~jzck/physics-notes

2d60cec521c1a7096ad248a0414b50eea16b786c — Jack Halford 1 year, 20 days ago 900e37b
publish html
75 files changed, 4549 insertions(+), 18 deletions(-)

M .gitignore
M Makefile
A docs/ammonia-maser.html
A docs/ammonia-maser.svg
A docs/angular-momentum.html
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A docs/central-field.html
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A docs/galileo-relativity.html
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graph
www-data
see_also

M Makefile => Makefile +12 -12
@@ 5,13 5,13 @@ SHELL := bash

MD	=	$(shell ls -1 md/*.md)
GRAPH	=	$(subst md/,graph/,$(MD:.md=))
SVG	=	$(subst md/,www-data/,$(MD:.md=.svg))
TEX	=	$(subst md/,www-data/,$(MD:.md=.tex))
HTML	=	$(subst md/,www-data/,$(MD:.md=.html)) www-data/index.html www-data/style.css
SVG	=	$(subst md/,docs/,$(MD:.md=.svg))
TEX	=	$(subst md/,docs/,$(MD:.md=.tex))
HTML	=	$(subst md/,docs/,$(MD:.md=.html)) docs/index.html docs/style.css

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

www-data/%.html: md/%.md | www-data/colors.css www-data
docs/%.html: md/%.md | docs/colors.css docs
> @printf '$@\n'
> pandoc	-s --filter tools/pf-filter.py \
		-H tools/header.html \


@@ 24,25 24,25 @@ www-data/%.html: md/%.md | www-data/colors.css www-data
		$< \
		-o $@ &

www-data/index.html: | www-data
docs/index.html: | docs
> cp tools/index.html $@
www-data/style.css: tools/style.css | www-data
docs/style.css: tools/style.css | docs
> cp tools/style.css $@
www-data/colors.css: | www-data
docs/colors.css: | docs
> ./tools/get bg_css >$@
www-data/%.svg: graph/% | www-data
docs/%.svg: graph/% | docs
> dot -Tsvg -o $@ <(./tools/generate-local-graph.sh $<) &
www-data/graph.svg: $(GRAPH) | www-data
docs/graph.svg: $(GRAPH) | docs
> dot -Tsvg -o $@ <(./tools/generate-graph.sh)

graph/%: md/%.md | graph
> @tools/get graph $(^F)>$@

www-data graph:
docs graph:
> @mkdir -p $@

clean:
> rm -rf www-data graph
> rm -rf docs graph

re: clean all


A docs/ammonia-maser.html => docs/ammonia-maser.html +96 -0
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<!DOCTYPE html>
<html xmlns="http://www.w3.org/1999/xhtml" lang="" xml:lang="">
<div align="left"><a href="index.html">back to graph</a></div>
<script src="https://cdn.jsdelivr.net/npm/mathjax@3/es5/tex-mml-chtml.js" type="text/javascript"></script>
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  <title>the ammonia maser</title>
  <style>
    code{white-space: pre-wrap;}
    span.smallcaps{font-variant: small-caps;}
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    div.column{display: inline-block; vertical-align: top; width: 50%;}
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    ul.task-list{list-style: none;}
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  <style>
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</head>
<div id="container">
<body class="_model">

<header id="title-block-header">
<h1 class="title"; style="text-align:left;float:left">the ammonia maser

<h5 style="text-align:right;float:right">model</h5>

</h1>

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


<figure>
<img src="../img/flp3-9-1.svg" alt="" /><figcaption>two base states of the ammonia molecule (source: FLP 3, fig 9-1)</figcaption>
</figure>
<h3 id="experimental-data">Experimental data</h3>
<p>In order to excite an electron inside an atom, it requires photons in the optical to ultraviolet range. In order to excite the vibrations on the molecules, it involves photons in the infrared. As for exciting rotations, it involves light in the far infrared. But the energy split <span class="math inline">\(2A\)</span> is much lower, we find experimentally that we can trigger this energy gap with microwave radiation, <span class="math inline">\(2A\approx 10^{-4}\text{eV}\)</span>.</p>
<h3 id="model">Model</h3>
<p>The geometry of the ammonia molecule <span class="math inline">\(NH3\)</span> is a tetrahedron, in this configuration, the hydrogen triangle can be either side of the nitrogren atom. Because of the tunnel effect, we can model the ammonia molecule as a two state system <span class="math inline">\(\left| \Psi \right&gt;=C_L\left| L \right&gt;+C_R\left| R \right&gt;\)</span>.</p>
<p>Because of the symmetry of the system, we can write the <a href="hamiltonian-operator.html">Hamiltonian</a> in the <span class="math inline">\(L,R\)</span> basis</p>
<p><span class="math display">\[ \hat{H}\vphantom{H}^{}=\begin{pmatrix} E&amp; -A\\ -A&amp; E \end{pmatrix} \]</span></p>
<p>where <span class="math inline">\(A\)</span> represent the probability for the molecule to flip between the two states. If the potential barrier between the two sides were infinite, we would have <span class="math inline">\(A=0\)</span>, no chance for the molecule to flip states.</p>
<h2 id="the-states-of-the-ammonia-molecule">The states of the ammonia molecule</h2>
<p>To find the eigenvalues (energies) we diagonalize the Hamiltonian matrix.</p>
<p><span class="math display">\[
\begin{aligned}
\det(H-\lambda\mathbb{1})&amp;=0\\
(E-\lambda)^2&amp;=A^2\\
\lambda_\pm&amp;=E\pm A
\end{aligned}
\]</span></p>
<p>We find that the 2 energy levels of our model are <span class="math inline">\(E\pm A\)</span>, i.e. they are seperated by an energy <span class="math inline">\(2A\)</span>.</p>
<p>Let us now find the first eigenfunction <span class="math inline">\(\left| + \right&gt;\)</span> corresponding to energy <span class="math inline">\(E_+=E+A\)</span>. We note <span class="math inline">\(\left| + \right&gt;=a\left| L \right&gt;+b\left| R \right&gt;\)</span></p>
<p><span class="math display">\[
\begin{aligned}
\hat{H}\vphantom{H}^{}\left| + \right&gt;&amp;=(E+A)\left| + \right&gt;\\
\begin{pmatrix} E&amp; -A\\ -A&amp; E \end{pmatrix}
\begin{pmatrix}a\\b\end{pmatrix}
&amp;=(E+A)
\begin{pmatrix}a\\b\end{pmatrix}\\
a&amp;=b\\
\end{aligned}
\]</span></p>
<p>Because the <span class="math inline">\(+\)</span> state has to be normalized we have</p>
<p><span class="math display">\[ \left| + \right&gt;=\textstyle\frac{1}{\sqrt{2}}\displaystyle(\left| L \right&gt;+\left| R \right&gt;) \]</span></p>
<p>Because <span class="math inline">\(+\)</span> and <span class="math inline">\(-\)</span> are orthogonal states, we must have</p>
<p><span class="math display">\[ \left| - \right&gt;=\textstyle\frac{1}{\sqrt{2}}\displaystyle(\left| L \right&gt;-\left| R \right&gt;) \]</span></p>
<p>In the diagonal basis, the wave equation <span class="math inline">\(i\hbar\frac{\partial\left| \Psi \right&gt;}{\partial t}=\hat{H}\vphantom{H}^{}\left| \Psi \right&gt;\)</span> is trivial to solve</p>
<p><span class="math display">\[
\left| \Psi \right&gt;=e^{-iEt/\hbar}
\left(
\left&lt; + \middle| \Psi(0) \right&gt;e^{-iAt/\hbar}\left| + \right&gt;
+\left&lt; - \middle| \Psi(0) \right&gt;e^{iAt/\hbar} \left| - \right&gt;
\right)
\]</span></p>
<p>The wave function <span class="math inline">\(\Psi\)</span> depends on the initial state of the system <span class="math inline">\(\Psi(0)\)</span>. The wave function includes a phase factor <span class="math inline">\(e^{iEt/\hbar}\)</span> which has no physical bearing on the isolated system. Appart from the 2 stationnary states we find that the probability to find the molecule in the <span class="math inline">\(L\)</span> state <span class="math inline">\(\left&lt; L \middle| \Psi \middle| L \right&gt;\)</span> (or <span class="math inline">\(R\)</span>) oscillates with frequency <span class="math inline">\(2A/\hbar\)</span></p>
<h2 id="todo-the-molecule-in-a-static-electric-field">TODO: The molecule in a static electric field</h2>

