~jzck/physics-notes

7c2475121c468df2948dd16db961fc8a0cbb37a4 — Jack Halford 1 year, 1 month ago d37073c
stuff
M Makefile => Makefile +5 -1
@@ 6,7 6,7 @@ SHELL := bash
MD	=	$(shell ls -1 md/*.md)
GRAPH	=	$(subst md/,graph/,$(MD:.md=))
SVG	=	$(subst md/,www-data/,$(MD:.md=.svg))
HTML	=	$(subst md/,www-data/,$(MD:.md=.html))
HTML	=	$(subst md/,www-data/,$(MD:.md=.html)) www-data/index.html www-data/style.css

all: $(HTML) $(SVG) www-data/graph.svg



@@ 19,6 19,10 @@ www-data/%.html: md/%.md | www-data/colors.css www-data
		$^ <(tools/get see_also $<) \
		-o $@ &
> @printf '$@\n'
www-data/index.html: | www-data
> cp tools/index.html $@
www-data/style.css: | www-data
> cp tools/style.css $@
www-data/colors.css: | www-data
> ./tools/get bg_css >$@
www-data/%.svg: graph/% | www-data

A img/flp3-9-1.svg => img/flp3-9-1.svg +251 -0
@@ 0,0 1,251 @@
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<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" width="410pt" height="289pt" viewBox="0 0 237 167" version="1.1">
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M md/ammonia-maser.md => md/ammonia-maser.md +1 -1
@@ 1,7 1,7 @@
---
title: the ammonia maser
type: model
sources: FLP3.9
sources: FLP3-9
---

![two base states of the ammonia molecule (source: FLP 3, fig 9-1)](../img/flp3-9-1.svg)

M md/angular-momentum.md => md/angular-momentum.md +1 -1
@@ 1,6 1,6 @@
---
title: angular momentum
type: object
type: theory
---

$$ \delta \mathbf{r}=\delta\mathbf{ϕ}\times\mathbf{r} $$

M md/central-field.md => md/central-field.md +8 -8
@@ 1,6 1,6 @@
---
title: central field
type: model
type: theory
---

We saw that we can reduce a [two body problem](two-body-problem.md) into a system of one particle and a potential $U(r)$ that only depends on the distance $r$ from some fixed point. This is called a central field.


@@ 11,30 11,30 @@ We have shown that the [angular-momentum](angular-momentum.md) $\mathbf{M}=\math

We use polar coordinates $r,\phi$ to describe the Lagrangian in this plane

$$ \lltag{1}{14.1} L=\mfrac{1}{2}m(\dot{r}^2+r^2\dot{\phi^2})-U(r) $$
$$ \ptag[LL1]{14.1} L=\mfrac{1}{2}m(\dot{r}^2+r^2\dot{\phi^2})-U(r) $$

As this function does not have the co-ordinate $\phi$ explicitly (Lagrangian is cyclic in $\phi$), we can greatly simplify the problem by using Lagrange's equation $(\dd{}/\dd{t})\partial L/\partial \dot{\phi}=\partial L/\partial \phi=0$, thus we get the conservation of the angular momentum by differentiating $L$ with respect to $\dot{\phi}$.

$$ \lltag{1}{14.2} \frac{\partial L}{\partial \dot{\phi}}=mr^2\dot{\phi}=\text{constant} (=M_z=p_\phi)$$
$$ \ptag[LL1]{14.2} \frac{\partial L}{\partial \dot{\phi}}=mr^2\dot{\phi}=\text{constant} (=M_z=p_\phi)$$

We know from polar coordinates that $\dd{A}=\mfrac{1}{2}\mathbf{r}^2\dd{\phi}$ is the area element of the trajectory. We can write this as

$$ \lltag{1}{14.3} 2m\dd{A}=M\dd{t} $$
$$ \ptag[LL1]{14.3} 2m\dd{A}=M\dd{t} $$

in other words, in equal times the radius vector of the particle sweeps out equal areas (Kepler's second law).

