=\frac{kQ}{\text{scripty}r},V_{\text{linear}}=-2k\lambda \ln \text{scripty}r,V_{\text{surfacial}}=-\frac{\rho \text{scripty}r}{2{ϵ}_0}\right]$ $\left[E_{\text{spherical}}=\frac{kQ}{\text{scripty}{r}^2},E_{\text{linear}}=\frac{2k\lambda }{\text{scripty}r},E_{\text{surfacial}}=\frac{\rho }{2{ϵ}_0}\right]$ {\bfseries Boundary Conditions:}\par [constraints] $\left[{\vec{E}}_{\text{above}}-{\vec{E}}_{\text{below}}=\frac{\sigma }{{ϵ}_0}\hat{n},V_{\text{above}}=V_{\text{below}}:{\partial }_n{V}_{\text{above}}-{\partial }_n{V}_{\text{below}}=-\frac{\sigma }{{ϵ}_0}\right]$ {\bfseries Special Techniques} \begin{enumerate} \item {[Laplace Equation]} $\left[{\vec{\nabla }}^{2}V=\frac{\rho }{{ϵ}_0}:[\rho =0],{\vec{\nabla }}^{2}V=0\right]$ $\text{[neighbourphile, extremephile]}$ $\text{[Uniqueness Theorem: Dirichlet’s, Neumann’s]}$ \item {[Method Of Images]}\par [“the potential is made to satisfy given boundary conditions by placing singularities outside its domain such that the additional singularities are an artifact needed to satisfy the prescribed boundary conditions”] $\left[\text{Plane}:q^{\prime }=-q:d^{\prime }=-d,\text{Sphere}:q^{\prime }=-\frac{R}{d}q:d^{\prime }=-\left(R-\frac{R^2}{d}\right)\right]$ \item {[Complex Picture]} $\left[\right]$ \end{enumerate} {\bfseries Inside Matter:}\par [creates approximate dipoles, force on dipoles if non uniform, torque on dipoles ($\vec{p}\times \vec{E},-\vec{p}\cdot \vec{E})$ so] $\left[V_{\text{induced dipoles}}=V_{\text{bound charges}}+V_{\text{volume charges}}=k\int \frac{[{\sigma }_b(r^{\prime })=\vec{P}.\text{uvec}n]}{\text{scripty}r}{d}^2{r}^{\prime }+k\int \frac{[{\rho }_b(r^{\prime })=-\vec{\nabla }.\vec{P}]}{\text{scripty}r}{d}^3{r}^{\prime }\right]$ [modifications] $\left[\vec{\nabla }\cdot ({ϵ}_0\vec{E}+\vec{P})={\rho }_f\left[\text{if linear: }\vec{\nabla }\cdot [{ϵ}_0(1+{\chi }_e)\vec{E}]={\rho }_f\right],\vec{\nabla }\times \vec{E}=0\right]$ [matter constraints] $\left[D_{\text{above}}^{⊢}-D_{\text{below}}^{⊢}={\sigma }_f,D_{\text{above}}^{||}-D_{\text{below}}^{||}=P_{\text{above}}^{||}-P_{\text{below}}^{||}\right]$ {\bfseries Capacitance:}\par [measure of tendency to aid accumulation of charges by virtue which stores energy in electric field] $\left[V_{\text{across conductors}}=C^{-1}{Q}_{\text{on conductors}}\right]$ \section{Moving Charges} {\bfseries Electric Current:}\par [motion of charges] $\left[I=\vec{I}(r^{\prime }),I=\int \vec{J}(r^{\prime })\cdot \overrightarrow{d^2{r}^{\prime }}:\vec{\nabla }\cdot \vec{J}+\frac{d\rho }{dt}=0\right]$ [governed as] $\left[\vec{J}=\sigma {\vec{f}}_s\sim \sigma \vec{E}\right]$ {\bfseries Resistance:}\par [a measure of tendency to oppose to the flow of electric current thus dissipating the energy] $\left[V_{\text{across conductors}}=R{I}_{\text{through conductors}}\right]$ {\bfseries Magnetostatics:}\par [field equations] $\left[\vec{B}(r)=k\int \frac{\vec{I}(r^{\prime })\times \text{uvec}\text{scripty}r}{\text{scripty}{r}^2}d{r}^{\prime }\left[\frac{{\mu }_{0}I}{2\pi s}\text{uvec}\varphi \right]\sim k\int \frac{\vec{J}(r^{\prime })\times \text{uvec}\text{scripty}r}{\text{scripty}{r}^2}{d}^3{r}^{\prime }:\vec{\nabla }\cdot \vec{B}=0,\vec{\nabla }\times \vec{B}={\mu }_0\vec{J}\right]$ [derives as] $\left[{\phi }_B=\oint \vec{B}(r)\cdot d\vec{a}(r)=\int \frac{{\mu }_{0}I}{2\pi s}\hat{\varphi }\cdot (sd\varphi dz\hat{s})=0\right]$

$\left[\vec{\nabla }\cdot \vec{B}(r)=k\int \vec{\nabla }\cdot \left(\frac{\vec{J}(r^{\prime })\times \widehat{\text{scripty}r}}{\text{scripty}{r}^2}\right)d^3{r}^{\prime }=0\right]$

$\left[{\lambda }_B=\oint \vec{B}(r)\cdot d\vec{l}(r)={\int }_a^b\frac{{\mu }_{0}I}{2\pi s}\hat{\varphi }\cdot (ds\hat{s}+sd\varphi \hat{\varphi }+dz\hat{z})={\mu }_{0}I\right]$

$\left[\vec{\nabla }\cdot \vec{B}(r)=k\int \vec{\nabla }\times \left(\frac{\vec{J}(r^{\prime })\times \widehat{\text{scripty}r}}{\text{scripty}{r}^2}\right)d^3{r}^{\prime }=k\int 4\pi {\delta }^3(\text{scripty}r)\vec{J}(r^{\prime })d^3{r}^{\prime }={\mu }_0\vec{J}(r)\right]$ [which allows] $\left[\vec{B}(r)=\vec{\nabla }\times \vec{A}(r)[\text{no significant physical meaning}]\right]$ [analogy extension] $\left[\vec{A}(r)=k\int \frac{\vec{I}(r^{\prime })}{\text{scripty}r}d{r}^{\prime }\sim k\int \frac{\vec{J}(r^{\prime })}{\text{scripty}r}{d}^3{r}^{\prime }={\sum }_{n=0}^{\infty }\frac{k}{r^{n+1}}\int {r^{\prime }}^n{P}_n(\cos {\theta }_{r-r^{\prime }}^{\prime })\vec{J}(r^{\prime })d^3{r}^{\prime }\right]$[individuals] $\left[{\vec{A}}_{\text{mon}}=0:{\vec{B}}_{\text{mon}}=0,{\vec{A}}_{\text{dip}}=k\frac{\vec{m}\cdot \text{uvec}r}{r^2}:{\vec{B}}_{\text{dip}}=k\frac{3(\vec{m}\cdot \text{uvec}r)\text{uvec}r-\vec{m}}{r^3}\left[\vec{m}=\frac{1}{2}\int {\vec{r}}^{\prime }\times \vec{J}(r^{\prime })d^3{r}^{\prime }=I\vec{A}\right]\right]$ {\bfseries Boundary Conditions:}\par [constraints] $\left[{\vec{B}}_{\text{above}}-{\vec{B}}_{\text{below}}={\mu }_0(\vec{K}\times \hat{n}),{\vec{A}}_{\text{above}}={\vec{A}}_{\text{below}}:{\partial }_n{\vec{A}}_{\text{above}}-{\partial }_n{\vec{A}}_{\text{below}}=-{\mu }_0\vec{K}\right]$ {\bfseries Inside Matter:}\par [creates approximate dipoles, force on dipoles if non uniform, torque on dipoles ($\vec{m}\times \vec{B},-\vec{m}\cdot \vec{B})$ so] $\left[{\vec{A}}_{\text{induced dipoles}}={\vec{A}}_{\text{bound currents}}+{\vec{A}}_{\text{vol currents}}=k\int \frac{[{\vec{K}}_b(r^{\prime })=\vec{M}\times \hat{n}]}{\text{scripty}r}{d}^2{r}^{\prime }+k\int \frac{[{\vec{J}}_b(r^{\prime })=\vec{\nabla }\times \vec{M}]}{\text{scripty}r}{d}^3{r}^{\prime }\right]$ [modifications] $\left[\vec{\nabla }\times \left(\frac{\vec{B}}{{\mu }_0}-\vec{M}\right)={\vec{J}}_f\left[\text{if linear: }\vec{\nabla }\times [\frac{\vec{B}}{{\mu }_0(1+{\chi }_b)}]={\vec{J}}_f\right],\vec{\nabla }\cdot \vec{B}=0\right]$ [matter constraints] $\left[H_{\text{above}}^{⊢}-H_{\text{below}}^{⊢}=-(M_{\text{above}}^{⊢}-M_{\text{below}}^{⊢}),H_{\text{above}}^{||}-H_{\text{below}}^{||}=|\overrightarrow{K_f}\times \hat{n}|\right]$ {\bfseries Electromotive Force:}\par [electrical action of non-electrical sources] $\left[\xi =\oint \vec{f}\cdot \overrightarrow{dl}-=\oint [{\vec{f}}_{\text{nE}}+\vec{E}]\cdot \overrightarrow{dl}=\oint {\vec{f}}_{\text{nE}}\cdot \overrightarrow{dl}\right]$ [other forces] \begin{enumerate} \item ~[A Battery]:\par $\left[\xi =\oint {\vec{f}}_{\text{nE}}\cdot \overrightarrow{dl}={\int }_{\text{S}}^{\text{E}}{\vec{f}}_{\text{nE}}\cdot \overrightarrow{dl}=-{\int }_{\text{S}}^{\text{E}}\vec{E}\cdot \overrightarrow{dl}=V_{\text{S-E}}\right]$ \item ~[Lorentz Force]: \par $\left[\xi =\oint [\vec{v}\times \vec{B}]\cdot \overrightarrow{dl}=-\frac{d{\phi }_B}{dt}\right]$ \item ~[Induced Electric Field]: $\left[\xi =\oint [?]\cdot \overrightarrow{dl}=-\frac{d{\phi }_B}{dt}:[?]={\vec{E}}_{\text{induced}},\vec{\nabla }\times \vec{E}=-\frac{\partial B}{\partial t}\right]$ $\text{[nature abhors change in flux]}$\end{enumerate} [formation of magnetic system, analogy] $\left[w=\frac{1}{2}\int \vec{J}(r^{\prime })\cdot \vec{A}(r)d^{3}r=\frac{1}{2{\mu }_0}\int {\vec{B}}^2{d}^{3}r\right]$ [other systems] $\left[B_{\text{linear}}=\frac{2kI}{\text{scripty}r},E_{\text{general}}=\frac{kI}{\text{scripty}r}[\sin {\alpha }_1+\sin {\alpha }_2]\right]$ {\bfseries Inductance:}\par [measure of tendency to oppose change in e-current by virtue which stores energy in magnetic field] $\left[V_{\text{across conductors}}=L{d}_t{I}_{\text{over conductors}}\right]$ \section{The Final Touch} {\bfseries The Mathematical Touch:}\par [imposition of mathematical consistency] $\left[\vec{\nabla }\cdot \vec{J}=-\frac{\partial \rho }{\partial t}=-\frac{\partial }{\partial t}(ϵ\vec{\nabla }\cdot \vec{E})=-\vec{\nabla }\cdot \left(ϵ\frac{\partial \vec{E}}{\partial t}\right):\vec{\nabla }\times \vec{B}={\mu }_0\vec{J}+{\mu }_0{ϵ}_0\frac{\partial \vec{E}}{\partial t}\right]$ {\bfseries The Matter:}\par [changing polarization] $\left[\vec{\nabla }\times \left(\frac{\vec{B}}{{\mu }_0}-\vec{M}\right)={\vec{J}}_f+\frac{\partial \vec{P}}{\partial t}+{ϵ}_0\frac{\partial \vec{E}}{\partial t}=\overrightarrow{J_f}+\frac{\partial }{\partial t}({ϵ}_0\vec{E}+\vec{P})\right]$ {\bfseries Conclusion:}\par [final set] $\left[\vec{\nabla }\cdot \vec{D}={\rho }_f,\vec{\nabla }\times \vec{E}=-\frac{\partial B}{\partial t},\vec{\nabla }\cdot \vec{B}=0,\vec{\nabla }\times \vec{H}={\vec{J}}_f+{ϵ}_0\frac{\partial \vec{D}}{\partial t}\right]$ {\bfseries Conservation Laws:}\par [energy] \left[\frac{dW}{dt}=\int(\vec{E}\cdot\vec{J})d^3r=-\frac{d}{dt}\int\left(\epsilon_0\vec{E}^2+\frac{\vec{B}^2}{\mu_0}\right)d^3r-\frac{1}{\mu_0}\oint(\vec{E}\times\vec{B})\cdot d\vec{a}\right]$$$$\left[ :\frac{d}{dt}(u_\text{mech} + u_\text{elec})=-\vec{\nabla}\cdot\vec{S}\right] [momentum] $\left[\vec{f}=\rho \vec{E}+\vec{J}\times \vec{B}={ϵ}_0[(\vec{\nabla }\cdot \vec{E})\vec{E}+(\vec{E}\cdot \vec{\nabla })\vec{E}]+\frac{1}{{\mu }_0}[(\vec{\nabla }\cdot \vec{B})\vec{B}+(\vec{B}\cdot \vec{\nabla })\vec{B}]-\frac{1}{2}\vec{\nabla }{u}_{\text{elec}}-{ϵ}_0{\mu }_0\frac{\partial }{\partial t}\vec{S}\right]$ \section{Electric Circuits} {\bfseries Magnetic Circuits}\par [mean lumping magnetic systems] $\left[{\varphi }_B=\frac{F}{R}:F=Ni,R=\frac{l_c}{{\mu }_c{A}_c},L=\frac{N^2}{R},E_{\text{ac}}=4.44fN{\varphi }_m\right]$ [ideal transforming action, no load and load] $\left[{\varphi }_m=\frac{V_1}{4.44f{N}_1}:\frac{E_1}{E_2}=\frac{N_1}{N_2}:E_1{I}_1=E_2{I}_2:Z_1=a^2{Z}_2\right]$ [full load lumping non ideal transformer] $\left[\text{Cu: }{R}_1,R_2;\text{Fe: }{G}_i;{\Phi }_B:X_1,X_2;\mu :B_m\right]$ [reference equivalency of the transformer] [efficiency] $\left[\eta =\frac{V_2{I}_2\cos {\theta }_2}{V_2{I}_2\cos {\theta }_2+P_i+I_2^2{R}_{\text{e2}}}\right]$ [remember the graph of efficiency and regulation]\par [also KVA size and ratings, how full load means the max current and not max load lmao] \section{Electromagnetic Waves} {\bfseries Lossless Dielectrics:}\par [coupled fields] $\left[{\nabla }^{2}E=ϵ\mu \frac{{\partial }^{2}E}{\partial t^2},{\nabla }^{2}B=ϵ\mu \frac{{\partial }^{2}B}{\partial t^2}\right]$ [which solves as] $\left[:\tilde{E}(r,t)={\tilde{E}}_0{e}^{i(r\cdot k-wt)}\hat{n},\tilde{B}(r,t)={\tilde{B}}_0{e}^{i(r\cdot k-wt)}(k\times \hat{n});\widetilde{B_0}=\frac{1}{v}(\hat{k}\times \widetilde{E_0})\right]$ {\bfseries Waves In Conductors:}\par [phasors] $\left[\nabla \cdot ϵ\tilde{E}=0,\nabla \times \tilde{E}=-jw\mu \tilde{H},\nabla \cdot \mu \tilde{H}=0,\nabla \times \tilde{H}=\sigma \tilde{E}+jwϵ\tilde{E}\right]$ $\left[\frac{{\partial }^2{E}_x}{\partial z^2}-{\gamma }^2{E}_x=0,\frac{{\partial }^2{H}_y}{\partial z^2}-{\gamma }^2{H}_y=0;(\alpha +i\beta {)}^2=-\mu ϵ{\omega }^2+i\mu \sigma \omega ;\alpha ,\beta =w\sqrt{\frac{ϵ\mu }{2}}{\left[\sqrt{1+{\left(\frac{\sigma }{ϵw}\right)}^2}\mp 1\right]}^{\frac{1}{2}}\right]$ $\left[\alpha ,\beta =0,w\sqrt{\mu ϵ};\sqrt{\frac{\mu \sigma w}{2}},\sqrt{\frac{\mu \sigma w}{2}}\right]$ $\left[:E_x=E_0{e}^{-\gamma z},H_y=\frac{E_0}{\eta }{e}^{-\gamma z};\eta =\frac{iw\mu }{\gamma }=\sqrt{\frac{iw\mu }{\sigma +iwϵ}}\right]$ $\left[|n|\angle \theta =\sqrt{\frac{\mu }{ϵ}},0;\sqrt{\frac{w\mu }{\sigma }},45;\frac{\sqrt{\frac{\mu }{ϵ}}}{{\left[1+{\left(\frac{\sigma }{wϵ}\right)}^2\right]}^{\frac{1}{4}}},\frac{1}{2}\arctan \frac{\sigma }{wϵ}\right]$ $\left[⟨S_z⟩=\frac{1}{2}Re(E_s\times H_s^{\ast })\right]$ [on boundaries] $\left[E_i=E_{i0}^{+}{e}^{-{\gamma }_{1}z},H_i=\frac{E_{i0}^{+}}{n_1}{e}^{-{\gamma }_{1}z};E_r=E_{r0}^{-}{e}^{+{\gamma }_{1}z},H_r=-\frac{E_{r0}^{-}}{n_1}{e}^{+{\gamma }_{1}z};E_t=E_{t0}^{+}{e}^{-{\gamma }_{2}z},H_t=\frac{E_{t0}^{+}}{n_2}{e}^{-{\gamma }_{2}z}\right]$ $\left[E_{i0}^{+}+E_{r0}^{-}=E_{t0}^{+};\frac{E_{i0}^{+}}{{\eta }_1}-\frac{E_{r0}^{-}}{{\eta }_1}=\frac{E_{t0}^{+}}{{\eta }_2}:\frac{E_{r0}^{-}}{E_{i0}^{+}}=\frac{{\eta }_2-{\eta }_1}{{\eta }_2+{\eta }_1}\left[\Gamma =|\Gamma |e^{j\varphi }\right];\frac{E_{t0}^{+}}{E_{i0}^{+}}=\frac{2{\eta }_2}{{\eta }_2+{\eta }_1}\right]$ [summing up] $\left[E_{ir}=(e^{-{\gamma }_{1}z}+\Gamma e^{{\gamma }_{1}z})E_0;H_{ir}=(e^{-{\gamma }_{1}z}-\Gamma e^{{\gamma }_{1}z})\frac{E_0}{{\eta }_1}\right]$ [on lossless] $\left[E_{ir}=(e^{-j{\beta }_{1}z}+\Gamma e^{j{\beta }_{1}z})E_0=(e^{-j{\beta }_{1}z}+|\Gamma |e^{j(\varphi +{\beta }_{1}z)})E_0\right]$ $\left[{\beta }_1{z}_{\text{max}}=-\frac{\varphi }{2}-\text{int}\pi ,{\beta }_1{z}_{\text{min}}=-\frac{\varphi }{2}-\text{odd}\frac{\pi }{2}\right]$ $\left[{\eta }_{\text{in}}={\eta }_1\frac{e^{j{\beta }_{1}l}+\Gamma e^{-j{\beta }_{1}l}}{e^{j{\beta }_{1}l}-\Gamma e^{-j{\beta }_{1}l}}={\eta }_1\frac{{\eta }_2+j{\eta }_1\tan {\beta }_{1}l}{{\eta }_1+j{\eta }_2\tan {\beta }_{1}l}\right]$ [on conductors] $\left[E_{ir}=-2j{E}_0\sin {\beta }_{1}z,H_{ir}=2\frac{E_0}{{\eta }_1}\cos {\beta }_{1}z\right]$ $\left[E_{irmax}=\pm 2E_0[{\beta }_{1}z=-\text{odd}\frac{\pi }{2}],E_{irmin}=0[{\beta }_{1}z=-\text{int}\pi ],{\eta }_{\text{in}}=j{\eta }_1\tan {\beta }_{1}l\right]$ [electric polarization] $\left[E_{xi}+E_{xr}=E_{xt}:E_i+E_r=E_t,H_{yi}+H_{yr}=H_{yt}:H_i\cos {\theta }_i-H_r\cos {\theta }_i=H_t\cos {\theta }_t\right]$ $\left[{\Gamma }_{\perp }=\frac{{\eta }_2\cos {\theta }_i-{\eta }_1\cos {\theta }_t}{{\eta }_2\cos {\theta }_i+{\eta }_1\cos {\theta }_t};{\tau }_{\perp }=\frac{2{\eta }_2\cos {\theta }_i}{{\eta }_2\cos {\theta }_i+{\eta }_1\cos {\theta }_t}\right]$ [magnetic polarization] $\left[H_{yi}+H_{yr}=H_{yt}:H_i+H_r=H_t,E_{xi}+E_{xr}=E_{xt}:E_i\cos {\theta }_i-E_r\cos {\theta }_i=E_t\cos {\theta }_t\right]$ $\left[{\Gamma }_{||}=\frac{{\eta }_2\cos {\theta }_t-{\eta }_1\cos {\theta }_i}{{\eta }_2\cos {\theta }_t+{\eta }_1\cos {\theta }_i};{\tau }_{||}=\frac{2{\eta }_2\cos {\theta }_i}{{\eta }_2\cos {\theta }_t+{\eta }_1\cos {\theta }_i}\right]$ \section{Potential Fields} {\bfseries Gauge Transformation:}\par [dynamic fields] $\left[{\vec{\nabla }}^{2}V+\frac{\partial }{\partial t}(\vec{\nabla }\cdot A)=-\frac{\rho }{{ϵ}_0},\left({\vec{\nabla }}^2\vec{A}-{ϵ}_0{\mu }_0\frac{{\partial }^2\vec{A}}{\partial t^2}\right)+\vec{\nabla }\left(\vec{\nabla }\cdot \vec{A}+{ϵ}_0{\mu }_0\frac{\partial V}{\partial t}\right)=-{\mu }_0\vec{J}\right]$ $\left[\vec{E}=-\nabla V-\frac{\partial \vec{A}}{\partial t},\vec{B}=\vec{\nabla }\times \vec{A}\right]$ [lorentz pickup] $\left[{\vec{A}}^{\prime }=\vec{A}+\vec{\nabla }\lambda ,V^{\prime }=V-\frac{\partial \lambda }{\partial t};\vec{\nabla }\cdot \vec{A}=-{ϵ}_0{\mu }_0\frac{\partial V}{\partial t}:{\vec{\square }}^{2}V=-\frac{\rho }{{ϵ}_0},{\vec{\square }}^{2}A=-{\mu }_0\vec{J}\right]$ [solves as] $\left[V(r,t)=\frac{1}{4\pi {ϵ}_0}\int \frac{\rho (r^{\prime },t_r)}{\text{scripty}r}{d}^3{r}^{\prime },\vec{A}(r,t)=\frac{{\mu }_0}{4\pi }\int \frac{\vec{J}(r^{\prime },t_r)}{\text{scripty}r}{d}^3{r}^{\prime };t_r=t-\frac{\text{scripty}r}{c}\right]$ $\left[\vec{E}(r,t)=k\int \left[\frac{\rho (r^{\prime },t_r)}{\text{scripty}{r}^2}\widehat{\text{scripty}r}+\frac{\dot{\rho }(r^{\prime },t_r)}{c\text{scripty}r}\widehat{\text{scripty}r}-\frac{\dot{\vec{J}}(r^{\prime },t_r)}{c^2\text{scripty}r}\widehat{\text{scripty}r}\right]d^3{r}^{\prime },\vec{B}(r,t)=k\int \left[\frac{\vec{J}(r^{\prime },t_r)}{c\text{scripty}{r}^2}+\frac{\dot{\vec{J}}(r^{\prime },t_r)}{c\text{scripty}r}\right]\times \text{scripty}r{d}^3{r}^{\prime }\right]$ [for point charges] $\left[V(r,t)=\frac{1}{4\pi {ϵ}_0}\frac{qc}{\text{scripty}rc-\overrightarrow{\text{scripty}r}\cdot \vec{v}},\vec{A}(r,t)=\frac{\vec{v}}{c^2}V(r,t)\right]$ $\left[\vec{E}(r,t)=\frac{q}{4\pi {ϵ}_0}\frac{\text{scripty}r}{(\overrightarrow{\text{scripty}r}\cdot \vec{u}{)}^3}[(c^2-v^2)\vec{u}+\overrightarrow{\text{scripty}r}\times (\vec{u}\times \vec{a})],\vec{B}(r,t)=\frac{1}{c}[\widehat{\text{scripty}r}\times \vec{E}(r,t)];\vec{u}=c\widehat{\text{scripty}r}-\vec{v}\right]$ \section{Radiation} [signature] $\left[P_{\text{rad}}={\lim }_{r\to \infty }=\frac{1}{{\mu }_0}\oint (\vec{E}\times \vec{B})\cdot d\vec{a}\right]$ [oscillating electric dipole] $\left[V=-\frac{p_0\omega }{4\pi {ϵ}_{0}c}\left(\frac{\cos \theta }{r}\right)\sin \left[\omega \left(t-\frac{r}{c}\right)\right],\vec{A}=-\frac{{\mu }_0{p}_0\omega }{4\pi r}\sin \left[\omega \left(t-\frac{r}{c}\right)\right]\hat{z}\right]$ $\left[\vec{E}=-\frac{{\mu }_0{p}_0{\omega }^2}{4\pi }\left(\frac{\sin \theta }{r}\right)\cos \left[\omega \left(t-\frac{r}{c}\right)\right]\hat{\theta },\vec{B}=-\frac{{\mu }_0{p}_0{\omega }^2}{4\pi }\left(\frac{\sin \theta }{r}\right)\cos \left[\omega \left(t-\frac{r}{c}\right)\right]\hat{\varphi }\right]$ $\left[⟨\vec{S}⟩=\left(\frac{{\mu }_0{p}_0^2{\omega }^4}{32{\pi }^{2}c}\right)\frac{{\sin }^2\theta }{r^2}\hat{r},⟨P_{\text{rad}}⟩=\frac{{\mu }_0{p}_0{\omega }^4}{12\pi c}\right]$ [oscillating magnetic dipole] $\left[V=0,\vec{A}=-\frac{{\mu }_0{m}_0\omega }{4\pi c}\left(\frac{\sin \theta }{r}\right)\sin \left[\omega \left(t-\frac{r}{c}\right)\right]\hat{\varphi }\right]$ $\left[\vec{E}=-\frac{{\mu }_0{m}_0{\omega }^2}{4\pi c}\left(\frac{\sin \theta }{r}\right)\cos \left[\omega \left(t-\frac{r}{c}\right)\right]\hat{\varphi },\vec{B}=-\frac{{\mu }_0{m}_0{\omega }^2}{4\pi c^2}\left(\frac{\sin \theta }{r}\right)\cos \left[\omega \left(t-\frac{r}{c}\right)\right]\hat{\theta }\right]$ $\left[⟨\vec{S}⟩=\left(\frac{{\mu }_0{m}_0^2{\omega }^4}{32{\pi }^2{c}^3}\right)\frac{{\sin }^2\theta }{r^2}\hat{r},⟨P_{\text{rad}}⟩=\frac{{\mu }_0{m}_0{\omega }^4}{12\pi c^3}\right]$ [arbitrary charge] $\left[V(r,t)=\frac{1}{4\pi {ϵ}_0}\left[\frac{Q}{r}+\frac{\hat{r}\cdot \vec{p}(t_0)}{r^2}+\frac{\hat{r}\cdot \dot{\vec{p}}(t_0)}{rc}\right],\vec{A}=\frac{{\mu }_0}{4\pi }\frac{\dot{\vec{p}}(t_0)}{r}\right]$ $\left[\vec{E}(r,t)=\frac{{\mu }_0}{4\pi r}[\hat{r}\times (\hat{r}\times \ddot{\vec{p}})]=\frac{{\mu }_0\ddot{p}(t_0)}{4\pi }\left(\frac{\sin \theta }{r}\right)\hat{\theta },\vec{B}(r,t)=-\frac{{\mu }_0}{4\pi rc}[\hat{r}\times \ddot{\vec{p}}]=\frac{{\mu }_0\ddot{p}(t_0)}{4\pi c}\left(\frac{\sin \theta }{r}\right)\hat{\varphi }\right]$ $\left[⟨\vec{S}⟩=\frac{{\mu }_0}{16{\pi }^{2}c}[\ddot{p}(t_0){]}^2\frac{{\sin }^2\theta }{r^2}\hat{r},⟨P_{\text{rad}}⟩=\frac{{\mu }_0{\ddot{p}}^2}{6\pi c}\right]$ [a point charge] $\left[P_{\text{rad}}=\frac{{\mu }_0{q}^2{a}^2}{6\pi c},{\vec{F}}_{\text{rad}}=\frac{{\mu }_0{q}^2}{6\pi c}\dot{\vec{a}}\right]$ $\left[\text{[on const ext force}]\vec{a}(t)=\left[\left(\frac{F}{m}+B\right)e^{\frac{t}{\tau }}\right],\left[\frac{F}{m}+B{e}^{\frac{t}{\tau }}\right],\left[\left(\frac{F}{m}{e}^{-\frac{T}{\tau }}+B\right)e^{\frac{t}{\tau }}\right]\right]$ \section{Relativity} [gedanken] ["\text{two events that are simultaneous in one inertial frame are not, in general, simultaneous in another}"]$$$$["\text{moving clock runs slow}"] $["\text{moving objects are shortened}"]$ $["\text{dimensions perpendicular to velocity are not contracted}"]$ {\bfseries Lorentz Transformation:}\par [geometry of relativity] $\left[\bar{x}=\gamma (x-vt),\bar{y}=y,\bar{z}=z,\bar{t}=\gamma \left(t-\frac{v}{c^2}x\right)\right]$ [fours]

\bar{x_0}\\ \bar{x_1}\\ \bar{x_2}\\ \bar{x_3}\\ \end{array}\right]=\left[\begin{array}{cccc} \gamma & -\gamma\beta & 0 & 0\\ -\gamma\beta & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \\ \end{array} \right]\left[\begin{array}{c} {x_0}\\ {x_1}\\ {x_2}\\ {x_3}\\ \end{array}\right],~[\bar{t}^{\mu\nu}=\Lambda^{\mu}_{\lambda}\Lambda^{\nu}_{\sigma}t^{\lambda\sigma}\right]$$ [invariant interval] $$\left[I=(\Delta x)_{\mu}(\Delta x)^{\mu}=-(\Delta x^0)^2+(\Delta x^1)^2+(\Delta x^2)^2+(\Delta x^3)^2=-c^2t^2+d^2;~(\Delta x)^{\mu}=x_A^{\mu}-x_{B}^{\mu}\right]$$ $$\text{[timelike, spacelike, lightlike]}$$ [minkowski spacetime] $$["\text{slopes, cones, hypers, worldlines, locus, causality, diagrams comparison}"]$$ $$[\text{"the invariant interval between casually related events is always timelike}$$ $$\text{ and their temporal ordering is the same for all intertial observers"}]$$ {\bfseries Mechanics:}\par [invariant timelike propers from spacetime] $$\left[d\tau={\gamma}^{-1}dt,~U=\gamma(c,u),~P={\gamma}(m_0c,m_0u)=\left(\frac{E}{c},p\right)[\text{conserved}],~K=\gamma\left(\frac{1}{c}\frac{dE}{dt},\frac{dp}{dt}\right)\right]$$ {\bfseries Electrodynamics:}\par [field transformation] $$\left[\bar{E_x}=E_x,~\bar{E_y}=\gamma(E_y-vB_z),~\bar{E_z}=\gamma(E_z+vB_y),~\bar{B_x}=B_x,~\bar{B_y}=\gamma\left(B_y+\frac{v}{c^2}E_z\right),~\bar{B_z}=\gamma\left(B_z-\frac{v}{c^2}E_y\right)\right]$$ $$\left[[\text{if }B=0],~\bar{B}=-\frac{1}{c^2}(v\times\bar{E}),~[\text{if }E=0],~\bar{E}=v\times\bar{B}\right]$$ [invariant quantities] $$\left[~ F^{\mu\nu}=\left[\begin{array}{cccc} 0 & bE_x{c} & bE_y & bE_z\\[0.4cm] -bE_x &0 & B_z & -B_y\\[0.4cm] -bE_y & -B_z & 0 & B_x\\[0.4cm] -bE_z & B_y & -B_x & 0 \\[0.4cm] \end{array} \right],~G^{\mu\nu}=\left[\begin{array}{cccc} 0 & B_x & B_y & B_z\\[0.4cm] -B_x &0 & -bE_z & bE_y\\[0.4cm] -{B_y} & bE_z & 0 & -bE_x\\[0.4cm] -B_z & -bE_y & bE_x & 0 \\[0.4cm] \end{array}\right]\right]$$ $$\left[J^\mu=(c\rho, J_x,J_y,J_z),~A^\mu=(bV,A_x,A_y,A_z)\right]$$ [field equations] $$\left[\frac{\partial J^\mu}{\partial x^\mu}=0,~\frac{\partial F^{\mu\nu}}{\partial x^{\nu}}=\mu_0J^{\mu},~\frac{\partial G^{\mu\nu}}{\partial x^\nu}=0,~K^\mu=q\eta_\nu F^{\mu\nu},~F^{\mu\nu}=\frac{\partial A^{\nu}}{\partial x_\mu}-\frac{\partial A^\mu}{\partial x_\nu},~\frac{\partial}{\partial x_{\nu}}\frac{\partial}{\partial x^{\mu}}A^{\mu}=-\mu_0J^{\mu}\right]$$ \chapter{Fundamentals} [in a damped mass on a spring driven by a sinusoidal external force, second law takes the form of] \begin{enumerate} \item {\bfseries [SHM:]} \par [Differential Equation:] $$\left[mx''+kx=0 \text{ or, }x''+\omega_0^2x=0\right]$$ [Solution:] $$\left[x=A\cos(\omega_0t)+B\sin(\omega_0t)\right]$$ \item {\bfseries [Damped:]} \par [Differential Equation:] $$\left[mx''+bx'+kx=0 \text{ or, }x''+2\gamma \omega_0x'+\omega_0^2x=0\right]$$ [Solution:] $$\left[x=e^{-\gamma\omega_0t}\left[A\cos\left(\sqrt{1-\gamma^2}\omega_0t\right)+B\sin\left(\sqrt{1-\gamma^2}\omega_0t\right)\right]\right]$$ \item {\bfseries [Forced:]} \par [Differential Equation:] $$\left[mx''+bx'+kx=F_0sc(\omega t+\Delta) \text{ or, }x''+2\gamma \omega_0x'+\omega_0^2x=F_0sc(\omega t+\Delta)\right]$$ [Steady Solution:] $$\left[x=\frac{F_0}{m\sqrt{(2\omega\omega_0\gamma)^2+(\omega_0^2-\omega^2)}}sc(\omega t+\Delta+\phi),~~\tan\phi=-\frac{2\omega\omega_0\gamma}{\omega_0^2-\omega^2}\right]$$ [Alternative:] $$\left[m\left[(i\omega)^2\uvec{x}+2\gamma \omega_0(i\omega)\uvec{x}+\omega_0^2\uvec{x}\right]e^{i\omega t}=\uvec{F}e^{i\omega t}\right]$$ [Solution:] $$\left[\uvec{x}=\frac{\uvec{F}}{m(\omega_0^2-\omega^2+i2\omega\omega_0\gamma)}=\uvec{R}\uvec{F}\right]$$ \item {\bfseries [Quality Factor:]}\par [dimensionless parameter that describes degree of underdamping of an oscillator or resonator] $$[\text{ratio of the stored energy in resonator to the dissipated energy per cycle by damping}]$$ $$[\text{ratio of a resonator's centre frequency to its bandwidth when subject to driving force}]$$ $$\left[Q=\frac{\text{[energy stored]}}{\text{[energy lost per cycle]}}~~[E=E_0e^{-2\gamma\omega_0t}]=\frac{1}{2\gamma}\equiv\frac{\text{[resonance frequency]}}{\text{[bandwidth]}} =\frac{\omega_0 m}{b}=\frac{k}{\omega_0b}\right]$$ \item {\bfseries [Electric Circuits:]}\par [circuit allows maximum current for a given frequency of the source of alternating supply as] $$\left[[m=L][b=R][k=\frac{1}{C}]\right]~~~~~~~~$$ $$\left[\uvec{I}=\frac{1}{\left[i\omega L+R-\frac{i}{\omega C}\right]}\uvec{V}\right]~~~~~~~~$$ [parallel follows duality] $$\left[I:V,~R:G,~X:B,~Z:Y\right]~~~~~~~~$$ $$\left[{R_s}=R_p\left(\frac{1}{1+Q^2}\right)\sim \frac{R_p}{Q^2}\sim 0,~{X_s}=X_p\left(1+\frac{1}{Q^2}\right)\sim X_p\right]~~~~~~~~$$ \item {\bfseries [Lumped Model:]}\par [attributes concentrated into idealized electrical compos- joined by perfectly conducting wires] $$\left[VS,CS;C,R,L;VCVS,VCCS,CCVS,CCCS\right]$$ \item {\bfseries [Net Impedances:]}\par [in series] $$\left[Z_{\text{eq}}=Z_1+Z_1+\ldots+Z_n;~V_1=\frac{Z_1}{Z_1+Z_2+Z_3}V;~I_1=I\right]$$ $$\left[[\text{voltage across self}]=\frac{[\text{self impedance}]}{[\text{self impedance}]+[\text{neighbour impedance}]}[\text{voltage across whole}]\right]$$ [in parallel] $$\frac{1}{Z_{\text{eq}}}=\frac{1}{Z_1}+\frac{1}{Z_2}+\ldots+\frac{1}{Z_n};~I_1=\frac{Z_2Z_3}{Z_1Z_2+Z_2Z_3+Z_3Z_1}I;~V_1=V$$ $$\left[[\text{current through self}]=\frac{[\text{neighbour impedance}]}{[\text{self impedance}]+[\text{neighbour impedance}]}[\text{current through whole}]\right]$$ \item {\bfseries [Star Delta:]} $$\left[R_A=\frac{R_{AB}R_{AC}}{R_{AB}+R_{AC}+R_{BC}},~R_B=\frac{R_{AB}R_{BC}}{R_{AB}+R_{AC}+R_{BC}},~R_C=\frac{R_{AC}R_{BC}}{R_{AB}+R_{AC}+R_{BC}}\right]$$ $$\left[R_{AB}=\frac{R_AR_B+R_AR_C+R_BR_C}{R_C},~R_{BC}=\frac{R_AR_B+R_AR_C+R_BR_C}{R_A},~R_{AC}=\frac{R_AR_B+R_AR_C+R_BR_C}{R_B}\right]$$ \item {\bfseries [Kirchhoff's Laws:]} $$["\text{for any node, the sum of currents flowing into is equal to the sum of currents flowing out of}"]$$ $$["\text{the directed sum of the potential differences around any closed loop is zero}"]$$ \end{enumerate} \chapter{Quantum Mechanics} \section{Stern Gerlach Experiments} \begin{enumerate} \item Moment Momentum $$\left[\vec{\mu}\propto \vec{S}\right]$$ \item Experiment 1 \begin{center} \begin{tikzpicture}[scale=0.2] \tikzstyle{every node}+=[inner sep=0pt] \draw [black] (27.8,-25.8) circle (3); \draw (27.8,-25.8) node {$SGz$}; \draw [black] (27.8,-25.8) circle (2.4); \draw (41.2,-22.8) node {$\left[\left.\middle|+z\right\rangle\right]$}; \draw (41.2,-28.7) node {$\left[\left.\middle|-z\right\rangle\right]$}; \draw [black] (53.3,-22.8) circle (3); \draw (53.3,-22.8) node {$SGz$}; \draw [black] (53.3,-22.8) circle (2.4); \draw (65.2,-22.8) node {$\left[\left.\middle|+z\right\rangle\right]$}; \draw [black] (20.6,-25.8) -- (24.8,-25.8); \fill [black] (24.8,-25.8) -- (24,-25.3) -- (24,-26.3); \draw [black] (30.73,-25.14) -- (38.27,-23.46); \fill [black] (38.27,-23.46) -- (37.38,-23.14) -- (37.6,-24.12); \draw [black] (30.73,-26.43) -- (38.27,-28.07); \fill [black] (38.27,-28.07) -- (37.59,-27.41) -- (37.38,-28.38); \draw [black] (44.2,-22.8) -- (50.3,-22.8); \fill [black] (50.3,-22.8) -- (49.5,-22.3) -- (49.5,-23.3); \draw (47.25,-22.3) node [above] {N0}; \draw [black] (56.3,-22.8) -- (62.2,-22.8); \fill [black] (62.2,-22.8) -- (61.4,-22.3) -- (61.4,-23.3); \draw (59.25,-22.3) node [above] {N0}; \end{tikzpicture} \end{center} \item Experiment 2 \begin{center} \begin{tikzpicture}[scale=0.2] \tikzstyle{every node}+=[inner sep=0pt] \draw [black] (27.8,-25.8) circle (3); \draw (27.8,-25.8) node {$SGz$}; \draw [black] (27.8,-25.8) circle (2.4); \draw (41.2,-22.8) node {$\left[\left.\middle|+z\right\rangle\right]$}; \draw (41.2,-28.7) node {$\left[\left.\middle|-z\right\rangle\right]$}; \draw [black] (53.3,-22.8) circle (3); \draw (53.3,-22.8) node {$SGx$}; \draw [black] (53.3,-22.8) circle (2.4); \draw (70.1,-19.6) node {$\left[\left.\middle|+x\right\rangle\right]$}; \draw (70.1,-25.8) node {$\left[\left.\middle|-x\right\rangle\right]$}; \draw [black] (20.6,-25.8) -- (24.8,-25.8); \fill [black] (24.8,-25.8) -- (24,-25.3) -- (24,-26.3); \draw [black] (30.73,-25.14) -- (38.27,-23.46); \fill [black] (38.27,-23.46) -- (37.38,-23.14) -- (37.6,-24.12); \draw [black] (30.73,-26.43) -- (38.27,-28.07); \fill [black] (38.27,-28.07) -- (37.59,-27.41) -- (37.38,-28.38); \draw [black] (44.2,-22.8) -- (50.3,-22.8); \fill [black] (50.3,-22.8) -- (49.5,-22.3) -- (49.5,-23.3); \draw (47.25,-22.3) node [above] {N0}; \draw [black] (56.25,-22.24) -- (67.15,-20.16); \fill [black] (67.15,-20.16) -- (66.27,-19.82) -- (66.46,-20.8); \draw (60.55,-20.16) node [above] {$\frac{\text{N}0}{2}$}; \draw [black] (56.25,-23.33) -- (67.15,-25.27); \fill [black] (67.15,-25.27) -- (66.45,-24.64) -- (66.27,-25.62); \draw (60.82,-25.35) node [below] {$\frac{\text{N}0}{2}$}; \end{tikzpicture} \end{center} \item Experiment 3 \begin{center} \begin{tikzpicture}[scale=0.2] \tikzstyle{every node}+=[inner sep=0pt] \draw [black] (7.1,-25.8) circle (3); \draw (7.1,-25.8) node {$SGz$}; \draw [black] (7.1,-25.8) circle (2.4); \draw (18.5,-22.8) node {$\left[\left.\middle|+z\right\rangle\right]$}; \draw (18.5,-28.9) node {$\left[\left.\middle|-z\right\rangle\right]$}; \draw [black] (29,-22.8) circle (3); \draw (29,-22.8) node {$SGx$}; \draw [black] (29,-22.8) circle (2.4); \draw (43.5,-19.8) node {$\left[\left.\middle|+x\right\rangle\right]$}; \draw (43.5,-25.8) node {$\left[\left.\middle|-x\right\rangle\right]$}; \draw [black] (57.3,-19.8) circle (3); \draw (57.3,-19.8) node {$SGz$}; \draw [black] (57.3,-19.8) circle (2.4); \draw (71.8,-15.8) node {$\left[\left.\middle|+z\right\rangle\right]$}; \draw (71.8,-23.9) node {$\left[\left.\middle|-z\right\rangle\right]$}; \draw [black] (-0.1,-25.8) -- (4.1,-25.8); \fill [black] (4.1,-25.8) -- (3.3,-25.3) -- (3.3,-26.3); \draw [black] (10,-25.04) -- (15.6,-23.56); \fill [black] (15.6,-23.56) -- (14.7,-23.28) -- (14.95,-24.25); \draw [black] (9.99,-26.59) -- (15.61,-28.11); \fill [black] (15.61,-28.11) -- (14.96,-27.42) -- (14.7,-28.39); \draw [black] (21.5,-22.8) -- (26,-22.8); \fill [black] (26,-22.8) -- (25.2,-22.3) -- (25.2,-23.3); \draw (23.2,-22.3) node [above] {N0}; \draw [black] (31.94,-22.19) -- (40.56,-20.41); \fill [black] (40.56,-20.41) -- (39.68,-20.08) -- (39.88,-21.06); \draw (34.75,-20.29) node [above] {$\frac{\text{N}0}{2}$}; \draw [black] (31.94,-23.41) -- (40.56,-25.19); \fill [black] (40.56,-25.19) -- (39.88,-24.54) -- (39.68,-25.52); \draw (34.75,-25.31) node [below] {$\frac{\text{N}0}{2}$}; \draw [black] (46.5,-19.8) -- (54.3,-19.8); \fill [black] (54.3,-19.8) -- (53.5,-19.3) -- (53.5,-20.3); \draw (50.4,-19.3) node [above] {$\frac{\text{N}0}{2}$}; \draw [black] (60.19,-19) -- (68.91,-16.6); \fill [black] (68.91,-16.6) -- (68,-16.33) -- (68.27,-17.29); \draw (62.86,-16.95) node [above] {$\frac{\text{N}0}{4}$}; \draw [black] (60.19,-20.62) -- (68.91,-23.08); \fill [black] (68.91,-23.08) -- (68.28,-22.38) -- (68.01,-23.35); \draw (62.76,-22.69) node [below] {$\frac{\text{N}0}{4}$}; \end{tikzpicture} \end{center} \item Polarization Analogy $$\left[\left.\middle|\pm x\right\rangle,~\left.\middle|\pm y\right\rangle,~\left.\middle|\pm z\right\rangle\right]$$ $$\left[\left.\middle|\pm x\right\rangle=\left.\middle|+z\right\rangle\frac{1}{\sqrt{2}}\pm \left.\middle|-z\right\rangle \frac{1}{\sqrt{2}};~\left.\middle|\pm y\right\rangle=\left.\middle|+z\right\rangle\frac{1}{\sqrt{2}}\pm\left.\middle|-z\right\rangle \frac{i}{\sqrt{2}}\right]$$ \end{enumerate} \subsection{BraKet Space} \begin{enumerate} \item Observables $$\left[\hat{A}\left.\middle|a'\right\rangle=a'\left.\middle|a'\right\rangle,~\hat{A}\left.\middle|a''\right\rangle=a''\left.\middle|a''\right\rangle,~\hat{A}\left.\middle|a'''\right\rangle=a'''\left.\middle|a'''\right\rangle,~\ldots\right]$$ \item Products $$\left[\left.\middle|\alpha\right\rangle\xleftrightarrow{DC}\left\langle\alpha\middle|\right.:~\left\langle\beta\middle|\alpha\right\rangle=\left\langle\alpha\middle|\beta\right\rangle^*\right]$$ $$\left[\hat{X}\left.\middle|\alpha\right\rangle \xleftrightarrow{DC}\left\langle\alpha\middle|\right.\hat{X}^{\bot}:~X=\left.\middle|\beta\right\rangle\left\langle\alpha\middle|\right.:~X^\bot=\left.\middle|\alpha\right\rangle\left\langle\beta\middle|\right.\right]$$ \item Associativity $$\left[(\left.\middle|\beta\right\rangle\left\langle\alpha\middle|\right.)\cdot \left.\middle|\gamma\right\rangle=\left.\middle|\beta\right\rangle \cdot (\left\langle\alpha\middle|\gamma\right\rangle):~(\left\langle\beta\middle|\right.)\cdot(\hat{X}\left.\middle|\alpha\right\rangle)=(\left\langle\beta\middle|\right. X)\cdot(\left.\middle|\alpha\right\rangle)\right]$$ \end{enumerate} \subsection{Representations} \begin{enumerate} \item Matrix $$\left[\hat{X}\dot{=}\left(\begin{array}{ccc} \left\langle a^{(1)}\middle|\hat{X}\middle|a^{(1)}\right\rangle & \left\langle a^{(1)}\middle|\hat{X}\middle|a^{(2)}\right\rangle & \ldots \\ \left\langle a^{(2)}\middle|\hat{X}\middle|a^{(1)}\right\rangle & \left\langle a^{(2)}\middle|\hat{X}\middle|a^{(2)}\right\rangle & \ldots \\ \ldots & \ldots & \vdots \end{array}\right)\right]$$ \item Column $$\left[\left.\middle|\alpha\right\rangle\dot{=}\left(\begin{array}{c} \left\langle a^{(1)}\middle|\alpha\right\rangle\\ \left\langle a^{(2)}\middle|\alpha\right\rangle\\ \left\langle a^{(3)}\middle|\alpha\right\rangle\\ \vdots \end{array}\right)\right]$$ \item Row $$\left[\left\langle\alpha\middle|\right.\dot{=}\left(\begin{array}{cccc} \left\langle \alpha\middle|a^{(1)}\right\rangle & \left\langle \alpha\middle|a^{(2)}\right\rangle & \left\langle \alpha\middle|a^{(3)}\right\rangle & \ldots \end{array}\right)\right]$$ \end{enumerate} \subsection{Hermitians} \begin{enumerate} \item Eigen Basekets $$\left[\left.\middle|\alpha\right\rangle=\sum_{a'}\left.\middle|a'\right\rangle\left\langle a'\middle|\alpha\right\rangle:~P(\left.\middle|\alpha\right\rangle\xrightarrow{M(A)}\left.