<hr>
<h3 class="sources">Sources</h3>
<ul>
  <li>FLP3-9</li>
</ul>
<hr>
<h3>See also</h3>
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</body>
</div>
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A docs/ammonia-maser.svg => docs/ammonia-maser.svg +36 -0
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<!DOCTYPE html>
<html xmlns="http://www.w3.org/1999/xhtml" lang="" xml:lang="">
<div align="left"><a href="index.html">back to graph</a></div>
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<body class="_theory">

<header id="title-block-header">
<h1 class="title"; style="text-align:left;float:left">angular momentum

<h5 style="text-align:right;float:right">theory</h5>

</h1>

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


<p><span class="math display">\[ \delta \mathbf{r}=\delta\mathbf{ϕ}\times\mathbf{r} \]</span> <span class="math display">\[ \delta \mathbf{v}=\delta\mathbf{ϕ}\times\mathbf{v} \]</span> <span class="math display">\[ \mathbf{M}\equiv\sum_a\mathbf{r}_a\times\mathbf{p}_a \]</span> <span class="math display">\[ \mathbf{M}=\mathbf{M}&#39;+a\times\mathbf{P} \]</span> <span class="math display">\[ \mathbf{M}=\mathbf{M}&#39;+\mu \mathbf{R}\times\mathbf{V} \]</span> <span class="math display">\[ \mathbf{M}=\mathbf{M}&#39;+\mathbf{R}\times\mathbf{P} \]</span> <span class="math display">\[ M_z=\sum_a\frac{\partial L}{\partial \dot{\phi_a}} \]</span> <span class="math display">\[ M_z=\sum_a m_ar_a^2\dot{\phi_a} \]</span></p>
<h2 id="assumptions">Assumptions</h2>
<ul>
<li><a href="mechanical-system.html">Classical mechanics</a></li>
</ul>

<hr>
<h3>See also</h3>
<center><object data=angular-momentum.svg type=image/svg+xml></object></center>
</body>
</div>
</html>

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<!DOCTYPE html>
<html xmlns="http://www.w3.org/1999/xhtml" lang="" xml:lang="">
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<header id="title-block-header">
<h1 class="title"; style="text-align:left;float:left">central field

<h5 style="text-align:right;float:right">theory</h5>

</h1>

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


<p>We saw that we can reduce a <a href="two-body-problem.html">two body problem</a> into a system of one particle and a potential <span class="math inline">\(U(r)\)</span> that only depends on the distance <span class="math inline">\(r\)</span> from some fixed point. This is called a central field.</p>
<p>This potential generates a radial force <span class="math inline">\(\mathbf{F}=-\frac{\partial U(r)}{\partial\mathbf{r}}=-\frac{\mathrm{d}^{}U}{\mathrm{d}^{}r}\frac{\mathbf{r}}{r}\)</span></p>
<p>We have shown that the <a href="angular-momentum.html">angular-momentum</a> <span class="math inline">\(\mathbf{M}=\mathbf{r}\times\mathbf{p}\)</span> of a system in a central field is conserved. Since <span class="math inline">\(\mathbf{M}\)</span> is perpendicular to <span class="math inline">\(r\)</span>, the constancy of <span class="math inline">\(\mathbf{M}\)</span> show that the radius vector <span class="math inline">\(\mathbf{r}\)</span> must remain must remain in the plane perpendicular to <span class="math inline">\(\mathbf{M}\)</span>.</p>
<p>We use polar coordinates <span class="math inline">\(r,\phi\)</span> to describe the Lagrangian in this plane</p>
<p><span class="math display">\[\label{LL1/14.1}\tag*{(14.1)}L=\textstyle\frac{1}{2}\displaystyle m(\dot{r}^2+r^2\dot{\phi^2})-U(r)\]</span></p>

<p>As this function does not have the co-ordinate <span class="math inline">\(\phi\)</span> explicitly (Lagrangian is cyclic in <span class="math inline">\(\phi\)</span>), we can greatly simplify the problem by using Lagrange’s equation <span class="math inline">\((\mathrm{d}^{}/\mathrm{d}^{}t)\partial L/\partial \dot{\phi}=\partial L/\partial \phi=0\)</span>, thus we get the conservation of the angular momentum by differentiating <span class="math inline">\(L\)</span> with respect to <span class="math inline">\(\dot{\phi}\)</span>.</p>
<p><span class="math display">\[\label{LL1/14.2}\tag*{(14.2)}\frac{\partial L}{\partial \dot{\phi}}=mr^2\dot{\phi}=\text{constant} (=M_z=p_\phi)\]</span></p>

<p>We know from polar coordinates that <span class="math inline">\(\mathrm{d}^{}A=\textstyle\frac{1}{2}\displaystyle\mathbf{r}^2\mathrm{d}^{}\phi\)</span> is the area element of the trajectory. We can write this as</p>
<p><span class="math display">\[\label{LL1/14.3}\tag*{(14.3)}2m\mathrm{d}^{}A=M\mathrm{d}^{}t\]</span></p>

<p>in other words, in equal times the radius vector of the particle sweeps out equal areas (Kepler’s second law).</p>
<p>To find the laws of motion, we start from the conservation of energy</p>
<p><span class="math display">\[\label{LL1/14.4}\tag*{(14.4)}E=\textstyle\frac{1}{2}\displaystyle m(\dot{r}^2+r^2\dot{\phi^2})+U(r)=\textstyle\frac{1}{2}\displaystyle m\dot{r}^2+\textstyle\frac{1}{2}\displaystyle M^2/mr^2+U(r)\]</span></p>

<p>we rewrite this as</p>
<p><span class="math display">\[\label{LL1/14.5}\tag*{(14.5)}\dot{\mathbf{r}}=\frac{\mathrm{d}^{}r}{\mathrm{d}^{}t}=\sqrt{\frac{2}{m}(E-U(r))-\frac{M^2}{m^2r^2}}\]</span></p>