To find the laws of motion, we start from the conservation of energy

$$ \lltag{1}{14.4} E=\mfrac{1}{2}m(\dot{r}^2+r^2\dot{\phi^2})+U(r)=\mfrac{1}{2}m\dot{r}^2+\mfrac{1}{2}M^2/mr^2+U(r)$$
$$ \ptag[LL1]{14.4} E=\mfrac{1}{2}m(\dot{r}^2+r^2\dot{\phi^2})+U(r)=\mfrac{1}{2}m\dot{r}^2+\mfrac{1}{2}M^2/mr^2+U(r)$$

we rewrite this as

$$ \lltag{1}{14.5} \dot{\mathbf{r}}=\frac{\dd{r}}{\dd{t}}=\sqrt{\frac{2}{m}(E-U(r))-\frac{M^2}{m^2r^2}} $$
$$ \ptag[LL1]{14.5} \dot{\mathbf{r}}=\frac{\dd{r}}{\dd{t}}=\sqrt{\frac{2}{m}(E-U(r))-\frac{M^2}{m^2r^2}} $$

by integrating

$$ \lltag{1}{14.6} t=\int\dd{r}\left(\frac{2}{m}(E-U(r))-\frac{M^2}{m^2r^2}\right)^{-1/2}+\text{constant} $$
$$ \ptag[LL1]{14.6} t=\int\dd{r}\left(\frac{2}{m}(E-U(r))-\frac{M^2}{m^2r^2}\right)^{-1/2}+\text{constant} $$

writing $(14.2)$ as $\dd{\phi}=\frac{M}{mr^2}\dd{t}$ we also find

$$ \lltag[LL1]{14.7} \phi=\int\frac{M\dd{r}/r^2}{\sqrt{\frac{2}{m}(E-U(r))-\frac{M^2}{m^2r^2}}}+\text{constant} $$
$$ \ptag[LL1]{14.7} \phi=\int\frac{M\dd{r}/r^2}{\sqrt{\frac{2}{m}(E-U(r))-\frac{M^2}{m^2r^2}}}+\text{constant} $$

M md/centre-of-mass.md => md/centre-of-mass.md +1 -1
@@ 1,6 1,6 @@
---
title: centre of mass
type: object
type: theory
---

\begin{equation} \mathbf{P}=\mathbf{P}'+\mathbf{V}\sum_a m_a \end{equation}

M md/density-matrix.md => md/density-matrix.md +1 -1
@@ 1,6 1,6 @@
---
title: density matrix
type: object
type: theory
---

paragraph 14

M md/energy.md => md/energy.md +1 -1
@@ 1,6 1,6 @@
---
title: energy
type: object
type: theory
---

$$\tag{6.1} E\equiv\sum_i\dot{q_i}\frac{\partial L}{\partial \dot{q_i}}-L $$

A md/erf.md => md/erf.md +41 -0
@@ 0,0 1,41 @@
---
title: exponential response function
type: math
sources: MIT 18.031x
---

## First order ERF

This trick works for finding the solution to $n^{th}$ order ODE with constant factors when the input is of exponential form.

Consider the following $n^th$ order constant factor ODE.

$$ P(D)z=e^{rt} $$

where $P(D)$ is a polynomial of the derivative operator $D$.

because of the property of exponential we have

$$ \tag{1} P(D)e^{rt}=P(r)e^{rt} $$

If $P(D)\neq 0$, by linearity of the operator $P(D)$ we have immediately

$$ P(D)\left(\frac{e^{rt}}{P(r)}\right)=e^{rt} $$

Therefore we have the particular solution to our ODE

$$ z_p=\frac{e^{rt}}{P(r)} $$

## Generalized ERF

When $P(r)=0$ can simply take the derivative of $(1)$ with respect to $r$.

$$ P(D)te^{rt}=P'(r)e^{rt} $$

or more generally if all the $P^{(i<m)}(r)=0$

$$ \tag{2} P(D)t^me^{rt}=P^{(m)}(r)e^{rt} $$

In which case our particular solution is

$$ z_p=\frac{t^{m}e^{rt}}{P^{(m)}(r)} $$

M md/galileo-relativity.md => md/galileo-relativity.md +1 -1
@@ 1,6 1,6 @@
---
title: Galileo's relativity
type: object
type: theory
---

\begin{equation} L=L(v^2) \end{equation}

M md/hamiltonian-operator.md => md/hamiltonian-operator.md +2 -2
@@ 1,6 1,6 @@
---
title: the Hamiltonian operator
type: object
type: theory
---

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


@@ 31,7 31,7 @@ Comparing this to the definition $\partial\Psi/\partial t=-i\qop{L}\Psi$ we have

As we know from classical mechanics, the derivative $-\partial S/\partial t$ is just Hamilton's function $H$ for a mechanical system. In quantum mechanics we call this the Hamiltonian [operator](quantum-operator.md), or the Hamiltonian of a system.

$$ \lltag{3}{8.1} i\hbar\frac{\partial\Psi}{\partial t}=\qop{H}\Psi $$
$$ \ptag[LL3]{8.1} i\hbar\frac{\partial\Psi}{\partial t}=\qop{H}\Psi $$

If the form of the Hamiltonian is known, equation $(8.1)$ determines the wave function of the physical system concerned. This fundamental equation of quantum mechanics is called the wave equation.