\middle|a'\right\rangle)=|\left\langle a'\middle|\alpha\right\rangle|^2:~\hat{A}=\sum_{a'}a'\Lambda_{a'}:~\langle \hat{A}\rangle_\alpha=\left\langle\alpha\middle|\hat{A}\middle|\alpha\right\rangle\right]$$ \item Unitary Relations $$\left[\hat{U}=\sum_{k}\left.\middle|b^{(k)}\right\rangle\left\langle a^{(k)}\middle|\right.:~\left.\middle|b\right\rangle=U^\bot\left.\middle|a\right\rangle:~X_b=U^\bot X_aU\right]$$ \item Commutatives $$\left[\hat{A}\left.\middle|a',b'\right\rangle=a'\left.\middle|a',b'\right\rangle,~\hat{B}\left.\middle|a',b'\right\rangle=b'\left.\middle|a',b'\right\rangle:~b'=\left\langle a'\middle|\hat{B}\middle|a'\right\rangle\right]$$ \end{enumerate} \subsection{Spin Half Applications} \begin{enumerate} \item Orthogonal XYs $$\left[\left.\middle|\pm x\right\rangle=\frac{1}{\sqrt{2}}\left.\middle|+ z\right\rangle\pm\frac{1}{\sqrt{2}}e^{i\delta_x}\left.\middle|- z\right\rangle,~\left.\middle|\pm y\right\rangle=\frac{1}{\sqrt{2}}\left.\middle|+ z\right\rangle\pm\frac{1}{\sqrt{2}}e^{i\delta_y}\left.\middle|- z\right\rangle:~\delta_y-\delta_x=\frac{\pi}{2}\right]$$ \item Observables $$\left[\hat{S}_z=\left.\middle| +z\right\rangle\left\langle+z\middle|\right.-\left.\middle|- z\right\rangle\left\langle-z\middle|\right.\right]$$ $$\left[\hat{S}_x=\left.\middle| +z\right\rangle\left\langle-z\middle|\right.+\left.\middle|- z\right\rangle\left\langle+z\middle|\right.,~\hat{S}_y=i\left.\middle| -z\right\rangle\left\langle+z\middle|\right.-i\left.\middle|+ z\right\rangle\left\langle-z\middle|\right.\right]$$ \item Unitary Rotators $$\left[{\hat{R}}(d\phi\hat{k})=1-\frac{i}{\hslash}{\hat{J}}_zd\phi:~\hat{J}_z=\hat{J}^\bot_z:~\hat{R}(\phi\hat{k})=\lim_{N\to\infty}\left[1-\frac{i}{\hslash}\hat{J}_z\left(\frac{\phi}{N}\right)\right]^N=e^\frac{-i\hat{J}_z\phi}{\hslash}\right]$$ $$\left[\hat{J}_z\left.\middle|\pm z\right \rangle=\pm\frac{\hslash}{2}\left.\middle|\pm z\right\rangle:~\hat{R}(\phi\hat{k})\left.\middle|\pm z\right \rangle=e^{\mp\frac{i\phi}{2}}\left.\middle|\pm z\right\rangle\right]$$ \end{enumerate} \section{Angular Momentum Theory} \subsection{Foundations} \begin{enumerate} \item Communication Spectral Theory $$\left[[\hat{J}_x,\hat{J}_y]=i\hslash \hat{J}_z:~[\hat{J}^2,\hat{J}_z]=0\right]$$ $$\left[\hat{J}^2\left.\middle|\lambda,m\right\rangle=\lambda\hslash^2\left.\middle|\lambda,m\right\rangle,~\hat{J}_z\left.\middle|\lambda,m\right\rangle=m\hslash\left.\middle|\lambda,m\right\rangle\right]$$ \item Laddering Operators $$\left[\hat{J}_\pm=\hat{J}_x\pm i\hat{J}_y:~[\hat{J}_z,\hat{J}_\pm]=\pm\hslash\hat{J}_\pm;~[\hat{J}^2,\hat{J}_\pm]=0\right]$$ $$\left[\hat{J}_z[\hat{J}_\pm\left.\middle|\lambda,m\right\rangle]=(m\pm 1)\hslash\hat{J}_\pm\left.\middle|\lambda,m\right\rangle:~\hat{J}^2[\hat{J}_\pm\left.\middle|\lambda,m\right\rangle]=\lambda\hslash^2\hat{J}_\pm\left.\middle|\lambda,m\right\rangle\right]$$ $$\left[\hat{J}_\pm\left.\middle|j,m\right\rangle=\sqrt{j(j+1)-m(m\pm 1)}\hslash\left.\middle|j,m\pm1\right\rangle\right]$$ \item Limits $$\left[m^2\le\lambda:~\hat{J}_-\hat{J}_+\left.\middle|\lambda,j\right\rangle=0=\hat{J}_+\hat{J}_-\left.\middle|\lambda,j'\right\rangle=0:~j=0,\frac{1}{2},1,\frac{3}{2},2,\ldots:~m=j,j-1,j-2,\ldots,-j+1,-j\right]$$ $$\left[\hat{J}_z\left.\middle|j,m\right\rangle=m\hslash\left.\middle|j,m\right\rangle:~\hat{J}^2\left.\middle|j,m\right\rangle=j(j+1)\hslash^2\left.\middle|j,m\right\rangle\right]$$ \end{enumerate} \subsection{Spin Half} \begin{enumerate} \item Foundation Elements $$\left[s=\frac{1}{2},~m=\frac{1}{2},-\frac{1}{2}\right]$$ $$\left[\hat{S}_+\xrightarrow{z}{\hslash}\left(\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right),~\hat{S}_-\xrightarrow{z}\hslash\left(\begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array}\right)\right]$$ \item Operator Elements $$\left[\hat{S}_z\xrightarrow{z}\frac{\hslash}{2}\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)\right]$$ $$\left[\hat{S}_x\xrightarrow{z}\frac{\hslash}{2}\left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right),~\hat{S}_y-\xrightarrow{z}\frac{\hslash}{2}\left(\begin{array}{cc} 0 & -i \\ i & 0 \end{array}\right)\right]$$ \item Eigen Lists $$\left[\hat{S}_n\left.\middle|\mu\right\rangle=[{S}_x\cos\phi+S_y\sin\phi]\left.\middle|\mu\right\rangle=\mu{\hslash}\left.\middle|\mu\right\rangle\right]$$ $$\left[\mu=\pm\frac{1}{2}:~\left.\middle|\mu= \pm\frac{1}{2}\right\rangle=\frac{1}{\sqrt{2}}\left.\middle|\frac{1}{2}, +\frac{1}{2}\right\rangle_z\pm\frac{e^{i\phi}}{\sqrt{2}}\left.\middle|\frac{1}{2}, -\frac{1}{2}\right\rangle_z\right]$$ $$\left[\left.\middle|\mu=+\frac{1}{2}\right\rangle=\cos\frac{\theta}{2}\left.\middle|+z\right\rangle+e^{i\phi}\sin\frac{\theta}{2}\left.\middle|-z\right\rangle:~\left.\middle|\mu=-\frac{1}{2}\right\rangle=\sin\frac{\theta}{2}\left.\middle|+z\right\rangle-e^{i\phi}\cos\frac{\theta}{2}\left.\middle|-z\right\rangle\right]$$ \end{enumerate} \subsection{Spin Full} \begin{enumerate} \item Foundation Elements $$\left[s={1},~m={1},0,{-1}\right]$$ $$\left[ \hat{S}_+\xrightarrow{z}{\hslash}\left(\begin{array}{ccc} 0 & \sqrt{2} & 0 \\ 0 & 0 & \sqrt{2} \\ 0 & 0 & 0 \end{array}\right),~\hat{S}_-\xrightarrow{z}\hslash\left(\begin{array}{ccc} 0 & 0 & 0 \\ \sqrt{2} & 0 & 0 \\ 0 & \sqrt{2} & 0 \end{array}\right)\right]$$ \item Operator Elements $$\left[\hat{S}_z\xrightarrow{z}{\hslash}\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{array}\right)\right]$$ $$\left[\hat{S}_x\xrightarrow{z}\frac{\hslash}{\sqrt{2}}\left(\begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & ~0 \end{array}\right),~\hat{S}_y\xrightarrow{z}\frac{\hslash}{\sqrt{2}}\left(\begin{array}{ccc} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0 \end{array}\right)\right]$$ \item Eigen Lists $$\left[\hat{S}_n\left.\middle|\mu\right\rangle=m{\hslash}\left.\middle|\mu\right\rangle\right]$$ $$\left[\mu=+1:~\left.\middle|\mu=+1\right\rangle=\frac{1}{2}\left.\middle|1, +1\right\rangle_z+\frac{e^{i\phi}}{\sqrt{2}}\left.\middle|1, 0\right\rangle_z+\frac{e^{2i\phi}}{2}\left.\middle|1, -1\right\rangle_z\right]$$ $$\left[\mu=0:~\left.\middle|\mu=0\right\rangle=\frac{1}{\sqrt{2}}\left.\middle|1, +1\right\rangle_z-\frac{e^{2i\phi}}{\sqrt{2}}\left.\middle|1, -1\right\rangle_z\right]$$ $$\left[\mu=-1:~\left.\middle|\mu=-1\right\rangle=\frac{1}{2}\left.\middle|1, +1\right\rangle_z-\frac{e^{i\phi}}{\sqrt{2}}\left.\middle|1, 0\right\rangle_z+\frac{e^{2i\phi}}{2}\left.\middle|1, -1\right\rangle_z\right]$$ \end{enumerate} \section{Time Evolution} \subsection{Foundations} \begin{enumerate} \item Independent Evolutors $$\left[\hat{U}(dt)=1-\frac{i}{\hslash}\hat{H}dt:~\hat{H}=\hat{H}^\bot:~\hat{U}(t)=\lim_{N\to\infty}\left[1-\frac{i}{\hslash}\hat{H}\left(\frac{t}{N}\right)\right]^N=e^\frac{-i\hat{H}t}{\hslash}\right]$$ \item Schrodingers' $$\left[i\hslash\frac{d}{dt}\hat{U}(t)=\hat{H}\hat{U}(t):~i\hslash\frac{d}{dt}\left.\middle|\Psi(t)\right\rangle=\hat{H}\left.\middle|\Psi(t)\right\rangle\right]$$ \item Expectation Rates $$\left[\frac{d}{dt}\langle \hat{A}\rangle_{\Psi(t)}=\frac{i}{\hslash}\left\langle\Psi(t)\middle|[\hat{H},\hat{A}]\middle|\Psi(t)\right\rangle+\left\langle\Psi(t)\middle|\frac{\partial\hat{A}}{\partial t}\middle|\Psi(t)\right\rangle\right]$$ \end{enumerate} \subsection{Some Hamiltonians} \begin{enumerate} \item zB Hamiltonian $$\left[\hat{H}=-\vec{\mu}\cdot\vec{B}=\omega_0\hat{S}_z:~\hat{U}(t)=e^{-\frac{i\omega_0\hat{S}_zt}{\hslash}}=\hat{R}(\omega_0t\hat{k})\right]$$ $$\left[\hat{H}\left.\middle|\pm z\right\rangle=\pm\frac{\hslash\omega_0}{2}\left.\middle|\pm z\right\rangle:~\hat{U}(t)\left.\middle|\pm z\right\rangle=e^{\mp\frac{i\omega_0 t}{2}}\left.\middle|\pm z\right\rangle\right]$$ $$\left[\hat{U}(t)\left.\middle|+x\right\rangle=e^{-\frac{i\hat{H}t}{\hslash}}\left.\middle|+x\right\rangle=\frac{e^{-\frac{i\omega_0t}{2}}}{\sqrt{2}}\left.\middle|+z\right\rangle+\frac{e^{\frac{i\omega_0t}{2}}}{\sqrt{2}}\left.\middle|-z\right\rangle\right]$$ $$\left[|\left\langle\pm x\middle|\Psi(t)\right\rangle|^2=\frac{1\pm\cos\omega_0t}{2}:~|\left\langle\pm y\middle|\Psi(t)\right\rangle|^2=\frac{1\pm \sin\omega_0t}{2}\right]$$ \item xB Hamiltonian $$\left[\hat{H}=-\vec{\mu}\cdot\vec{B}=\omega_0\hat{S}_z+\omega_1(\cos\omega t)\hat{S}_x\right]$$ $$\left[\hat{U}(t)\left.\middle|+z\right\rangle=\cos\left(\frac{\omega_1t}{4}\right)e^{-\frac{i\omega_0t}{2}}\left.\middle|+z\right\rangle-i\sin\left(\frac{\omega_1t}{4}\right)e^{\frac{i\omega_0t}{2}}\left.\middle|-z\right\rangle\right]$$ $$\left[|\left\langle+z\middle|\Psi\right\rangle|^2=\cos^2\left(\frac{\omega_1t}{4}\right),~|\left\langle-z\middle|\Psi\right\rangle|^2=\sin^2\left(\frac{\omega_1t}{4}\right)\right]$$ \item Energy Uncertainty $$\left[\Delta H\Delta t\ge\frac{\hslash}{2}\right]$$ \end{enumerate} \section{Spin Half Systems} \subsection{Foundations} \begin{enumerate} \item Basises $$\left[\left.\middle|\pm z,\pm z\right\rangle=\left.\middle|\pm z\right\rangle_1\otimes\left.\middle|\pm z\right\rangle_2:~\left.\middle|1\right\rangle=\left.\middle|+z,+z\right\rangle,~\left.\middle|2\right\rangle=\left.\middle|+z,-z\right\rangle,~\left.\middle|3\right\rangle=\left.\middle|-z,+z\right\rangle,~\left.\middle|4\right\rangle=\left.\middle|-z,-z\right\rangle\right]$$ \item Operators $$\left[\hat{H}=\frac{2A}{\hslash^2}{\hat{S}_1}\cdot{\hat{S}_2}:~\hat{S}=\hat{S}_1\otimes 1+1\otimes\hat{S}_2\right]$$ \item Eigen Lists $$\left[E=\frac{A}{2}:~\left.\middle|E_1,E_2,E_3\right\rangle=\left.\middle|1\right\rangle,~\frac{1}{\sqrt{2}}\left.\middle|2\right\rangle+\frac{1}{\sqrt{2}}\left.\middle|3\right\rangle,~\left.\middle|4\right\rangle::~E=-\frac{3A}{2}:~\left.\middle|E_4\right\rangle=\frac{1}{\sqrt{2}}\left.\middle|2\right\rangle-\frac{1}{\sqrt{2}}\left.\middle|3\right\rangle\right]$$ $$\left[s=1:~\left.\middle|1,1\right\rangle=\left.\middle|1\right\rangle;~\left.\middle|1,0\right\rangle=\frac{1}{\sqrt{2}}\left.\middle|2\right\rangle+\frac{1}{\sqrt{2}}\left.\middle|3\right\rangle;~\left.\middle|1,-1\right\rangle=\left.\middle|4\right\rangle::~s=0:~\left.\middle|0,0\right\rangle=\frac{1}{\sqrt{2}}\left.\middle|2\right\rangle-\frac{1}{\sqrt{2}}\left.\middle|3\right\rangle\right]$$ \end{enumerate} \subsection{Paradoxes} \begin{enumerate} \item Spin Zero $$\left[\left.\middle|0,0\right\rangle=\frac{1}{\sqrt{2}}\left.\middle|+z,-z\right\rangle-\frac{1}{\sqrt{2}}\left.\middle|-z,+z\right\rangle=-\frac{1}{\sqrt{2}}\left.\middle|+x,-x\right\rangle+\frac{1}{\sqrt{2}}\left.\middle|-x,+x\right\rangle\right]$$ \item Experimental Quests $$\left[P(A;B:\hat{a},\hat{b},\hat{c})\ge\frac{1}{3}?,~=\frac{1}{4}?\right]$$ \item Bell's Inequality $$\left[N_3+N_4\le (N_2+N_4)+(N_3+N_7):~P(+a;+b)\le P(+a,+c)+P(+c,+a)\right]$$ $$\left[\sin^2\left(\frac{\theta_{ab}}{2}\right)\le \sin^2\left(\frac{\theta_{ac}}{2}\right)+\sin^2\left(\frac{\theta_{cb}}{2}\right)\right]$$ \end{enumerate} \subsection{Teleportations} \begin{enumerate} \item Utility Entanglement $$\left[\left.\middle|\Psi_{23}^{(-)}\right\rangle=\frac{1}{\sqrt{2}}\left.\middle|+z\right\rangle_2\left.\middle|-z\right\rangle_3-\frac{1}{\sqrt{2}}\left.\middle|-z\right\rangle_2\left.\middle|+z\right\rangle_3\right]$$ $$\left[\left.\middle|\Psi_{123}\right\rangle=[\left.\middle|\Psi_{1}\right\rangle=a\left.\middle|+z\right\rangle_1+b\left.\middle|-z\right\rangle_1]\left[\left.\middle|\Psi_{23}^{(-)}\right\rangle=\frac{1}{\sqrt{2}}\left.\middle|+z\right\rangle_2\left.\middle|-z\right\rangle_3-\frac{1}{\sqrt{2}}\left.\middle|-z\right\rangle_2\left.\middle|+z\right\rangle_3\right]\right]$$ \item Bell Basis 12 $$\left[\left.\middle|\Psi_{12}^{(\pm)}\right\rangle=\frac{1}{\sqrt{2}}\left.\middle|+z\right\rangle_1\left.\middle|-z\right\rangle_2\pm\frac{1}{\sqrt{2}}\left.\middle|-z\right\rangle_1\left.\middle|+z\right\rangle_2\right]$$ $$\left[\left.\middle|\Phi_{12}^{(\pm)}\right\rangle=\frac{1}{\sqrt{2}}\left.\middle|+z\right\rangle_1\left.\middle|+z\right\rangle_2\pm\frac{1}{\sqrt{2}}\left.\middle|-z\right\rangle_1\left.\middle|-z\right\rangle_2\right]$$ \item Bell Measurement $$\left[\left.\middle|\Psi_{123}\right\rangle=\right]$$ $$\left[\frac{1}{2}\left.\middle|\Psi_{12}^{(-)}\right\rangle(-a\left.\middle|+z\right\rangle_3-b\left.\middle|-z\right\rangle_3)+\frac{1}{2}\left.\middle|\Psi_{12}^{(+)}\right\rangle(-a\left.\middle|+z\right\rangle_3+b\left.\middle|-z\right\rangle_3)\right]$$ $$\left[+\frac{1}{2}\left.\middle|\Phi_{12}^{(-)}\right\rangle(b\left.\middle|+z\right\rangle_3+a\left.\middle|-z\right\rangle_3)+\frac{1}{2}\left.\middle|\Phi_{12}^{(+)}\right\rangle(-b\left.\middle|+z\right\rangle_3+a\left.\middle|-z\right\rangle_3)\right]$$ \end{enumerate} \section{One Dimensional Mechanics} \subsection{Continuums} \begin{enumerate} \item Eigen Basekets $$\left[\left.\middle|\alpha\right\rangle=\int_{-\infty}^{\infty}dx\left.\middle|x\right\rangle\left\langle x\middle|\alpha\right\rangle\right]$$ \item Translators $$\left[\hat{T}(dx)=1-\frac{i}{\hslash}\hat{p}_xdx:~\hat{p}_x=\hat{p}_x^\bot:~\hat{T}(a)=\lim_{N\to\infty}\left[1-\frac{i}{\hslash}\hat{p}_x\left(\frac{a}{N}\right)\right]^N=e^{-\frac{i\hat{p}_xa}{\hslash}}\right]$$ $$\left[\hat{T}(\delta x)\left.\middle|\Psi\right\rangle=\left.\middle|\Psi\right\rangle-\delta x\int dx\left.\middle|x\right\rangle\frac{\partial}{\partial x}\left\langle x\middle|\Psi\right\rangle\right]$$ \item Wave Function $$\left[\left\langle x\middle|p\right\rangle=\frac{1}{\sqrt{2\pi\hslash}}e^{\frac{ipx}{\hslash}}:~\lambda=\frac{h}{p}\right]$$ $$\left[\left\langle p\middle|\Psi\right\rangle=\int dx\frac{1}{\sqrt{2\pi\hslash}}e^{-\frac{ipx}{\hslash}}\left\langle x\middle|\Psi\right\rangle:~\left\langle x\middle|\Psi\right\rangle=\int dp\frac{1}{\sqrt{2\pi\hslash}}e^{-\frac{ipx}{\hslash}}\left\langle p\middle|\Psi\right\rangle\right]$$ \end{enumerate} \subsection{Gaussian Wave Packets} \begin{enumerate} \item Communications $$\left[ [\hat{x},\hat{p}_x]=i\hslash\right]$$ \item PosiMomenta Space $$\left[\left\langle x\middle|\Psi\right\rangle=\frac{1}{\sqrt{\sqrt{\pi}a}}e^{-\frac{x^2}{2a^2}}:~\Delta x=\frac{a}{\sqrt{2}}\right]$$ $$\left[\left\langle p\middle|\Psi\right\rangle=\sqrt{\frac{a}{\hslash\sqrt{\pi}}}e^{-\frac{p^2a^2}{2\hslash^2}}:~\Delta p_x=\frac{\hslash}{\sqrt{2}a}\right]$$ \item Free Evolution $$\left[\hat{H}=\frac{\hat{p}_x^2}{2m}:~\left.\middle|\Psi(x,t)\right\rangle=\frac{1}{\sqrt{\sqrt{\pi}[a+\frac{i\hslash t}{ma}]}}e^{-\frac{x^2}{2a^2\left[1+\frac{i\hslash t}{ma^2}\right]}}:~\Delta x=\frac{a}{\sqrt{2}}\left[1+\frac{\hslash^2t^2}{m^2a^4}\right]^\frac{1}{2}\right]$$ \end{enumerate} \subsection{General Solutions} \begin{enumerate} \item Time Dependent $$\left[i\hslash\frac{\partial}{\partial t}\Psi(x,t)=\left[-\frac{\hslash^2}{2m}\frac{\partial^2}{\partial x^2}+V(x)\right]\Psi(x,t)\right]$$ \item Eigens State $$\left[\Psi_E(x,t)=\left\langle x\middle|E\right\rangle e^{-\frac{iEt}{\hslash}}:~\frac{d^2\Psi}{dx^2}=-\frac{2m(E-V_0)}{\hslash^2}\Psi\right]$$ \item Potential Well $$\left[\Psi(x)=A\sin kx+B\cos kx:~\Psi_n(x)=\sqrt{\frac{2}{a}}\cos\frac{n\pi x}{a},~n=1,3,5,\ldots:~\sqrt{\frac{2}{a}}\sin\frac{n\pi x}{a},n=2,4,6\right]$$ $$\left[E_n=\frac{\hslash^2\pi^2n^2}{2ma^2}:~[n=1,2,3,\ldots]\right]$$ \end{enumerate} \subsection{Scattering} \begin{enumerate} \item Probability Current $$\left[\frac{\partial}{\partial t}|\Psi|^2=-\frac{\partial j_x}{\partial x}:~j_x=\frac{\hslash}{2mi}\left[\Psi^*\frac{\partial \Psi}{\partial t}-\Psi\frac{\partial \Psi^*}{\partial t}\right]\right]$$ $$\left[\frac{d}{dt}\int_a^bdx|\Psi|^2=-j_x(b,t)+j_x(a,t)\right]$$ \item Step Potential $$\left[\Psi(x)=Ae^{ikx}+Be^{-ikx}:~j_x=j_\text{inc}-j_\text{ref}=\frac{\hslash k}{m}[|A|^2-|B|^2]\right]$$ $$\left[E>V_0:~\Psi(x)=Ce^{ik_0x}:~j_\text{trans}=\frac{\hslash k_0}{m}|C|^2:~R=\frac{(k-k_0)^2}{(k+k_0)^2}:~T=\frac{4kk_0}{(k+k_0)^2}\right]$$ \item Cutting Off $$\left[T=\frac{1}{1+ \left(\frac{k^2+q^2}{2kq}\right)^2\sinh^2qa}\right]$$ \end{enumerate} \subsection{Harmonic Oscillations} \begin{enumerate} \item Oscillations $$\left[\hat{H}=\frac{\hat{p}_x^2}{2m}+\frac{1}{2}m\omega^2\hat{x}^2\right]$$ \item Laddering Operators $$\left[\hat{a}=\sqrt{\frac{m\omega}{2\hslash}}\left(\hat{x}+\frac{i}{m\omega}\hat{p}_x\right):~[\hat{a},\hat{a}^\bot]=1:~\hat{H}=\hslash\omega\left(\hat{a}\hat{a}^\bot+\frac{1}{2}\right):~[\hat{N},\hat{a}]=-\hat{a}:~[\hat{N},\hat{a}^\bot]=\hat{a}^\bot\right]$$ $$\left[Na^\bot\left.\middle|\eta\right\rangle=(n+1)a^\bot\left.\middle|\eta\right\rangle:~Na\left.\middle|\eta\right\rangle=(n-1)a\left.