<p>by integrating</p>
<p><span class="math display">\[\label{LL1/14.6}\tag*{(14.6)}t=\int\mathrm{d}^{}r\left(\frac{2}{m}(E-U(r))-\frac{M^2}{m^2r^2}\right)^{-1/2}+\text{constant}\]</span></p>

<p>writing <span class="math inline">\((14.2)\)</span> as <span class="math inline">\(\mathrm{d}^{}\phi=\frac{M}{mr^2}\mathrm{d}^{}t\)</span> we also find</p>
<p><span class="math display">\[\label{LL1/14.7}\tag*{(14.7)}\phi=\int\frac{M\mathrm{d}^{}r/r^2}{\sqrt{\frac{2}{m}(E-U(r))-\frac{M^2}{m^2r^2}}}+\text{constant}\]</span></p>

<hr>
<h3>See also</h3>
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A docs/centre-of-mass.html => docs/centre-of-mass.html +51 -0
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<hr style="clear:both;"/>


<p><span class="math display">\[\begin{equation} \mathbf{P}=\mathbf{P}&#39;+\mathbf{V}\sum_a m_a \end{equation}\]</span> <span class="math display">\[\begin{equation} \mathbf{V}=\mathbf{P}/\sum_a m_a=\sum_a m_a\mathbf{v}_a/\sum_a m_a \end{equation}\]</span> <span class="math display">\[\begin{equation} \mathbf{R}\equiv\sum_a m_a\mathbf{r}_a/\sum_a m_a \end{equation}\]</span> <span class="math display">\[\begin{equation} E=\frac{1}{2}\mu V^2+E_i \end{equation}\]</span> <span class="math display">\[\begin{equation} E=E&#39;+\mathbf{V}\cdot\mathbf{P}&#39;+\frac{1}{2}\mu V^2 \end{equation}\]</span></p>
<h2 id="model">Model</h2>
<ul>
<li><a href="mechanical-system.html">Classical mechanics</a></li>
</ul>

<hr>
<h3>See also</h3>
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<p>Find the <a href="lagrangian.html">Lagrangian</a> of a coplanar double pendulum</p>
<details><summary><b>Solution</b></summary>
<div class="solution">
<p><span class="math display">\[ L=\textstyle\frac{1}{2}\displaystyle(m_1+m_2)l_1^2\phi_1^2
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<hr>
<h3>See also</h3>
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<p>paragraph 14</p>

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<h3>See also</h3>
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</header>
<hr style="clear:both;"/>


<p><span class="math display">\[\begin{equation} \epsilon=E_i-E_{1i}-E_{2i} \end{equation}\]</span> <span class="math display">\[\begin{equation} \epsilon=\frac{1}{2}p_0^2\left(\frac{1}{m_1}+\frac{1}{m_2}\right)=\frac{p_0^2}{2m} \end{equation}\]</span> <span class="math display">\[\begin{equation} v^2+V^2-2vV\cos\theta=v_0^2 \end{equation}\]</span> <span class="math display">\[\begin{equation} \sin\theta_\text{max}=v_0/V \end{equation}\]</span> <span class="math display">\[\begin{equation} \tan\theta=v_0\sin\theta_0/(v_0\cos\theta_0+V) \end{equation}\]</span> <span class="math display">\[\begin{equation} \cos\theta_0=-\textstyle\frac{V}{v_0}\displaystyle\sin^2\theta\pm\cos\theta\sqrt{1-\textstyle\frac{V^2}{v_0}\displaystyle\sin^2\theta} \end{equation}\]</span> <span class="math display">\[\begin{equation} \frac{1}{2}\sin\theta_0d\theta_0 \end{equation}\]</span> <span class="math display">\[\begin{equation} \frac{1}{2}mv_0VdT \end{equation}\]</span> <span class="math display">\[\begin{equation} T_{10,\text{max}}=(M-m_1)\epsilon/M \end{equation}\]</span></p>
<h2 id="assumptions">Assumptions</h2>
<ul>
<li><a href="two-body-problem.html">Two body problem</a></li>
</ul>

<hr>
<h3>See also</h3>
<center><object data=disintegration.svg type=image/svg+xml></object></center>
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</div>
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<h5 style="text-align:right;float:right">model</h5>

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<hr style="clear:both;"/>


<p><span class="math display">\[ \mathbf{v}_{10}&#39;=m_2v\mathbf{n}_0/(m_1+m_2),\qquad \mathbf{v}_{20}&#39;=-m_1v\mathbf{n}_0/(m_1+m_2) \]</span></p>
<p><span class="math display">\[\begin{align}
\mathbf{v}_{1}&#39;&amp;=m_2v\mathbf{n}_0/(m_1+m_2)+(m_1\mathbf{v}_1+m_2\mathbf{v}_2)/(m_1+m_2),\\
\mathbf{v}_{2}&#39;&amp;=-m_1v\mathbf{n}_0/(m_1+m_2)+(m_1\mathbf{v}_1+m_2\mathbf{v}_2)/(m_1+m_2)
\end{align}\]</span></p>
<p><span class="math display">\[\begin{align}
\mathbf{p}_{1}&#39;&amp;=mv\mathbf{n}_0+m_1(\mathbf{p}_1+\mathbf{p}_2)/(m_1+m_2),\\
\mathbf{p}_{2}&#39;&amp;=-mv\mathbf{n}_0+m_2(\mathbf{p}_1+\mathbf{p}_2)/(m_1+m_2)
\end{align}\]</span></p>
<p><span class="math display">\[\begin{equation} \tan\theta_1=\frac{m_2\sin\chi}{m_1+m_2\cos\chi},\qquad \theta_2=\textstyle\frac{1}{2}\displaystyle(\textstyle\frac{\tau}{2}\displaystyle-\chi) \end{equation}\]</span> <span class="math display">\[\begin{align}
v_1&#39;&amp;=\frac{v}{m_1+m_2}\sqrt{m_1^2+m_2^2+2m_1m_2\cos\chi},\\
v_2&#39;&amp;=\frac{2m_1v}{m_1+m_2}\sin\textstyle\frac{1}{2}\displaystyle\chi
\end{align}\]</span> <span class="math display">\[\begin{equation} \mathbf{v}_1&#39;=\frac{m_1-m_2}{m_1+m_2}\mathbf{v},\qquad \mathbf{v}_2&#39;=\frac{2m_1}{m_1+m_2}\mathbf{v} \end{equation}\]</span> <span class="math display">\[\begin{equation} {{E_2}&#39;}_\text{max}=\frac{1}{2}m_2{{{v_2}&#39;}_\text{max}}^2=\frac{4m_1m_2}{(m_1+m_2)^2}E_1 \end{equation}\]</span> <span class="math display">\[\begin{equation} \sin\theta_\text{max}=\text{OC/OA}=m_2/m_1 \end{equation}\]</span> <span class="math display">\[\begin{equation} \theta_1=\textstyle\frac{1}{2}\displaystyle\chi,\qquad \theta_2=\textstyle\frac{1}{2}\displaystyle(\textstyle\frac{\tau}{2}\displaystyle-\chi) \end{equation}\]</span> <span class="math display">\[\begin{equation} v_1&#39;=v\cos\textstyle\frac{1}{2}\displaystyle\chi,\qquad v_2&#39;=v\sin\textstyle\frac{1}{2}\displaystyle\chi \end{equation}\]</span></p>
<h2 id="system">System</h2>
<ul>
<li><a href="two-body-problem.html">Two body problem</a></li>
</ul>