M md/inertia-tensor.md => md/inertia-tensor.md +1 -1
@@ 1,6 1,6 @@
---
title: inertia tensor
type: object
type: theory
---

\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}

A md/integration-factor.md => md/integration-factor.md +31 -0
@@ 0,0 1,31 @@
---
title: integration factor
type: math
sources: MIT 18.031x
---

The are many ways to solve first order linear differential equations, but this trick is quite neat.

the general form of a first order linear DE is

$$ \dot{y}+p(x)y=q(x) $$

We are going to multiply through by the "integration factor" $e^{\int p\dd{x}}$

$$ \dot{y}e^{\int p\dd{x}}+pye^{\int p\dd{x}}=qe^{\int p\dd{x}} $$

Here we use the identity $\frac{\dd{}}{\dd{x}}\int p\dd{x}=p(x)$ and we recognize the derivative of a product

$$ (ye^{\int p\dd{x}})'=qe^{\int p\dd{x}} $$

Thus, integration on both sides we have

$$ y=e^{-\int p\dd{x}}\int qe^{\int p\dd{x}}\dd{x}+\text{constant} $$

Which is a general solution for $y$

### Notes

The trick works with any primitive $P(x)$ of $p(x)$, so we don't need to bother with the constant in the integration factor $e^{\int p\dd{x}}$.

Other methods for solving the same equation involve solving the homogeneous equaton $(q(x)=0)$, and then varying the constant of the homegenous solution. I find the integration factor more elegant.

M md/lagrangian.md => md/lagrangian.md +1 -1
@@ 1,6 1,6 @@
---
title: Lagrangian
type: object
type: theory
---

The Lagrangian is defined by the principle of least action

M md/matrix-mechanics.md => md/matrix-mechanics.md +1 -1
@@ 1,6 1,6 @@
---
title: matrix mechanics
type: object
type: theory
---

paragraph 11

M md/mechanical-similarity.md => md/mechanical-similarity.md +1 -1
@@ 1,6 1,6 @@
---
title: mechanical similarity
type: object
type: theory
---

\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}

M md/momentum-operator.md => md/momentum-operator.md +14 -14
@@ 1,6 1,6 @@
---
title: the momentum operator
type: object
type: theory
---

Considering a system of particles not in an external field. The system should behave the same if we make an identical parallel displacement of each particle.


@@ 13,7 13,7 @@ $$ \qop{H}\qop{O}-\qop{O}\qop{H}=0 $$

Since the unit operator and the multiplication by $\delta r$ both commute with $\qop{H}$ we are left with

$$ \lltag{3}{15.1} (\sum_a\nabla_a)\qop{H}-\qop{H}(\sum_a\nabla_a)=0 $$
$$ \ptag[LL3]{15.1} (\sum_a\nabla_a)\qop{H}-\qop{H}(\sum_a\nabla_a)=0 $$

As we know the commutation with $\qop{H}$ mean that the corresponding physical quantity is conserved. The conserved quantity that follows from the homogoneity of space is the momentum $p$.



@@ 23,7 23,7 @@ $$ \qop{p}\Psi=(i/\hbar)cae^{(i/\hbar)S}\nabla S=c(i/\hbar)\Psi\nabla S $$

We know from classical mechanics that $p=\nabla S$. Therefore $c=-i\hbar$. Thus we have the momentum operator $\qop{p}=-i\hbar\nabla$, or in components

$$ \lltag{3}{15.2}
$$ \ptag[LL3]{15.2}
\qop{p}_x=i\hbar\partial/\partial x,\qquad
\qop{p}_y=i\hbar\partial/\partial y,\qquad
\qop{p}_z=i\hbar\partial/\partial z


@@ 37,7 37,7 @@ and this is the condition that the operator should be Hermitian.

Since differentiating with respect to two different variables is independant of order, it is clear that the operators commute with one another.

$$ \lltag{3}{15.3}
$$ \ptag[LL3]{15.3}
\qop{p}_x\qop{p}y-\qop{p}_y\qop{p}x=0,\qquad
\qop{p}_x\qop{p}z-\qop{p}_z\qop{p}x=0,\qquad
\qop{p}_y\qop{p}z-\qop{p}_z\qop{p}y=0


@@ 47,7 47,7 @@ This means that all 3 components can have definite values simultaneously, which 

To find the eigenfunctions and eigenvalues of the momentum operators we have to solve

$$ \lltag{3}{15.4}
$$ \ptag[LL3]{15.4}
-i\hbar\partial\psi/\partial x=p_x\psi,\qquad
-i\hbar\partial\psi/\partial y=p_y\psi,\qquad
-i\hbar\partial\psi/\partial z=p_z\psi


@@ 61,15 61,15 @@ Thus the eigenvalues form a continuous spectrum from $-\infty$ to $+\infty$.

