1a061c9fbe824a468f6aa9652a43b43771af8b78 — Jack Halford 1 year, 3 months ago
first commit
A  => .gitignore +3 -0
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A  => Makefile +43 -0
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SHELL := bash
.SHELLFLAGS := -eu -o pipefail -c

MD	=	$(shell ls -1 md/*.md)
GRAPH	=	$(subst md/,graph/,$(MD))
REFS	=	$(subst md/,external_refs/,$(MD))
HTML	=	$(subst md/,html/,$(MD:.md=.html))

all: $(HTML) html/colors.css html/graph.svg $(REFS)

html/%.html: md/%.md external_refs/%.md
> @mkdir -p html
> @pandoc	--filter tools/pf-filter.py \
		-H tools/header.html \
		--template tools/template.html \
		--mathjax -f markdown \
		tools/latex-includes.yaml \
		$^ -o $@ &
> @printf '$@\n'

external_refs/%.md: graph/%.md
> @mkdir -p external_refs
> @[ -s $^ ] && echo '## See also' > $@
> @awk '{printf "- [%s](%s)\n", $$1, $$1;}' $^>>$@

graph/%.md: md/%.md
> @mkdir -p graph
> @grep $(notdir $@) md/*.md | awk -F: '{print $$1}' | uniq | sed s:md/::g | grep -v lexicon > $@ || true

html/graph.svg: $(GRAPH)
> dot -Tsvg -o $@ <(./tools/generate-graph.sh)

html/colors.css: tools/colors
> ./tools/generate-bg-css.sh >$@

> rm -rf html graph external_refs

re: clean all

.PHONY: clean all re

A  => md/angular-momentum.md +16 -0
@@ 1,16 @@
title: angular momentum
type: object

$$ \delta \mathbf{r}=\delta\mathbf{ϕ}\times\mathbf{r} $$
$$ \delta \mathbf{v}=\delta\mathbf{ϕ}\times\mathbf{v} $$
$$ \mathbf{M}\equiv\sum_a\mathbf{r}_a\times\mathbf{p}_a $$
$$ \mathbf{M}=\mathbf{M}'+a\times\mathbf{P} $$
$$ \mathbf{M}=\mathbf{M}'+\mu \mathbf{R}\times\mathbf{V} $$
$$ \mathbf{M}=\mathbf{M}'+\mathbf{R}\times\mathbf{P} $$
$$ M_z=\sum_a\frac{\partial L}{\partial \dot{\phi_a}} $$
$$ M_z=\sum_a m_ar_a^2\dot{\phi_a} $$

## Assumptions
 - [Classical mechanics](mechanical-system.md)

A  => md/centre-of-mass.md +13 -0
@@ 1,13 @@
title: centre of mass
type: object

\begin{equation} \mathbf{P}=\mathbf{P}'+\mathbf{V}\sum_a m_a \end{equation}
\begin{equation} \mathbf{V}=\mathbf{P}/\sum_a m_a=\sum_a m_a\mathbf{v}_a/\sum_a m_a \end{equation}
\begin{equation} \mathbf{R}\equiv\sum_a m_a\mathbf{r}_a/\sum_a m_a \end{equation}
\begin{equation} E=\frac{1}{2}\mu V^2+E_i \end{equation}
\begin{equation} E=E'+\mathbf{V}\cdot\mathbf{P}'+\frac{1}{2}\mu V^2 \end{equation}

## Model
 - [Classical mechanics](mechanical-system.md)

A  => md/coplanar-double-pendulum.md +16 -0
@@ 1,16 @@
title: coplanar double pendulum
type: problem

## Solution

$$ L=\mfrac{1}{2}(m_1+m_2)l_1^2\phi_1^2

## Model
 - [Classical mechanics](mechanical-system.md)

A  => md/disintegration.md +17 -0
@@ 1,17 @@
title: disintegration
type: model

\begin{equation} \epsilon=E_i-E_{1i}-E_{2i} \end{equation}
\begin{equation} \epsilon=\frac{1}{2}p_0^2\left(\frac{1}{m_1}+\frac{1}{m_2}\right)=\frac{p_0^2}{2m} \end{equation}
\begin{equation} v^2+V^2-2vV\cos\theta=v_0^2 \end{equation}
\begin{equation} \sin\theta_\text{max}=v_0/V \end{equation}
\begin{equation} \tan\theta=v_0\sin\theta_0/(v_0\cos\theta_0+V) \end{equation}
\begin{equation} \cos\theta_0=-\mfrac{V}{v_0}\sin^2\theta\pm\cos\theta\sqrt{1-\mfrac{V^2}{v_0}\sin^2\theta} \end{equation}
\begin{equation} \frac{1}{2}\sin\theta_0d\theta_0 \end{equation}
\begin{equation} \frac{1}{2}mv_0VdT \end{equation}
\begin{equation} T_{10,\text{max}}=(M-m_1)\epsilon/M \end{equation}