<hr>
<h3>See also</h3>
<center><object data=elastic-collisions.svg type=image/svg+xml></object></center>
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<h5 style="text-align:right;float:right">theory</h5>

</h1>

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


<p><span class="math display">\[\label{LL1/6.1}\tag*{(6.1)}E\equiv\sum_i\dot{q_i}\frac{\partial L}{\partial \dot{q_i}}-L\]</span></p>

<p><span class="math display">\[\label{LL1/6.2}\tag*{(6.2)}E=T(q,\dot{q})+U(q)\]</span></p>

<p><span class="math display">\[\label{LL1/6.3}\tag*{(6.3)}E=\sum_a\frac{1}{2}m_av_a^2+U(\mathbf{r}_1,\mathbf{r}_2,\ldots)\]</span></p>

<h2 id="model">Model</h2>
<ul>
<li><a href="mechanical-system.html">Classical mechanics</a></li>
</ul>

<hr>
<h3>See also</h3>
<center><object data=energy.svg type=image/svg+xml></object></center>
</body>
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<h5 style="text-align:right;float:right">math</h5>

</h1>

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


<h2 id="first-order-erf">First order ERF</h2>
<p>This trick works for finding a particular solution to <span class="math inline">\(n^{th}\)</span> order ODE with constant factors when the input is of exponential form.</p>
<p>Consider the following <span class="math inline">\(n^{th}\)</span> order constant factor ODE.</p>
<p><span class="math display">\[ P(D)z=e^{rt} \]</span></p>
<p>where <span class="math inline">\(P(D)\)</span> is a polynomial of the derivative operator <span class="math inline">\(D\)</span>, and <span class="math inline">\(r\)</span> is any complex number.</p>
<p>because of the property of exponential we have</p>
<p><span class="math display">\[ \tag{1} P(D)e^{rt}=P(r)e^{rt} \]</span></p>
<p>If <span class="math inline">\(P(D)\neq 0\)</span>, by linearity of the operator <span class="math inline">\(P(D)\)</span> we have immediately</p>
<p><span class="math display">\[ P(D)\left(\frac{e^{rt}}{P(r)}\right)=e^{rt} \]</span></p>
<p>Therefore we have the particular solution to our ODE</p>
<p><span class="math display">\[ z_p=\frac{e^{rt}}{P(r)} \]</span></p>
<h2 id="higher-order-erf">Higher order ERF</h2>
<p>When <span class="math inline">\(P(r)=0\)</span> can simply take the derivative of <span class="math inline">\((1)\)</span> with respect to <span class="math inline">\(r\)</span>.</p>
<p><span class="math display">\[ P(D)te^{rt}=P&#39;(r)e^{rt} \]</span></p>
<p>or more generally if all the <span class="math inline">\(P^{(i&lt;m)}(r)=0\)</span></p>
<p><span class="math display">\[ \tag{2} P(D)t^me^{rt}=P^{(m)}(r)e^{rt} \]</span></p>
<p>In which case our particular solution is</p>
<p><span class="math display">\[ z_p=\frac{t^{m}e^{rt}}{P^{(m)}(r)} \]</span></p>

<hr>
<h3 class="sources">Sources</h3>
<ul>
  <li>MIT 18.031x</li>
</ul>
<hr>
<h3>See also</h3>
<center><object data=erf.svg type=image/svg+xml></object></center>
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<h1 class="title"; style="text-align:left;float:left">forced oscillations

<h5 style="text-align:right;float:right">model</h5>

</h1>

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


<p>We consider <a href="small-oscillations.html">small oscillations</a> on which a variable force <span class="math inline">\(F(t)\)</span> acts, weak enough not to make the oscillations too large.</p>
<hr>
<p>Additionally to the 
        <span class="tooltip" class="math inline">
        \((21.2)\)
        <span class="tooltiptext"><span><span class="math inline">\(U(x)=\textstyle\frac{1}{2}\displaystyle kx^2\)</span></span>
</span>
        </span>
         potential, the system now has the potential energy <span class="math inline">\(U_e(x, t)\)</span> resulting from the force. We expand this as a series of the small quantity <span class="math inline">\(x\)</span>: <span class="math inline">\(U_e(x,t)\approx U(0,t)+x[\partial U_e/\partial x]_{x=0}\)</span>. We can omit the first term because it is a function of time only, and thus is a total derivative of a function of time, which doesn’t affect the Lagrangian. The second term is the the external alone is <span class="math inline">\(x[\partial U_e/\partial x]_{x=0}=-xF(t)\)</span>. Thus the Lagrangian of the system is</p>
<p><span class="math display">\[\label{LL1/22.1}\tag*{(22.1)}L=\textstyle\frac{1}{2}\displaystyle m\dot{x}^2-\textstyle\frac{1}{2}\displaystyle kx^2+xF(t)\]</span></p>

<p>According to 
        <span class="tooltip" class="math inline">
        \((2.6)\)
        <span class="tooltiptext"><span><span class="math inline">\(\frac{\mathrm{d}^{}}{\mathrm{d}^{}t}\frac{\partial L}{\partial \dot{q_i}}-\frac{\partial L}{\partial q_i}=0\qquad(i=1,2,...,s)\)</span></span>
</span>
        </span>
         we have the equation of motion</p>
<p><span class="math display">\[\label{LL1/22.2}\tag*{(22.2)}\ddot{x}+\omega^2 x=F(t)/m\]</span></p>

<h2 id="periodic-force">Periodic force</h2>
<p>We consider the special case of a periodic force of frequency <span class="math inline">\(\gamma\)</span></p>
<p><span class="math display">\[\label{LL1/22.3}\tag*{(22.3)}F(t)=f\cos(\gamma t+\beta)=\text{re}\left[Fe^{i\gamma t}\right]\]</span></p>

<p>We can easily find a particular solution to the equation of motion 
        <span class="tooltip" class="math inline">
        \((22.2)\)
        <span class="tooltiptext"><span><span class="math inline">\(\ddot{x}+\omega^2 x=F(t)/m\)</span></span>
</span>
        </span>
         using the <a href="erf.html">exponential response function</a>, which gives us</p>
<p><span class="math display">\[\label{LL1/22.4a}\tag*{(22.4a)}x_p=\text{re}\left[\frac{Fe^{i\gamma t}}{m(\omega^2-\gamma^2)}\right]=\frac{f\cos(\gamma t+\beta)}{m(\omega^2-\gamma^2)}\]</span></p>