There is a common solution to all three equations, which correspond to a state where the momentum $\v{p}$ is entirely defined.

$$ \lltag{3}{15.5} \psi=Ce^{(i/\hbar)\v{p}\cdot\v{r}} $$
$$ \ptag[LL3]{15.5} \psi=Ce^{(i/\hbar)\v{p}\cdot\v{r}} $$

This is a completely determined wave function, therefore the momentum $\v{p}=(p_x,p_y,p_z)$ forms a complete basis. Let us now find the normalization factor $C$. The rule for normalization is

$$ \lltag{3}{15.6} \int\Psi_{p'}^*\Psi_p\dd{v}=\delta(\v{p'}-\v{p}) $$
$$ \ptag[LL3]{15.6} \int\Psi_{p'}^*\Psi_p\dd{v}=\delta(\v{p'}-\v{p}) $$

The integration can be effected with the formula

$$ \lltag{3}{15.7} \mfrac{1}{\tau}\int_{-\infty}^{+\infty}e^{i\alpha x}\dd{x}=\delta(\alpha) $$
$$ \ptag[LL3]{15.7} \mfrac{1}{\tau}\int_{-\infty}^{+\infty}e^{i\alpha x}\dd{x}=\delta(\alpha) $$

We have



@@ 80,18 80,18 @@ We have

Therefore we mut have $C^2h^3=1$. Thus the normalized eigenfunction $\psi_p$ is

$$ \lltag{3}{15.8} \psi=h^{-3/2}e^{i\tau(\v{p}\cdot\v{r}/h)} $$
$$ \ptag[LL3]{15.8} \psi=h^{-3/2}e^{i\tau(\v{p}\cdot\v{r}/h)} $$

We can expand any wave function $\psi(\v{r})$ in terms of the eigenfunctions $\psi_\v{p}$, as a Fourier integral

$$ \lltag{3}{15.9} \psi(\v{r})=
$$ \ptag[LL3]{15.9} \psi(\v{r})=
\int a(\v{p})\psi_\v{p}(\v{r})\dd[3]{p}=
\int a(\v{p})e^{i\tau(\v{p}\cdot\v{r}/h)}\dd[3]{p}
$$

(where $\dd[3]{p}=\dd{p_x}\dd{p_y}\dd{p_z}$). The expansion coefficients $a(\v{p})$ are, according to formula $(5.3)$

$$ \lltag{3}{15.10} a(\v{p})=
$$ \ptag[LL3]{15.10} a(\v{p})=
\int \psi(\v{r})\psi_\v{p}^*(\v{r})\dd{V}=
h^{-3/2}\int\psi(\v{r})e^{i\tau(\v{p}\cdot\v{r}/h)}\dd{V}
$$


@@ 106,7 106,7 @@ TODO 15.11 and 15.12

Let us now derive the commutation rules for the operators we have found. Since partial differentation of one cartesian variable doesn't affect the others, we have directly the commutation rule

$$ \lltag{3}{16.1}
$$ \ptag[LL3]{16.1}
\qop{p}_xy-y\qop{p}_x=0,\qquad
\qop{p}_xz-z\qop{p}_x=0,\qquad
\qop{p}_yz-z\qop{p}_y=0


@@ 122,7 122,7 @@ For the commutation of $\qop{p}_x$ and $x$ we write
We find that the commutator reduces to a multiplication by $-i\hbar$, the same is true for $y$ and $z$


$$ \lltag{3}{16.2}
$$ \ptag[LL3]{16.2}
\qop{p}_xx-x\qop{p}_x=
\qop{p}_yy-y\qop{p}_y=
\qop{p}_zz-z\qop{p}_z=


@@ 131,7 131,7 @@ $$

or they can be rewritten in the form

$$ \lltag{3}{16.3} \qop{p}_ik-k\qop{p}_i=-i\hbar\delta_{ki}\qquad(i,k=x,y,z) $$
$$ \ptag[LL3]{16.3} \qop{p}_ik-k\qop{p}_i=-i\hbar\delta_{ki}\qquad(i,k=x,y,z) $$

These relations show that the co-ordinate of a particle can have a definite value at the same time as the components of the momentum along the other two axes. However, the components of co-ordinate and momentum along th same axis cannot exists simultaneously. 