\middle|\eta\right\rangle\right]$$ $$\left[\hat{a}^\bot\left.\middle|n\right\rangle=\sqrt{n+1}\left.\middle|n+1\right\rangle:~\hat{a}\left.\middle|n\right\rangle=\sqrt{n}\left.\middle|n-1\right\rangle\right]$$ \item Limits $$\left[\eta\ge 0:~a^\bot a\left\middle|\eta_m\right\rangle=0:~\eta_m=0\right]$$ $$\left[N\left.\middle|n\right\rangle=n\left.\middle|n\right\rangle,~n=0,1,2,\ldots:~H\left.\middle|n\right\rangle=\hslash\omega\left(n+\frac{1}{2}\right)=E_n\left.\middle|n\right\rangle,~n=0,1,2,\ldots\right]$$ \end{enumerate} \subsection{Harmonic Waves} \begin{enumerate} \item Zero Waves $$\left[\left\langle x\middle|0\right\rangle=\left(\frac{m\omega}{\pi\hslash}\right)^\frac{1}{4}e^{-\frac{m\omega x^2}{2\hslash}}\right]$$ \item n Waves $$\left[\left\langle x\middle|n\right\rangle=\frac{1}{\sqrt{n!}}\left\langle x\middle|(a^\bot)^n\middle|0\right\rangle\right]$$ \item Uncertainities $$\left[\Delta x=\sqrt{\left(n+\frac{1}{2}\right)\frac{\hslash}{m\omega}}:~\Delta p_x=\sqrt{\left(n+\frac{1}{2}\right)\hslash m\omega}\right]$$ \end{enumerate} \subsection{Fore postulatiosn} \begin{enumerate} \item INSPIRATION: Experiment shows that quantum phenomena are probabilistic but probabilities are like waves i.e. able to cancel each other but if you measure hen follows the classical probability theory. The weirdness of qm seems to challenge some expectations from the universe like locality and local realism \item STATE OF QUANTUM OBJECT: A pure state is represented by a normalized vector in hilbert space. yeha superposiiot bhanera superopsoitno of classical staets whle the quantum particles do not need to obey the callicaslness so they are actually just one. For eg. the photon is just a single state quantum particle but it can also be interpreted as combination of electron and positorn and whatever is allowed. We take them as a superopositon. For two vectors in that space their superpositoin is also is in that space. Since just multiplying by $e^{i\theta}$ does not change to the normalization so no physical difference there. \item MEASUREMENT: any physical quantity (i.e., observable) a, there exists a corresponding Hermitian operator A acting on H. When we make a measurement of a, we obtain one of the eigenvalues of the operator A. Let $\lambda 1$ and $\lambda 2$ be two eigenvalues of A $$A\left.\middle|\lambda_i\right\rangle=\lambda_i\left.\middle|\lambda_i\right\rangle$$ Suppose the system is in a superposition state $c_1\left.\middle|\lambda_1\right\rangle$+$c_2\left.\middle|\lambda_2\right\rangle$. If we measure 'a' in this state, then the state undergoes an abrupt change to one of the eigenstates corresponding to the observed eigenvalue: If the observed eigenvalue is $\lambda_1(\lambda_2)$ the system undergoes a wave function collapse as follows: $$c_1\left.\middle|\lambda_1\right\rangle+c_2\left.\middle|\lambda_2\right\rangle\rightarrow \left.\middle|\lambda_1\right\rangle(\left.\middel|\lambda_2\right\rangle)$$ and the state imme- diately after the measurement is one of them. The probabiltiy of collapsing to one of the state is squraed of that coefficent $c^2$\par Statistically the expectation value of the obserable A on the state psi will be $$\left\langle\Psi\middle|A\middle|\Psi\right\rangle$$ For if we have back to back two observable, whose hermitian do ont commutate, then we have what we call heisenberg unceantaintly principle \item EVOLUTION: The time dependecne of a system is $$i\hbar \frac{\partial\left.\middle|\Psi\right\rangle}{\partial t}=H\left.\middle|\Psi\right\rangle$$ where the H is hamilotonian, the hermitian operator corresponding to the energy of the system which can be solved for H being time independent $$\left.\middle|\Psi\right\rangle = e^{-\frac{iHt}{\hbar}}\left.\middle|0\right\rangle$$ the hamiltonians are like instantenous time genrators what give unitary transformations. The exp blob is the unitary operator to prove that for diagonalizable matrix $e^A=Ue^DU^*$\par Hence gien a hermitan we can always find the unitary gien by that euqaiotn over there. Now the reverse is also true bit given a unitary can arise lots of hamiltoninan from it. \item MORE ON HAMILTONIAN: The eigen values of hamiltoanians (are real) called energies system can attain. (any state can be represented in superpositn of those energy sates) Now the eigen vectors corresponding to those eigen values are also the eigen vectors of the unitary. Hence when you apply a uniteray to the eigen value then we get the same state ie unitary cannot change them. Also the most states will like to be in their ground states (second law of thermodynaics bitch) so the quantum shits happing is those states that dont sit on their ground state. \item THE FIRST NP PROBLEM: the order of hermitian and unitary operators acting on hilbert space of dimension n have size n*n. When you have many energies ating on a system, like knieti, potential, elector and what not then hamiltonians will be made up ofS $$H=H_0+H_1+\ldots+H_{m-1}$$ so getting the ground sate by diagonalization could be hard, the queison is can i get ground state by diagonaliation of those sijmple hamiltonians. This is an NP hard problem to find gound state of H. \end{enumerate} \subsection{Qubit specials} \begin{enumerate} \item STATES: We will talk all about qubits even for higher diemnsional hilbert spaces. A qubit lives in two dimensional hilbert space, while n qubits will live in 2$^n$. Our default basis in the qubit systems are labeled $\left.\middle|0\right\rangle$ and 1. Some other useful states that can also be used as basis (but are not) are +, -, i and -i . The states with real amplitudes (when expressed in 0,1 bassi) can be represented in unit circle like the coordinates so the states: 0, 1, + and - lie there bbut i and -i dont. \par REMEMBER THE COLUMN MATRIX OF EACH IN 0,1 BASIS \item HERMITIANS: the measurmente will always be in the computational basis, if we require to do measurement in other basis then we use another technique (coming later) \item UNITARIES: the transformations include \begin{enumerate} \item identity gate \item X gate, Z gate, Y gate \item H gate \item Phase gate (caes: S and T gate) \item RY gate, RX gate, RZ gate \item U gate \end{enumerate} REMEMBER THE RECTANGULAR OF EACH IN 0,1 BASIS \item WORD ON HADAMARD: it maps 0,1 to +,- and vice versa: $$\left[H\left.\middle|x\right\rangle=\frac{1}{\sqrt{2}}(\left.\middle|0\right\rangle+(-1)^x\left.\middle|1\right\rangle)\right]$$ NOTE: MEasurement in basis some basis v,w other than 0,1 is making a unitary transformation then measuring in 0,1. basis. For eg if you want to measure in basis +,- then applly the hadarmard and then measure in 0,1 basis whereby 0 corresponds to + and 1 corresponds to -. \item PHYSICAL IMPLICATIONS: 1) as unitary matrices are reversible the information cannot be desroyed, unvierse is essentially reversible 2) deterministic: there is nothing probabilisitic in the unitary evolution process itslef 3)continuous: unitar transformations take place over intervals of time which can always be broekn into samller and smaller intervals, eg if a process was applied for a second, then the same process can be splitted as if two of it are applied for half half second, but it is sometimes imposible to get the real elements in those half matrices. On the other hand, the measuremetn break all these thres rules: irreversible, proailistic, instanteaneous \item QUANTUM ZENO EFFECT: \par DIFFERENT BASE MEASUREMENT: if two parties have same quibit but different bases say you have basis 0,1 and another basis $\theta,\theta+\pi/2$ (rotated by some real angle) then the probability of getting the bits in order is $cos^2\theta$\par that a system at 0 could be moved to 1 without unitary by just measuirng it frquency in the basis that is rotates just a bit, and another vice versa is that a system can be frequentyly measured to make it stable like shit. For the math Involved: keep in mind the unit circle (and the word on measurement by me), the probability that the state ends up on the basis that is nearer to the basis it was rotate from is cos2e (pretty good for small e), so each time we change the basis by epsilon then in about pi/(2e) measurments the state will be at 1. Probability of failure is at most (by union formula) pi/2e*sin2e. \par Same applied every time you make the measurement on the basis (0,1) to make the state stay at 0 when the state drift by some small angle e \item BOTH ZENOS EFFECT: its like sending a partilcle, if we actually send particle then it checks the bomb and if the bomb is good it explodes else the particle is returned as if the mirror. Let the bomb be quantum.\par What we do is we send the state (being absent and partile being present in superposition where being absent if significant). $$\text{particle absent}\rightarrow (\text{rotate by e})\rightarrow \text{cose being absent+sine being present}$$ Now the bomb measures it and with very high probability finds the state (being absent) and returns us that only, else sends the whole state (superimposed state). We then apply the same rotation to the state without doing anything which after pi/(2*e) times either will stay at 0 only or will turn to 1 state. If the bomb is working then his anti zeno effect will take us out (giving the 0 state) on the other hand if bomb is not working our zeno effect will beat him (getting the 1 state). The max (upperbound) proability that the bomb explode is: pi/2e * sin2e \end{enumerate} \subsection{Bipartite systems} \begin{enumerate} \item STATE: Supppose a sytem made up of two qubits one living in $H_1$(dim 2) another in $H_2$(dim 2) then the system as a whole lives in $H_1\otimes H_2$ whose dimension will be 4 and the multiplicaiton goes on like for three its 8. There would be some states in H called a separable states if can be written as tensor product of two vectors each in H1 and H2. And others which cannot be are entangled. There are many more entangled states than there are separable. Almost all pure states are entangled ie probability of finding the entangled states is one (just like finding integers in real number line), \par To defferentiate them? When separable, when entangled? A vector in H = H1 o H2 can be decomposed as $$\left.\middle|\Psi\right\rangle = \sum_{i=1}^r\sqrt{s_i}\left.\middle|f_{1,i}\right\rangle\otimes\left.\middle|f_{2,i}\right\rangle$$ where the f's are the complete set in each hilbert space H1 and H2 This is such a way that one of the basis vector in space H1 comes only with another and only another in space H2 and this basis make the vector writable in the given form. Here the r is schmidth number.\par In other words, the general vector of the combined hilbert space can be written as a matrix of order 2 by 2 to get which resize the state into a matrix) then this matrix can be SVDed. The number of non zero singular values is the schmidth number\par In 2 quit state 4 imp states which are what called maximally entangled. \par REMEMBER THEIR COLUMN MATRICES \item MEASUREMENT: in the multipartite system one sytem could be on earth another on andromeda galaxy, you can make a measurement on earth that will collapse the system here while another system will adjust just like that (called partial measruemnet rule) $$a\left.\middle|00\right\rangle+b\left.\middle|01\right\rangle+a\left.\middle|10\right\rangle+d\left.\middle|11\right\rangle\rightarrow\text{ make measurement on first qubit}$$ $$\left.\middle|0\right\rangle\otimes\frac{a\left.\middle|0\right\rangle+b\left.\middle|1\right\rangle}{\sqrt{a^2+b^2}}$$ \item UNITARIES: some multi gates that perform a action on one qbuit and something other on another then those matrices can be dreived by tensor producting. While something like CNOT cannot be written as a tensor product. \begin{itemize} \item Tensored gates \item C-version of every other gate mentioned above (to get the matrix of which just put identity on second quandard and the matrix on fourth) \item Swap gate \item ADDING THE 3 QUBIT GATES HERE: fredkin and taffoli are those two \end{itemize} REMEMBER THE MATRICES \item WORD ON BELL CREATING GATES: But before that let me show you something sexy about hadamards: $$H^{\otimes n}\left.\middle|x\right\rangle_n=\frac{1}{\sqrt{2^n}}\sum_{y=0}^{2^n-1}(-1)^{x\cdot y}\left.\middle|y\right\rangle_n$$ Now the following circuit produces what we earlied called the maximally entanged states, note what it does to 00[00+11], 01[01+10], 10[00-11], and 11[01-10] \begin{center} \begin{tikzpicture} \node[scale=1.0] { \begin{quantikz} \left.\middle|i\right\rangle&&\gate{H}\slice{0}&\ctrl{1}\slice{1}&\qw\\ \left.\middle|j\right\rangle&&\qw&\targ{}&\qw \end{quantikz} } \end{tikzpicture} \end{center} That bell states are maximally entangled in any other orthonormal basis ie can be written as just like that 00+11 where 0 and 1 are replcaed by some others. Also the one bell state can be converted into another just by manipulation qubit of one system. \item PHYSICAL IMPLICATION: In entangled state, when alice makes a measurement, bob's result are fixed but since alice does not have contorl over his results they cannot communitacte. Hence, engtanglement is like classical coorelation where informaiton does not travel by speed of light but you will konw sutffs like that newspaper thing so Correlation is bassically like the stuff that learning someting about one bit says about anohter bit. But einstein has a questions? where he found that alice could do more than just mearuing the qubits. What if she measures in hadarmadrd basis? so if she measures +, bob collapses to + and if she sess 1 bob collapses to - ie. bob will know which basis she measured faster than light bro? But no, becuase there is not wway in whcih bob knows which basis alice makes her measuremet in (through the formalism of mixed states) known as the no communication theoerem: during measurement of an entangled quantum state, it is not possible for one observer, by making a measurement of a subsystem of the total state, to communicate information to another observer. \item GENERATIONLIZATIONS OF BOMB EFFECT. Now let there be a controlled gate either I gate or an X gate. To know which we could send 1 in upper (control bit) and 0 in target bit. Then depending on output we know which gate it was. But if it was a X gate then we would be in problem.\par In this case we do is we put 0 to the target bit and rotate it a little and send through the controlled gate. Then measure the second qubit. After doing this for pi/2e times we check the upper bit. If it was an identity gate then our zeno effect will win so we measure 1 and if it was a X gate then the antizeno will win so the result would be 0. The probabilities are same as before.\par Also on the other hand, a variation is if the qubit is unknown and the controlled X is known then we have vice versa way of knowing whether the known bit was + or not (if the possibilities were only 0, 1, + and -). Similarly to know if it was - or not we use -X controlled gate. \item QUANTUM MONEY: prevents money from counterfeiting. A money would contain a m digit long binary serial number and m number of qubits (each possible in 0, 1, +, -) state and a function that maps them, none is entangled. By guessing, each bit has probability of 3/4 so getting all correct is (3/4)**n. The possibility is the varied bomb attack. \end{enumerate} \subsection{Another postulations} \begin{enumerate} \item STATE: A quantum state can also be represented by the desnity matrix, but in additon we can also insert our ignorance in there. A state in 1qubit system (2 by 2) in 2qubitsystem (4by4( and in 3qubitsytem/8dimhilbertspace (8 by 8) and so on. So from n numbers we have not n by n numbers. The matrix is given by $$\rho = \sum_i p_i\left.\middle|\Psi_i\right\rangle\left\langle\Psi_i\middle|\right.$$ (Note that the density matrix is a SEMI DEIFNITE HERMITIAN with trace 1, which is also the requried condition for amatrix to be a density matrix, ALso the eigen values are non negative)\par The diagonal elements of the matrix give the probabilities (not amplitute) of getting the state in basis in which density matrix was formed. (when density matrix was formed what basis was those states in). The off diagonal signify that your system non-zero off-diagonal signify that your system features a quantum superposition between the basis vectors that you chose for rho ie your state is not only a statistical mixture of your basis states but an actual quantum superposition in which those basis elements are "coexisting"\par Since this is the square matrix we have the following properties of matrix: in general rank tells you about the purity of the state, if its 1 then pure state\par Also note that if a state is called maximally mixed if the density matrix is proportional to the identity matrix, Physically, it may be interpreted as a uniform mixture of states in an orthonormal basis. (regardless of what basis we measure in we the outcome will always be equally likely in any basis).