<h3 id="resonance">Resonance</h3>
<p>in the special case of resonance <span class="math inline">\(\gamma=\omega\)</span>, we use a second order <a href="erf.html">exponential reponse</a></p>
<p><span class="math display">\[\label{LL1/22.5a}\tag*{(22.5a)}x_p=t\cdot\text{re}\left[\frac{Fe^{i\gamma t}}{2im\omega}\right]=t\cdot\frac{f\sin(\gamma t+\beta)}{2m\omega}\]</span></p>

<p>Thus, the amplitude of oscillations increases linearly with time, until the oscillations are large enough that the current model of small oscillations becomes invalid.</p>
<h3 id="beats">Beats</h3>
<p>Let’s consider the case close to resonance <span class="math inline">\(\gamma=\omega+\epsilon\)</span>, with <span class="math inline">\(\epsilon\ll\omega\)</span>. The solution is a linear combition of the homogenous solitution 
        <span class="tooltip" class="math inline">
        \((21.11)\)
        <span class="tooltiptext"><span><span class="math inline">\(x=\text{re}\left[Ae^{i\omega t}\right]\)</span></span>
</span>
        </span>
         and the particular solution 
        <span class="tooltip" class="math inline">
        \((22.4a)\)
        <span class="tooltiptext"><span><span class="math inline">\(x_p=\text{re}\left[\frac{Fe^{i\gamma t}}{m(\omega^2-\gamma^2)}\right]=\frac{f\cos(\gamma t+\beta)}{m(\omega^2-\gamma^2)}\)</span></span>
</span>
        </span>
        </p>
<p><span class="math display">\[\label{LL1/22.6}\tag*{(22.6)}x=\text{re}\left[Ae^{i\omega t}+Be^{i(\omega+\epsilon)t}\right]=\text{re}\left[[A+Be^{i\epsilon t}]e^{i\omega t}\right]\]</span></p>

<p>We can consider the term <span class="math inline">\(C=[A+Be^{i\epsilon t}]\)</span> as an amplitude varying much slower than the factor <span class="math inline">\(e^{i\omega t}\)</span>, because the period is much larger <span class="math inline">\(\textstyle\frac{2\pi}{\epsilon}\displaystyle\gg\textstyle\frac{2\pi}{\omega}\displaystyle\)</span>. We find the real amplitude</p>
<p><span class="math display">\[\label{LL1/22.7}\tag*{(22.7)}c^2=a^2+b^2+2ab\cos(\epsilon t+\beta-\alpha)\]</span></p>

<p>Thus the amplitude varies periodically with frequency <span class="math inline">\(\epsilon\)</span> between the limits <span class="math inline">\(|a-b|\leq c\leq|a+b|\)</span>. This phenomenon is called <em>beats</em>.</p>
<h2 id="arbitrary-force">Arbitrary force</h2>
<p>We rewrite 
        <span class="tooltip" class="math inline">
        \((22.2)\)
        <span class="tooltiptext"><span><span class="math inline">\(\ddot{x}+\omega^2 x=F(t)/m\)</span></span>
</span>
        </span>
         as</p>
<p><span class="math display">\[ \frac{\mathrm{d}^{}}{\mathrm{d}^{}t}(\dot{x}+i\omega x)-i\omega(\dot{x}+i\omega x)=\textstyle\frac{1}{m}\displaystyle F(t) \]</span></p>
<p>or</p>
<p><span class="math display">\[\label{LL1/22.8}\tag*{(22.8)}\dot{\xi}-i\omega\xi=\textstyle\frac{1}{m}\displaystyle F(t)\]</span></p>

<p>where</p>
<p><span class="math display">\[\label{LL1/22.9}\tag*{(22.9)}\xi=\dot{x}+i\omega x\]</span></p>

<p>is a complex equation. We can solve 
        <span class="tooltip" class="math inline">
        \((22.8)\)
        <span class="tooltiptext"><span><span class="math inline">\(\dot{\xi}-i\omega\xi=\textstyle\frac{1}{m}\displaystyle F(t)\)</span></span>
</span>
        </span>
         using the <a href="integration-factor.html">integration factor</a> trick. In our case the integration factor is <span class="math inline">\(\int p\mathrm{d}^{}t=-i\omega t\)</span>, such that</p>
<p><span class="math display">\[\label{LL1/22.10}\tag*{(22.10)}\xi=e^{i\omega t}\left[\int_0^t\textstyle\frac{1}{m}\displaystyle F(t)e^{-i\omega t}+\xi_0 \right]\]</span></p>

<p>The function <span class="math inline">\(x(t)\)</span> is given by the imaginary part of 
        <span class="tooltip" class="math inline">
        \((22.10)\)
        <span class="tooltiptext"><span><span class="math inline">\(\xi=e^{i\omega t}\left[\int_0^t\textstyle\frac{1}{m}\displaystyle F(t)e^{-i\omega t}+\xi_0 \right]\)</span></span>
</span>
        </span>
        , divided by <span class="math inline">\(\omega\)</span>.</p>
<p>In forced oscillations, the ernegy is of the system is not conserved, because of the energy gained from the external field. The energy of the system is, from 
        <span class="tooltip" class="math inline">
        \((6.2)\)
        <span class="tooltiptext"><span><span class="math inline">\(E=T(q,\dot{q})+U(q)\)</span></span>
</span>
        </span>
        </p>
<p><span class="math display">\[\label{LL1/22.11}\tag*{(22.11)}E=\textstyle\frac{1}{2}\displaystyle m(\dot{x}^2+\omega^2x^2)=\textstyle\frac{1}{2}\displaystyle m|\xi|^2\]</span></p>

<p>To get the total amount of transferred energy we put the lower bound at <span class="math inline">\(-\infty\)</span> and we integrate until <span class="math inline">\(t=\infty\)</span>, we get</p>
<p><span class="math display">\[ |\xi|^2=\textstyle\frac{1}{m^2}\displaystyle\left|\int_{-\infty}^{\infty}F(t)e^{-i\omega t}\right|^2 \]</span></p>
<p>Subsituting, we obtain</p>
<p><span class="math display">\[\label{LL1/22.12}\tag*{(22.12)}E=\textstyle\frac{1}{2m}\displaystyle\left|\int_{-\infty}^{\infty}F(t)e^{-i\omega t}\right|^2\]</span></p>

<p>which is the squared Fourier component of the force at the intrisic frequency <span class="math inline">\(\omega\)</span> of the system.</p>
<h2 id="short-acting-force">Short acting force</h2>
<p>if <span class="math inline">\(F(t)\)</span> act during a short amount of time compared to <span class="math inline">\(1/\omega\)</span>, then we can put <span class="math inline">\(e^{-i\omega t}\approx 1\)</span>. The transferered energy is simply</p>
<p><span class="math display">\[ E=\textstyle\frac{1}{2m}\displaystyle\left|\int_{-\infty}^{\infty}F(t)\right|^2 \]</span></p>
<p>This result means that a force of short duration gives the system momentum <span class="math inline">\(\int F\mathrm{d}^{}t\)</span> without bringing a perceptible displacement.</p>

<hr>
<h3>See also</h3>
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<h1 class="title"; style="text-align:left;float:left">Galileo’s relativity

