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Diffstat (limited to '')
| -rw-r--r-- | buch/papers/punktgruppen/Makefile.inc | 12 | ||||
| -rw-r--r-- | buch/papers/punktgruppen/main.tex | 58 | ||||
| -rw-r--r-- | buch/papers/punktgruppen/packages.tex | 5 | ||||
| -rw-r--r-- | buch/papers/punktgruppen/teil0.tex | 22 | ||||
| -rw-r--r-- | buch/papers/punktgruppen/teil1.tex | 55 | ||||
| -rw-r--r-- | buch/papers/punktgruppen/teil2.tex | 40 | ||||
| -rw-r--r-- | buch/papers/punktgruppen/teil3.tex | 40 | ||||
| -rw-r--r-- | vorlesungen/punktgruppen/.gitignore | 20 | ||||
| -rw-r--r-- | vorlesungen/punktgruppen/Makefile | 18 | ||||
| -rw-r--r-- | vorlesungen/punktgruppen/crystals.py | 611 | ||||
| -rw-r--r-- | vorlesungen/punktgruppen/media/images/nosignal.jpg | bin | 0 -> 711846 bytes | |||
| -rw-r--r-- | vorlesungen/punktgruppen/poetry.lock | 743 | ||||
| -rw-r--r-- | vorlesungen/punktgruppen/pyproject.toml | 15 | ||||
| -rw-r--r-- | vorlesungen/punktgruppen/script.pdf | bin | 0 -> 44991 bytes | |||
| -rw-r--r-- | vorlesungen/punktgruppen/script.tex | 214 | ||||
| -rw-r--r-- | vorlesungen/punktgruppen/slides.pdf | bin | 0 -> 790926 bytes | |||
| -rw-r--r-- | vorlesungen/punktgruppen/slides.tex | 895 |
17 files changed, 2552 insertions, 196 deletions
diff --git a/buch/papers/punktgruppen/Makefile.inc b/buch/papers/punktgruppen/Makefile.inc index 7c6e70d..629abca 100644 --- a/buch/papers/punktgruppen/Makefile.inc +++ b/buch/papers/punktgruppen/Makefile.inc @@ -3,12 +3,8 @@ # # (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule # -dependencies-punktgruppen = \ - papers/punktgruppen/packages.tex \ - papers/punktgruppen/main.tex \ - papers/punktgruppen/references.bib \ - papers/punktgruppen/teil0.tex \ - papers/punktgruppen/teil1.tex \ - papers/punktgruppen/teil2.tex \ - papers/punktgruppen/teil3.tex +dependencies-punktgruppen = \ + papers/punktgruppen/packages.tex \ + papers/punktgruppen/main.tex \ + papers/punktgruppen/references.bib diff --git a/buch/papers/punktgruppen/main.tex b/buch/papers/punktgruppen/main.tex index fc91913..603f293 100644 --- a/buch/papers/punktgruppen/main.tex +++ b/buch/papers/punktgruppen/main.tex @@ -3,34 +3,38 @@ % % (c) 2020 Hochschule Rapperswil % -\chapter{Thema\label{chapter:punktgruppen}} -\lhead{Thema} +\chapter{Crystal M\rotatebox[origin=c]{180}{a}th\label{chapter:punktgruppen}} +\lhead{Crystal M\rotatebox[origin=c]{180}{a}th} \begin{refsection} -\chapterauthor{Hans Muster} - -Ein paar Hinweise für die korrekte Formatierung des Textes -\begin{itemize} -\item -Absätze werden gebildet, indem man eine Leerzeile einfügt. -Die Verwendung von \verb+\\+ ist nur in Tabellen und Arrays gestattet. -\item -Die explizite Platzierung von Bildern ist nicht erlaubt, entsprechende -Optionen werden gelöscht. -Verwenden Sie Labels und Verweise, um auf Bilder hinzuweisen. -\item -Beginnen Sie jeden Satz auf einer neuen Zeile. -Damit ermöglichen Sie dem Versionsverwaltungssysteme, Änderungen -in verschiedenen Sätzen von verschiedenen Autoren ohne Konflikt -anzuwenden. -\item -Bilden Sie auch für Formeln kurze Zeilen, einerseits der besseren -Übersicht wegen, aber auch um GIT die Arbeit zu erleichtern. -\end{itemize} - -\input{papers/punktgruppen/teil0.tex} -\input{papers/punktgruppen/teil1.tex} -\input{papers/punktgruppen/teil2.tex} -\input{papers/punktgruppen/teil3.tex} +\chapterauthor{Tim T\"onz, Naoki Pross} + +%% TODO: remove +%% Some ideas to motivate the topic: +%% - Physics in a crystal lattice structure +%% - Birifrencenge and scattering of light / Xray in Crystals +%% - Electron density function in a lattice +%% - Heat diffusion with lattice model +%% - Ising model for ferromagnetism (?? => H.D. Lang) +%% +%% - Homomorphic encryption (or lattice based cryptography) +%% + Q: Is it possible to edit encrypted data without decrypting it first? + +%% TODO: translated and move into a file {{{ + +\section{Motivation} +% birifrengence + +\section{Math} +% lattice group +% symmetry +% space group + +\section{Physics} +\subsection{Electromagnetic Waves} +\subsection{Crystal Lattice} + + +%% }}} \printbibliography[heading=subbibliography] \end{refsection} diff --git a/buch/papers/punktgruppen/packages.tex b/buch/papers/punktgruppen/packages.tex index 971bcfe..9953339 100644 --- a/buch/papers/punktgruppen/packages.tex +++ b/buch/papers/punktgruppen/packages.tex @@ -4,7 +4,4 @@ % (c) 2019 Prof Dr Andreas Müller, Hochschule Rapperswil % -% if your paper needs special packages, add package commands as in the -% following example -%\usepackage{packagename} - +\usepackage{tikz-3dplot} diff --git a/buch/papers/punktgruppen/teil0.tex b/buch/papers/punktgruppen/teil0.tex deleted file mode 100644 index 5a8278e..0000000 --- a/buch/papers/punktgruppen/teil0.tex +++ /dev/null @@ -1,22 +0,0 @@ -% -% einleitung.tex -- Beispiel-File für die Einleitung -% -% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil -% -\section{Teil 0\label{punktgruppen:section:teil0}} -\rhead{Teil 0} -Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam -nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam -erat, sed diam voluptua \cite{punktgruppen:bibtex}. -At vero eos et accusam et justo duo dolores et ea rebum. -Stet clita kasd gubergren, no sea takimata sanctus est Lorem ipsum -dolor sit amet. - -Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam -nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam -erat, sed diam voluptua. -At vero eos et accusam et justo duo dolores et ea rebum. Stet clita -kasd gubergren, no sea takimata sanctus est Lorem ipsum dolor sit -amet. - - diff --git a/buch/papers/punktgruppen/teil1.tex b/buch/papers/punktgruppen/teil1.tex deleted file mode 100644 index 228af33..0000000 --- a/buch/papers/punktgruppen/teil1.tex +++ /dev/null @@ -1,55 +0,0 @@ -% -% teil1.tex -- Beispiel-File für das Paper -% -% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil -% -\section{Teil 1 -\label{punktgruppen:section:teil1}} -\rhead{Problemstellung} -Sed ut perspiciatis unde omnis iste natus error sit voluptatem -accusantium doloremque laudantium, totam rem aperiam, eaque ipsa -quae ab illo inventore veritatis et quasi architecto beatae vitae -dicta sunt explicabo. -Nemo enim ipsam voluptatem quia voluptas sit aspernatur aut odit -aut fugit, sed quia consequuntur magni dolores eos qui ratione -voluptatem sequi nesciunt -\begin{equation} -\int_a^b x^2\, dx -= -\left[ \frac13 x^3 \right]_a^b -= -\frac{b^3-a^3}3. -\label{punktgruppen:equation1} -\end{equation} -Neque porro quisquam est, qui dolorem ipsum quia dolor sit amet, -consectetur, adipisci velit, sed quia non numquam eius modi tempora -incidunt ut labore et dolore magnam aliquam quaerat voluptatem. - -Ut enim ad minima veniam, quis nostrum exercitationem ullam corporis -suscipit laboriosam, nisi ut aliquid ex ea commodi consequatur? -Quis autem vel eum iure reprehenderit qui in ea voluptate velit -esse quam nihil molestiae consequatur, vel illum qui dolorem eum -fugiat quo voluptas nulla pariatur? - -\subsection{De finibus bonorum et malorum -\label{punktgruppen:subsection:finibus}} -At vero eos et accusamus et iusto odio dignissimos ducimus qui -blanditiis praesentium voluptatum deleniti atque corrupti quos -dolores et quas molestias excepturi sint occaecati cupiditate non -provident, similique sunt in culpa qui officia deserunt mollitia -animi, id est laborum et dolorum fuga \eqref{000tempmlate:equation1}. - -Et harum quidem rerum facilis est et expedita distinctio -\ref{punktgruppen:section:loesung}. -Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil -impedit quo minus id quod maxime placeat facere possimus, omnis -voluptas assumenda est, omnis dolor repellendus -\ref{punktgruppen:section:folgerung}. -Temporibus autem quibusdam et aut officiis debitis aut rerum -necessitatibus saepe eveniet ut et voluptates repudiandae sint et -molestiae non recusandae. -Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis -voluptatibus maiores alias consequatur aut perferendis doloribus -asperiores repellat. - - diff --git a/buch/papers/punktgruppen/teil2.tex b/buch/papers/punktgruppen/teil2.tex deleted file mode 100644 index b48e785..0000000 --- a/buch/papers/punktgruppen/teil2.tex +++ /dev/null @@ -1,40 +0,0 @@ -% -% teil2.tex -- Beispiel-File für teil2 -% -% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil -% -\section{Teil 2 -\label{punktgruppen:section:teil2}} -\rhead{Teil 2} -Sed ut perspiciatis unde omnis iste natus error sit voluptatem -accusantium doloremque laudantium, totam rem aperiam, eaque ipsa -quae ab illo inventore veritatis et quasi architecto beatae vitae -dicta sunt explicabo. Nemo enim ipsam voluptatem quia voluptas sit -aspernatur aut odit aut fugit, sed quia consequuntur magni dolores -eos qui ratione voluptatem sequi nesciunt. Neque porro quisquam -est, qui dolorem ipsum quia dolor sit amet, consectetur, adipisci -velit, sed quia non numquam eius modi tempora incidunt ut labore -et dolore magnam aliquam quaerat voluptatem. Ut enim ad minima -veniam, quis nostrum exercitationem ullam corporis suscipit laboriosam, -nisi ut aliquid ex ea commodi consequatur? Quis autem vel eum iure -reprehenderit qui in ea voluptate velit esse quam nihil molestiae -consequatur, vel illum qui dolorem eum fugiat quo voluptas nulla -pariatur? - -\subsection{De finibus bonorum et malorum -\label{punktgruppen:subsection:bonorum}} -At vero eos et accusamus et iusto odio dignissimos ducimus qui -blanditiis praesentium voluptatum deleniti atque corrupti quos -dolores et quas molestias excepturi sint occaecati cupiditate non -provident, similique sunt in culpa qui officia deserunt mollitia -animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis -est et expedita distinctio. Nam libero tempore, cum soluta nobis -est eligendi optio cumque nihil impedit quo minus id quod maxime -placeat facere possimus, omnis voluptas assumenda est, omnis dolor -repellendus. Temporibus autem quibusdam et aut officiis debitis aut -rerum necessitatibus saepe eveniet ut et voluptates repudiandae -sint et molestiae non recusandae. Itaque earum rerum hic tenetur a -sapiente delectus, ut aut reiciendis voluptatibus maiores alias -consequatur aut perferendis doloribus asperiores repellat. - - diff --git a/buch/papers/punktgruppen/teil3.tex b/buch/papers/punktgruppen/teil3.tex deleted file mode 100644 index 94abd74..0000000 --- a/buch/papers/punktgruppen/teil3.tex +++ /dev/null @@ -1,40 +0,0 @@ -% -% teil3.tex -- Beispiel-File für Teil 3 -% -% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil -% -\section{Teil 3 -\label{punktgruppen:section:teil3}} -\rhead{Teil 3} -Sed ut perspiciatis unde omnis iste natus error sit voluptatem -accusantium doloremque laudantium, totam rem aperiam, eaque ipsa -quae ab illo inventore veritatis et quasi architecto beatae vitae -dicta sunt explicabo. Nemo enim ipsam voluptatem quia voluptas sit -aspernatur aut odit aut fugit, sed quia consequuntur magni dolores -eos qui ratione voluptatem sequi nesciunt. Neque porro quisquam -est, qui dolorem ipsum quia dolor sit amet, consectetur, adipisci -velit, sed quia non numquam eius modi tempora incidunt ut labore -et dolore magnam aliquam quaerat voluptatem. Ut enim ad minima -veniam, quis nostrum exercitationem ullam corporis suscipit laboriosam, -nisi ut aliquid ex ea commodi consequatur? Quis autem vel eum iure -reprehenderit qui in ea voluptate velit esse quam nihil molestiae -consequatur, vel illum qui dolorem eum fugiat quo voluptas nulla -pariatur? - -\subsection{De finibus bonorum et malorum -\label{punktgruppen:subsection:malorum}} -At vero eos et accusamus et iusto odio dignissimos ducimus qui -blanditiis praesentium voluptatum deleniti atque corrupti quos -dolores et quas molestias excepturi sint occaecati cupiditate non -provident, similique sunt in culpa qui officia deserunt mollitia -animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis -est et expedita distinctio. Nam libero tempore, cum soluta nobis -est eligendi optio cumque nihil impedit quo minus id quod maxime -placeat facere possimus, omnis voluptas assumenda est, omnis dolor -repellendus. Temporibus autem quibusdam et aut officiis debitis aut -rerum necessitatibus saepe eveniet ut et voluptates repudiandae -sint et molestiae non recusandae. Itaque earum rerum hic tenetur a -sapiente delectus, ut aut reiciendis voluptatibus maiores alias -consequatur aut perferendis doloribus asperiores repellat. - - diff --git a/vorlesungen/punktgruppen/.gitignore b/vorlesungen/punktgruppen/.gitignore new file mode 100644 index 0000000..3633a3d --- /dev/null +++ b/vorlesungen/punktgruppen/.gitignore @@ -0,0 +1,20 @@ +# directories +__pycache__ +media/Tex + +media/images/crystal +media/images/freezeframes + +media/videos +media/audio + +media/Punktgruppen +media/Punktgruppen.mp4 + +build + +# files +script.log +slides.log +slides.vrb +missfont.log diff --git a/vorlesungen/punktgruppen/Makefile b/vorlesungen/punktgruppen/Makefile new file mode 100644 index 0000000..302e976 --- /dev/null +++ b/vorlesungen/punktgruppen/Makefile @@ -0,0 +1,18 @@ +TEX=xelatex +TEXARGS=--output-directory=build --halt-on-error --shell-escape + +all: slides.pdf script.pdf media + +.PHONY: clean +clean: + @rm -rfv build + +%.pdf: %.tex + mkdir -p build + $(TEX) $(TEXARGS) $< + $(TEX) $(TEXARGS) $< + cp build/$@ . + +media: + poetry install + poetry run manim -ql crystals.py diff --git a/vorlesungen/punktgruppen/crystals.py b/vorlesungen/punktgruppen/crystals.py new file mode 100644 index 0000000..4a9836a --- /dev/null +++ b/vorlesungen/punktgruppen/crystals.py @@ -0,0 +1,611 @@ +from manim import * + +import math as m +import numpy as np +import itertools as it + +# configure style +config.background_color = '#202020' +config.tex_template.add_to_preamble( + r"\usepackage[p,osf]{scholax}" + r"\usepackage{amsmath}" + r"\usepackage[scaled=1.075,ncf,vvarbb]{newtxmath}" +) + +# scenes +class Geometric2DSymmetries(Scene): + def construct(self): + self.wait(5) + + self.intro() + self.cyclic() + self.dihedral() + self.circle() + + def intro(self): + # create square + square = Square() + square.set_fill(PINK, opacity=.5) + self.play(SpinInFromNothing(square)) + self.wait() + + # the action of doing nothing + action = MathTex(r"\mathbb{1}") + self.play(Write(action)) + self.play(ApplyMethod(square.scale, 1.2)) + self.play(ApplyMethod(square.scale, 1/1.2)) + self.play(FadeOut(action)) + self.wait() + + # show some reflections + axis = DashedLine(2 * LEFT, 2 * RIGHT) + sigma = MathTex(r"\sigma") + sigma.next_to(axis, RIGHT) + + self.play(Create(axis)) + self.play(Write(sigma)) + self.play(ApplyMethod(square.flip, RIGHT)) + + for d in [UP + RIGHT, UP]: + self.play( + Rotate(axis, PI/4), + Rotate(sigma, PI/4, about_point=ORIGIN)) + + self.play(Rotate(sigma, -PI/4), run_time=.5) + self.play(ApplyMethod(square.flip, d)) + + self.play(FadeOutAndShift(sigma), Uncreate(axis)) + + # show some rotations + dot = Dot(UP + RIGHT) + figure = VGroup(square, dot) + + rot = MathTex(r"r") + self.play(Write(rot), Create(dot)) + + last = rot + for newrot in map(MathTex, [r"r", r"r^2", r"r^3"]): + self.play( + ReplacementTransform(last, newrot), + Rotate(figure, PI/2, about_point=ORIGIN)) + self.wait(.5) + last = newrot + + self.play(Uncreate(dot), FadeOut(square), FadeOut(last)) + + + def cyclic(self): + # create symmetric figure + figure = VGroup() + prev = [1.5, 0, 0] + for i in range(1,6): + pos = [ + 1.5*m.cos(2 * PI/5 * i), + 1.5*m.sin(2 * PI/5 * i), + 0 + ] + + if prev: + line = Line(prev, pos) + figure.add(line) + + dot = Dot(pos, radius=.1) + if i == 5: + dot.set_fill(RED) + + prev = pos + figure.add(dot) + + group = MathTex(r"G = \langle r \rangle") + self.play(Write(group), run_time = 2) + self.wait(3) + + self.play(ApplyMethod(group.to_edge, UP)) + + actions = map(MathTex, [ + r"\mathbb{1}", r"r", r"r^2", + r"r^3", r"r^4", r"\mathbb{1}", r"r"]) + + action = next(actions, MathTex(r"r")) + + self.play(Create(figure)) + self.play(Write(action)) + self.wait() + + for i in range(5): + newaction = next(actions, MathTex(r"r")) + self.play( + ReplacementTransform(action, newaction), + Rotate(figure, 2*PI/5, about_point=ORIGIN)) + action = newaction + + self.wait() + newaction = next(actions, MathTex(r"r")) + self.play( + ReplacementTransform(action, newaction), + Rotate(figure, 2*PI/5, about_point=ORIGIN)) + action = newaction + self.wait(2) + + self.play(Uncreate(figure), FadeOut(action)) + + whole_group = MathTex( + r"G = \langle r \rangle" + r"= \left\{\mathbb{1}, r, r^2, r^3, r^4 \right\}") + + self.play(ApplyMethod(group.move_to, ORIGIN)) + self.play(ReplacementTransform(group, whole_group)) + self.wait(5) + + cyclic = MathTex( + r"C_n = \langle r \rangle" + r"= \left\{\mathbb{1}, r, r^2, \dots, r^{n-1} \right\}") + + cyclic_title = Tex(r"Zyklische Gruppe") + cyclic_title.next_to(cyclic, UP * 2) + + cyclic.scale(1.2) + cyclic_title.scale(1.2) + + self.play(ReplacementTransform(whole_group, cyclic)) + self.play(FadeInFrom(cyclic_title, UP)) + + self.wait(5) + self.play(FadeOut(cyclic), FadeOut(cyclic_title)) + + def dihedral(self): + # create square + square = Square() + square.set_fill(PINK, opacity=.5) + + # generator equation + group = MathTex( + r"G = \langle \sigma, r \,|\,", + r"\sigma^2 = \mathbb{1},", + r"r^4 = \mathbb{1},", + r"(\sigma r)^2 = \mathbb{1} \rangle") + + self.play(Write(group), run_time = 2) + self.wait(5) + + self.play(ApplyMethod(group.to_edge, UP)) + self.play(FadeIn(square)) + self.wait() + + # flips + axis = DashedLine(2 * LEFT, 2 * RIGHT) + sigma = MathTex(r"\sigma^2 = \mathbb{1}") + sigma.next_to(axis, RIGHT) + self.play(Create(axis), Write(sigma)) + self.play(ApplyMethod(square.flip, RIGHT)) + self.play(ApplyMethod(square.flip, RIGHT)) + self.play(Uncreate(axis), FadeOut(sigma)) + + # rotations + dot = Dot(UP + RIGHT) + rot = MathTex(r"r^4 = \mathbb{1}") + rot.next_to(square, DOWN * 3) + + figure = VGroup(dot, square) + + self.play(Write(rot), Create(dot)) + for i in range(4): + self.play(Rotate(figure, PI/2)) + self.play(FadeOut(rot), Uncreate(dot)) + + # rotation and flip + action = MathTex(r"(\sigma r)^2 = \mathbb{1}") + action.next_to(square, DOWN * 5) + + dot = Dot(UP + RIGHT) + axis = DashedLine(2 * LEFT, 2 * RIGHT) + self.play(Create(dot), Create(axis), Write(action)) + + figure = VGroup(dot, square) + + for i in range(2): + self.play(Rotate(figure, PI/2)) + self.play(ApplyMethod(figure.flip, RIGHT)) + self.wait() + + self.play(Uncreate(dot), Uncreate(axis), FadeOut(action)) + self.play(FadeOut(square)) + + # equation for the whole + whole_group = MathTex( + r"G &= \langle \sigma, r \,|\," + r"\sigma^2 = r^4 = (\sigma r)^2 = \mathbb{1} \rangle \\" + r"&= \left\{" + r"\mathbb{1}, r, r^2, r^3, \sigma, \sigma r, \sigma r^2, \sigma r^3" + r"\right\}") + + self.play(ApplyMethod(group.move_to, ORIGIN)) + self.play(ReplacementTransform(group, whole_group)) + self.wait(2) + + dihedral = MathTex( + r"D_n &= \langle \sigma, r \,|\," + r"\sigma^2 = r^n = (\sigma r)^2 = \mathbb{1} \rangle \\" + r"&= \left\{" + r"\mathbb{1}, r, r^2, \dots, \sigma, \sigma r, \sigma r^2, \dots" + r"\right\}") + + dihedral_title = Tex(r"Diedergruppe: Symmetrien eines \(n\)-gons") + dihedral_title.next_to(dihedral, UP * 2) + + dihedral.scale(1.2) + dihedral_title.scale(1.2) + + self.play(ReplacementTransform(whole_group, dihedral)) + self.play(FadeInFrom(dihedral_title, UP)) + + self.wait(5) + self.play(FadeOut(dihedral), FadeOut(dihedral_title)) + + def circle(self): + circle = Circle(radius=2) + dot = Dot() + dot.move_to(2 * RIGHT) + + figure = VGroup(circle, dot) + group_name = MathTex(r"C_\infty") + + # create circle + self.play(Create(circle)) + self.play(Create(dot)) + + # move it around + self.play(Rotate(figure, PI/3)) + self.play(Rotate(figure, PI/6)) + self.play(Rotate(figure, -PI/3)) + + # show name + self.play(Rotate(figure, PI/4), Write(group_name)) + self.wait() + self.play(Uncreate(figure)) + + nsphere = MathTex(r"C_\infty \cong S^1 = \left\{z \in \mathbb{C} : |z| = 1\right\}") + nsphere_title = Tex(r"Kreisgruppe") + nsphere_title.next_to(nsphere, 2 * UP) + + nsphere.scale(1.2) + nsphere_title.scale(1.2) + + self.play(ReplacementTransform(group_name, nsphere)) + self.play(FadeInFrom(nsphere_title, UP)) + + self.wait(5) + self.play(FadeOut(nsphere_title), FadeOut(nsphere)) + self.wait(2) + + +class Geometric3DSymmetries(ThreeDScene): + def construct(self): + self.improper_rotation() + self.tetrahedron() + + def improper_rotation(self): + # changes the source of the light and camera + self.renderer.camera.light_source.move_to(3*IN) + self.set_camera_orientation(phi=0, theta=0) + + # initial square + square = Square() + square.set_fill(PINK, opacity=.5) + + self.play(SpinInFromNothing(square)) + self.wait(2) + + for i in range(4): + self.play(Rotate(square, PI/2)) + self.wait(.5) + + self.move_camera(phi= 75 * DEGREES, theta = -80 * DEGREES) + + # create rotation axis + axis = Line3D(start=[0,0,-2.5], end=[0,0,2.5]) + + axis_name = MathTex(r"r \in C_4") + # move to yz plane + axis_name.rotate(PI/2, axis = RIGHT) + axis_name.next_to(axis, OUT) + + self.play(Create(axis)) + self.play(Write(axis_name)) + self.wait() + + # create sphere from slices + cyclic_slices = [] + for i in range(4): + colors = [PINK, RED] if i % 2 == 0 else [BLUE_D, BLUE_E] + cyclic_slices.append(ParametricSurface( + lambda u, v: np.array([ + np.sqrt(2) * np.cos(u) * np.cos(v), + np.sqrt(2) * np.cos(u) * np.sin(v), + np.sqrt(2) * np.sin(u) + ]), + v_min=PI/4 + PI/2 * i, + v_max=PI/4 + PI/2 * (i + 1), + u_min=-PI/2, u_max=PI/2, + checkerboard_colors=colors, resolution=(10,5))) + + self.play(FadeOut(square), *map(Create, cyclic_slices)) + + cyclic_sphere = VGroup(*cyclic_slices) + for i in range(4): + self.play(Rotate(cyclic_sphere, PI/2)) + self.wait() + + new_axis_name = MathTex(r"r \in D_4") + # move to yz plane + new_axis_name.rotate(PI/2, axis = RIGHT) + new_axis_name.next_to(axis, OUT) + self.play(ReplacementTransform(axis_name, new_axis_name)) + + # reflection plane + self.play(FadeOut(cyclic_sphere), FadeIn(square)) + plane = ParametricSurface( + lambda u, v: np.array([u, 0, v]), + u_min = -2, u_max = 2, + v_min = -2, v_max = 2, + fill_opacity=.3, resolution=(1,1)) + + plane_name = MathTex(r"\sigma \in D_4") + # move to yz plane + plane_name.rotate(PI/2, axis = RIGHT) + plane_name.next_to(plane, OUT + RIGHT) + + self.play(Create(plane)) + self.play(Write(plane_name)) + self.wait() + + self.move_camera(phi = 25 * DEGREES, theta = -75 * DEGREES) + self.wait() + + condition = MathTex(r"(\sigma r)^2 = \mathbb{1}") + condition.next_to(square, DOWN); + + self.play(Write(condition)) + self.play(Rotate(square, PI/2)) + self.play(Rotate(square, PI, RIGHT)) + + self.play(Rotate(square, PI/2)) + self.play(Rotate(square, PI, RIGHT)) + self.play(FadeOut(condition)) + + self.move_camera(phi = 75 * DEGREES, theta = -80 * DEGREES) + + # create sphere from slices + dihedral_slices = [] + for i in range(4): + for j in range(2): + colors = [PINK, RED] if i % 2 == 0 else [BLUE_D, BLUE_E] + dihedral_slices.append(ParametricSurface( + lambda u, v: np.array([ + np.sqrt(2) * np.cos(u) * np.cos(v), + np.sqrt(2) * np.cos(u) * np.sin(v), + np.sqrt(2) * np.sin(u) + ]), + v_min=PI/2 * j + PI/4 + PI/2 * i, + v_max=PI/2 * j + PI/4 + PI/2 * (i + 1), + u_min=-PI/2 if j == 0 else 0, + u_max=0 if j == 0 else PI/2, + checkerboard_colors=colors, resolution=(10,5))) + + dihedral_sphere = VGroup(*dihedral_slices) + + self.play(FadeOut(square), Create(dihedral_sphere)) + + for i in range(2): + self.play(Rotate(dihedral_sphere, PI/2)) + self.play(Rotate(dihedral_sphere, PI, RIGHT)) + self.wait() + + self.wait(2) + self.play(*map(FadeOut, [dihedral_sphere, plane, plane_name, new_axis_name]), FadeIn(square)) + self.wait(3) + self.play(*map(FadeOut, [square, axis])) + self.wait(3) + + def tetrahedron(self): + tet = Tetrahedron(edge_length=2) + self.play(FadeIn(tet)) + + self.move_camera(phi = 75 * DEGREES, theta = -100 * DEGREES) + self.begin_ambient_camera_rotation(rate=.1) + + axes = [] + for coord in tet.vertex_coords: + axes.append((-2 * coord, 2 * coord)) + + lines = [ + Line3D(start=s, end=e) for s, e in axes + ] + + self.play(*map(Create, lines)) + self.wait() + + for axis in axes: + self.play(Rotate(tet, 2*PI/3, axis=axis[1])) + self.play(Rotate(tet, 2*PI/3, axis=axis[1])) + + self.wait(5) + self.stop_ambient_camera_rotation() + self.wait() + self.play(*map(Uncreate, lines)) + self.play(FadeOut(tet)) + self.wait(5) + + +class AlgebraicSymmetries(Scene): + def construct(self): + self.wait(5) + self.cyclic() + # self.matrices() + + def cyclic(self): + # show the i product + product = MathTex( + r"1", r"\cdot i &= i \\", + r"i \cdot i &= -1 \\", + r"-1 \cdot i &= -i \\", + r"-i \cdot i &= 1") + product.scale(1.5) + + for part in product: + self.play(Write(part)) + self.wait() + + self.play(ApplyMethod(product.scale, 1/1.5)) + + # gather in group + group = MathTex(r"G = \left\{ 1, i, -1, -i \right\}") + self.play(ReplacementTransform(product, group)) + self.wait(2) + + # show C4 + grouppow = MathTex( + r"G &= \left\{ 1, i, i^2, i^3 \right\} \\", + r"C_4 &= \left\{ \mathbb{1}, r, r^2, r^3 \right\}") + self.play(ReplacementTransform(group, grouppow[0])) + self.wait(2) + + self.play(Write(grouppow[1])) + self.wait(4) + + self.play(ApplyMethod(grouppow.to_edge, UP)) + + # define morphisms + morphism = MathTex(r"\phi: C_4 \to G \\") + morphism.shift(UP) + self.play(Write(morphism)) + self.wait() + + # show an example + mappings = MathTex( + r"\phi(\mathbb{1}) &= 1 \\", + r"\phi(r) &= i \\", + r"\phi(r^2) &= i^2 \\", + r"\phi(r^3) &= i^3 \\") + mappings.next_to(morphism, 2 * DOWN) + + self.play(Write(mappings)) + self.wait(3) + self.play(FadeOutAndShift(mappings, DOWN)) + + # more general definition + homomorphism = MathTex( + r"\phi(r\circ \mathbb{1}) &= \phi(r)\cdot\phi(\mathbb{1}) \\", + r"&= i\cdot 1") + homomorphism.next_to(morphism, DOWN).align_to(morphism, LEFT) + for part in homomorphism: + self.play(Write(part)) + self.wait() + + hom_bracegrp = VGroup(morphism, homomorphism) + + self.play( + ApplyMethod(grouppow.shift, 3 * LEFT), + ApplyMethod(hom_bracegrp.shift, 3 * LEFT)) + + hom_brace = Brace(hom_bracegrp, direction=RIGHT) + hom_text = Tex("Homomorphismus").next_to(hom_brace.get_tip(), RIGHT) + hom_text_short = MathTex(r"\mathrm{Hom}(C_4, G)").next_to(hom_brace.get_tip(), RIGHT) + + self.play(Create(hom_brace)) + self.play(Write(hom_text)) + self.wait() + self.play(ReplacementTransform(hom_text, hom_text_short)) + self.wait() + + # add the isomorphism part + isomorphism = Tex(r"\(\phi\) ist bijektiv") + isomorphism.next_to(homomorphism, DOWN).align_to(homomorphism, LEFT) + self.play(Write(isomorphism)) + + iso_bracegrp = VGroup(hom_bracegrp, isomorphism) + + iso_brace = Brace(iso_bracegrp, RIGHT) + iso_text = Tex("Isomorphismus").next_to(iso_brace.get_tip(), RIGHT) + iso_text_short = MathTex("C_4 \cong G").next_to(iso_brace.get_tip(), RIGHT) + + self.play( + ReplacementTransform(hom_brace, iso_brace), + ReplacementTransform(hom_text_short, iso_text)) + self.wait() + + self.play(ReplacementTransform(iso_text, iso_text_short)) + self.wait() + + # create a group for the whole + morphgrp = VGroup(iso_bracegrp, iso_brace, iso_text_short) + + self.play( + ApplyMethod(grouppow.to_edge, LEFT), + ApplyMethod(morphgrp.to_edge, LEFT)) + + # draw a complex plane + plane = ComplexPlane(x_range = [-2.5, 2.5]) + coordinates = plane.get_coordinate_labels(1, -1, 1j, -1j) + + roots = list(map(lambda p: Dot(p, fill_color=PINK), ( + [1, 0, 0], [0, 1, 0], [-1, 0, 0], [0, -1, 0] + ))) + + arrow = CurvedArrow( + 1.5 * np.array([m.cos(10 * DEGREES), m.sin(10 * DEGREES), 0]), + 1.5 * np.array([m.cos(80 * DEGREES), m.sin(80 * DEGREES), 0])) + arrowtext = MathTex("\cdot i") + arrowtext.move_to(2 / m.sqrt(2) * (UP + RIGHT)) + + square = Square().rotate(PI/4).scale(1/m.sqrt(2)) + square.set_fill(PINK).set_opacity(.4) + + figuregrp = VGroup(plane, square, arrow, arrowtext, *coordinates, *roots) + figuregrp.to_edge(RIGHT) + + self.play(Create(plane)) + self.play( + *map(Create, roots), + *map(Write, coordinates)) + self.wait() + self.play(FadeIn(square), Create(arrow), Write(arrowtext)) + + for _ in range(4): + self.play(Rotate(square, PI/2)) + self.wait(.5) + + self.play( + *map(FadeOut, (square, arrow, arrowtext)), + *map(FadeOut, coordinates), + *map(FadeOut, roots)) + self.play(Uncreate(plane)) + self.play( + FadeOutAndShift(grouppow, RIGHT), + FadeOutAndShift(morphgrp, RIGHT)) + + modulo = MathTex( + r"\phi: C_4 &\to (\mathbb{Z}/4\mathbb{Z}, +) \\" + r"\phi(\mathbb{1} \circ r^2) &= 0 + 2 \pmod 4").scale(1.5) + self.play(Write(modulo)) + self.wait(2) + + self.play(FadeOut(modulo)) + self.wait(3) + + def matrices(self): + question = MathTex( + r"D_n &\cong \,? \\" + r"S_n &\cong \,? \\" + r"A_n &\cong \,?").scale(1.5) + + answer = MathTex( + r"D_n &\cong \,?