## Assumptions
 - [Two body problem](two-body-problem.md)

A  => md/elastic-collisions.md +30 -0
@@ 1,30 @@
title: elastic collisions
type: model

$$ \mathbf{v}_{10}'=m_2v\mathbf{n}_0/(m_1+m_2),\qquad \mathbf{v}_{20}'=-m_1v\mathbf{n}_0/(m_1+m_2) $$



\begin{equation} \tan\theta_1=\frac{m_2\sin\chi}{m_1+m_2\cos\chi},\qquad \theta_2=\mfrac{1}{2}(\mfrac{\tau}{2}-\chi) \end{equation}
\begin{equation} \mathbf{v}_1'=\frac{m_1-m_2}{m_1+m_2}\mathbf{v},\qquad \mathbf{v}_2'=\frac{2m_1}{m_1+m_2}\mathbf{v} \end{equation}
\begin{equation} {{E_2}'}_\text{max}=\frac{1}{2}m_2{{{v_2}'}_\text{max}}^2=\frac{4m_1m_2}{(m_1+m_2)^2}E_1 \end{equation}
\begin{equation} \sin\theta_\text{max}=\text{OC/OA}=m_2/m_1 \end{equation}
\begin{equation} \theta_1=\mfrac{1}{2}\chi,\qquad \theta_2=\mfrac{1}{2}(\mfrac{\tau}{2}-\chi) \end{equation}
\begin{equation} v_1'=v\cos\mfrac{1}{2}\chi,\qquad v_2'=v\sin\mfrac{1}{2}\chi \end{equation}

## System
 - [Two body problem](two-body-problem.md)

A  => md/energy.md +11 -0
@@ 1,11 @@
title: energy
type: object

$$\tag{6.1} E\equiv\sum_i\dot{q_i}\frac{\partial L}{\partial \dot{q_i}}-L $$
$$\tag{6.2} E=T(q,\dot{q})+U(q) $$
$$\tag{6.3} E=\sum_a\frac{1}{2}m_av_a^2+U(\mathbf{r}_1,\mathbf{r}_2,\ldots) $$

## Model
 - [Classical mechanics](mechanical-system.md)

A  => md/galileo-relativity.md +8 -0
@@ 1,8 @@
title: Galileo's relativity
type: object

\begin{equation} L=L(v^2) \end{equation}
\begin{equation} \mathbf{v}=\text{constant} \end{equation}
\begin{equation} t=t' \end{equation}

A  => md/hamiltonian-operator.md +12 -0
@@ 1,12 @@
title: the hamiltonian operator
type: object

The [wave function](wave-function.md) completely determines the system, and also the future states of the system. This means that the derivative $\partial\Psi/\partial t$ must be determined by the function itself at an instant, and, by the principle of superpostion, the relationship must be linear

$$ i\partial\Psi/\partial t=\qop{L}\Psi $$

TODO blablabla

$$ \lltag{3}{8.1} i\hbar\partial\Psi/\partial t=\qop{H}\Psi $$

A  => md/hermitian-eigenfunctions.md +22 -0
@@ 1,22 @@
type: problem
title: hermitian eigenfunctions

We have seen that real physical quantities are represented by hermitian linear integral [operators](quantum-operator.md) $(3.15)$. Show that the eingenfunctions of these operators are orthogonal.

## solution

Let $f_n$ and $f_m$ be two different eigenvalues of the quantity $f$, and $\Psi_n$, $\Psi_m$ the corresponding eigenfunctions

$$ \hat{f}\Psi_n=f_n\Psi_n,\qquad \hat{f}\Psi_m=f_m\Psi_m $$

Multiplying the first equation by $\Psi_m*$, and multiplying the conjugate of the second by $\Psi_n$ we have

$$ \Psi_m^*\hat{f}\Psi_n-\Psi_n\hat{f}\vphantom{f}^*\Psi_m^*=(f_n-f_m)\Psi_n\Psi_m^* $$

integrating over q, and using the fact that $\hat{f}$ is Hermitian $\hat{f}\equiv\hat{f}\vphantom{f}^*$ so that the left side is zero, we have

$$ (f_n-f_m)\int\Psi_n\Psi_m\vphantom{f}^*\dd{q}=0 $$

and since $f_n\neq f_m$, we obtain the orthogonality property.

A  => md/inertia-tensor.md +38 -0
@@ 1,38 @@
title: inertia tensor
type: object

\begin{equation}\tag{32.1} T=\mfrac{1}{2}\mu V^2+\mfrac{1}{2}\sum m\left[\Omega^2r^2-(\mathbf{\Omega}\cdot\mathbf{r})^2\right] \end{equation}
\begin{equation}\tag{32.2} I_{ik}=\sum m\left(x_l^2\delta_{ik}-x_ix_k\right) \end{equation}
\begin{equation}\tag{32.3} T=\mfrac{1}{2}\mu V^2+\mfrac{1}{2}I_{ik}\Omega_i\Omega_k \end{equation}
\begin{equation}\tag{32.4} L=\mfrac{1}{2}\mu V^2+\mfrac{1}{2}I_{ik}\Omega_i\Omega_k-U \end{equation}
\begin{equation}\tag{32.5} I_{ik}=I_{ki} \end{equation}
\begin{equation}\tag{32.6} I_{ik}=\left[\begin{matrix}
	    \sum m(y^2+z^2) & -\sum mxy & -\sum mxz\\
	    -\sum mxy & \sum m(x^2+z^2) & -\sum myz\\
	    -\sum mxz & -\sum myz & \sum m(x^2+y^2)\\

\begin{equation}\tag{32.7} I_{ik}=\int \rho\left(x_l^2\delta_{ik}-x_ix_k\right)\dd{V} \end{equation}

\begin{equation}\tag{32.12} I'_{ik}=I_{ik}+\mu\left(a^2\delta_{ik}-a_ia_k\right) \end{equation}