<h5 style="text-align:right;float:right">model</h5>

</h1>

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<p><span class="math display">\[\begin{equation} L=L(v^2) \end{equation}\]</span> <span class="math display">\[\begin{equation} \mathbf{v}=\text{constant} \end{equation}\]</span> <span class="math display">\[\begin{equation} t=t&#39; \end{equation}\]</span></p>

<hr>
<h3>See also</h3>
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<!-- the ammonia maser -->
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<title>the ammonia maser</title>
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<h1 class="title"; style="text-align:left;float:left">the Hamiltonian operator

<h5 style="text-align:right;float:right">theory</h5>

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<p>The <a href="wave-function.html">wave function</a> completely determines the system, and also the future states of the system. This means that the derivative <span class="math inline">\(\partial\Psi/\partial t\)</span> must be determined by the function itself at an instant, and, by the principle of superpostion, the relationship must be linear</p>
<p><span class="math display">\[ i\partial\Psi/\partial t=\hat{L}\vphantom{L}^{}\Psi \]</span></p>
<p>We can derive some properties for <span class="math inline">\(\hat{L}\vphantom{L}^{}\)</span>, since the integral <span class="math inline">\(\left&lt; \Psi \middle| \Psi \right&gt;=\int\Psi^*\Psi\mathrm{d}^{}q\)</span> is a constant independant of time we have</p>
<p><span class="math display">\[ \frac{\mathrm{d}^{}}{\mathrm{d}^{}t}\int|\Psi|^2\mathrm{d}^{}q=
\int\Psi\frac{\partial\Psi^*}{\partial t}\mathrm{d}^{}q+
\int\frac{\partial\Psi}{\partial t}\Psi^*\mathrm{d}^{}q=0 \]</span></p>
<p>Substituting <span class="math inline">\(\partial\Psi/\partial t=-i\hat{L}\vphantom{L}^{}\Psi\)</span> and <span class="math inline">\(\partial\Psi^*/\partial t=i\hat{L}\vphantom{L}^{*}\Psi^*\)</span> and using</p>
<p><span class="math display">\[ \int\Psi\hat{L}\vphantom{L}^{*}\Psi^*\mathrm{d}^{}q-\int\Psi^*\hat{L}\vphantom{L}^{}\Psi\mathrm{d}^{}q=\int\Psi^*(\hat{L}\vphantom{L}^{t*}-\hat{L}\vphantom{L}^{})\Psi\mathrm{d}^{}q=0 \]</span></p>
<p>Since this must hold for an arbitrary <span class="math inline">\(\Psi\)</span>, we find that <span class="math inline">\(\hat{L}\vphantom{L}^{}\)</span> must be Hermitian</p>
<p><span class="math display">\[ \hat{L}\vphantom{L}^{t}=\hat{L}\vphantom{L}^{*} \]</span></p>
<p>Let us find the classical quantity to which <span class="math inline">\(\hat{L}\vphantom{L}^{}\)</span> corresponds, we differentiate 
        <span class="tooltip" class="math inline">
        \((6.1)\)
        <span class="tooltiptext"><span><span class="math inline">\(\Psi=ae^{iS/\hbar}\)</span></span>
</span>
        </span>
        </p>
<p><span class="math display">\[ \frac{\partial\Psi}{\partial t}=\frac{i}{\hbar}\frac{\partial S}{\partial t}\Psi \]</span></p>
<p>the slowly varying amplitude <span class="math inline">\(a\)</span> need not be differentiated.</p>
<p>Comparing this to the definition <span class="math inline">\(\partial\Psi/\partial t=-i\hat{L}\vphantom{L}^{}\Psi\)</span> we have <span class="math inline">\(\hat{L}\vphantom{L}^{}=-(1/\hbar)\partial S/\partial t\)</span>.</p>
<p>As we know from classical mechanics, the derivative <span class="math inline">\(-\partial S/\partial t\)</span> is just Hamilton’s function <span class="math inline">\(H\)</span> for a mechanical system. In quantum mechanics we call this the Hamiltonian <a href="quantum-operator.html">operator</a>, or the Hamiltonian of a system.</p>
<p><span class="math display">\[\label{LL3/8.1}\tag*{(8.1)}i\hbar\frac{\partial\Psi}{\partial t}=\hat{H}\vphantom{H}^{}\Psi\]</span></p>

<p>If the form of the Hamiltonian is known, equation 
        <span class="tooltip" class="math inline">
        \((8.1)\)
        <span class="tooltiptext"><span><span class="math inline">\(i\hbar\frac{\partial\Psi}{\partial t}=\hat{H}\vphantom{H}^{}\Psi\)</span></span>
</span>
        </span>
         determines the wave function of the physical system concerned. This fundamental equation of quantum mechanics is called the wave equation.</p>

<hr>
<h3>See also</h3>
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<p>We have seen that real physical quantities are represented by hermitian linear integral <a href="quantum-operator.html">operators</a> <span class="math inline">\((3.15)\)</span>. Show that the eingenfunctions of these operators are orthogonal.</p>
<h2 id="solution">solution</h2>
<p>Let <span class="math inline">\(f_n\)</span> and <span class="math inline">\(f_m\)</span> be two different eigenvalues of the quantity <span class="math inline">\(f\)</span>, and <span class="math inline">\(\Psi_n\)</span>, <span class="math inline">\(\Psi_m\)</span> the corresponding eigenfunctions</p>
<p><span class="math display">\[ \hat{f}\Psi_n=f_n\Psi_n,\qquad \hat{f}\Psi_m=f_m\Psi_m \]</span></p>
<p>Multiplying the first equation by <span class="math inline">\(\Psi_m*\)</span>, and multiplying the conjugate of the second by <span class="math inline">\(\Psi_n\)</span> we have</p>
<p><span class="math display">\[ \Psi_m^*\hat{f}\Psi_n-\Psi_n\hat{f}\vphantom{f}^*\Psi_m^*=(f_n-f_m)\Psi_n\Psi_m^* \]</span></p>
<p>integrating over q, and using the fact that <span class="math inline">\(\hat{f}\)</span> is Hermitian <span class="math inline">\(\hat{f}\equiv\hat{f}\vphantom{f}^*\)</span> so that the left side is zero, we have</p>
<p><span class="math display">\[ (f_n-f_m)\int\Psi_n\Psi_m\vphantom{f}^*\mathrm{d}^{}q=0 \]</span></p>
<p>and since <span class="math inline">\(f_n\neq f_m\)</span>, we obtain the orthogonality property.</p>

<hr>
<h3>See also</h3>
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</header>
<hr style="clear:both;"/>


<p><span class="math display">\[\label{LL1/32.1}\tag*{(32.1)}T=\textstyle\frac{1}{2}\displaystyle\mu V^2+\textstyle\frac{1}{2}\displaystyle\sum m\left[\Omega^2r^2-(\mathbf{\Omega}\cdot\mathbf{r})^2\right]\]</span></p>

<p><span class="math display">\[\label{LL1/32.2}\tag*{(32.2)}I_{ik}=\sum m\left(x_l^2\delta_{ik}-x_ix_k\right)\]</span></p>