M md/momentum.md => md/momentum.md +1 -1
@@ 1,6 1,6 @@
---
title: momentum
type: object
type: theory
---

$$ \sum_a\partial L/\partial \mathbf{r}_a=0 $$

M md/quantum-operator.md => md/quantum-operator.md +34 -34
@@ 1,23 1,23 @@
---
type: object
type: theory
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.

$$ \lltag{3}{3.1} \int|\Psi_n|^2\dd{q}=1 $$
$$ \ptag[LL3]{3.1} \int|\Psi_n|^2\dd{q}=1 $$

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).

$$ \lltag{3}{3.2} \Psi=\sum a_n\Psi_n $$
$$ \ptag[LL3]{3.2} \Psi=\sum a_n\Psi_n $$

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.

$$ \lltag{3}{3.3} \sum_n|a_n|^2=1 $$
$$ \ptag[LL3]{3.3} \sum_n|a_n|^2=1 $$

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

$$ \lltag{3}{3.4} \sum_n a_na_n^*=\int\Psi\Psi^*\dd{q} $$
$$ \ptag[LL3]{3.4} \sum_n a_na_n^*=\int\Psi\Psi^*\dd{q} $$

we can find a simple form for $a_n$



@@ 27,71 27,71 @@ we can find a simple form for $a_n$
\sum_n a_na_n^*&=\sum a_n^*\int\Psi_n^*\Psi\dd{q} &&\text{replacing with $(3.4)$}
\end{align}

$$ \lltag{3}{3.5} a_n=\int\Psi\Psi_n^*\dd{q} $$
$$ \ptag[LL3]{3.5} a_n=\int\Psi\Psi_n^*\dd{q} $$

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

$$ \lltag{3}{3.6} \int\Psi_m\Psi_n^*\dd{q}=\delta_{nm} $$
$$ \ptag[LL3]{3.6} \int\Psi_m\Psi_n^*\dd{q}=\delta_{nm} $$

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

$$ \lltag{3}{3.7} \overline{f}=\sum_nf_n|a_n|^2 $$
$$ \ptag[LL3]{3.7} \overline{f}=\sum_nf_n|a_n|^2 $$

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

$$ \lltag{3}{3.8} \overline{f}=\int \Psi^*(\qop{f}\Psi)\dd{q} $$
$$ \ptag[LL3]{3.8} \overline{f}=\int \Psi^*(\qop{f}\Psi)\dd{q} $$

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

$$ \lltag{3}{3.9} (\qop{f}\Psi)=\sum_nf_na_n\Psi_n $$
$$ \ptag[LL3]{3.9} (\qop{f}\Psi)=\sum_nf_na_n\Psi_n $$

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

$$ \lltag{3}{3.10} (\qop{f}\Psi)=\int K(q,q')\Psi(q')\dd{q} $$
$$ \ptag[LL3]{3.10} (\qop{f}\Psi)=\int K(q,q')\Psi(q')\dd{q} $$

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

$$ \lltag{3}{3.11} K(q,q')=\sum_n f_n\Psi_n^*(q')\Psi_n(q) $$
$$ \ptag[LL3]{3.11} K(q,q')=\sum_n f_n\Psi_n^*(q')\Psi_n(q) $$

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

$$ \lltag{3}{3.12} \qop{f}\Psi=f_n\Psi_n $$
$$ \ptag[LL3]{3.12} \qop{f}\Psi=f_n\Psi_n $$

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

$$ \lltag{3}{3.13} \int\Psi^*(\qop{f}\Psi)\dd{q}=\int\Psi(\qop[*]{f}\Psi^*)\dd{q} $$
$$ \ptag[LL3]{3.13} \int\Psi^*(\qop{f}\Psi)\dd{q}=\int\Psi(\qop[*]{f}\Psi^*)\dd{q} $$

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

$$ \lltag{3}{3.14} \int\Psi(\qop{f}\Phi)\dd{q}\equiv\int\Phi(\qop[t]{f}\Psi)\dd{q} $$
$$ \ptag[LL3]{3.14} \int\Psi(\qop{f}\Phi)\dd{q}\equiv\int\Phi(\qop[t]{f}\Psi)\dd{q} $$

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

$$ \lltag{3}{3.15} \qop[t]{f}=\qop[*]{f} $$
$$ \ptag[LL3]{3.15} \qop[t]{f}=\qop[*]{f} $$

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

$$ \lltag{3}{3.16} \qop[+]{f}=\qop[*t]{f} $$
$$ \ptag[LL3]{3.16} \qop[+]{f}=\qop[*t]{f} $$