### Principle moments of inertia

\begin{equation}\tag{32.8} T_\text{rot}=\mfrac{1}{2}\left(I_1\Omega_1^2+I_2\Omega_2^2+I_3\Omega_3^2\right) \end{equation}
\begin{equation}\tag{32.9} I_1+I_2=\sum m(x_1^2+x_2^2+2x_3^2) \geq \sum m(x_1^2+x_2^2)=I_3 \end{equation}

#### Coplanar system of particles
\begin{equation}\tag{32.10} I_3=I_1+I_2 \end{equation}

#### Colinear system of particles
\begin{equation}\tag{32.11} I_1=I_2=\sum m{x_3}^2,\quad I_3=0 \end{equation}

## Model
 - [Rigid body](rigid-body.md)

## Properties
 - rank 2 tensor
 - antisymetric 

A  => md/lagrangian.md +18 -0
@@ 1,18 @@
title: Lagrangian
type: object

The Lagrangian is defined by the principle of least action

$$\tag{2.1} S=\int_{t_1}^{t_2}L(q,\dot{q},t)\dd{t} $$
$$\tag{2.2} q(t)+\delta q(t) $$
$$\tag{2.3} \delta q(t_1)=\delta q(t_2) = 0 $$
$$\tag{2.4} \delta S=\delta \int_{t_1}^{t_2}L(q,\dot{q},t)\dd{t}=0 $$
$$\tag{2.5} \delta S=\left[\frac{\partial L}{\partial \dot{q}}\delta q\right]_{t_1}^{t_2}+\int_{t_1}^{t_2}\left(\frac{\partial L}{\partial q}-\frac{\dd{}}{\dd{t}}\frac{\partial L}{\partial \dot{q}}\right)\delta q \dd{t}=0 $$
$$\tag{2.6} \frac{\dd{}}{\dd{t}}\frac{\partial L}{\partial \dot{q_i}}-\frac{\partial L}{\partial q_i}=0\qquad(i=1,2,...,s) $$
$$\tag{2.7} \lim L=L_A+L_B $$
$$\tag{2.8} L'(q,\dot{q},t)=L(q,\dot{q},t)+\frac{\dd{}}{\dd{t}}f(q,t) $$
## Model
 - [Classical mechanics](mechanical-system.md)

A  => md/lexicon.md +9 -0
@@ 1,9 @@
title: Lexicon

degrees of freedom: the number of co-ordinates needed to _completely_ define the position of a system.  
generalised co-ordinates: Any $s$ quantities $q_1, q_2, ..., q_s$ which completely define the system the position of a system are  
generalised velocities: The derivatives $\dot{q}_1, \dot{q}_2, ..., \dot{q}_s$ of the _generalised co-ordinates_   
particle: a body whose dimension is neglected.  
rigid body: system of [particles](particle.md) such that the distances between the particles does not vary.  

A  => md/mechanical-similarity.md +22 -0
@@ 1,22 @@
title: mechanical similarity
type: object

\begin{equation} U(\alpha\mathbf{r}_1,\alpha\mathbf{r}_2,\ldots,\alpha\mathbf{r}_n)=\alpha^kU(\mathbf{r}_1,\mathbf{r}_2,\ldots,\mathbf{r}_n) \end{equation}
\begin{equation} t'/t=(l/l)^{1-k/2} \end{equation}


\begin{equation} 2T=\sum_a\mathbf{p}_a\mathbf{v}_a=\frac{\dd{}}{\dd{t}}(\sum_a\mathbf{p}_a\mathbf{r}_a)-\sum_a\mathbf{r}_a\dot{\mathbf{p}}_a \end{equation}
\begin{equation} 2\overline{T}=\overline{\sum_a\mathbf{r}_a\cdot\partial U/\partial \mathbf{r}_a} \end{equation}
\begin{equation} 2\overline{T}=k\overline{U} \end{equation}
\begin{equation} \overline{U}=2E/(k+2),\qquad \overline{T}=kE/(k+2) \end{equation}

## Assumptions
 - The potential energy is a _homogenous_ function of co-ordinates
 - [Classical mechanics](mechanical-system.md)