<p><span class="math display">\[\label{LL1/32.3}\tag*{(32.3)}T=\textstyle\frac{1}{2}\displaystyle\mu V^2+\textstyle\frac{1}{2}\displaystyle I_{ik}\Omega_i\Omega_k\]</span></p>

<p><span class="math display">\[\label{LL1/32.4}\tag*{(32.4)}L=\textstyle\frac{1}{2}\displaystyle\mu V^2+\textstyle\frac{1}{2}\displaystyle I_{ik}\Omega_i\Omega_k-U\]</span></p>

<p><span class="math display">\[\label{LL1/32.5}\tag*{(32.5)}I_{ik}=I_{ki}\]</span></p>

<p><span class="math display">\[\label{LL1/32.6}\tag*{(32.6)}I_{ik}=\left[\begin{matrix}
        \sum m(y^2+z^2) &amp; -\sum mxy &amp; -\sum mxz\\
        -\sum mxy &amp; \sum m(x^2+z^2) &amp; -\sum myz\\
        -\sum mxz &amp; -\sum myz &amp; \sum m(x^2+y^2)\\
\end{matrix}\right]\]</span></p>

<p><span class="math display">\[\label{LL1/32.7}\tag*{(32.7)}I_{ik}=\int \rho\left(x_l^2\delta_{ik}-x_ix_k\right)\mathrm{d}^{}V\]</span></p>

<p><span class="math display">\[\label{LL1/32.12}\tag*{(32.12)}I&#39;_{ik}=I_{ik}+\mu\left(a^2\delta_{ik}-a_ia_k\right)\]</span></p>

<h3 id="principle-moments-of-inertia">Principle moments of inertia</h3>
<p><span class="math display">\[\label{LL1/32.8}\tag*{(32.8)}T_\text{rot}=\textstyle\frac{1}{2}\displaystyle\left(I_1\Omega_1^2+I_2\Omega_2^2+I_3\Omega_3^2\right)\]</span></p>

<p><span class="math display">\[\label{LL1/32.9}\tag*{(32.9)}I_1+I_2=\sum m(x_1^2+x_2^2+2x_3^2) \geq \sum m(x_1^2+x_2^2)=I_3\]</span></p>

<h4 id="coplanar-system-of-particles">Coplanar system of particles</h4>
<p><span class="math display">\[\label{LL1/32.10}\tag*{(32.10)}I_3=I_1+I_2\]</span></p>

<h4 id="colinear-system-of-particles">Colinear system of particles</h4>
<p><span class="math display">\[\label{LL1/32.11}\tag*{(32.11)}I_1=I_2=\sum m{x_3}^2,\quad I_3=0\]</span></p>

<h2 id="model">Model</h2>
<ul>
<li><a href="rigid-body.html">Rigid body</a></li>
</ul>
<h2 id="properties">Properties</h2>
<ul>
<li>rank 2 tensor</li>
<li>antisymetric</li>
</ul>

<hr>
<h3>See also</h3>
<center><object data=inertia-tensor.svg type=image/svg+xml></object></center>
</body>
</div>
</html>

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<h1 class="title"; style="text-align:left;float:left">integration factor

<h5 style="text-align:right;float:right">math</h5>

</h1>

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


<p>The are many ways to solve first order linear differential equations, but this trick is quite neat.</p>
<p>the general form of a first order linear DE is</p>
<p><span class="math display">\[ \dot{y}+p(x)y=q(x) \]</span></p>
<p>We are going to multiply through by the “integration factor” <span class="math inline">\(e^{\int p\mathrm{d}^{}x}\)</span></p>
<p><span class="math display">\[ \dot{y}e^{\int p\mathrm{d}^{}x}+pye^{\int p\mathrm{d}^{}x}=qe^{\int p\mathrm{d}^{}x} \]</span></p>
<p>Here we use the identity <span class="math inline">\(\frac{\mathrm{d}^{}}{\mathrm{d}^{}x}\int p\mathrm{d}^{}x=p(x)\)</span> and we recognize the derivative of a product</p>
<p><span class="math display">\[ (ye^{\int p\mathrm{d}^{}x})&#39;=qe^{\int p\mathrm{d}^{}x} \]</span></p>
<p>Thus, integration on both sides we have</p>
<p><span class="math display">\[ y=e^{-\int p\mathrm{d}^{}x}\left[\int qe^{\int p\mathrm{d}^{}x}\mathrm{d}^{}x+\text{constant}\right] \]</span></p>
<p>Which is a general solution for <span class="math inline">\(y\)</span></p>
<h3 id="notes">Notes</h3>
<p>The trick works with any primitive <span class="math inline">\(P(x)\)</span> of <span class="math inline">\(p(x)\)</span>, so we don’t need to bother with the constant in the integration factor <span class="math inline">\(e^{\int p\mathrm{d}^{}x}\)</span>.</p>
<p>Other methods for solving the same equation involve solving the homogeneous equaton <span class="math inline">\((q(x)=0)\)</span>, and then varying the constant of the homegenous solution. I find the integration factor more elegant.</p>

<hr>
<h3 class="sources">Sources</h3>
<ul>
  <li>MIT 18.031x</li>
</ul>
<hr>
<h3>See also</h3>
<center><object data=integration-factor.svg type=image/svg+xml></object></center>
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</h1>

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<p>hello <a href="central-field.html">central field</a> TODO LL1 p15</p>

<hr>
<h3>See also</h3>
<center><object data=keplers-problem.svg type=image/svg+xml></object></center>
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<h5 style="text-align:right;float:right">theory</h5>

</h1>

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


<p>The Lagrangian is defined by the principle of least action</p>
<p><span class="math display">\[\label{LL1/2.1}\tag*{(2.1)}S=\int_{t_1}^{t_2}L(q,\dot{q},t)\mathrm{d}^{}t\]</span></p>

<p><span class="math display">\[\label{LL1/2.2}\tag*{(2.2)}q(t)+\delta q(t)\]</span></p>

<p><span class="math display">\[\label{LL1/2.3}\tag*{(2.3)}\delta q(t_1)=\delta q(t_2) = 0\]</span></p>

<p><span class="math display">\[\label{LL1/2.4}\tag*{(2.4)}\delta S=\delta \int_{t_1}^{t_2}L(q,\dot{q},t)\mathrm{d}^{}t=0\]</span></p>

<p><span class="math display">\[\label{LL1/2.5}\tag*{(2.5)}\delta S=\left[\frac{\partial L}{\partial \dot{q}}\delta q\right]_{t_1}^{t_2}+\int_{t_1}^{t_2}\left(\frac{\partial L}{\partial q}-\frac{\mathrm{d}^{}}{\mathrm{d}^{}t}\frac{\partial L}{\partial \dot{q}}\right)\delta q \mathrm{d}^{}t=0\]</span></p>

<p><span class="math display">\[\label{LL1/2.6}\tag*{(2.6)}\frac{\mathrm{d}^{}}{\mathrm{d}^{}t}\frac{\partial L}{\partial \dot{q_i}}-\frac{\partial L}{\partial q_i}=0\qquad(i=1,2,...,s)\]</span></p>