### 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

$$ \lltag{3}{4.1} \overline{f+g}=\overline{f}+\overline{g} $$
$$ \ptag[LL3]{4.1} \overline{f+g}=\overline{f}+\overline{g} $$

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

$$ \lltag{3}{4.2} (f+g)_0\geq f_0+g_0 $$
$$ \ptag[LL3]{4.2} (f+g)_0\geq f_0+g_0 $$

(need Hilbert algebra for proof of this)



@@ 99,7 99,7 @@ $$ \lltag{3}{4.2} (f+g)_0\geq f_0+g_0 $$

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 $\qop{g}\qop{f}$.

$$ \lltag{3}{4.3} \qop{f}\qop{g}-\qop{g}\qop{f}=0 $$
$$ \ptag[LL3]{4.3} \qop{f}\qop{g}-\qop{g}\qop{f}=0 $$

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).



@@ 118,30 118,30 @@ $$

from which we have directly

$$ \lltag{3}{4.4} (\qop{f}\qop{g})^t=\qop[t]{g}\qop[t]{f} $$
$$ \ptag[LL3]{4.4} (\qop{f}\qop{g})^t=\qop[t]{g}\qop[t]{f} $$

Taking the complex conjugate on both sides we have

$$ \lltag{3}{4.5} (\qop{f}\qop{g})^+=\qop[+]{g}\qop[+]{f} $$
$$ \ptag[LL3]{4.5} (\qop{f}\qop{g})^+=\qop[+]{g}\qop[+]{f} $$


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

$$ \lltag{3}{4.6} \mfrac{1}{2}(\qop{f}\qop{g}+\qop{g}\qop{f}) $$
$$ \ptag[LL3]{4.6} \mfrac{1}{2}(\qop{f}\qop{g}+\qop{g}\qop{f}) $$

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

$$ \lltag{3}{4.6} i(\qop{f}\qop{g}-\qop{g}\qop{f}) $$
$$ \ptag[LL3]{4.6} i(\qop{f}\qop{g}-\qop{g}\qop{f}) $$

For brevity we note the commutator

$$ \lltag{3}{4.7} \{\qop{f},\qop{g}\}=\qop{f}\qop{g}-\qop{g}\qop{f} $$
$$ \ptag[LL3]{4.7} \{\qop{f},\qop{g}\}=\qop{f}\qop{g}-\qop{g}\qop{f} $$

It is easily seen that 

$$ \lltag{3}{4.8} \{\qop{f}\qop{g},\qop{h}\}=\{\qop{f},\qop{h}\}\qop{g}-\qop{f}\{\qop{g},\qop{h}\} $$
$$ \ptag[LL3]{4.8} \{\qop{f}\qop{g},\qop{h}\}=\{\qop{f},\qop{h}\}\qop{g}-\qop{f}\{\qop{g},\qop{h}\} $$

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).



@@ 149,13 149,13 @@ And we notice that if $\{\qop{f},\qop{h}\}=0$ and $\{\qop{g},\qop{h}\}=0$, it do

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.

$$ \lltag{3}{5.1} \Psi(q)=\int a_f\Psi_f(q)\dd{f} $$
$$ \ptag[LL3]{5.1} \Psi(q)=\int a_f\Psi_f(q)\dd{f} $$

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

$$ \lltag{3}{5.2} \int|a_f|^2\dd{f}=1 $$
$$ \ptag[LL3]{5.2} \int|a_f|^2\dd{f}=1 $$

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



@@ 166,7 166,7 @@ In the same way we found $(3.5)$, we can write

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

$$ \lltag{3}{5.3} a_f=\int\Psi(q)\Psi_f^*(q)\dd{q} $$
$$ \ptag[LL3]{5.3} a_f=\int\Psi(q)\Psi_f^*(q)\dd{q} $$

in exact analogy to $(3.5)$.



@@ 176,7 176,7 @@ $$ a_f=\int a_{f'}(\Psi_{f'}\Psi_f^*\dd{q})\dd{f'} $$

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

$$ \lltag{3}{5.4} \int\Psi_{f'}\Psi_f^*\dd{q}=\delta(f'-f) $$
$$ \ptag[LL3]{5.4} \int\Psi_{f'}\Psi_f^*\dd{q}=\delta(f'-f) $$

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



@@ 184,11 184,11 @@ $$ \Psi(q)=\int\Psi(q')\left(\int\Psi_f^*(q')\Psi_f(q)\dd{f}\right)\dd{q'} $$

From which we deduce immediately

$$ \lltag{3}{5.11} \int\Psi_f^*(q')\Psi_f(q)\dd{f}=\delta(q-q') $$
$$ \ptag[LL3]{5.11} \int\Psi_f^*(q')\Psi_f(q)\dd{f}=\delta(q-q') $$

The analogous for a discrete spectrum is

$$ \lltag{3}{5.12} \sum_n\Psi_n^*(q')\Psi_n(q)\dd{f}=\delta(q-q') $$
$$ \ptag[LL3]{5.12} \sum_n\Psi_n^*(q')\Psi_n(q)\dd{f}=\delta(q-q') $$

$(5.1)$ 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 $\dd{f}$.