\par REMEMBER THE FOLLOWING TWO DENSITIES: $$I/2 \text{equal mixure of 0 and 1 for superposiiton: }1/2*1111 \text{ wala matrix}$$ $$\left(\begin{array}{cc} 0.5 & 0 \\ 0 & 0.5 \end{array}\right)\left(\begin{array}{cc} 0.5 & 0.5 \\ 0.5 & 0.5 \end{array}\right)$$ \item HERMITIAN: when a meawsurement is taken (given by the hermiitan matrix) then the state has to abrupty change into one of the eigen states, with the corresponding eigen value. As already said for standard basis measurement that probabilites are given by the diagonal. While for general this takes place with the probability of $tr(P_a\rho)$ where $P_a =\left.\middle|\lambda_a\right\rangle\left\langle\lambda_a\middle|\right.$ Also the expectation values is given by such $$\left\langle A\right\rangle = tr(\rho A)$$ \item UNITARY: The temporal evolution of the desnity matrix is given by the liovuille von nuermmall eqna $$i\hbar\frac{d\rho}{dt}=[H,\rho]$$ now the unitary operator U that applies to the vector states, when applied to the desnity matrix maps it to $U\rho U^t$ \par REMEMBER THE FORMULA (COZ NO MATRIX) \end{enumerate} \subsection{Bipartite Desnties} \begin{enumerate} \item STATES: There are also density matrix for 2 qubit hibert space (4 dim), just like the separable and entangled states there is the three types of division in density operators. They will be separable if it is written in the form $$\left[\rho = \sum_i~~(p_i)~~\rho_{1,i}\otimes \rho_{2,i}\right]$$ where pi's sum to 1, else inseparable of course.\par For 2 by 2 systems, let $\rho$ be a bipartite state nd define the partial transpose of $\rho^{pt}$ of $\rho$ with respect to the second Hilbert space as $$\rho_{ij,kl}\rightarrow \rho_{il,kj}$$ ignoring all these mess it is just each block of the matrix is spearated rotated, the four quadrants are transposed individually. If it was separable then it would still be a desntiy matrix form, Hence forms the condition that a separate matrix's partial transpose is also a density matrix. Whose vice versa is also true for 2*2 rey. So check the trace and for semi positive hermitianity. \par The density matrix of the bell states is $$\left[00\pm 11=\frac{1}{2}\left(\begin{array}{cccc} 1&0&0&\pm 1 \\ 0&0&0&0 \\ 0&0&0&0 \\ \pm 1&0&0&1 \\ \end{array}\right)\right]$$ $$\left[01\pm 10=\frac{1}{2}\left(\begin{array}{cccc} 0&0&0&0 \\ 0&1&\pm 1&0 \\ 0&\pm 1&1&0 \\ 0&0&0&0 \\ \end{array}\right)\right]$$ REMEMBER THE GIVEN MATRICES AT LEAST \item HERMITIANS: ? \item UNITARY: ? \item PARTIAL TRACKINGS: Say you have a desnity matrix of some pure two qubit system (might be entangled and might not be), so they probably depend on each other, now you have no control over the another qubit so you add the information contained in it as the ignorance for your qubit. The process of going from a pure state of a composite system, to the mixed state of part of the system (whcih made that composite system, like going from shared state of alice and bob to single of bob becuase when you go from known pure composite state to bob's state then there is some ignorance on the bob's side because alice could measure anything), is called tracing out. Suppose we are interested only in the first system and have no access to the second system. Then the partial trace allows us to “forget” about the second system. In other words, the partial trace quantifies our ignorance of the second sytem.\par We say you obtain reducced density matrix $\rho_A$ by tracing over the rhoAB matrix with respect to B\par To find so $\rho_A$, first you divide the whole matrix $\rho_{AB}$ in four quandrant, do $\rho_a= tr_b(\rho)$ say it loudly: trace of density matrix with respect to b. Taking b as a second system $$tr_b = \text{sum the first, second, third, forth quadrant}$$ $$tr_a=\left(\begin{array}{cccc} \rho_{11} & \rho_{12} & \rho_{13} & \rho_{14} \\ \rho_{21} & \rho_{22} & \rho_{23} & \rho_{24} \\ \rho_{31} & \rho_{32} & \rho_{33} & \rho_{34} \\ \rho_{41} & \rho_{42} & \rho_{43} & \rho_{44} \end{array}\right) = \left(\begin{array}{cc} \rho_{11}+\rho_{33} & \rho_{12}+\rho_{34} \\ \rho_{21}+\rho_{43} & \rho_{22}+\rho_{44} \end{array}\right)$$ Reverse is possible and is called purification but you see it is not unqiue at all, in fact infinitely many possibility \item FINALLY ALICE BOB: When alive and bob share entagneld system there is nothing alice can do to change bobs desnity sub system. You already have the tools to prove this: just calculate Bob’s reduced density matrix, then apply a unitary transformation to Alice’s side, then see if Bob’s density matrix changes. Or have Alice measure her side, and see if Bob’s reduced density matrix changes. about a quantum system. TO show: bob will trace over the density matrix of bell state with respect to alice which by above dallo formula gives: I/2 so a maximally mixed state. \end{enumerate} \subsection{BLOCHENDRA} \begin{enumerate} \item The bloch sphere is for visualization of all about one qubit system. The pure states live in the surface and the mixed states live inside the sphere. The z axis, x axis, y axis are for the three special vectors we talked about. Centre is the special point whose matrix is I/2. \item Notice you can draw a line that touches two points on the surface ie two pure states, while many mixed states are touched. To see which qubits make that mixed state draw the lines. Hence mixed states in one qubit state are made from two pure states. \item Every quantum gate can be described as a 3D rotation by some angle theta about some axis defined by the eigenvectors of the gate. \begin{table}[H] \centering \begin{tabular}{|c|c|c|} \hline Gates & Angle of Rotation & Axis of Rotation\\[0.5cm] I & - & -\\[0.5cm] X(NOT) & ${\pi}$ & x-\\[0.5cm] Z(FLIP) & ${\pi}$ & z-\\[0.5cm] Y & ${\pi}$ & y-\\[0.5cm] H & $\pi$ & $\frac{[x-]+[z-]}{\sqrt{2}}$\\[0.5cm] P & - & -\\[0.5cm] S [ROOT2Z] & $\frac{\pi}{2}$ & z-\\[0.5cm] T [ROOT4Z] & $\frac{\pi}{4}$ & z-\\[0.5cm] RY & $\theta$ & y- \\[0.5cm] RX & $\phi$ & x- \\[0.5cm] RZ & $\lambda$ & z- \\[0.5cm] U & - & -\\ \hline \end{tabular} \end{table} \end{enumerate} \subsection{Finally} \begin{enumerate} \item NO CLONING THOERM: if bob could make many copies of his state he could know which basis alice used for measurement. "if a state is given which is otherwise unknown you cannot make a copy of it". To prove which copy a quibt $a\left.\middle|0\right\rangle+b\left.\middle|1\right\rangle$ we send this bit and another bit through a two-input gate and the result gate should give us two of those states $$\left[\left(a\left.\middle|0\right\rangle+b\left.\middle|1\right\rangle\right)\otimes\left.\middle|0\right\rangle\rightarrow a\left.\middle|0\right\rangle+b\left.\middle|1\right\rangle\otimes \left.\middle|0\right\rangle+b\left.\middle|1\right\rangle\right]$$ in matrix notation $$[a^2~ab~ab~b^2]=U[a~0~b~0]$$ which wants U to be nonlinear (not possible) \item QUANTUM KEY DISTRIBUTION: classical private key encryption is not handy while public key requires some computational assumption on the crackers. While quantum has solution \begin{itemize} \item alice randomly chooses a pair of strings x,y **n binary \item alice uses y string to determine which bases to encode her qubits: 0 for 0,1 and 1 for +,-; and elements of x string to determine which state the qubit likes in 0->0,+, 1->1,- \item alice sends that qubit to bob. \item Now bob picks up a random string y' to decide which basis to measure in and sore the result in x'. AFter bob declares that he has measure they do foloowing \item they share the y and y', now whichever does not match they discard those bits in x and remaining will be their shared key. \item If Eve measures the qubit in middle, let the bit was +, and she measured in wrong basis and get 0. Then even if bob measures in right basis then he has 50 50 probability of getting the right bit. Then they can decide if the required bits are matched by discussing over the unsecured channel. So at last of the bits in x' (say k number) they publicly share k/2 and check if they are matched. if check passes they do some cryptostuff. \end{itemize} \item DENSECODING: shannon says by sending n bits you cannot communicate more than n bits of information. But can share 2 bits of info by sending one qubit on the presimise that they already share a entangled pair. Alice can get three different states, all of them orthogonal to the original Bell pair and to each other, by applying the following gates to her qubit, these three are orthonormal to orignal bell state $$(X\oplus I)\left(\frac{\left.\middle|00\right\rangle+\left.\middle|11\right\rangle}{\sqrt{2}}\right)=\left(\frac{\left.\middle|01\right\rangle+\left.\middle|10\right\rangle}{\sqrt{2}}\right)$$ $$(Z\oplus I)\left(\frac{\left.\middle|00\right\rangle+\left.\middle|11\right\rangle}{\sqrt{2}}\right)=\left(\frac{\left.\middle|00\right\rangle-\left.\middle|11\right\rangle}{\sqrt{2}}\right)$$ $$(Z\oplus I)(X\oplus I)\left(\frac{\left.\middle|00\right\rangle+\left.\middle|11\right\rangle}{\sqrt{2}}\right)=\left(\frac{\left.\middle|01\right\rangle-\left.\middle|10\right\rangle}{\sqrt{2}}\right)$$ Now if she wants to send two bits x,y then for $x=1$ he applies the $x o I$ gate and for $z = 1$ he applies $zoI$. Bob then measures by reversing the bell states ie applying a not gate and hadamard gate as follows. ie his own qubit will be the controlling and alice's will be target. His qubit will be the y and alices will be the x. \begin{center} \begin{tikzpicture} \node[scale=1.0] { \begin{quantikz} \qw&\targ{}&\qw&\qw\\ \qw&\ctrl{-1}&\gate{H}&\qw \end{quantikz} } \end{tikzpicture} \end{center} But the thing is if they were sharing more pre entanglement still only 2 bits of information $$1 qubit\text{ [the entangled one] } + 1 ebit >= 2bit$$ ie by having one preshared entangled bits and sending one qubit we cuold share 2 bits of inormatiion while the vice versa is also possible that by having one preshared entanglement and sending two bits of information, one qubit could be sent. Hence total two qubits involved. \item TELEPORATION: The reverse is possible $$1ebit + 2 bits >= 1 qubit$$ ie to transmit one qubit you need 1 entangled bit and 2 classical bits. The circuit that does this is the following: \begin{center} \begin{tikzpicture} \node[scale=1.0] { \begin{quantikz} \qw&\ctrl{1}&\gate{H}&\qw&meter\\ \qw&\targ{}&\qw&\qw&meter \\ \qw&\qw&\qw&\qw& \end{quantikz} } \end{tikzpicture} \end{center} The above two lines are of alice (hence last two are entangled bits). The first one is that needs to be sent which is destroyed at alices side and the last bit will get that information, this last qubit belongs to bobs.\par Alice uses the qubit to teleport as the controller of controlled not gate whose target is her shared part of bell state. After hadamarding the gate she measures both of her qubit and according to four possible results sends two 2 clbits to bob. Bob performs one of the four operations. If first bit sent is 1 then applies Z and if also second bit is 1 then applies X. \item ANY PHYSICS: Now the dilemma is teleportation is also like encoding but we have sent the information to make the qubit in another place now suppose we encode the whole works of shakespeare into the coefficients of a qubit(ie the amplitudes) then we transfer the qubit like easy but this information cannot be retrieved by using just that qubit no copy available. Also the no cloning theorem saves us here because the received qubit could be copied multiple items to see the probability densities. \item FIRST VERSION OF SWAPPING: if bob and alice are sharing a entagled pair. and bob goes to mike and swap qubits with then now alice and mike will be having the entangled pair while bob will not be. See this as the triangle thing. Now if nobody wants to move then bob and mike must share a entangled pair ie two sides of triangle must have entangled pair then by teleporting they do not need to move. Hence from two sides entanglement is shifted to one. \item SECOND ENTANGLEMENT SWAPPING: Consider a square where the edges that form a C represents the entangled bits. Now the upper and lower edges half bell pairs gets teleported to the corresponding sides. As a result the right corners qubits are entangled even though no contact has been made. \end{enumerate} \subsection{More on Entanglement} \begin{enumerate} \item FIND ENTANGLEMENT: Impressed to entanglement huh? How do you know if two states (here classical or quantum does not matter) are entangled (or correlated for classical case)? (no matrix of state). Given that you have multiple (very large) copies of those states in the hand. and you go on measuring all of them. If they show some kind of correlation they are entangled.\par Now imagine a scenario with three bits distributed to three people, each being either 0 or 1. Now when they three get together and see their bits (actually they will also need many copies of those) they realize a classical correlation. This would be true even if two of them meet and they will realize that they two are having correlation. But are these correlations correspond to the entanglements in quantum world? \par Let the three players share a GHZ state (000+111/root2). With all three of them together they can see that the state is entangled, (How? by just seeing the correlations) but what if Charlie is gone? Can Alice and Bob see that they’re entangled with each other? (they are not sure if charlie measured or not) No. To see this observe that by the No-Communication Theorem, Charlie could’ve measured without Alice and Bob knowing. Even though they donot know if charlie measured the clearly have classical correlation either both 0’s (if Charlie got the measurement outcome |0〉) or both 1 (if Charlie got 1) or any one of them if not measured.\par Hence the GHZ state proves that entanglement is something more than just classical correlations\par To make GHZ statue is similar like bell state but you use two cnot gates, second cNOT connecting the middle and last wires \par Another is the W State: While the W states 001+010+100 state is entangled even if someone measures them so correlation just giving entanglement. Hence somehow weakly entangled. \item MONOGAMY OF ENTANGLEMENT: if Alice has a qubit that is maximally entangled with Bob’s qubit, then that qubit can’t also be maximally entangled with Charlie’s qubit. in order for two qubits A and B to be maximally entangled, they must not be entangled with any third qubit C whatsoever. Just like GHZ state in which they are entangled with each other in some manner but not maximally entangled with any of each other. \end{enumerate} \subsection{Measure of entanglement} \begin{enumerate} \item Most measure of entanglement take it as a resource. For the special case of an entangled pure states (not entanglement of mixed states) shared by two parties, it turns out that there’s a single measure of entanglement which counts the number of Bell pairs needed to form the state, or equivalently the number of Bell pairs that can be extracted from it. \item Von Neumann Entropy: When you diagonalise the density matrix it represents the probability over n possible outcomes, these probability distribution over n outcomes gives the von nuemman entropy $$S(\rho)=\sum_{i=0}^{n-1}\gamma_i\log_2\left(\frac{1}{\gamma_i}\right)$$ We know that the diagonal elements of density matrix give the probability of finding the state in given basis (in which density matrix is represented). Now if we take density matrix to all the possible basis and now each diagonal will contain the probability of finding the given state in one of the bases. Calculate shannon entropy in each of these. The minimum of those would be the von neumann entropy $$S(\rho) = \text{min running over all possbile U's~}H(\text{diag}(U\rho U^*)$$ where H represents tha shannon entropy \par Also note that the veon neummann entorpy is invariable fron change of basis ie Urho U* has von neumman as that of rho\par For pure states, in one of the running matrices, there will be a one case in which all the diagonal are zero and only one is 1. This would be the basis in the pure state itself lies. Hence von neumman entropy for pure states is 0. On the other hand the von neumann entropy will be 1 for mmaximmaly entangled state I/2 (single qubit). On the other hand, for the n dimensinoal hilbert space maximally mixed state it will be n; \item ENTANGLEMENT MEASUREMENT: the entanglement entropy of pure biparititle state is given by $$S(\text{that state}=S(\rho^A)=S(\rho^B)=H(\lambda_0^2,...,\lambda_{n-1}^2)$$ where the lambdas are probability disrubiton correspoindg to schmidth form of alice and bobs shared state. ie you find the scmidth decompositon (or SVD decomposiiton) and the diagonal elements are the heros) \par This definition has the property that the entanglement of separable sate has entropy 0 and of bell pair (the first one hence all others have) is 1.