\\" + r"S_4 &\cong \mathrm{Aut}(Q_8) \\" + r"A_5 &\cong \mathrm{PSL}_2 (5)").scale(1.5) + + self.play(Write(question)) + self.wait() + self.play(ReplacementTransform(question, answer)) + + self.wait(3) diff --git a/vorlesungen/punktgruppen/media/images/nosignal.jpg 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a/vorlesungen/punktgruppen/script.pdf b/vorlesungen/punktgruppen/script.pdf Binary files differnew file mode 100644 index 0000000..91993fb --- /dev/null +++ b/vorlesungen/punktgruppen/script.pdf diff --git a/vorlesungen/punktgruppen/script.tex b/vorlesungen/punktgruppen/script.tex new file mode 100644 index 0000000..bc50e21 --- /dev/null +++ b/vorlesungen/punktgruppen/script.tex @@ -0,0 +1,214 @@ +\documentclass[a4paper]{article} + +\usepackage{amsmath} +\usepackage{amssymb} + +\usepackage[cm]{manuscript} +\usepackage{xcolor} + +\newcommand{\scene}[1]{\par\noindent[ #1 ]\par} +\newenvironment{totranslate}{\color{blue!70!black}}{} + +\begin{document} +\section{Das sind wir} +\scene{Camera} + +\section{Ablauf} +Zuerst werden wir Symmetrien in 2 Dimensionen anschauen, dann \"uberlegen wir +kurz was es heisst f\"ur eine Symmetrie ``algebraisch'' zu sein. Von da aus +kommt die dritte Dimension hinzu, die man besser mit Matrizen verstehen kann. +Mit der aufgebauten Theorie werden wir versuchen Kristalle zu klassifizieren. +Und zum Schluss kommen wir zu Anwendungen, welche f\"ur Ingenieure von +Interesse sind. + +\section{intro} +\scene{Spontan} + +\section{2D Geometrie} +\scene{Intro} +Wir fangen mit den 2 dimensionalen Symmetrien an, da man sie sich am +einfachsten vorstellen kann. Eine Symmetrie eines Objektes beschreibt eine +Aktion, welche nachdem sie auf das Objekt wirkt, das Objekt wieder gleich +aussehen l\"asst. + +\scene{Viereck} +Die einfachste Aktion, ist das Viereck zu nehmen, und wieder hinzulegen. +Eine andere Aktion k\"onnte sein, das Objekt um eine Achse zu spiegeln, +oder eine Rotation um 90 Grad. + +\scene{Zyklische Gruppe} +Fokussieren wir uns auf die einfachste Klassen von Symmetrien: diejenigen die +von einer reinen Drehung generiert werden. Wir sammeln diese in einer Gruppe +\(G\), und notieren das sie von eine Rotation \(r\) generiert worden sind, mit +diesen spitzen Klammern. + +Nehmen wir als Beispiel dieses Pentagon. Wenn wir \(r\) 5-mal anwenden, ist es +dasselbe als wenn wir nichts gemacht h\"atten. Wenn wir es noch ein 6. mal +drehen, entspricht dies dasselbe wie \(r\) nur 1 mal zu nutzen. + +\scene{Notation} +So, die Gruppe setzt sich zusammen aus dem neutralen Element, und den Potenzen +1 bis 4 von \(r\). Oder im allgemein Gruppen mit dieser Struktur, in welcher die +Aktion \(n-1\) mal angewendet werden kann, heissen ``Zyklische Gruppe''. + +\scene{Diedergruppe} +Nehmen wir nun auch noch die Spiegeloperation \(\sigma\) dazu. Weil wir jetzt 2 +Operationen haben, m\"ussen wir auch im Generator schreiben wie sie +zusammenh\"angen. Schauen wir dann uns genauer diesen Ausdr\"uck an. Zweimal +Spielegeln ist \"aquivalent zum neutralen Element, sowie 4 mal um 90 Grad +drehen und 2 Drehspiegelungen, welche man auch Inversion nennt. + +\scene{Notation} +Daraus k\"onnen wir wieder die ganze Gruppe erzeugen, die im allgemeinen den +Symmetrien eines \(n\)-gons entsprechen. + +\scene{Kreisgruppe} +Bis jetzt hatten wir nur diskrete Symmetrien, was nicht zwingend der Fall sein +muss. Ein Ring kann man kontinuierlich drehen, und sieht dabei immer gleich +aus. + +Diese Symmetrie ist auch als Kreisgruppe bekannt, die man sch\"on mit dem +komplexen Einheitskreis definieren kann. + +\section{Algebra} +\scene{Produkt mit \(i\)} +\"Uberlegen wir uns eine spezielle algebraische Operation: Multiplikation mit +der imagin\"aren Einheit. \(1\) mal \(i\) ist gleich \(i\). Wieder mal \(i\) +ist \(-1\), dann \(-i\) und schliesslich kommen wir z\"uruck auf \(1\). Diese +fassen wir in eine Gruppe \(G\) zusammen. Oder sch\"oner geschrieben:. Sieht das +bekannt aus? + +\scene{Morphismen} +Das Gefühl, dass es sich um dasselbe handelt, kann wie folgt formalisiert +werden. Sei \(\phi\) eine Funktion von \(C_4\) zu \(G\) und ordnen wir zu +jeder Symmetrieoperation ein Element aus \(G\). Wenn man die Zuordnung richtig +definiert, dann sieht man die folgende Eigenschaft: Eine Operation nach eine +andere zu nutzen, und dann die Funktion des Resultats zu nehmen, ist gleich wie +die Funktion der einzelnen Operazionen zu nehmen und die Resultate zu +multiplizieren. Dieses Ergebnis ist so bemerkenswert, dass es in der Mathematik +einen Namen bekommen hat: Homorphismus, von griechisch "homos" dasselbe und +"morphe" Form. Manchmal auch so geschrieben. Ausserdem, wenn \(\phi\) eins zu +eins ist, heisst es \emph{Iso}morphismus: "iso" gleiche Form. Was man +typischerweise mit diesem Symbol schreibt. + +\scene{Animation} +Sie haben wahrscheinlich schon gesehen, worauf das hinausläuft. Dass die +zyklische Gruppe \(C_4\) und \(G\) isomorph sind ist nicht nur Fachjargon der +mathematik, sondern sie haben wirklich die selbe Struktur. + +\scene{Modulo} +Das Beispiel mit der komplexen Einheit, war wahrscheinlich nicht so +\"uberraschend. Aber was merkw\"urdig ist, ist das Beziehungen zwischen +Symmetrien und Algebra auch in Bereichen gefunden werden, welche auf den ersten +Blick, nicht geomerisch erscheinen. Ein R\"atsel für die Neugierigen: die Summe +in der Modulo-Arithmetik. Als Hinweis: Um die Geometrie zu finden denken Sie +an einer Uhr. + +\section{3D Geometrie} +2 Dimensionen sind einfacher zu zeichnen, aber leider leben wir im 3 +dimensionalen Raum. + +\scene{Zyklische Gruppe} +Wenn wir unser bekanntes Viereck mit seiner zyklischer Symmetrie in 3 +Dimensionen betrachten, k\"onnen wir seine Drehachse sehen. + +\scene{Diedergruppe} +Um auch noch die andere Symmetrie des Rechteckes zu sehen, ben\"otigen wir eine +Spiegelachse \(\sigma\), die hier eine Spiegelebene ist. + +\scene{Transition} +Um die Punktsymmetrien zu klassifizieren orientiert man sich an einer Achse, um +welche sich die meisten Symmetrien drehen. Das geht aber nicht immer, wie beim +Tetraeder. + +\scene{Tetraedergruppe} +Diese Geometrie hat 4 gleichwertige Symmetrieachsen, die eben eine +Symmetriegruppe aufbauen, welche kreativer weise Tetraedergruppe genannt wird. +Vielleicht fallen Ihnnen weitere Polygone ein mit dieser Eigenschaft, bevor wir +zum n\"achsten Thema weitergehen. + +\section{Matrizen} +\scene{Titelseite} +Nun gehen wir kurz auf den Thema unseres Seminars ein: Matrizen. Das man mit +Matrizen Dinge darstellen kann, ist keine Neuigkeit mehr, nach einem +Semester MatheSeminar. Also überrascht es wohl auch keinen, das man alle +punktsymmetrischen Operationen auch mit Matrizen Formulieren kann. + +\scene{Matrizen} + +Sei dann \(G\) unsere Symmetrie Gruppe, die unsere abstrakte Drehungen und +Spiegelungen enth\"ahlt. Die Matrix Darstellung dieser Gruppe, ist eine +Funktion gross \(\Phi\), von \(G\) zur orthogonalen Gruppe \(O(3)\), die zu +jeder Symmetrie Operation klein \(g\) eine Matrix gross \(\Phi_g\) zuordnet. + +Zur Erinnerung, die Orthogonale Gruppe ist definiert als die Matrizen, deren +transponierte auch die inverse ist. Da diese Volumen und Distanzen erhalten, +natuerlich nur bis zu einer Vorzeichenumkehrung, macht es Sinn, dass diese +Punksymmetrien genau beschreiben. + +Nehmen wir die folgende Operationen als Beispiele. Die Matrix der trivialen +Operation, dass heisst nichts zu machen, ist die Einheitsmatrix. Eine +Spiegelung ist dasselbe aber mit einem Minus, und Drehungen sind uns schon +dank Herrn M\"uller bekannt. + +\section{Kristalle} +\scene{Spontan} + +\section{Piezo} +\scene{Spontan} + +\section{Licht} +Als Finale, haben wir ein schwieriges Problem aus der Physik. Das Ziel dieser +Folie ist nicht jedes Zeichen zu versehen, sondern zu zeigen wie man von hier +weiter gehen kann. Wir mochten sehen wie sich Licht in einem Kristall