@@ 198,7 198,7 @@ $(5.1)$ and $(5.3)$ are analogous: $\Psi(q)$ can be expanded in terms of the fun

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 $\bar{f}$

$$ \lltag{3}{9.1} \overline{\dot{f}}=\dot{\overline{f}} $$
$$ \ptag[LL3]{9.1} \overline{\dot{f}}=\dot{\overline{f}} $$

by definition we have the mean as $\overline{f}=\int\Psi^*\qop{f}\Psi\dd{q}$,



@@ 226,6 226,6 @@ $$

Since, by definition of the mean value, the operator inside the parenthesis must be the derivative operator of the quantity $f$

$$ \lltag{3}{9.2} \qop{\dot{f}}=\frac{\partial\qop{f}}{\partial t}+\frac{i}{\hbar}\left(\qop[]{H}\qop{f}-\qop{f}\qop{H}\right) $$
$$ \ptag[LL3]{9.2} \qop{\dot{f}}=\frac{\partial\qop{f}}{\partial t}+\frac{i}{\hbar}\left(\qop[]{H}\qop{f}-\qop{f}\qop{H}\right) $$

We notice that if $f$ does not depend explicitly on time, then the derivative $\qop{\dot{f}}$ is, apart from a constant factor, just a commutation of $f$ with the Hamiltonian $\qop{H}$.

M md/solid-angular-momentum.md => md/solid-angular-momentum.md +1 -1
@@ 1,6 1,6 @@
---
title: rigid angular momentum
type: object
type: theory
---

## Model

M md/two-body-problem.md => md/two-body-problem.md +1 -1
@@ 3,7 3,7 @@ title: two body problem
type: model
---

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

M md/wave-function.md => md/wave-function.md +6 -6
@@ 1,25 1,25 @@
---
type: object
type: theory
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 $\Psi^*$.

$$ \lltag{3}{2.1} \int\int\Psi(q)\Psi^*(q')\phi(q,q')\dd{q}\dd{q'} $$
$$ \ptag[LL3]{2.1} \int\int\Psi(q)\Psi^*(q')\phi(q,q')\dd{q}\dd{q'} $$

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

$$ \lltag{3}{2.2} \int|\Psi|^2\dd{q}=1 $$
$$ \ptag[LL3]{2.2} \int|\Psi|^2\dd{q}=1 $$

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

$$ \lltag{3}{2.3} \Psi_{12}(q_1,q_2)=\Psi_1(q_1)\Psi_2(q_2) $$
$$ \ptag[LL3]{2.3} \Psi_{12}(q_1,q_2)=\Psi_1(q_1)\Psi_2(q_2) $$

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

$$ \lltag{3}{2.4} \Psi_{12}(q_1,q_2,t)=\Psi_1(q_1,t)\Psi_2(q_2,t) $$
$$ \ptag[LL3]{2.4} \Psi_{12}(q_1,q_2,t)=\Psi_1(q_1,t)\Psi_2(q_2,t) $$

### Passage to classical mechanics



@@ 29,6 29,6 @@ $$ \hbar=1.054\times10^{-34} \text{J$\cdot$s} $$

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

$$ \lltag{3}{6.1} \Psi=ae^{iS/\hbar} $$
$$ \ptag[LL3]{6.1} \Psi=ae^{iS/\hbar} $$

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 $\hbar\to 0$.