\par You can think of the entanglement entropy of a state as either the number of Bell pairs it would take to create it or as the number of Bell pairs that you can extract from it. It’s not immediately obvious that these numbers are the same, but for pure states they are. For mixed states they need not be. For eg if the entorpy is .942 then if they shared 1000 copies of that state, then they would be able to releport 942 qubits. But how? \item MIXED STATE ENTANGLEMENT: For bipartite mixed states, there are entanglemetn measures values: The Entanglement of Formation and Distillable Entanglement: $$\text{EF is the no of ebits needed per copy of the state as the no of copies gets large}$$ $$\text{ED is the no of ebits that Alice and Bob can extract per copy of the rho state in limit}$$ Obviously $E_F>=E_D$ but also for some pure bipartitie states $E_F>>E_D$ which means takes lots of entanglement ot make but only a little can be extracted (by using LOOCs). For those whose Ed is zero are called bound enangled states, some pure states are like this. They are very weak form of entanglemnet. \item Now after measuing entanglement you might like to emasur eht closeness of two systems to know how close a system is to a entangled system or in general measure the closenss of two systems. For two desniisty matrices of sames space $$F(\rho_1,\rho_2)=tr\sqrt{\sqrt{\rho_1}\rho_2\sqrt{\rho_1}}$$ which equals 1 for same density matrices and <1 for different density matrices, is symmetric though. For pure states it breaks down to just square of their inner product \end{enumerate} \subsection{INTERPRETATIONS} \begin{enumerate} \item what is quantum mechanics telling about our reality? \item The opposite nature of unitary and measuriment raises a questions: How does the universe know when to apply unitary evolution and when to apply measurement? \item COPENHAGEN: that the unvirse has the two parts, or two sides or two peeks with fuzzy boundaries: quantum wherer untieray stuffs happens and claiiscal where our stuffs happens and they are baically connectd by measuremnet. Something nice: at the end of the day QM works so why getting philosophical bro. SUAC. \item Problem with copenhagen interpretaion: Schrodingers cat and wigner's friend showed that the idea of having boundary between qwuantum and claissical realm is absurd. Winger thinks of food he wantts to eat (tea or coffee). Wingner friend who can read his mind (quantum mechanically) reads his mmind now his mind is in entanglement (winger friend reads tea, winger wants tea and winger friend reads coffe wiger wants coffee). Now what happens when anoterh firned come and anohter. Seems like the event is something for someone and somting for other. \item MANY WORLDS INTERPRETATION: Looks like this measurement thingy is causing real problem. Lets remove it such that entire universe is a single state (probably the tensor prodcuct) undergoing unitary transformation. Then mmeasurement or collapse is a result of inside quantum systems becoming entangled with each other when they interact so $$\frac{0+1}{\sqrt{2}}\rightarrow \frac{0~you_0+1~you_1}{\sqrt{2}}$$ now you get branched into two poissibies one wher you observed 0 and another 1. these branches are very uniqukely to intearact with each ohters. In general the universe gets branched very strageely each time a microscopic quantum system gets amplified to have macroscopic effect (each branch equally real). The branches unlikeldy to interfere could be literally seen as the second law of thermodynamcis. Hene there is a unitary transoration that induces a branching corespoinds to the measurment \item Problem with MWI: what about he born's rule, why probabilies there when each universe is equally likely? \item Dynamic Collapse: That when we zoom out from the quantum world ie when we try to exceed some type of complexity then the quantum effect leaves the universe. This doesnot interpet quantum mechanics but rther a rival theory. In case of cat, when the quantumness of decarying aprtcle takes over the cat, it at some point gets classical. \par Penrose says that superposiitns collapse when we have sufficiently large mass and siatne due to some graviational phenomena. \item Prolem with dynamics collaps: tha experiments been showing double slit experiment for even 6000 atoms. And other quantum exerpiemnts been going on. Now this is a perfeclty testifiable thoery. which always says that no quantum computer is possible becuase once the cmoplexity is crossed the wquantuumness goes away. Also probably does not solve the measurement problem. \item SUPTERDETEMINISM: still another way out is that the basis of measurment observes were chossing were not indenepent, since the choice of measurment were predetermined, outocmes at one place coulld affect anohter \item Rpboelmm with Superedtermis: is that this has the capacity of collapsing the scicneve because sceince has been based on freedom of experimentalists \end{enumerate} \subsection{Escape the craziness} \begin{enumerate} \item Let hidden variables are typically additional degrees of freedom which we either don’t have access to or haven’t yet discovered. The unverse has already set the result of measurement while the superpositnin is a way of making prediciotn \item Bohmian mechancis which is pretty tough to understand is non lcoal hidden variable theory. The poblem is ofcourse nonlocality \item Einsteins' local hidden variable possiblity (is local realism essentially as you will come to know) ie hidden variables are localized near poitns and are affected by things happening close to them. ie Each qubit will carry the answer which state to give when measured in every possible basis. When entangelds they are separated they take their own local copy of answrs. so essentially the iiidea is that they are classically correlated in a way that if anyone asks we will be 0. \item Bell's Game: The idea is Alice and Bob are placed in separate rooms and are both given a challenge bit (x and y, respectively) by a referee, Charlie. The challenge bits are chosen uniformly at random, and independently of each other. Alice sends an answer bit, a, back to the referee and Bob sends back an answer bit b. Alice and Bob “win” the game iff $a + b = xy (mod 2)$ ie to win: 11 pathauda they need to send different bits else aru case ma they need to send the same bits. \item Alice and Bob are allowed to agree on a strategy in advance (in other words correlate their answers) and to share random bits for probabilistic strategy. Solution: Any probabilistic strategy is equivalent to a mixture of deterministic ones. As such, the win probability for a probabilistic strategy is the weighted average over all the deterministic strategies the probabilistic strategy employs. There must be some deterministic strategy in the mixture that does at least as well as the average, so we can derandomize the protocol without negatively impacting the overall success probability; this is called a convexity argument \item Tabulating the deterministic strategy like sending 0 both, sending 1 by both, sending bits same as inputs, opposite as inputs. The first two has probability of 75 while the last two has 25. \item Bell statement is just that above. WTF BRO THAT BROING. ie bell statement just says that if classical correlation is all that there is then the chances will be 75 only because for the statistic probability we just iterated over all the possibilities. \item while if they were sharig a entangled pair (first bell) then it can be increased to cos squared pi/8 = 85 percentae \item The strategy will be: Upon receiving their challenge bits from the referee, Charlie, Alice and Bob then measure their respective qubits in different bases depending on whether their input bits x and y are 0 or 1. Based on the results of their measurements, Alice and Bob then output bits a and b based on the outcomes \item intoducing 4 more states, two of each forming an orthonormal basis (remember the 2D circle), $$\{\left.\middle|\pi/8\right\rangle,\left.\middle|5\pi/8\right\rangle\},~\{\left.\middle|-\pi/8\right\rangle,\left.\middle|3\pi/8\right\rangle\}$$ the angles are just giving the degree of rotation matrix applied to 0 state. And we can measure in these bases by respectively applying Rpi/8 and R-pi/8 and then measuring in 0,1 basis. (beause remember they way of emasuring in other basis by appling a transformaiton) \item Strategy: \begin{itemize} \item Alice takes measurement in eigther 0,1 or +,- basis depending on whether she reviews 0 or 1 respectiely \item then she sends 0 or 1 if she emasures to be the first state or seond state respectivley. \item the same applies to bob in given sates in that same order \end{itemize} \item Analysis: (by analysing each possiblites) \begin{itemize} \item Both gets 0, 0 so need to send same bits (non communcation allows us to) Alice measures first. \item let she gets 0 (prob 1/2) then bob will also collapse to 0. but bob will measure in pi/8, 5pi/8 basis and he will get pi/8 and 5pi/8 with probabiilties given by $$P(\left.\middle|\pi/8\right\rangle = |\left\lange 0\middle|\pi/8\right\rangle|^2=\cos^2\pi/8\text{ and }\cos^25\pi/8=\sin^2\pi/8$$ Here Alice returns 0, bob that cos probability to return 0 and sin probabiity to return 1. \item Also if she gets 1 (prob 1/2) then they measure in same basis and get $$\sin^2\pi/8 \text{ and }\cos^2\pi/8$$ Here ALice will return 1, bob will return 0 for sin probabiliy and 1 for cos probability \item hence total probability of winning, (as each time cos will give us the same bits. $$\cos^2\frac{\pi}{8}$$ \item again in other cases that they get like 0,1 and 1,0. they need to send the same bits so the analysis is same \item while for getting 1,1 they need to send different bits. \item DO SOME HARD WORK and use your brain that a bell state can be written as properly entangled in any of the orthonormal bases. \end{itemize} \item INTERPRETAIOTNS: if the correlations between qubits were just like if anyone measures we will be 0 then this iwll make winnin probailty 75 becuase for classical corelations the solution is bound to 75percent thats what bell pointed out but thats not what happens. so realism goes donw. So what? are they communicating if so? its also not possible locality is still not allowing communicaiton. we are in between. \item Weird number 85 percent huh? thats called tsirelson's inequality whose proof is tough as fuck. So theres been serach for theroy to violate tsirelson inequaality wihtout violating speed of light but unscuess so far. More shring or etnanglemetn cannot help \item General approach to game: \begin{itemize} \item PROBLEM STATEMENT: The two parties are asked a questions Q(x) = tell me the answer (0 or 1) related to the input x which will satisfy this conditino (this condition not only depends upon your answer but also on your mates answer). The input comes from the set of possible input symbols. So the question would be essentially to when to generate parity and when not to generate parity bits\par In this game input x could be 0 or 1 so possible permutation (since two people) are 00,01,10,11. The condition says that except the last one they have to generate parity bits. \item SOLUTION: In this case the optimum classical is they could just give the answer of parity that is demanded by the set of permutation having higher numbers. While the quantum one is \par There will be n number of basis for both alice and bob where n is the number of possible input sysbols. So initially fix the basis on which alice will measure when he receives the possible inputs. The order of the measurement result basis decides which bit he will seend to the refree. Then we decide on bob's basis by some analysis, his basis will simply be some angle from those of alice, the angle is to be determined\par In this game you run through all the possiblity and find the angle theta that maximizes it, i couldnot do it \end{itemize} \end{enumerate} \subsection{Some More Games} \begin{enumerate} \item Lets see what entanglement can further help us with? \item ODD CYCLE: alice and bob argue that they can 2 color the cycle of length old. \item the refree said he will do consistency checks as follows \begin{itemize} \item he asks them if the color of a random vertex, if they say same then pass \item he asks color of a vertex to alice and the color of another adjacent vertex to bob if different they pass \end{itemize} \item He takes both in separte rooms and asks one question with equal probability. if he were to ask multiple times each time with differing vertices then they would surely fail. But as he just aks only one question they probaility of success is 1-1/2n by just doing coloring way and letting one node underterminate when asked about dat node they could say anything \item QUANTUM STARAGETY: they share a bell pair anad make the following stargety \begin{itemize} \item Note there are n possible input symbols and as usual two outputs they can give. \item Lets fix the n alice basis the angle varying from pi/(2n) in the unit circle. Hence for first vertex we have 0,1 basis while for last vertex we have pi(n-1)/(2n). \item Now to find the angle for the basis of bob we find that the angle required is 0 ie the same basis can be used for the bob as well. \item Now on depending on their answer which is for first of their basis ie if measuring in 0,1 basis gives then 0 then they say blue else red. same thing for other basis also. \item To see, when asked about the same vertex, they measure in same basis (again remember that bell state can be repsensed as bell sate in any other orhonoma lbasis) so they give same answer. Now when asked about adjacent vertices then the proability that their result is opposite is cosine squared the angle between their measurement basis ie $$P(win) = \cos^2\pi/2n$$ so you could find probabiliy of losing which will be O(1/n**2) \end{itemize} \end{enumerate} \subsection{Quantum Computers} \begin{enumerate} \item david from interpretation and feynman from simulation of qm, could a computer handle semantically different sates in superposintnon, and could the superposition power be used for good? feynman was a n qubit sate wuold require $2^n$ to simulate it thats tough how about. will qc solve any problem classical cant ofc not becoa a classical will simulate just good so halting problem not be soled but time improvements coluld be there. \item So we will be using superpoisiotns right? what about those superpoisition are so cool that will probably solve them.. ummm lets think \item INTERFERENCE: The probablies are amplitiues bro how about we use them? so we design or choreogrpah such that the wrong answers destruct themselves. If there was no interfrence its just likec lassical computers with some bit of randomness lmao \item ENTANGLEMENT: ah ha there is entanglemetn also, an unentalged state of n qubit will requie 2n amplities while so there would be no speed up wihtout entanglement because the entangled state require $2^n$ sates. \item The circuit model of computation is essentially the boolean functions implemented by gates $$f(x_n)_m$$ wich means that takes in bits of length n and output the length m.