A  => md/mechanical-system.md +20 -0
@@ 1,20 @@
type: model
title: mechanical system

$$ L=\sum_a\frac{1}{2}m_av_a^2-U(\mathbf{r}_1,\mathbf{r}_2,\ldots) $$
$$ \frac{\dd{}}{\dd{t}}\frac{\partial L}{\partial \mathbf{v}_a}=\frac{\partial L}{\partial \mathbf{r}_a} $$
$$ m_a\dd{\mathbf{v}_a}/\dd{t}=-\partial U/\partial \mathbf{r}_a $$
$$ \mathbf{F}=-\partial U/\partial \mathbf{r}_a $$
$$ L=\sum_{i,k}a_{ik}(q)\dot{q_i}\dot{q_k}-U(q) $$
$$ L=\frac{1}{2}mv^2-U(\mathbf{r}, t) $$
$$ m\dot{\mathbf{v}}=-\partial U/\partial \mathbf{r} $$
$$ U=-\mathbf{F}\cdot\mathbf{r} $$

## Assumption
 - [The Principle of Least Action](principle-of-least-action.md)
 - [Galileo's Relativity](galileo-relativity.md)

## Model
 - [particles](particle.md)

A  => md/momentum.md +15 -0
@@ 1,15 @@
title: momentum
type: object

$$ \sum_a\partial L/\partial \mathbf{r}_a=0 $$
$$ \mathbf{P}\equiv\sum_a\partial L/\partial \mathbf{v}_a $$
$$ \mathbf{P}=\sum_am_a\mathbf{v}_a $$
$$ \sum_a\mathbf{F}_a=0 $$
$$ p_i=\partial L/\partial \dot{q}_i $$
$$ F_i=\partial L/\partial q_i $$
$$ \dot{p}_i=F_i $$

## Model
 - [Classical mechanics](mechanical-system.md)

A  => md/particle.md +16 -0
@@ 1,16 @@
title: particle
type: model

- A system of $N$ particles has $3N$ degrees of freedom.
- If all the co-ordinates and velocities are specified, the mechanical state of the system is determined.

### For a free particle

$$\tag{4.1} L=\frac{1}{2}mv^2 $$
$$\tag{4.2} L=\sum\frac{1}{2}m_av_a^2 $$
$$\tag{4.3} v=(\dd{l}/\dd{t})^2=(\dd{l})^2/(\dd{t})^2 $$
$$\tag{4.4} \dd{l}^2=\frac{1}{2}(\dot{x}^2+\dot{y}^2+\dot{z}^2) $$
$$\tag{4.5} \dd{l}^2=\frac{1}{2}(\dot{r}^2+r^2\dot{\phi}^2+\dot{z}^2) $$
$$\tag{4.6} \dd{l}^2=\frac{1}{2}(\dot{r}^2+r^2\dot{\theta}^2+r^2\dot{\phi}^2\sin^2\theta) $$

A  => md/quantum-operator.md +204 -0
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type: object
title: quantum operators

Real physical quantities $f$ are contained in the [wave function](wave-function.md)  $\Psi$, measurement is done by the application of an operator $(\qop{f}\Psi)$. The values that are taken by a physical quantities are eigenvalues $f_n$ of it's operator. The set of eigenvalues form a "spectrum". The spectrum can be discrete (e.g. energy) or continous (e.g. position) or a mix of both (see anharmonic oscillator). To each eingenvalue is associate a eigen-wavefunction $\Psi_n$ which is also normalized.

$$ \lltag{3}{3.1} \int|\Psi_n|^2\dd{q}=1 $$

If a measurement $\qop{f}$ is carried out on a system $\Psi$, the result will be one of $f_n$. The wavefunction must be a linear combination of the eigen-wavefunctions. (see principle of superposition).

$$ \lltag{3}{3.2} \Psi=\sum a_n\Psi_n $$

A wavefunction can be expanded in terms of eigenfunction of any physical quantity. These $a_n$ represent the probability of a given eigenfunction which are $|a_n|^2$. We also must have unit probability over the eigenfunction.

$$ \lltag{3}{3.3} \sum_n|a_n|^2=1 $$

This relation doesn't hold if $\Psi$ is not normalized. The sum $\sum|a_n|^2$ must be bilinear in $\Psi, \Psi^*$, and become unit when $\Psi$ is normalized, thus we have

$$ \lltag{3}{3.4} \sum_n a_na_n^*=\int\Psi\Psi^*\dd{q} $$

we can find a simple form for $a_n$

\Psi^*&=\sum a_n^*\Psi_n^* &&\text{conjugate of $(3.2)$}\\ 
\int\Psi\Psi^*\dd{q}&=\sum a_n^*\int\Psi_n^*\Psi\dd{q} &&\text{by substituting $\Psi^*$}\\
\sum_n a_na_n^*&=\sum a_n^*\int\Psi_n^*\Psi\dd{q} &&\text{replacing with $(3.4)$}

$$ \lltag{3}{3.5} a_n=\int\Psi\Psi_n^*\dd{q} $$

if we substitute $(3.2)$ we find $a_n=\sum a_m\int\Psi_m\Psi_n^*\dd{q}$, from which it is evident the eigenfunctions must satisfy be an orthogonal set

$$ \lltag{3}{3.6} \int\Psi_m\Psi_n^*\dd{q}=\delta_{nm} $$

Thus the spectrum of eigenfunctions is orthonormal.