verhaltet. +Genauer, wir m\"ochten die Amplitude einer +elektromagnetischer Welle in einem Kristall beschreiben. + +Das Beispiel richtet sich mehr an Elektrotechnik Studenten, aber die Theorie +ist die gleiche bei mechanischen Wellen in Materialien mit einer +Spannungstensor wie dem, den wir letzte Woche gesehen haben. +% Ganz grob gesagt, ersetzt man E durch Xi und epsilon durch das Sigma. + +Um eine Welle zu beschreiben, verwenden wir die Helmholtz-Gleichung, die einige +von uns bereits in anderen Kursen gel\"ost haben. Schwierig wird aber dieses +Problem, wenn der Term vor der Zeitableitung ein Tensor ist (f\"ur uns eine Matrix). + +Zur Vereinfachung werden wir eine ebene Welle verwenden. Setzt man dieses E in +die Helmholtz-Gleichung ein, erhält man folgendes zurück: ein Eigenwertproblem. + +Physikalisch bedeutet dies, dass die Welle in diesem Material ihre Amplitude in +Abhängigkeit von der Ausbreitungsrichtung ändert. Und die Eigenwerte sagen +aus, wie stark die Amplitude der Welle in jeder Richtung skaliert wird. + +Ich sagte, in jede Richtung skaliert, aber welche Richtungen genau? +Physikalisch hängt das von der kristallinen Struktur des Materials ab, aber +mathematisch können wir sagen: in Richtung der Eigenvektoren! Aber diesen +Eigenraum zu finden, in dem die Eigenvektoren wohnen, ist beliebig schwierig. + +Hier kommt unsere Gruppentheorie zu Hilfe. Wir können die Symmetrien unseres +Kristalls zur Hilfe nehmen. Zu jeder dieser Symmetrien lässt sich bekanntlich eine +einfache Matrix finden, deren Eigenraum ebenfalls relativ leicht zu finden ist. +Zum Beispiel ist der Eigenraum der Rotation \(r\), die Rotationsachse, für die +Reflexion \(\sigma\) eine Ebene, und so weiter. + +Nun ist die Frage, ob man diese Eingenraume der Symmetrienoperationen +kombinieren kann um den Eigenraum des physikalisches Problems zu finden. + +Aber leider ist meine Zeit abgelaufen in der Recherche, also müssen Sie mir 2 +Dingen einfach glauben, erstens dass es einen Weg gibt, und zweitens dass eher +nicht so schlimm ist, wenn man die Notation einmal gelernt hat. + +Nachdem wir an, wir haben den Eigenraum U gefunden, dann können wir einen +(Eigen)Vektor E daraus nehmen und in ihm direkt lambda ablesen. Das sagt uns, +wie die Amplitude der Welle, in diese Richtung gedämpft wurde. + +Diese Methode ist nicht spezifisch für dieses Problem, im Gegenteil, ich habe +gesehen, dass sie in vielen Bereichen eingesetzt wird, wie z.B.: +Kristallographie, Festkörperphysik, Molekülschwingungen in der Quantenchemie +und numerische Simulationen von Membranen. + +\section{Outro} +\scene{Camera} + +\end{document} +% vim:et ts=2 sw=2: diff --git a/vorlesungen/punktgruppen/slides.pdf b/vorlesungen/punktgruppen/slides.pdf Binary files differnew file mode 100644 index 0000000..e1769f8 --- /dev/null +++ b/vorlesungen/punktgruppen/slides.pdf diff --git a/vorlesungen/punktgruppen/slides.tex b/vorlesungen/punktgruppen/slides.tex new file mode 100644 index 0000000..cd3d7d7 --- /dev/null +++ b/vorlesungen/punktgruppen/slides.tex @@ -0,0 +1,895 @@ +\documentclass[12pt, xcolor, aspectratio=169, handout]{beamer} + +% language +\usepackage{polyglossia} +\setmainlanguage{german} + +% pretty drawings +\usepackage{tikz} +\usepackage{tikz-3dplot} + +\usetikzlibrary{positioning} +\usetikzlibrary{arrows.meta} +\usetikzlibrary{shapes.misc} +\usetikzlibrary{calc} + +\usetikzlibrary{external} +\tikzexternalize[ + mode = graphics if exists, + figure list = true, + prefix=build/ +] + +% Theme +\beamertemplatenavigationsymbolsempty + +% set look +\usetheme{default} +\usecolortheme{fly} +\usefonttheme{serif} + +%% Set font +\usepackage[p,osf]{scholax} +\usepackage{amsmath} +\usepackage[scaled=1.075,ncf,vvarbb]{newtxmath} + +% set colors +\definecolor{background}{HTML}{202020} + +\setbeamercolor{normal text}{fg=white, bg=background} +\setbeamercolor{structure}{fg=white} + +\setbeamercolor{item projected}{use=item,fg=background,bg=item.fg!35} + +\setbeamercolor*{palette primary}{use=structure,fg=white,bg=structure.fg} +\setbeamercolor*{palette secondary}{use=structure,fg=white,bg=structure.fg!75} +\setbeamercolor*{palette tertiary}{use=structure,fg=white,bg=structure.fg!50} +\setbeamercolor*{palette quaternary}{fg=white,bg=background} + +\setbeamercolor*{block title}{parent=structure} +\setbeamercolor*{block body}{fg=background, bg=} + +\setbeamercolor*{framesubtitle}{fg=white} + +\setbeamertemplate{section page} +{ + \begin{center} + \Huge + \insertsection + \end{center} +} +\AtBeginSection{\frame{\sectionpage}} + +% Macros +\newcommand{\ten}[1]{#1} + +% Metadata +\title{\LARGE \scshape Punktgruppen und Kristalle} +\author[N. Pross, T. T\"onz]{Naoki Pross, Tim T\"onz} +\institute{Hochschule f\"ur Technik OST, Rapperswil} +\date{10. Mai 2021} + +% Slides +\begin{document} +\frame{ + \titlepage + \vfill + \begin{center} + \small \color{gray} + Slides: \texttt{s.0hm.ch/ctBsD} + \end{center} +} +\frame{\tableofcontents} + +\frame{ + \begin{itemize} + \item Was heisst \emph{Symmetrie} in der Mathematik? \pause + \item Wie kann ein Kristall modelliert werden? \pause + \item Aus der Physik: Licht, Piezoelektrizit\"at \pause + \end{itemize} + \begin{center} + \begin{tikzpicture} + \begin{scope}[ + node distance = 0cm + ] + \node[ + rectangle, fill = gray!40!background, + minimum width = 3cm, minimum height = 2cm, + ] (body) {\(\vec{E}_p = \vec{0}\)}; + + \node[ + draw, rectangle, thick, white, fill = red!50, + minimum width = 3cm, minimum height = 1mm, + above = of body + ] (pos) {}; + + \node[ + draw, rectangle, thick, white, fill = blue!50, + minimum width = 3cm, minimum height = 1mm, + below = of body + ] (neg) {}; + + \draw[white, very thick, -Circle] (pos.east) to ++ (1,0) node (p) {}; + \draw[white, very thick, -Circle] (neg.east) to ++ (1,0) node (n) {}; + + \draw[white, thick, ->] (p) to[out = -70, in = 70] node[midway, right] {\(U = 0\)} (n); + \end{scope} + \begin{scope}[ + node distance = 0cm, + xshift = 7cm + ] + \node[ + rectangle, fill = gray!40!background, + minimum width = 3cm, minimum height = 1.5cm, + ] (body) {\(\vec{E}_p = \vec{0}\)}; + + \node[ + draw, rectangle, thick, white, fill = red!50, + minimum width = 3cm, minimum height = 1mm, + above = of body + ] (pos) {}; + + \node[ + draw, rectangle, thick, white, fill = blue!50, + minimum width = 3cm, minimum height = 1mm, + below = of body + ] (neg) {}; + + \draw[orange, very thick, <-] (pos.north) to node[near end, right] {\(\vec{F}\)} ++(0,1); + \draw[orange, very thick, <-] (neg.south) to node[near end, right] {\(\vec{F}\)} ++(0,-1); + + \draw[white, very thick, -Circle] (pos.east) to ++ (1,0) node (p) {}; + \draw[white, very thick, -Circle] (neg.east) to ++ (1,0) node (n) {}; + + \draw[white, thick, ->] (p) to[out = -70, in = 70] node[midway, right] {\(U > 0\)} (n); + \end{scope} + \end{tikzpicture} + \end{center} +} + +\section{2D Symmetrien} +%% Made in video +{ + \usebackgroundtemplate{ + \includegraphics[height=\paperheight]{media/images/nosignal}} + \frame{} +} + +\section{Algebraische Symmetrien} +%% Made in video +\frame{ + \begin{columns}[T] + \begin{column}{.5\textwidth} + Produkt mit \(i\) + \begin{align*} + 1 \cdot i &= i \\ + i \cdot i &= -1 \\ + -1 \cdot i &= -i \\ + -i \cdot i &= 1 + \end{align*} + \pause + % + Gruppe + \begin{align*} + G &= \left\{ + 1, i, -1, -i + \right\} \\ + &= \left\{ + 1, i, i^2, i^3 + \right\} \\ + C_4 &= \left\{ + \mathbb{1}, r, r^2, r^3 + \right\} + \end{align*} + \pause + \end{column} + \begin{column}{.5\textwidth} + Darstellung \(\phi : C_4 \to G\) + \begin{align*} + \phi(\mathbb{1}) &= 1 & \phi(r^2) &= i^2 \\ + \phi(r) &= i & \phi(r^3) &= i^3 + \end{align*} + \pause + % + Homomorphismus + \begin{align*} + \phi(r \circ \mathbb{1}) &= \phi(r) \cdot \phi(\mathbb{1}) \\ + &= i \cdot 1 + \end{align*} + \pause + % + \(\phi\) ist bijektiv \(\implies C_4 \cong G\) + \pause + % + \begin{align*} + \psi : C_4 &\to (\mathbb{Z}/4\mathbb{Z}, +) \\ + \psi(\mathbb{1}\circ r^2) &= 0 + 2 \pmod{4} + \end{align*} + \end{column} + \end{columns} +} + +\section{3D Symmetrien} +%% Made in video +{ + \usebackgroundtemplate{ + \includegraphics[height=\paperheight]{media/images/nosignal}} + \frame{} +} + +\section{Matrizen} +\frame{ + \begin{columns}[T] + \begin{column}{.5\textwidth} + Symmetriegruppe + \[ + G = \left\{\mathbb{1}, r, \sigma, \dots \right\} + \] + \pause + Matrixdarstellung + \begin{align*} + \Phi : G &\to O(3) \\ + g &\mapsto \Phi_g + \end{align*} + \pause + Orthogonale Gruppe + \[ + O(n) = \left\{ Q : QQ^t = Q^tQ = I \right\} + \] + \end{column} + \pause + \begin{column}{.5\textwidth} + \begin{align*} + \Phi_\mathbb{1} &= \begin{pmatrix} + 1 & 0 & 0 \\ + 0 & 1 & 0 \\ + 0 & 0 & 1 + \end{pmatrix} = I \\[1em] + \Phi_\sigma &= \begin{pmatrix} + 1 & 0 & 0 \\ + 0 & -1 & 0 \\ + 0 & 0 & 1 + \end{pmatrix} \\[1em] + \Phi_r &= \begin{pmatrix} + \cos \alpha & -\sin \alpha & 0 \\ + \sin \alpha & \cos \alpha & 0 \\ + 0 & 0 & 1 + \end{pmatrix} + \end{align*} + \end{column} + \end{columns} +} + +\section{Kristalle} +\begin{frame}[fragile]{} + \begin{columns} + \onslide<1->{ + \begin{column}{.5\textwidth} + \begin{center} + \begin{tikzpicture}[ + dot/.style = { + draw, circle, thick, white, fill = gray!40!background, + minimum size = 2mm, + inner sep = 0pt, + outer sep = 1mm, + }, + ] + + \begin{scope} + \clip (-2,-2) rectangle (3,4); + \foreach \y in {-7,-6,...,7} { + \foreach \x in {-7,-6,...,7} { + \node[dot, xshift=3mm*\y] (N\x\y) at (\x, \y) {}; + } + } + \end{scope} + \draw[white, thick] (-2, -2) rectangle (3,4); + + \draw[red!80!background, thick, ->] + (N00) to node[midway, below] {\(\vec{a}_1\)} (N10); + \draw[cyan!80!background, thick, ->] + (N00) to node[midway, left] {\(\vec{a}_2\)} (N01); + \end{tikzpicture} + \end{center} + \end{column} + } + \begin{column}{.5\textwidth} + \onslide<2->{ + Kristallgitter: + \(n_i \in \mathbb{Z}\), + } + \onslide<3->{ + \(\vec{a}_i \in \mathbb{R}^3\) + } + \onslide<2->{ + \[ + \vec{r} = n_1 \vec{a}_1 + n_2 \vec{a}_2 \onslide<3->{+ n_3 \vec{a}_3} + \] + } + \vspace{1cm} + + \onslide<4->{ + Invariant unter Translation + \[ + Q_i(\vec{r}) = \vec{r} + \vec{a}_i + \] + } + \end{column} + \end{columns} +\end{frame} + +\begin{frame}[fragile]{} + \begin{columns}[T] + \begin{column}{.5\textwidth} + \onslide<1->{ + Wie kombiniert sich \(Q_i\) mit der anderen Symmetrien? + } + \begin{center} + \begin{tikzpicture}[ + dot/.style = { + draw, circle, thick, white, fill = gray!40!background, + minimum size = 2mm, + inner sep = 0pt, + outer sep = 1mm, + }, + ] + + \onslide<2->{ + \node[dot] (A1) at (0,0) {}; 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& + \node[classcirc] (C4h) {} node[classlabel] {\(C_{4h}\)}; & + \node[classcirc] (C6h) {} node[classlabel] {\(C_{6h}\)}; & + + \node[classcirc, dashed] (Th) {} node[classlabel] {\(T_{h}\)}; & + \node[classcirc] (C4v) {} node[classlabel] {\(C_{4v}\)}; & + \node[classcirc] (C6v) {} node[classlabel] {\(C_{6v}\)}; & + \node[classcirc, dashed] (Td) {} node[classlabel] {\(T_{d}\)}; \\ + + \node[classcirc] (D2d) {} node[classlabel] {\(D_{2d}\)}; & + \node[classcirc] (D3h) {} node[classlabel] {\(D_{3h}\)}; & + \node[classcirc, dashed] (O) {} node[classlabel] {\(O_{}\)}; & + \node[classcirc] (D4) {} node[classlabel] {\(D_{4}\)}; & + + \node[classcirc] (D6) {} node[classlabel] {\(D_{6}\)}; & + \node[classcirc, dashed] (Oh) {} node[classlabel] {\(O_{h}\)}; & + \node[classcirc] (D4h) {} node[classlabel] {\(D_{4h}\)}; & + \node[classcirc] (D6h) {} node[classlabel] {\(D_{6h}\)}; \\ + }; + + + \node[cross] at ($(C1)+(4mm,0)$) {}; + + + \node[cross] at ($(C2)+(4mm,0)$) {}; + \node[cross] at ($(C2)-(4mm,0)$) {}; + + + \node[cross] at ($(C3)+( 0:4mm)$) {}; + \node[cross] at ($(C3)+(120:4mm)$) {}; + \node[cross] at ($(C3)+(240:4mm)$) {}; + + + \node[cross] at ($(Ci)+(4mm,0)$) {}; + \node[round] at ($(Ci)-(4mm,0)$) {}; + + + \node[cross] at ($(Cs)+(4mm,0)$) {}; + \node[round] at ($(Cs)+(4mm,0)$) {}; + + + \node[cross] at ($(C3i)+( 0:4mm)$) {}; + \node[cross] at ($(C3i)+(120:4mm)$) {}; + \node[cross] at ($(C3i)+(240:4mm)$) {}; + \node[round] at ($(C3i)+( 60:4mm)$) {}; + \node[round] at ($(C3i)+(180:4mm)$) {}; + \node[round] at ($(C3i)+(300:4mm)$) {}; + + + \node[cross] at ($(C2h)+(4mm,0)$) {}; + \node[cross] at ($(C2h)-(4mm,0)$) {}; + \node[round] at ($(C2h)+(4mm,0)$) {}; + \node[round] at ($(C2h)-(4mm,0)$) {}; + + + \node[cross] at ($(D2)+( 20:4mm)$) {}; + \node[cross] at ($(D2)+(200:4mm)$) {}; + \node[round] at ($(D2)+(160:4mm)$) {}; + \node[round] at ($(D2)+(340:4mm)$) {}; + + + \foreach \x in {0, 120, 240} { + \node[cross] at ($(D3d)+({\x+15}:4mm)$) {}; + \node[cross] at ($(D3d)+({\x-15}:4mm)$) {}; + } + + + \foreach \x in {0, 180} { + \node[cross] at ($(C2v)+({\x+15}:4mm)$) {}; + \node[cross] at ($(C2v)+({\x-15}:4mm)$) {}; + } + + + \foreach \x in {0, 180} { + \node[cross] at ($(D2h)+({\x+15}:4mm)$) {}; + \node[cross] at ($(D2h)+({\x-15}:4mm)$) {}; + \node[round] at ($(D2h)+({\x+15}:4mm)$) {}; + \node[round] at ($(D2h)+({\x-15}:4mm)$) {}; + } + + + \foreach \x in {0, 120, 240} { + \node[cross] at ($(D3)+({\x+15}:4mm)$) {}; + \node[round] at ($(D3)+({\x-15}:4mm)$) {}; + } + + + \foreach \x in {0, 90, 180, 270} { + \node[cross] at ($(C4)+(\x:4mm)$) {}; + } + + + \foreach \x in {0, 60, 120, 180, 240, 300} { + \node[cross] at ($(C6)+(\x:4mm)$) {}; + } + + + \foreach \x in {0, 120, 240} { + \node[cross] at ($(D3dP)+({\x+15}:4mm)$) {}; + \node[cross] at ($(D3dP)+({\x-15}:4mm)$) {}; + \node[round] at ($(D3dP)+({\x+15+60}:4mm)$) {}; + \node[round] at ($(D3dP)+({\x-15+60}:4mm)$) {}; + } + + + \node[cross] at ($(S4)+(4mm,0)$) {}; + \node[cross] at ($(S4)-(4mm,0)$) {}; + \node[round] at ($(S4)+(0,4mm)$) {}; + \node[round] at ($(S4)-(0,4mm)$) {}; + + + \foreach \x in {0, 120, 240} { + \node[cross] at ($(S3)+(\x:4mm)$) {}; + \node[round] at ($(S3)+(\x:4mm)$) {}; + } + + + %% TODO: T + + + \foreach \x in {0, 90, 180, 270} { + \node[cross] at ($(C4h)+(\x:4mm)$) {}; + \node[round] at ($(C4h)+(\x:4mm)$) {}; + } + + + \foreach \x in {0, 60, 120, 180, 240, 300} { + \node[cross] at ($(C6h)+(\x:4mm)$) {}; + \node[round] at ($(C6h)+(\x:4mm)$) {}; + } + + + %% TODO: Th + + + \foreach \x in {0, 90, 180, 270} { + \node[cross] at ($(C4v)+(\x+15:4mm)$) {}; + \node[cross] at ($(C4v)+(\x-15:4mm)$) {}; + } + + + + \foreach \x in {0, 60, 120, 180, 240, 300} { + \node[cross] at ($(C6v)+(\x+10:4mm)$) {}; + \node[cross] at ($(C6v)+(\x-10:4mm)$) {}; + } + + + %% TODO: Td + + + \foreach \x in {0, 180} { + \node[cross] at ($(D2d)+({\x+15}:4mm)$) {}; + \node[round] at ($(D2d)+({\x-15}:4mm)$) {}; + + \node[round] at ($(D2d)+({\x+15+90}:4mm)$) {}; + \node[cross] at ($(D2d)+({\x-15+90}:4mm)$) {}; + } + + + \foreach \x in {0, 120, 240} { + \node[cross] at ($(D3h)+({\x+15}:4mm)$) {}; + \node[cross] at ($(D3h)+({\x-15}:4mm)$) {}; + \node[round] at ($(D3h)+({\x+15}:4mm)$) {}; + \node[round] at ($(D3h)+({\x-15}:4mm)$) {}; + } + + + %% TODO: O + + + \foreach \x in {0, 90, 180, 270} { + \node[cross] at ($(D4)+({\x+15}:4mm)$) {}; + \node[round] at ($(D4)+({\x-15}:4mm)$) {}; + } + + \foreach \x in {0, 60, 120, 180, 240, 300} { + \node[cross] at ($(D6)+({\x+10}:4mm)$) {}; + \node[round] at ($(D6)+({\x-10}:4mm)$) {}; + } + + + % TODO Oh + + + \foreach \x in {0, 90, 180, 270} { + \node[cross] at ($(D4h)+(\x+15:4mm)$) {}; + \node[cross] at ($(D4h)+(\x-15:4mm)$) {}; + \node[round] at ($(D4h)+(\x+15:4mm)$) {}; + \node[round] at ($(D4h)+(\x-15:4mm)$) {}; + } + + + \foreach \x in {0, 60, 120, 180, 240, 300} { + \node[cross] at ($(D6h)+({\x+10}:4mm)$) {}; + \node[cross] at ($(D6h)+({\x-10}:4mm)$) {}; + \node[round] at ($(D6h)+({\x+10}:4mm)$) {}; + \node[round] at ($(D6h)+({\x-10}:4mm)$) {}; + } + \end{tikzpicture} + } + \begin{frame}[fragile]{} + \end{frame} +} + +\section{Anwendungen} +\begin{frame}[fragile]{} + \centering + \begin{tikzpicture}[ + box/.style = { + rectangle, thick, draw = white, fill = darkgray!50!background, + minimum height = 1cm, outer sep = 2mm, + }, + ] + + \matrix [nodes = {box, align = center}, column sep = 1cm, row sep = 1.5cm] { + & \node (A) {32 Kristallklassen}; \\ + \node (B) {11 Mit\\ Inversionszentrum}; & \node (C) {21 Ohne\\ Inversionszentrum}; \\ + & \node[fill=red!20!background] (D) {20 Piezoelektrisch}; & \node (E) {1 Nicht\\ piezoelektrisch}; \\ + }; + + \draw[thick, ->] (A.west) to[out=180, in=90] (B.north); + \draw[thick, ->] (A.south) to (C); + \draw[thick, ->] (C.south) to (D.north); + \draw[thick, ->] (C.east) to[out=0, in=90] (E.north); + \end{tikzpicture} +\end{frame} + +\begin{frame}[fragile]{} + \begin{tikzpicture}[ + overlay, xshift = 1.5cm, yshift = 1.5cm, + node distance = 2mm, + charge/.style = { + circle, draw = white, thick, + minimum size = 5mm + }, + positive/.style = { fill = red!50 }, + negative/.style = { fill = blue!50 }, + ] + + \node[font = {\large\bfseries}, align = center] (title) at (5.5,0) {Mit und Ohne\\ Symmetriezentrum}; + \pause + + \begin{scope} + \matrix[nodes = { charge }, row sep = 8mm, column sep = 8mm] { + \node[positive] {}; & \node[negative] (N) {}; & \node [positive] {}; \\ + \node[negative] (W) {}; & \node[positive] {}; & \node [negative] (E) {}; \\ + \node[positive] {}; & \node[negative] (S) {}; & \node [positive] {}; \\ + }; + \draw[gray, dashed] (W) to (N) to (E) to (S) to (W); + \end{scope} + \pause + + \begin{scope}[xshift=11cm] + \foreach \x/\t [count=\i] in {60/positive, 120/negative, 180/positive, 240/negative, 300/positive, 360/negative} { + \node[charge, \t] (C\i) at (\x:1.5cm) {}; + } + + \draw[white] (C1) to (C2) to (C3) to (C4) to (C5) to (C6) to (C1); + \node[circle, draw=gray, fill=gray, outer sep = 0, inner sep = 0, minimum size = 3mm] {}; + % \draw[gray, dashed] (C2) to (C4) to (C6) to (C2); + \end{scope} + \pause + + %% + \node[below = of title] {Polarisation Feld \(\vec{E}_p\)}; + + %% hex with vertical pressure + \begin{scope}[xshift=11cm, yshift=-4.5cm] + \node[charge, positive, yshift=-2.5mm] (C1) at ( 60:1.5cm) {}; + \node[charge, negative, yshift=-2.5mm] (C2) at (120:1.5cm) {}; + \node[charge, positive, xshift=-2.5mm] (C3) at (180:1.5cm) {}; + \node[charge, negative, yshift= 2.5mm] (C4) at (240:1.5cm) {}; + \node[charge, positive, yshift= 2.5mm] (C5) at (300:1.5cm) {}; + \node[charge, negative, xshift= 2.5mm] (C6) at (360:1.5cm) {}; + + \draw[white] (C1) to (C2) to (C3) to (C4) to (C5) to (C6) to (C1); + % \draw[gray, dashed] (C2) to (C4) to (C6) to (C2); + + \foreach \d in {C1, C2} { + \draw[orange, very thick, <-] (\d) to ++(0,.7); + } + + \foreach \d in {C4, C5} { + \draw[orange, very thick, <-] (\d) to ++(0,-.7); + } + + \node[white] (E) {\(\vec{E}_p\)}; + \begin{scope}[node distance = .5mm] + \node[red!50, right = of E] {\(+\)}; + \node[blue!50, left = of E] {\(-\)}; + \end{scope} + % \draw[gray, thick, dotted] (E) to ++(0,2); + % \draw[gray, thick, dotted] (E) to ++(0,-2); + \end{scope} + \pause + + %% square with vertical pressure + \begin{scope}[yshift=-4.5cm] + \matrix[nodes = { charge }, row sep = 5mm, column sep = 1cm] { + \node[positive] (NW) {}; & \node[negative] (N) {}; & \node [positive] (NE) {}; \\ + \node[negative] (W) {}; & \node[positive] {}; & \node [negative] (E) {}; \\ + \node[positive] (SW) {}; & \node[negative] (S) {}; & \node [positive] (SE) {}; \\ + }; + + \foreach \d in {NW, N, NE} { + \draw[orange, very thick, <-] (\d) to ++(0,.7); + } + + \foreach \d in {SW, S, SE} { + \draw[orange, very thick, <-] (\d) to ++(0,-.7); + } + + \draw[gray, dashed] (W) to (N) to (E) to (S) to (W); + \end{scope} + \pause + + %% hex with horizontal pressure + \begin{scope}[xshift=5.5cm, yshift=-4.5cm] + \node[charge, positive, yshift= 2.5mm] (C1) at ( 60:1.5cm) {}; + \node[charge, negative, yshift= 2.5mm] (C2) at (120:1.5cm) {}; + \node[charge, positive, xshift= 2.5mm] (C3) at (180:1.5cm) {}; + \node[charge, negative, yshift=-2.5mm] (C4) at (240:1.5cm) {}; + \node[charge, positive, yshift=-2.5mm] (C5) at (300:1.5cm) {}; + \node[charge, negative, xshift=-2.5mm] (C6) at (360:1.5cm) {}; + + \draw[white] (C1) to (C2) to (C3) to (C4) to (C5) to (C6) to (C1); + % \draw[gray, dashed] (C2) to (C4) to (C6) to (C2); + + \draw[orange, very thick, <-] (C6) to ++(.7,0); + \draw[orange, very thick, <-] (C3) to ++(-.7,0); + + \node[white] (E) {\(\vec{E}_p\)}; + \begin{scope}[node distance = .5mm] + \node[blue!50, right = of E] {\(-\)}; + \node[red!50, left = of E] {\(+\)}; + \end{scope} + % \draw[gray, thick, dotted] (E) to ++(0,2); + % \draw[gray, thick, dotted] (E) to ++(0,-2); + \end{scope} + \pause + + + \end{tikzpicture} +\end{frame} + +\frame{ + \frametitle{Licht in Kristallen} + \begin{columns}[T] + \begin{column}{.45\textwidth} + \onslide<2->{ + Helmholtz Wellengleichung + \[ + \nabla^2 \vec{E} = \ten{\varepsilon}\mu + \frac{\partial^2}{\partial t^2} \vec{E} + \] + } + \onslide<3->{ + Ebene Welle + \[ + \vec{E} = \vec{E}_0 \exp\left[i + \left(\vec{k}\cdot\vec{r} - \omega t \right)\right] + \] + } + \onslide<4->{ + Anisotropisch Dielektrikum + \[ + (\ten{K}\ten{\varepsilon})\vec{E} + = \frac{k^2}{\mu \omega^2} \vec{E} + \implies + \Phi \vec{E} = \lambda \vec{E} + \] + } + \end{column} + \begin{column}{.55\textwidth} + \onslide<5->{ + Eingenraum + \begin{align*} + U_\lambda &= \left\{ v : \Phi v = \lambda v \right\} + = \mathrm{null}\left(\Phi - \lambda I\right) + \end{align*} + }\onslide<6->{ + Symmetriegruppe und Darstellung + \begin{align*} + G &= \left\{\mathbb{1}, r, \sigma, \dots \right\} \\ + &\Phi : G \to O(n) + \end{align*} + }\onslide<7->{ + Kann man \(U_\lambda\) von \(G\) herauslesen? + \only<7>{ + \[ + U_\lambda \stackrel{?}{=} f\left(\bigoplus_{g \in G} \Phi_g\right) + \] + }\only<8>{ + \begin{align*} + \mathrm{Tr}\left[\Phi_r(g)\right] + &= \sum_i n_i \mathrm{Tr}\left[\Psi_i(g)\right] \\ + |G| &= \sum_i\mathrm{Tr}\left[\Psi_i(\mathbb{1})\right] + \end{align*} + } + } + \end{column} + \end{columns} +} + +% \begin{frame}[fragile] +% \centering +% \tdplotsetmaincoords{70}{110} +% \begin{tikzpicture}[scale=2, tdplot_main_coords] +% \node[draw=white, thick, minimum size = 3cm, circle] {}; +% % \foreach \x in {0, 120, 240} { +% % } +% \end{tikzpicture} +% \end{frame} + + +\end{document} |