\par The circuit model is essentialy the mulltisets of \{NAND\}, \{NOR\}, \{FREDKIN\}, \{TAFFOLI\}, two of them being reversible and two being non reversible \par for quantum circuit model the function would essentially be reversible like this $$f_q(|x_n,0_m>) = |x_n,f(x)_m>$$ but in general we do not think of implementing functions but rather unitary transformations \item INTERCONVERSITON: To realize quantum power, is to do following: given a non reversible circuit implementing the boolean function how do you get the reversible circuit implementing the unitary function. \item To do such conversion we have what we need quantum computation model which turns out to be multiset of taffoli gate. Here fredkin does not work. \item But this conversion requires exponential qubits so we allow some error into the circuit implementation and see if we can achieve polynomiality, this is sort of weak model, first lets see which gates cannot give us weak universality. Some of them are cnot, Rpi/8, S and also taffoli, hadamard, S make some examples. \end{enumerate} \subsection{Lets do it} \begin{enumerate} \item Query complexity: In circuit model, the algorithm are complexied on the basis of query complexity. The black box function (circuits) will be given to us then in an algorithm we count the number of calls it makes to the oracle function. In case of quantum, the oracles gotta be reversible so for the function that takes in n bits and give out m bits the oracle would be a XOR version of the oracle $$\left[U_f\left.\middle|x\right\rangle_n\left.\middle|y\right\rangle_m=\left.\middle|x\right\rangle_n\left.\middle|f(x)\oplus y\right\rangle_m\right]$$ Also used oracle is the phase oracle given by $$\left[U_f\left.\middle|x\right\rangle_n\left.\middle|b\right\rangle_m=(-1)^{f(x)\cdot b}\left.\middle|x\right\rangle_n\left.\middle|b\right\rangle_m\right]$$ To convert XOR to phase we have for m = 1\par \begin{center} \begin{tikzpicture} \node[scale=1.0] { \begin{quantikz} \left.\middle|x\right\rangle_n&&\gate[wires=2]{XOR}&\qw&\left.\middle|x\right\rangle_n\\ \left.\middle|y\right\rangle&&&\qw&\left.\middle|y\oplus f(x)\right\rangle \end{quantikz} } \end{tikzpicture} \begin{tikzpicture} \node[scale=1.0] { \begin{quantikz} \left.\middle|x\right\rangle_n&&\gate[wires=2]{Phase}&\qw&(-1)^{f(x)}\left.\middle|x\right\rangle_n\\ \frac{1}{\sqrt{2}}(\left.\middle|0\right\rangle-\left.\middle|1\right\rangle)&&&\qw&\frac{1}{\sqrt{2}}(\left.\middle|0\right\rangle-\left.\middle|1\right\rangle) \end{quantikz} } \end{tikzpicture} \end{center} \item Now for practicality of the algorithm how difficult is it to make such oracles. If you dont care about no of qubits you use, you can use total universality or else if u conservative then use weak universality. Now the thing is converting the two input gates into three qubit gates (like taffoli) introduces some ancilla bits, these bits at the outcomes are garbage that they could be entangled with the answer we care about so. for eg for one bit output function: $$|x_n 0....0> \rightarrow |x_n, gar, f(x_n)_1>$$ Thus we can always turn a irreversible gate into a reversible gate, at the expense of having a larger gate: a gate with “garbage bits” which aren’t necessarily used in the irreversible circu\par \item To remove the garbage we use techniques like first applying the C gate, also the wire comes with input 0, then we CNOT the answer into that safe wire. then the inverse C applies on the top side. here the CNOT copies because the f(x) will be the general classical states ie having no superposiotin.\par \begin{center} \begin{tikzpicture} \node[scale=1.0] { \begin{quantikz} \text{n bits input}&&\gate[wires=3]{C}&\qw&\text{n bits input}\\ \text{k bits garbage 0}&&&\qw&\text{k bits garbage g(x)}\\ \text{m bits 0}&&&\qw&\text{m bit f(x)} \end{quantikz} } \end{tikzpicture} \end{center} \begin{center} \begin{tikzpicture} \node[scale=1.0] { \begin{quantikz} \text{n bits input}&&\gate[wires=3]{C}&\qw&\gate[wires=3]{C-1}&\qw&\text{n bits input}\\ \text{k bits garbage 0}&&&\qw&\qw&\qw&\text{k bits garbage 0}\\ \text{m bits 0}&&&\ctrl{1}&\qw&\qw&\text{m bit 0}\\ \text{m bits 0}&&\qw&\targ{}&\qw&\qw&\text{m bits f(x)} \end{quantikz} } \end{tikzpicture} \end{center} \item Our problem involves some function, f (x), that we want to learn some property of f(x) in a way that only requires evaluating f(x) on various inputs, while ignoring the details of how f is computed. So, for example, you might want to learn: \begin{itemize} \item is there some input x such that f(x) = 1?, Does f (x) = 1 for the majority of x’s? \item Does f satisfy some global symmetry, such as periodicity, or is it far from any function satisfying the symmetry \end{itemize} We then want to know how many queries to f are needed to learn this property. In this model we ignore the cost of any quantum gates that are needed before or after the queries \end{enumerate} \subsection{The Real Algorihtms} \begin{enumerate} \item The Deutsh Parity Checks (to know whether the output is of same parity or different, classically you would send 0 and 1 both to check), the idea is to check parity of the phase oracle. To do so we convert the default 0 bits to superposition of 0 and 1 then if parity were same then the oracle will convert to + state if different then - state so we measure in +,- basis. \begin{center} \begin{tikzpicture} \node[scale=1.0] { \begin{quantikz} \left.\middle|0\right\rangle&\slice{0}&\gate{H}\slice{1}&\gate[wires=1]{Phased_f}\slice{2}&\gate{H}\slice{3}&\meter{}&\qw\end{quantikz} } \end{tikzpicture} \end{center} In one qubit system, the phased oracle are either identity or Z gates (phase not the xors or what). In general if you wanted to check the parity of n bits then you have to make n/2 queries twice less the queries. Here the hadamards gate enable quantum parallelism. \item Jonza Constatnty checks: a given function taking n bit string (so has possible of $2^n$ combination) and outputting 0 or 1 is known to be either constant or balanced. Then to know which you have to make $2^{n-1}+1$ queries. \par We send all the possible inputs in superposition so send $2^n$ inputs in superposed form they you send it through the phase oracle and if the function was constant we could take all commons out and we will have + in all the states but if balanced then we get some combination of +,- in the states. So we measure in hadamard basis. if got ++++ then cool else done.\par One way to make that XOR oracle for constant is the series of identity. While for balacned is to use the all the qubits as controlled bits that control the X on some output string. Then feed the output string with 0-1root2 state which makes it a phase oracle. The last output qubit can be ignored now \item Bernstein Vazirani: constant or balance ta alli bhayeran bro tetro dherai bits xa lets get some good info from them. Classically, make at least n queries. Means, What if we not get +++++ do we just ignore the bits that we get, they must imply something yes they do, the above guys found that they correspond to the bits of s if the function f(x) was of the given form (+ for 0 and - for 1) which takes a string of n bits and give 0 or 1 $$f(x) = s\cdot x~(\text{mod 2})$$ note that you get the bits of s in the same order you apply the function to the input bits for eg if the topmost is considered msb then same propagates to the output\par To implement the XOR oracle we use the qubits where the s bit's has one as controllers that control X one output qubit wire. As always that ouput wire is fed 0-1root2. \item Simon's: given a function taking n bits and thowirng n bits such that there is a secret string s $$f(x_n)=f(x_n\oplus s_n) \text{ or }f(x_1)=f(x_2)\leftrightarrow x_1=x_2\oplus s$$ [the functions nature is not specificed meaning] where s is not 000.., which could be called xor periodic, meaning that if you flip (flip all or none) bits of the strings here s has 1's will give you same output. So there would be only A PAIR for which the function will give same value. Actually two pairs can give same value so there would be two pairs and just like that any multiple of two will be giving same value. BUT I THINK WE ASSUME THAT THEY ARE UNIQUE. To solve it you could tabalize $2^n-1+1$ values and XOR out all the inputs that map to same value\par The probabilistic classical attack to this is by birthday problemm where the birthday function supposedly random throws out birthday of people (365 range) will have half chance of collide when you check about 23 people (which scales like the square root of the size of range). Now in this case having range $2^n$ we will have to make $O(2^{n/2}$ queries. Better way is not avaialble.\par First note we cannot use the oracled form of the function so we have to use its reversible version. \begin{center} \begin{tikzpicture} \node[scale=1.0] { \begin{quantikz} \left.\middle|0\right\rangle_n&\slice{0}&\gate{H^{\otimes n}}\slice{1}&\gate[wires=2]{f}&\qw&\slice{2}&\meter{}\\ \left.\middle|0\right\rangle_n&&\qw&\qw&\qw&\meter{} \end{quantikz} } \end{tikzpicture} \end{center} Upto here we did not have input and output register we just had one becuase we were using phase oracle (they didnot get entangled and it was okay). But here the result at the line 2 will be entangled. Now we measure and collapse the entangolement \item First say we measure the output (answer) register and pops out some value. Then can you see what happens to the upper register? they get collapsed to the superposition of values which correspond to the measured value. For eg if we mesured say w then the upper register will contain the states $x_1+x_2/\sqrt{2}$ for which $f(x_1)=f(x_2)$ in equal superposition. \item Now we will be needing those two states (our inputs for which the answer was found) which were in superpositoin so you probably think one way to do is we repeat the expiermnt two times, but we will get different w on the output register. \par Improved version: what if we measure the upper input reigster in hadarmard basis. \begin{center} \begin{tikzpicture} \node[scale=1.0] { \begin{quantikz} \left.\middle|0\right\rangle_n&\slice{0}&\gate{H^{\otimes n}}\slice{1}&\gate[wires=2]{f}\slice{2}&\gate{H^{\otimes n}}\slice{3}&\meter{}\\ \left.\middle|0\right\rangle_n&&\qw&\qw&\qw&\meter{} \end{quantikz} } \end{tikzpicture} \end{center} then the observed z is found to satisfy the following: $$s\cdot z = 0\text{ mod 2}$$ Now we repaet the experiment multiple times (about n+c times) and get a little more than n different z's. There are about $2^{n-1}$ such z's. We need little more than n becuase if we get just n then the solution might be just 0. $$s\cdot z _1=0, s\cdot z_2=0\ldots s\cdot z_{n+c}=0\text{ moded 2}$$ \item NOw that we have slight more than n equations we will with overwhelming probability get exactly two solutions one being 0 and another the s, we ignore 0. So with O(n) queires we did it. \item One way of making such XOR oracle is first copy the input register into ancilla register as by controlling NOTing the corresponding qubits, Then find the smallest index where the b's or s's bit value is 1. Then the qubit at that position will controll the ancilla qubits which correspond to b's 1s. See githubSS \end{enumerate} \subsection{Find Periods} \begin{enumerate} \item A few words about the function $a^x \text{ mod N}$ which we would like to find period of. Here we like a and N to be coprime. Why you ask?. Lets see when they are not. \begin{itemize} \item if they have a common factor the function will turn into a constant after some value of x \item while if they are coprime then it becomes periodic from the beginning, we need to find the smallest period. Now the function will take out values from 0 to N-1 so for N inputs (values from 0 to N-1) one output is bound to be repeated. If N is not prime it would be more lesser. \end{itemize} \item How many qubit (in inputs and outputs) to use is the question. Probably >=lgN qubits right? But just like that in simon's algorithm we would be needing at least two input values for which the output is found. So, the inputs should go from 0 to N-1 to 2N-2 >=log(2N-1) qubits right? Yes that would work. However for some better reasons we would demand the number of qubits to be >=lg($N^2$). Hence the basis of the first register will be q. The output qubits on the other hand could be just >=lg(N) \item Now just like in simon's we hadamard the inputs and send them to queries \item Its very important to visualize these superposiots as the peaks of deltas throughout these algoriihtms. In each and every step. \item now the state sent to the the black box will be $$\frac{1}{\sqrt{Q}}\sum_{r=0}^{q-1}\left.\middle|r\right\rangle\left.\middle|0\right\rangle$$ where Q is out max limit from the input qubits, and the black box sends us this $$\frac{1}{\sqrt{Q}}\sum_{r=0}^{q-1}\left.\middle|r\right\rangle\left.\middle|f(r) \right\rangle$$ Now we measure the output register and fix it to some value say f(r) such thta the upper register collapse to the equal superosooit of those values which will correspond to the emasured function value. $$\frac{\left.\middle|r\right\rangle+\left.\middle|r+s\right\rangle+\left.\middle|r+2s\right\rangle+\ldots+\left.\middle|r+(L-1)s\right\rangle}{\sqrt{L}}$$ where scott says L is the approximately chooosen normalization constant \item just like in simon's algorithm we could measure the states and hopefully get the desired s but the problem is r could vary like shit. So do we hadamard it or what? \item Before we dwelve in lets talk out our XOR oracle here only. The xor oracle can be implemented by what we call the modula exponentian technique. i.e to implement $a^x\text{ mod N}$ we need to manipulate $ax \text{ mod N}$. Let the input's ie x's bits goes like $x_{n-1},\ldots,x_0$ then if (n-1)th bit is 1 then we apply that ax function $2^{n-1}$ times and so on. \item While the act of implementing ax is now different for different N's. We will talk about N = 15 here. \end{enumerate} \subsection{Fourier transform} \begin{enumerate} \item any time you have a periodic function and you want to extrat its period its tranform time. From the insight driven from Descrete analysis. FFT is just hte matrice consisting of roots of unity like $$F_{2^1}=\frac{1}{\sqrt{2}}\begin{array}{cc} 1 & 1 \\ 1 & \omega \end{array}, ~F_{2^2}=\frac{1}{2}\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 1 & \omega & \omega^2 & \omega^3 \\ 1 & \omega^2 & \omega^4 & \omega^6 \\ 1 & \omega^3 & \omega^6 &\omega^9 \end{array}$$ As these are unitary you can apply them in quantum circuit. For n qubits you would need $F_{2^n}$. \item By the DDFT princple we will have can write the fouriter transform for n qubit in terms of transform of n-1 qubit as follows $$F_Q=\frac{1}{\sqrt{2}}\left(\begin{array}{cc} F_{Q/2} & B_{Q/2}F_{Q/2} \\ F_{Q/2} & -B_{Q/2}F_{Q/2} \end{array}\right)$$ where each element of the matrix is one quadarant of the old matrix. ;Here Bq/2 is the diagonal matrix of 1, w, w2, .. upto q/2-1. Now how to apply this matrix is the quetion? \item the above matrix iwll be broken down into. $$=\frac{1}{\sqrt{2}}\left(\begin{array}{cc} I & I\\ I & -I \end{array}\right)\left(\begin{array}{cc} I & 0\\ 0 & B_{q/2} \end{array}\right)\left(\begin{array}{cc} F_{q/2} & 0\\ 0 & F_{q/2} \end{array}\right)$$ \begin{itemize} \item The first matrix is saying to apply Fq/2 to each half of the vector so that would probably be applying to n-1 qubits leaving the msb out. \item its like applying I to upper half and lower half apply Bq/2, hence esssentially the signiicant bits control the shift of the insignificant. ie the top significant bits that are left out will control just one below their level. \item the final is the hadamard gate to the going to be left out bit \item i know the langugae been rought but lets destroy the recursion now. \end{itemize} \begin{center} \begin{tikzpicture} \node[scale=1.0] { \begin{quantikz} q_4&\gate[wires=4]{F_{q/2}}&\gate[wires=4]{CB}&\qw&\qw\\ q_3&&&\qw&\qw\\ q_2&&&\qw&\qw\\ q_1&&&\qw&\qw\\ q_0&\qw&\qw&\gate{H}&\qw \end{quantikz} } \end{tikzpicture} \begin{tikzpicture} \node[scale=1.0] { \begin{quantikz} q_4&\gate[wires=4]{F_{q/2}}&\qw&\qw&\qw&\qw\\ q_3&&\qw&\qw&\qw&\qw\\ q_2&&\qw&\qw&\qw&\qw\\ q_1&&\qw\slice{.}&\gate{S\left(\frac{2\pi}{2^2}\right)}\slice{.}&\qw&\qw\\ q_0&\qw&\qw&\ctrl{-1}&\gate{H}&\qw \end{quantikz} } \end{tikzpicture} \end{center} \item expanding futher: \begin{center} \begin{tikzpicture} \node[scale=1.0] { \begin{quantikz} q_4\slice{.