The mean value $\overline{f}$ of a physical quantity $f$, with the usual definition using wighted probabilities we get

$$ \lltag{3}{3.7} \overline{f}=\sum_nf_n|a_n|^2 $$

Let $(\qop{f}\Psi)$ be the result of the operator acting on the function $\Psi$. We define $\qop{f}$ in such a way that

$$ \lltag{3}{3.8} \overline{f}=\int \Psi^*(\qop{f}\Psi)\dd{q} $$

In general $\qop{f}$ is linear, if we plug $(3.5)$ in $(3.7)$ we find $\overline{f}=\sum_n f_na_na_n^*=\int\Psi(\sum_n a_nf_n\Psi_n)\dd{q}$

Comparing with $(3.8)$ we get that the effect of $\qop{f}$ on $\Psi$ is

$$ \lltag{3}{3.9} (\qop{f}\Psi)=\sum_nf_na_n\Psi_n $$

If we substitute in $(3.5)$ for $a_n$, we find that $\qop{f}$ is an integral operator of the form

$$ \lltag{3}{3.10} (\qop{f}\Psi)=\int K(q,q')\Psi(q')\dd{q} $$

where the kernel function $K(q,q')$ is

$$ \lltag{3}{3.11} K(q,q')=\sum_n f_n\Psi_n^*(q')\Psi_n(q) $$

Thus, for every physical quantity there is a definite linear integral operator. From $(3.9)$ we see that if $\Psi$ is one the the eigenfunction $\Psi_n$, then

$$ \lltag{3}{3.12} \qop{f}\Psi=f_n\Psi_n $$

The values taken by physical quantities are necessarily real, hence the mean must also be real, in any state. And if the mean is real, then all of the eigenvalues must be real, because the mean values coincidence with the eigenvalues in the states $\Psi_n$. Because the mean is real, we can equate $(3.8)$ to it's conjugate form

$$ \lltag{3}{3.13} \int\Psi^*(\qop{f}\Psi)\dd{q}=\int\Psi(\qop[*]{f}\Psi^*)\dd{q} $$

This doens't hold in general for any linear operator $\qop{f}$, it is a restriction for physical quantities.

For an arbitrary $\qop{f}$ we can find the transposed operator $\qop{f}\qop[t]{f}$ such that

$$ \lltag{3}{3.14} \int\Psi(\qop{f}\Phi)\dd{q}\equiv\int\Phi(\qop[t]{f}\Psi)\dd{q} $$

where $\Psi$ and $\Phi$ are different, if we take $\Phi=\Psi^*$ then from $(3.13)$ we have

$$ \lltag{3}{3.15} \qop[t]{f}=\qop[*]{f} $$

Thus, the operators that represent physical quantities must be Hermitian.

If we consider a complex physical quantity $f$ (need example here), it's complex conjugate is $f^*$, whose eigenvalues are the complex conjugate of those of $f$. We denote $\qop[+]{f}$ the operator corresponding to the physical quantity $f^*$. It is the Hermitian conjugate of $\qop{f}$ and in general is different from $\qop[*]{f}$: from the condition $\overline{f^*}=\overline{f}^*$ we find that

$$ \lltag{3}{3.16} \qop[+]{f}=\qop[*t]{f} $$

### Addition of operators

Let $f$ and $g$ be two physical quantities. The eigenvalues of the sum $f+g$ are equal to the sums of the eigenvalues of $f$ and $g$. The quantity $f+g$ is represented by the operator $\qop{f}+\qop{g}$. Sometimes $f$ and $g$ can't take definite values at the same time, in this case we define the mean of the sum

$$ \lltag{3}{4.1} \overline{f+g}=\overline{f}+\overline{g} $$

In this case, the eigenvalues of the new operator $\qop{f}+\qop{g}$ are real valued, but they don't bear any more relation to those of the quantities $f$ and $g$ separately.

Let $f_0$ and $g_0$ be the smallest eigenvalues of the quantities $f$ and $g$, and $(f+g)_0$ of the quantity $f+g$, then

$$ \lltag{3}{4.2} (f+g)_0\geq f_0+g_0 $$

(need Hilbert algebra for proof of this)

### Product of operators

Let $f$ and $g$ be quantities that can be measured simultaneously. The product $\qop{f}\qop{g}$, it is the successive application of $\qop{g}$, then $\qop{f}$. If $\Psi_n$ are eigenfunctions common to $\qop{f}$ and $\qop{g}$ we hav $\qop{f}\qop{g}\Psi=\qop{f}g_n\Psi_n=g_n\qop{f}\Psi_n=g_nf_n\Psi_n$, we could have equally taken the operator $\qop{g}\qop{f}$. Sinec $\Psi$ can always be written as a linear combination of $\Psi_n$, it follows that $\qop{f}\qop{g}$ is the same as $\qop{g}\qop{f}$.

$$ \lltag{3}{4.3} \qop{f}\qop{g}-\qop{g}\qop{f}=0 $$

Thus, we arrive at this important result: if two quantities $f$ and $g$ can simultaneously take definite values, then their operators $\qop{f}$ and $\qop{g}$ commute. (there is also proof of the converse in par11).