M tools/generate-graph.sh => tools/generate-graph.sh +1 -1
@@ 5,7 5,7 @@ GET=tools/get
generate_one_node() {
    node=$(basename ${1%.md})
    
    [ "$($GET type $node)" == "problem" ] && return
    # [ "$($GET type $node)" == "problem" ] && return
    $GET graph_node $node
    title=$($GET title $node)
    [ -f graph/$node ] && for line in $(cat graph/$node); do

M tools/get => tools/get +5 -3
@@ 1,9 1,11 @@
#!/bin/bash

declare -A colors
colors[model]="#abffab"		#greeen
colors[object]="#a8deff"	#blue
colors[experiment]="#000000"
colors[problem]="#faf884"	#yellow
colors[model]="#abffab"		#greeen
colors[theory]="#a8deff"	#blue
colors[math]="#ffd1dc"		#pink
colors[DEFAULT]="#ffd1dc"	#pink

_or_default() { grep ^ || echo DEFAULT; }


@@ 14,7 16,7 @@ graph_node() { printf '    "%s"[href="%s.html", target="_parent", fillcolor="%s"
graph() { grep "($1.md)" md/*.md | awk -F: '{print $1}' | uniq | sed s:md/::g | sed s:.md::g | grep -v lexicon || true; }
bg_css() {
    for type in ${!colors[@]}; do
	    echo ".$type{ background-color: ${colors[$type]}; }"
	    echo "._$type{ background-color: ${colors[$type]}; }"
    done
}
see_also() {

M tools/header.html => tools/header.html +0 -10
@@ 1,13 1,3 @@
<link rel="stylesheet" type="text/css" href="colors.css">
<style>
body {
  display: block;
  margin: 0 auto;
  margin-top: 50px;
  max-width: 800px;
  text-align: justify;
}
h1 {
  text-transform: capitalize;
}
</style>

A tools/index.html => tools/index.html +10 -0
@@ 0,0 1,10 @@
<link rel="stylesheet" type="text/css" href="colors.css">
<link rel="stylesheet" type="text/css" href="style.css">

<!-- <div id="container"> -->
<h1>Main Graph</h1>
<!-- </div"> -->

<body>
<div style="margin:auto"><object data=graph.svg type=image/svg+xml></object></div>
</body>

M tools/latex-includes.yaml => tools/latex-includes.yaml +1 -1
@@ 11,7 11,7 @@ header-includes: |
  \newcommand{\qop}[2][]{\hat{#2}\vphantom{#2}^{#1}}
  \newcommand{\mfrac}[2]{\textstyle\frac{#1}{#2}\displaystyle}

  \newcommand{\lltag}[2][]{\tag*{(#2)$_{\text{#1}}$}}
  \newcommand{\ptag}[2][]{\tag*{(#2)$_{\text{#1}}$}}

  \newenvironment{nalign}{
      \begin{equation}

A tools/style.css => tools/style.css +14 -0
@@ 0,0 1,14 @@
#container {
  display: block;
  margin: 0 auto;
  margin-top: 50px;
  max-width: 800px;
  text-align: justify;
}
h1 {
  text-transform: capitalize;
}
hr { display: block; height: 1px;
    border: 0; border-top: 1px solid #000000;
    margin: 1em 0; padding: 0;
}

M tools/template.html => tools/template.html +25 -18
@@ 1,5 1,8 @@
<!DOCTYPE html>
<html xmlns="http://www.w3.org/1999/xhtml" lang="$lang$" xml:lang="$lang$"$if(dir)$ dir="$dir$"$endif$>
<div align="left"><a href="index.html">back to graph</a></div>
<link rel="stylesheet" type="text/css" href="colors.css">
<link rel="stylesheet" type="text/css" href="style.css">
<head>
  <meta charset="utf-8" />
  <meta name="generator" content="pandoc" />


@@ 23,31 26,27 @@ $endfor$
$if(math)$
  $math$
$endif$
  <!--[if lt IE 9]>
    <script src="//cdnjs.cloudflare.com/ajax/libs/html5shiv/3.7.3/html5shiv-printshiv.min.js"></script>
  <![endif]-->
$for(header-includes)$
  $header-includes$
$endfor$
</head>
<body class="$type$">
$for(include-before)$
$include-before$
$endfor$
<div id="container">
<body class="_$type$">

$if(title)$
<header id="title-block-header">
<h1 class="title">$title$</h1>
$if(subtitle)$
<p class="subtitle">$subtitle$</p>
$endif$
$for(author)$
<p class="author">$author$</p>
$endfor$
$if(date)$
<p class="date">$date$</p>
<h1 class="title"; style="text-align:left;float:left">$title$

$if(type)$
<h5 style="text-align:right;float:right">$type$</h5>
$endif$

</h1>

</header>
<hr style="clear:both;"/>
$endif$

$if(toc)$
<nav id="$idprefix$TOC" role="doc-toc">
$if(toc-title)$


@@ 56,9 55,17 @@ $endif$
$table-of-contents$
</nav>
$endif$

$body$
$for(include-after)$
$include-after$

$if(sources)$
<h3 class="sources">Sources</h3>
<ul>
$for(sources)$
  <li>$sources$</li>
$endfor$
</ul>
$endif$
</body>
</div>
</html>