}&\gate[wires=3]{F_{q/2}}&\gate[wires=3]{CB}&\qw&\qw&\qw&\qw\\ q_3&&\qw&\qw&\qw&\qw&\qw\\ q_2&&\qw&\qw&\qw&\qw&\qw\\ q_1&\qw&\qw&\gate{H}\slice{.}&\gate{S\left(\frac{2\pi}{2^2}\right)}&\qw&\qw\\ q_0&\qw&\qw&\qw&\ctrl{-1}&\gate{H}&\qw \end{quantikz} } \end{tikzpicture} \end{center} whcih agian on replacing the CB \begin{center} \begin{tikzpicture} \node[scale=1.0] { \begin{quantikz} q_4&\gate[wires=3]{F_{q/2}}\slice{.}&\qw&\qw&\qw&\qw&\qw&\qw\\ q_3&&\qw&\qw&\qw&\qw&\qw&\qw\\ q_2&&\gate{S\left(\frac{2\pi}{2^2}\right)}&\gate{S\left(\frac{2\pi}{2^3}\right)}\slice{.}&\qw&\qw&\qw&\qw\\ q_1&\qw&\ctrl{-1}&\qw&\gate{H}&\gate{S\left(\frac{2\pi}{2^2}\right)}&\qw&\qw\\ q_0&\qw&\qw&\ctrl{-2}&\qw&\ctrl{-1}&\gate{H}&\qw \end{quantikz} } \end{tikzpicture} \end{center} \item finally the bits at the last are reordered \end{enumerate} \subsection{Nicely done now what result do you get after applying FT} \begin{enumerate} \item out origiral stae was which goes into $$\sum_{l=0}^{L-1}\frac{\left.\middle|r+ls\right\rangle}{\sqrt{L}}\rightarrow\frac{1}{\sqrt{LQ}}\sum_{k=0}^{Q-1}\sum_{l=0}^{L-1}\omega^{(r+ls)k}\left.\middle|k\right\rangle$$ which is the superpositon of the same numbers going from 0 to Q-1 \item looks like a hard mess but lets for easeness assume s is the divisor of Q then the relevant coefficint of k by (takes out the gloal phase). Here the L will be Q/s in this case. $$\sum_{l=0}^{L-1}\omega^{lsk}=1+\omega^{sk}+\omega^{2sk}+\omega^{3sk}+\ldots+\omega^{(\frac{Q}{s}-1)sk}$$ Notice this is the sum of Q/s numer of roots of unity. So the phases of thse witll be $$0,~2\pi\frac{sk}{Q},~2\pi\frac{2sk}{Q},~2\pi\frac{3sk}{Q}\ldots ,~2\pi\frac{(Q/s-1)sk}{Q}$$ Since we know $$( \text{any Qth root of unity})^Q=1$$ so when ks is the multiple of Q then they survive (also check that the phase all point on the x axis) \par in actual values, the probability amplitude of obtaining one of such k's is $$\frac{1}{\sqrt{LQ}}*L={\frac{\sqrt{s}}{Q}}*\frac{Q}{s}=\frac{1}{\sqrt{s}}\rightarrow \frac{1}{s}$$ HOw many k's are there such that $k=\text{some constant}\frac{Q}{s}$ there are s of them becuase the maximum number k can go is from 0 to Q. \item Hence the total probaiblity of getting such k's is unity. ALl other states are destructed. \item Hence whatever remains in the superposiotn are the k's who are the multiple of Q/s, so by collecting a bunch of them we will with high probability get Q/s itself by gcding those bad bitches. \end{enumerate} \subsection{Hard Part} \begin{enumerate} \item Previously the probability map was straight deltas but not when s does not divide Q then we will have non perfect constructiive and estricve probabiblies. \itemm then now whole range of k's will be there in superpsoiton, but those K's which are near to Q/s (Q/s is not the integer remember bitch) \item let us suppose we got $$k=c\frac{Q}{s}\pm\epsilon\rightarrow \frac{k}{Q}=\frac{c}{s}\pm\frac{\epsilon}{Q}\rightarrow \left|\frac{k}{Q}-\frac{c}{s}\right|\le \frac{\epsilon}{Q}$$ Here since we chose $N^2$ we have good approximation $\frac{c}{s}$ to $\frac{k}{Q}$ \item Now we approximate the rational number k/Q to the fraction c/s, the denominator will be s. \item we obtain some more c/s and then the LCM of s's will give us the s. \end{enumerate} \subsection{The Delta Visualization} \begin{enumerate} \item We divide our whole qubits into two namely the QR and ANCILLAs \item So initially hadamarding the qrs give us equal deltas on QR and 0's on ANCILLAS \item After quering we have same QRs and now ANCILLas will dpepend on the function \item You measure the ancella and QR collapse into a series of deltas separated by period we need to find. How many deltas are there is unknown? \item Now after doing the fourier transform, the QRs will distribute such that the values near Q/s will peak very hard while others will be very less. \item Now since our N squared approximation k/Q will approximate the fraction c/s very goodly \item On the other the QPE variation seems to peak at last values of Q/s ie Q(s-1)/s so it becomes easy \end{enumerate} \subsection{Grover's Algorith} \begin{enumerate} \item it's just the unstructured search algorithm that deals with the black box quetsions like given a function (0,...N-1)->(0,1) is there a x such that f(x)=1 and what x is that? \item a function is made from the database such that the required to be searched elements give 1 to our result and others give us 0. you can you need about linear quieres to go. lets begin by assuming that solution exists and is uniuqe \item ALGORITHM: we will implement the functions as a phase oracle. Now let the function takes (0,1)**n to (0,1). Then we creaet a superposiotn of all these staes hence will be the superpsoiotn of |0>,|1>,|2> to |N-1>. So all together N basis. Remember the deltas \item Now when you send this to the oracle, you will have your to serach delta downed. Ie the amplitude corespoinding to x0 will donwloards (but equal) and others will be upwards \item now groer realized a sexy matrix $$\left(\begin{array}{cccc} \frac{2}{N}-1 & \frac{2}{N} & \ldots & \frac{2}{N} \\ \frac{2}{N} & \frac{2}{N}-1 & \ldots & \frac{2}{N}\\ \vdots & \vdots & \ddots & \vdots\\ \frac{2}{N}&\frac{2}{N}&\ldots&\frac{2}{N}-1 \end{array}\right)$$ such a matrix of order N by N whose diagoranla are as shown and other elements are also shown. \item What this does is flippling all the N amplitudtes about the mean aplitude. $$\text{one amplitude}\rightarrow 2\text{mean of all the N amplitudes}-\text{that one amplitude}$$ \item Now to visuaize this as the deltas, you see one delta is down at neg 1/root(N) and otehras are at the positive ie others are up The mean will on the uppper deltas, just a bit below their top bar. (N-2)/[Nroot(N)] \item Now the result will be that (not focusing on actual values) the lower one goes up higher while the gangs will be lower than their preivous values. you can do the math. But they will be on same side. Apply the function again \item apply it again and again and you see that the gang amplitudes lower to the shit. In about O(rootN) you have very high proabaiblty \item Now about he diffusion operator: Probalby grover converted the matrix in hadamard basis (you know how to do it right by that UU* wala tarika) then he saw that the matrix is easy as fuck. ie it comes diagonal $$\left(\begin{array}{cccc} 1 & 0 & \ldots & 0 \\ 0 & -1 & \ldots & 0\\ 0 & \vdots & \ddots & \vdots\\ 0 & 0 &\ldots&-1 \end{array}\right)$$ \item So to implement the diffusion operator you do is apply the series of hadamard then this diagonal then again series of hadamard. THen done measuring in computational basis works. This diagonal piece of shit is just series of X gates+multicontrolled z gate+series of x gate \end{enumerate} \subsection{Goodness in Grover's} \begin{enumerate} \item Another way of thinking of the gover's algorihtm is thorugh the geomtery. Bause the diffusion operator is: lets make a subspace made by the basis, the search stte and all other unsearched staes in superpooisiton along y axis and x asix respecitlvey. \item Then first hadamarding our staes will be in slight angle with the x axis. when we send it to the funcion it will just reflcet on the x axis and go down making the same angle in opposite diection \item now what the diffusion operator (int computational basis) does is reflects that state about hte axis made by the original state before reflection. Hence if it was making theta (arcsin 1/rootN) before now will make 3thta after one iteration. \item after t iteratoin the prob of success is $$\sin^2(\frac{2t+1}{\sqrt{N}})$$ \item put t = about $\frac{\pi}{4}\sqrt{N}$ then we will get nearly to 1, we could overshoot as well, notice how if we do the algoirhtm many times we could get back to where we staretd. \item Also if you take your chance by syaing bro i htink i did enough of the iteraiton and get the wrong angswer then you have to start from the benining \end{enumerate} \subsection{let's get to more items} \begin{enumerate} \item what if K (we know what K is) items out of N are marked ie they give 1 from the funciton. \item then the peak at after $\pi/4\sqrt{\frac{N}{K}}$ ie we would get one of the K's after we measure them. \item what if K is unknow, then we could lose the peak, so what we do is we do not do the given iiteration instead we pick random number from 0 to root(N) and run iteration that many times. IN this case we have constant probability 40 or 50 or 60 of measuring the answer. wwe repeat the ALGORITHM aabout O(1) times unitll we emasure the marked itms. So the bound of rootN is maintained \item what if we have not marked items and we dont know that? then each sates have same probabilty. We measure them and we see wtf bro this is not marked? how long do we go. Again as i said if we repeat the algorithm O(1) we are with very sure to find the marked items. If we dont then we will be almost sure that there are no items amrked. Mainatining the bouon \item See the difference, we will know one marked items (if there is) in O(rootN/K) runs if we know K, while O(rooN) if we dont know K. \item Now we could do better by guessing the values of K. \begin{itemize} \item Guess K = N, then apply it pi/4 times and measure, found? great? no? \item let k = N/2 then apply the calculate number of imtes found? no? \item then let k = N/4, N/8 and so on \item either we will find or we will do unsuccessfully for K =1 \end{itemize} \end{enumerate} \subsection{Apppliaitons} \begin{enumerate} \item so now the BBBV theorem says the rootN is optimal. proof is damn looking complicated so move on. In more gneral that those which ignore the strucute of NP complete problems can ahiceve max of gorver speed up. \item ORs of ANDs: given a table of size N, consisting of roder root(N) by root(N). The problem is to determine whether there is a row of all bits 1. why callled ors of ands: because you have a rows in OR fashion and you need those rows to have ANDs \item Classically you would go about checking each row until you find the row with all 1s. For each row you check the bits found 0 then stop. so that would be O(rootN*rootN), ie N for searching in each row (size of a row is rootN bro) and rootN for number of rows hence O(N) \item While quantumically you could search for 0 in each row by root(rootN)=$N^{0.25}$ or if you want better then $N^{0.25}$logN (if you want a probability of surence) so for each row row you would have to multiply by rootN, ie $N^{0.75}logN$ \item ANOTHER GROVENESS: is to have a outer grover algorithm. which has a function that gives 1 for the rows (ie decide them marked) when if and only if classical algorithm (which we run as the outer loop) finds no zero in it. But as you could see there would be the same complexity because of the classical ness \item Naturally the next idea is to run Grover’s algorithm recursively, where the outer Grover (over the rows) will count a given row as being marked if and only if the inner Grover failed to find a zero in that row. So, $N^{0.25}$ for each outer and inner grover and logN for error we got the damn $N^{0.5}logN$ (by some tricks peopeld remove logNs) \item Now we could generalize that to ORs of ANDs of ORs or to any level,you could do in rootN. Wow damn how? idk bro first do your assingment \end{enumerate} \chapter{Quantum Computing} \begin{enumerate} \item QPEs \begin{center} \begin{tikzpicture} \node[scale=1.0] { \begin{quantikz} \left.\middle|0\right\rangle&\slice{0}&\gate{H}\slice{1}&\ctrl{1}\slice{2}&\gate{H}\slice{3}&\qw\\ \left.\middle|\Psi\right\rangle_k&&\qw&\gate{U_f}&\qw&\qw \end{quantikz} } \end{tikzpicture} \end{center} $$\left[\left.\middle|0\right\rangle\left.\middle|\Psi\right\rangle,~\frac{1}{\sqrt{2}}(\left.\middle|0\right\rangle+\left.\middle|1\right\rangle)\left.\middle|\Psi\right\rangle,~\frac{1}{\sqrt{2}}(\left.\middle|0\right\rangle+e^{i\theta_\Psi}\left.\middle|1\right\rangle)\left.\middle|\Psi\right\rangle,~\frac{1}{{2}}((1+e^{i\theta_\Psi})\left.\middle|0\right\rangle+(1-e^{i\theta_\Psi})\left.\middle|1\right\rangle)\left.\middle|\Psi\right\rangle\right]$$ \item Precise QPEs \begin{center} \begin{tikzpicture} \node[scale=1.0] { \begin{quantikz} \left.\middle|0\right\rangle_n&\slice{0}&\gate{H^{\otimes n}}\slice{1}&\ctrl{1}&\ctrl{1}&\ctrl{1}&\ctrl{1}&\gate{QFT^{-1}}\slice{6}&\qw\\ \left.\middle|\Psi\right\rangle&&\qw&\gate{U_f^{2^{n-1}}}\slice{2}&\gate{U_f^{2^{n-2}}}\slice{3}&\gate{\ldots}\slice{4}&\gate{U_f^{2^0}}\slice{5}&\qw&\qw \end{quantikz} } \end{tikzpicture} \end{center} $$\left[\left.\middle|0\right\rangle_n,~\frac{1}{{2}^\frac{n}{2}}(\left.\middle|0\right\rangle+\left.\middle|1\right\rangle)^{\otimes n},~\frac{1}{{2}^\frac{n}{2}}(\left.\middle|0\right\rangle+e^{i\theta_\Psi2^{n-1}}\left.\middle|1\right\rangle)\otimes(\left.\middle|0\right\rangle+e^{i\theta_\Psi2^{n-2}}\left.\middle|1\right\rangle)\otimes\ldots\otimes(\left.\middle|0\right\rangle+e^{i\theta_\Psi2^0}\left.\middle|1\right\rangle)\right]$$ $$\left[\theta_\Psi=2\pi\left(\frac{\theta_x}{2^n}\right)=2\pi\left(\frac{\{0,1,2,\ldots,2^n-1\}}{2^n}\right)\right]$$ \item Period Finders $$\left[U_f\left.\middle|\Psi\right\rangle=\left.\middle|a\Psi~\% N\right\rangle:~\left.\middle|u_s\right\rangle=\frac{1}{\sqrt{r}}\sum_{k=0}^{r-1}e^{-\frac{2\pi i sk}{r}}\left.\middle|a^k~\%N\right\rangle:~U_f\left.\middle|u_s\right\rangle=e^{\frac{2\pi i s}{r}}\left.\middle|u_s\right\rangle:~\frac{1}{\sqrt{r}}\sum_{s=0}^{r-1}\left.\middle|u_s\right\rangle=\left.\middle|1\right\rangle\right]$$ \end{enumerate} \end{document} \chapter{Not yet there} \item Single Qubit Decompositions $$\left[Q=\left(\begin{array}{cc} +e^{i(\delta+\alpha+\gamma)}\cos\beta & +e^{i(\delta+\alpha-\gamma)}\sin\beta \\ -e^{i(\delta-\alpha+\gamma)}\sin\beta & +e^{i(\delta-\alpha-\gamma)}\cos\beta \end{array}\right)=K(\delta)T(\alpha)R(\beta)T(\gamma)\right]$$ \begin{center} \begin{tikzpicture} \node[scale=1.0] { \begin{quantikz} \qw&\gate{T\left(-\frac{\delta}{2}\right)}&\gate{K\left(\frac{\delta}{2}\right)}&\qw&\ctrl{1}&\qw&\ctrl{1}&\qw&\qw \\ \qw&\qw&\qw&\gate{T\left(\frac{\gamma-\alpha}{2}\right)}&\targ{}&\gate{R\left(-\frac{\beta}{2}\right)T\left(\frac{-\gamma-\alpha}{2}\right)}&\targ{}&\gate{T\left(\alpha\right)R\left(\frac{\beta}{2}\right)}&\qw \end{quantikz} } \end{tikzpicture} \end{center} \item Two Qubit Decompositions\par \begin{tikzpicture} \node[scale=1.0] { \begin{quantikz} \qw&\ctrl{1}&\ctrl{1}&\ctrl{2}\qw&\qw&\ctrl{2}&\qw&\ctrl{2}&\qw\\ \qw&\gate{T\left(-\frac{\delta}{2}\right)}&\gate{K\left(\frac{\delta}{2}\right)}&\qw&\ctrl{1}&\qw&\ctrl{1}&\qw&\qw \\ \qw&\qw&\qw&\gate{T\left(\frac{\gamma-\alpha}{2}\right)}&\targ{}&\gate{R\left(-\frac{\beta}{2}\right)T\left(\frac{-\gamma-\alpha}{2}\right)}&\targ{}&\gate{T\left(\alpha\right)R\left(\frac{\beta}{2}\right)}&\qw \end{quantikz} } \end{tikzpicture} \item Universal Gates $$\left[T\left.\middle|1,1,x\right\rangle=\left.\middle|1,1,x'\right\rangle:~T\left.\middle|x,y,0\right\rangle=\left.\middle|x,y,x\&y\right\rangle:~T\left.\middle|1,x,y\right\rangle=\left.\middle|1,x,x\oplus y\right\rangle:~T\left.\middle|x,y,1\right\rangle=\left.\middle|x,y,(x\&y)'\right\rangle\right]$$ $$\left[F\left.\middle|x,0,1\right\rangle=\left.\middle|x,x,x'\right\rangle:~T\left.\middle|x,y,1\right\rangle=\left.\middle|x,y+x,y+x'\right\rangle:~T\left.\middle|x,0,y\right\rangle=\left.\middle|x,y\&x,y\&x'\right\rangle\right]$$ \item Simon's Algorithm \begin{center} \begin{tikzpicture} \node[scale=1.0] { \begin{quantikz} \left.\middle|0\right\rangle_n&&\gate{H^{\otimes n}}\slice{0}&\gate[wires=2]{U_f}\slice{1}&\qw&\gate{H^{\otimes n}}\slice{3}&\meter{}&\qw\\ \left.\middle|0\right\rangle_{n-1}&&\qw&&\meter{}\slice{2}&\qw&\qw&\qw\end{quantikz} } \end{tikzpicture} \end{center} $$\left[\left.\middle|\Psi_0\right\rangle=\frac{1}{2^\frac{n}{2}}\sum_{x=0}^{2^n-1}\left.\middle|x\right\rangle_n\otimes\left.\middle|0\right\rangle_{n-1}\right]$$ $$\left[\left.\middle|\Psi_1\right\rangle=\frac{1}{2^\frac{n}{2}}\sum_{x=0}^{2^n-1}\left.\middle|x\right\rangle_n\otimes\left.\middle|f(x)\right\rangle_{n-1}\right]$$ $$\left[\left.\middle|\Psi_2\right\rangle=\left[\frac{1}{\sqrt{2}}\left.\middle|x_0\right\rangle_n+\frac{1}{\sqrt{2}}\left.\middle|x_0\oplus a\right\rangle_n\right]\otimes \left.\middle|f_0\right\rangle_{n-1}\right]$$ $$\left[\left.\middle|\Psi_3\right\rangle=\left[\frac{1}{\sqrt{2}\sqrt{2^n}}\sum_{y=0}^{2^n-1}\left[(-1)^{x_0\cdot y}+(-1)^{(x_0\oplus a)\cdot y}\right]\left.\middle|y\right\rangle_n\right]\otimes \left.\middle|f_0\right\rangle_{n-1}\right]$$