Let us consider an operator raised to a power $p$. From the above we deduce that the eigenvalues of $\qop[p]{f}$ are $f_n^p$. Any function $\Phi(\qop{f})$ of an operator can be defined as an operator whose eigenvalues are equal to the same function $\Phi(f)$ of the eigenvalues of the operator $\qop{f}$.

The inverse operator $\qop[-1]{f}$ is such that $\qop{f}\qop[-1]{f}=\qop[-1]{f}\qop{f}=1$.

If the quantities $f$ and $g$ cannot simultaneously take definite values, the product cannot be defined as above. This appears in the fact that $\qop{f}\qop{g}$ is not self-conjugate, so it cannot represent a physical quantity. By definition of the transpose we have


from which we have directly

$$ \lltag{3}{4.4} (\qop{f}\qop{g})^t=\qop[t]{g}\qop[t]{f} $$

Taking the complex conjugate on both sides we have

$$ \lltag{3}{4.5} (\qop{f}\qop{g})^+=\qop[+]{g}\qop[+]{f} $$

If both $\qop{f}$ and $\qop{g}$ are Hermitian, then $(\qop{f}\qop{g})^+=\qop{g}\qop{f}$. It follows that the operator $\qop{f}\qop{g}$ is Hermitian if and only if the factors $\qop{f}$ and $\qop{g}$ commute.

We note that for non commuting operators, we can form an Hermitian operator by taking the symmetrical combination

$$ \lltag{3}{4.6} \mfrac{1}{2}(\qop{f}\qop{g}+\qop{g}\qop{f}) $$

The difference $\qop{f}\qop{g}-\qop{g}\qop{f}$ is an anti-Hermitian operator. It can be made Hermitian by mutlplying by $i$

$$ \lltag{3}{4.6} i(\qop{f}\qop{g}-\qop{g}\qop{f}) $$

For brevity we note the commutator

$$ \lltag{3}{4.7} \{\qop{f},\qop{g}\}=\qop{f}\qop{g}-\qop{g}\qop{f} $$

It is easily seen that 

$$ \lltag{3}{4.8} \{\qop{f}\qop{g},\qop{h}\}=\{\qop{f},\qop{h}\}\qop{g}-\qop{f}\{\qop{g},\qop{h}\} $$

And we notice that if $\{\qop{f},\qop{h}\}=0$ and $\{\qop{g},\qop{h}\}=0$, it does not in general follow that $\qop{f}$ and $\qop{g}$ commute, i.e. commutation is not transitive. (math example is easy to find, need an example here with a physical meaning).

### The continuous spectrum

We can generalize the above results for operator an operator $\qop{f}$ with a continuous spectrum. We shall denote it's eigenvalues $f$, without suffix, because they take a continuous range of values. We note $\Psi_f$ the eigenfunction corresponding to the eigenvalue $f$. Just as $\Psi$ can be exanded in a series $(3.2)$ of eingenfunctions, it can also be expanded in terms of of the complete set of eigenfunctions of a quantity with a continuous spectrum as an integral.

$$ \lltag{3}{5.1} \Psi(q)=\int a_f\Psi_f(q)\dd{f} $$

where the integration is taken over the whole range of values that can be taken by the quantity $f$.

The subject of normalisation of the eigenfunctions of a continuous spectrum is more complex than that of a discrete spectrum. We don't try to normalize the square modulus of the wavefunction, instead we normalize so that the $|a_f|^2\dd{f}$ is the probability that the physical quantity has a value between $f$ and $f+\dd{f}$. Since the sum of all probabilities must equal unity, we have

$$ \lltag{3}{5.2} \int|a_f|^2\dd{f}=1 $$

In the same way we found $(3.5)$, we can write

\int\Psi\Psi^*\dd{q}&=\int\int a_f^*\Psi_f^*\Psi\dd{f}\dd{q} &&\text{by substituting $\Psi^*$ from $(5.1)$}

By comparing these two we find the expression for the expansion coefficients

$$ \lltag{3}{5.3} a_f=\int\Psi(q)\Psi_f^*(q)\dd{q} $$

in exact analogy to $(3.5)$.

To derive the normalisation condition, we substitute $(5.1)$ in $(5.3)$

$$ a_f=\int a_{f'}(\Psi_{f'}\Psi_f^*\dd{q})\dd{f'} $$

This relation must hold for any $a_f$. The only solution is

$$ \lltag{3}{5.4} \int\Psi_{f'}\Psi_f^*\dd{q}=\delta(f'-f) $$

This gives the normalisation rule for the eigenfunctions; replacing condition $(3.6)$. Similarly, we have $\Psi_f$ and $\Psi_{f'}$ orthogonal for $f\neq f'$. However the integrals of $|\Psi_f|^2$ diverge for a continuous system. The eigenfunctions satisfy another relation by subsituting the other way around

$$ \Psi(q)=\int\Psi(q')\left(\int\Psi_f^*(q')\Psi_f(q)\dd{f}\right)\dd{q'} $$

From which we deduce immediately

$$ \lltag{3}{5.11} \int\Psi_f^*(q')\Psi_f(q)\dd{f}=\delta(q-q') $$

The analogous for a discrete spectrum is

$$ \lltag{3}{5.12} \sum_n\Psi_n^*(q')\Psi_n(q)\dd{f}=\delta(q-q') $$

$(5.1)$ and $(5.3)$ are analogous: $\Psi(q)$ can be expanded in terms of the functions $\Psi_f(q)$ with expansion coefficients $a_f$, or otherwise we can expand $a_f\equiv a(f)$ in terms of the functions $\Psi_f^*(q)$ while $\Psi(q)$ play the expansion coefficients. The function $a(f)$, like $\Psi(q)$ completely determines the state of the system. Just as $|\Psi(q)|^2$ determines the probability for the system to have coordinates lying in an interval $\dd{q}$, so $|a(f)|^2$ determines the probability for the values of the quantity $f$ to lie in a given interval $\dd{f}$.

(TODO: 5.13 -> 5.17)

### Differentiating operators with respect to time

The classical definition of a time derivative doesn't hold in quantum mechanics. In quantum mechanics, if a quantity has a value at one instant, it does not in general have a definite value at subsequent instants. However it is still natural to define the derivative $\dot{f}$ of a quantity $f$ as the quantity whose mean value is equal to the derivative of the mean value $\bar{f}$

$$ \lltag{3}{9.1} \overline{\dot{f}}=\dot{\overline{f}} $$

by definition we have the mean as $\overline{f}

A  => md/rigid-body.md +24 -0
@@ 1,24 @@
title: rigid body
type: model

the study of motion of a rigid body


## Assumptions
 - [Particles](particle.md) within solids are attached firmly, no deformation.
 - [Particles](particle.md) within solids move at the same instant, no delay.

## Model
 - [Particles](particle.md)

A  => md/scattering.md +16 -0
@@ 1,16 @@
title: scattering
type: model

\begin{equation} \chi=|\mfrac{\tau}{2}-2\phi_0| \end{equation}
\begin{equation} \phi_0=\int\limits_{r_\text{min}}^\infty\frac{(M/r^2)\dd{r}}{\sqrt{2m[E-U(r)]-M^2/r^2}} \end{equation}
\begin{equation} E=\mfrac{1}{2}mv_\infty^2,\qquad M=m\rho v_\infty \end{equation}
\begin{equation} \phi_0=\int\limits_{r_\text{min}}^\infty\frac{(\rho/r^2)\dd{r}}{\sqrt{1-(\rho^2/r^2)-(2U/mv_\infty^2)}} \end{equation}
\begin{equation} \dd{\sigma}=\dd{N}/n \end{equation}
\begin{equation} \dd{\sigma}=\tau\rho\dd{\rho} \end{equation}
\begin{equation} \dd{\sigma}=\tau\rho\left|\frac{\dd{\rho}}{\dd{\chi}}\right|\dd{\chi} \end{equation}
\begin{equation} \dd{\sigma}=\frac{\rho}{\sin\chi}\left|\frac{\dd{\rho}}{\dd{\chi}}\right|\dd{o} \end{equation}

## Assumptions
 - [Two body problem](two-body-problem.md)

A  => md/solid-angular-momentum.md +7 -0
@@ 1,7 @@
title: rigid angular momentum
type: object

## Model
 - [Rigid bodies](rigid-body.md)

A  => md/translation-operator.md +8 -0
@@ 1,8 @@
title: translation operator
type: problem

Express the [operator](quantum-operator.md) $\hat{T_a}$ of a parallel displacement over a finite distance $a$ in terms of the momentum operator.

## solution

A  => md/two-body-problem.md +13 -0
@@ 1,13 @@
type: model
title: two body problem

$$\tag{13.1} L=\frac{1}{2}m\dot{\mathbf{r}}_1+\frac{1}{2}m\dot{\mathbf{r}}_2-U(|\mathbf{r}_1-\mathbf{r}_2|) $$
$$\tag{13.2} \mathbf{r}_1=m_2\mathbf{r}/(m_1+m_2),\qquad \mathbf{r}_2=-m_1\mathbf{r}/(m_1+m_2),\qquad $$
$$\tag{13.3} L=\frac{1}{2}m\dot{\mathbf{r}}^2-U(r) $$
$$\tag{13.4} m=m_1m_2/(m_1+m_2) $$

## Theory
 - [Mechanical System](mechanical-system.md)
 - [Centre of mass](centre-of-mass.md)

A  => md/wave-function.md +34 -0
@@ 1,34 @@
type: object
title: the wave function

The wave function describes a quantum-mechanical system. From the wave function we can calculate the probability of a given event (see uncertainty principle) with a form bilinear in $\Psi$ and $\Psi^*$.

$$ \lltag{3}{2.1} \int\int\Psi(q)\Psi^*(q')\phi(q,q')\dd{q}\dd{q'} $$

The sum of the probabilities of all possible values of the co-ordinates of the system must be unity.

$$ \lltag{3}{2.2} \int|\Psi|^2\dd{q}=1 $$

Sometimes $|\Psi|^2$ diverges, in this case it does not represent the absolute values of probability, but rather the relative probability between two events in the co-ordinate system.

If we know the wave function for two systems, then the state of the whole system is

$$ \lltag{3}{2.3} \Psi_{12}(q_1,q_2)=\Psi_1(q_1)\Psi_2(q_2) $$

This relation stands as long as the two systems don't interact. (see entanglement)

$$ \lltag{3}{2.4} \Psi_{12}(q_1,q_2,t)=\Psi_1(q_1,t)\Psi_2(q_2,t) $$

### Passage to classical mechanics

TODO read LL2 to grok eikonal stuff

$$ \hbar=1.054\times10^{-34} \text{J$\cdot$s} $$

The wave function of an "almost classical" system has the form

$$ \lltag{3}{6.1} \Psi=ae^{iS/\hbar} $$

Planck's constant $\hbar$ plays the role of the "extent of quantistion", the passage from quantum to classical mechanics, corresponding to large phase, can be formally described as a passage to the limit $\hbar\to 0$.

A  => notes.md +7 -0
@@ 1,7 @@
# Notation

I note the transposed operator ^t instead of a tilde hat

## Mistakes

LL3. between 4.6 and 4.7, I think that fg-gf is anti-Hermitian iff f and g are Hermitian

A  => tools/colors +5 -0
@@ 1,5 @@
declare -A colors
colors[model]="#abffab"		#greeen
colors[object]="#a8deff"	#blue
colors[problem]="#faf884"	#yellow

A  => tools/generate-bg-css.sh +6 -0
@@ 1,6 @@

. tools/colors
for type in ${!colors[@]}; do
	echo ".$type{ background-color: ${colors[$type]}; }"

A  => tools/generate-graph.sh +27 -0
@@ 1,27 @@

. tools/colors

title_of() { awk -F': ' '/title/ {print $2}' md/$1; }
type_of() { awk -F': ' '/type/ {print $2}' md/$1; }
color_of() {
    type=$(type_of $1)
    echo ${colors[$type]}

echo 'digraph {'
echo 'node  [style="rounded,filled", shape=box]'
echo 'rankdir=LR'
for node in md/*.md; do
    node=$(basename $node)
    [ "$(type_of $node)" == "problem" ] && continue
    title=$(title_of $node)
    color=$(color_of $node)
    echo "    \"$title\"[href=\"../html/$(basename ${node/.md/.html} | tr -d '_')\", fillcolor=\"$color\"]"
    [ -f graph/$node ] && for line in $(cat graph/$node); do
    echo "    \"$(title_of $line)\" -> \"$title\""
echo "}"

A  => tools/header.html +14 -0
@@ 1,14 @@
<link rel="stylesheet" type="text/css" href="colors.css">
<script src="https://cdn.jsdelivr.net/npm/mathjax@3/es5/tex-mml-chtml.js" type="text/javascript"></script>
body {
  display: block;
  margin: 0 auto;
  margin-top: 50px;
  max-width: 800px;
  text-align: justify;
h1 {
  text-transform: capitalize;

A  => tools/latex-includes.yaml +16 -0
@@ 1,16 @@
header-includes: |
  \newcommand{\lltag}[2]{\tag*{LL#1 (#2)}}

A  => tools/pf-filter.py +21 -0
@@ 1,21 @@
import panflute as pf

def prepare(doc):
    # doc.content.insert(0, pf.RawBlock("<h1>title</h1"))
    # doc.content.insert(1, pf.RawBlock('<body class="theory">'))

def action(elem, doc):
    if isinstance(elem, pf.Code):
        doc.content.insert(0, pf.RawBlock(f"HELLO"))

    # .md to .html
    if isinstance(elem, pf.Link) and elem.url.endswith('.md'):
        elem.url = elem.url[:-3] + '.html'
        return elem

if __name__ == '__main__':
    pf.run_filter(action, prepare=prepare)

A  => tools/template.html +64 -0
@@ 1,64 @@
<!DOCTYPE html>
<html xmlns="http://www.w3.org/1999/xhtml" lang="$lang$" xml:lang="$lang$"$if(dir)$ dir="$dir$"$endif$>
  <meta charset="utf-8" />
  <meta name="generator" content="pandoc" />
  <meta name="viewport" content="width=device-width, initial-scale=1.0, user-scalable=yes" />
  <meta name="author" content="$author-meta$" />
  <meta name="dcterms.date" content="$date-meta$" />
  <meta name="keywords" content="$for(keywords)$$keywords$$sep$, $endfor$" />
  <title>$if(title-prefix)$$title-prefix$ – $endif$$pagetitle$</title>
  <link rel="stylesheet" href="$css$" />
  <!--[if lt IE 9]>
    <script src="//cdnjs.cloudflare.com/ajax/libs/html5shiv/3.7.3/html5shiv-printshiv.min.js"></script>
<body class="$type$">
<header id="title-block-header">
<h1 class="title">$title$</h1>
<p class="subtitle">$subtitle$</p>
<p class="author">$author$</p>
<p class="date">$date$</p>
<nav id="$idprefix$TOC" role="doc-toc">
<h2 id="$idprefix$toc-title">$toc-title$</h2>