<p><span class="math display">\[\label{LL1/2.7}\tag*{(2.7)}\lim L=L_A+L_B\]</span></p>

<p><span class="math display">\[\label{LL1/2.8}\tag*{(2.8)}L&#39;(q,\dot{q},t)=L(q,\dot{q},t)+\frac{\mathrm{d}^{}}{\mathrm{d}^{}t}f(q,t)\]</span></p>

<h2 id="model">Model</h2>
<ul>
<li><a href="mechanical-system.html">Classical mechanics</a></li>
</ul>

<hr>
<h3>See also</h3>
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<h2 id="indeterminate-form-00">indeterminate form 0/0</h2>
<p>if <span class="math inline">\(f(x)\xrightarrow[a]{}0\)</span> and <span class="math inline">\(g(x)\xrightarrow[a]{}0\)</span></p>
<p>and the functions <span class="math inline">\(f\)</span> and <span class="math inline">\(g\)</span> are differentiable near the point <span class="math inline">\(x=a\)</span>, then</p>
<p><span class="math display">\[ \lim_{x\to a}{\frac{f(x)}{g(x)}} =
\lim_{x\to a}{\frac{f&#39;(x)}{g&#39;(x)}}
\]</span></p>
<h2 id="indeterminate-form-inftyinfty">indeterminate form <span class="math inline">\(\infty/\infty\)</span></h2>
<p>if <span class="math inline">\(f(x)\xrightarrow[a]{}\infty\)</span> and <span class="math inline">\(g(x)\xrightarrow[a]{}\infty\)</span></p>
<p>and the functions <span class="math inline">\(f\)</span> and <span class="math inline">\(g\)</span> are differentiable near the point <span class="math inline">\(x=a\)</span>, then</p>
<p><span class="math display">\[ \lim_{x\to a}{\frac{f(x)}{g(x)}} =
\lim_{x\to a}{\frac{f&#39;(x)}{g&#39;(x)}}
\]</span></p>
<h2 id="notes">notes</h2>
<p>We can replace <span class="math inline">\(a\)</span> with <span class="math inline">\(a^+\)</span> or <span class="math inline">\(a^-\)</span> and the results still hold.<br />
We can replace <span class="math inline">\(a\)</span> with <span class="math inline">\(\pm\infty\)</span> and the results still hold.</p>
<h2 id="other-indeterminate-forms">other indeterminate forms</h2>
<p><span class="math inline">\(0\cdot\infty\)</span><br />
<span class="math inline">\(\infty-\infty\)</span><br />
<span class="math inline">\(0^0\)</span><br />
<span class="math inline">\(1^\infty\)</span><br />
<span class="math inline">\(\infty^0\)</span></p>
<p>Should be rearranged to of the form <span class="math inline">\(0/0\)</span> or <span class="math inline">\(\infty/\infty\)</span> in order to apply l’Hôpital’s rule.</p>

<hr>
<h3 class="sources">Sources</h3>
<ul>
  <li>MIT 18.01x</li>
</ul>
<hr>
<h3>See also</h3>
<center><object data=lhopital.svg type=image/svg+xml></object></center>
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<p>paragraph 11</p>

<hr>
<h3>See also</h3>
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<p>We have found that the equations of motion are unchanged by multiplication of the <a href="lagrangian.html">Lagrangian</a> by any constant. This allows us to determine some preperties of motion without necessarily solving the equations.</p>
<p>Let us consider a homogeneous potential <span class="math inline">\(U(q)\)</span></p>
<p><span class="math display">\[\label{LL1/10.1}\tag*{(10.1)}U(\alpha\mathbf{r}_1,\alpha\mathbf{r}_2,\ldots,\alpha\mathbf{r}_n)=\alpha^kU(\mathbf{r}_1,\mathbf{r}_2,\ldots,\mathbf{r}_n)\]</span></p>

<p>where <span class="math inline">\(\alpha\)</span> is any constant and <span class="math inline">\(k\)</span> is the degree of homogeneity of the function.</p>
<p>Let us carry out a transformation where the co-ordinates are changed by a factor <span class="math inline">\(\alpha\)</span> and the time by a factor <span class="math inline">\(\beta\)</span></p>
<p><span class="math display">\[
\mathbf{r}_a\to\alpha\mathbf{r&#39;}_a,\qquad
t\to\beta t&#39;
\]</span></p>
<p>Then all velocities <span class="math inline">\(\mathbf{v}_a=\mathrm{d}^{}\mathbf{r}_a/\mathrm{d}^{}t\)</span> are changed by a factor <span class="math inline">\(\alpha/\beta\)</span>, therefore the kinetic energy by a factor <span class="math inline">\(\alpha^2/\beta^2\)</span>. The potential energy is multiplied by a factor <span class="math inline">\(\alpha^k\)</span>.</p>
<p><span class="math display">\[
T\to\alpha^2/\beta^2T&#39;,\qquad
U\to\alpha^kU&#39;
\]</span></p>
<p>If these factor are the same, i.e. <span class="math inline">\(\beta=\alpha^{1-k/2}\)</span>, then the motion is unchanged. (todo link to paragraph 2 explanation). A change of co-ordinates of the particle by the same factor corresponds to the replacement of the paths by other paths, geometrically similar but differing in size. By definition of <span class="math inline">\(\alpha\)</span> and <span class="math inline">\(\beta\)</span>, the times of the motion between corresponding points are in the ratio</p>
<p><span class="math display">\[\label{LL1/10.2}\tag*{(10.2)}t&#39;/t=(l&#39;/l)^{1-k/2}\]</span></p>

<p>where <span class="math inline">\(l&#39;/l\)</span> is the ratio of linear dimensions of the two paths. We can find similar relations for other characteristics of motion.</p>
<p><span class="math display">\[\label{LL1/10.3}\tag*{(10.3)}v&#39;/v=(l&#39;/l)^{k/2},\quad
E&#39;/E=(l&#39;/l)^{k},\quad
M&#39;/M=(l&#39;/l)^{1+k/2}\quad\]</span></p>

<h2 id="some-consequences">Some consequences</h2>
<h3 id="small-oscillations-k2">Small oscillations <span class="math inline">\(k=2\)</span></h3>
<p>For <span class="math inline">\(k=2\)</span>, we find the motion is 
        <span class="tooltip" class="math inline">
        \((21.11)\)
        <span class="tooltiptext"><span><span class="math inline">\(x=\text{re}\left[Ae^{i\omega t}\right]\)</span></span>
</span>
        </span>
        , which verifies that the period of oscillations is independant of the amplitude.</p>
<h3 id="free-fall-k1">Free fall <span class="math inline">\(k=1\)</span></h3>
<p>In a uniform field, the potential energy is a linear function of co-ordinates (see 
        <span class="tooltip" class="math inline">
        \((5.8)\)
        <span class="tooltiptext"><span><span class="math inline">\(\)</span></span>
</span>
        </span>
        ), <span class="math inline">\(k=1\)</span>. We then have <span class="math inline">\(t&#39;/t=\sqrt{l&#39;/l}\)</span>. Hence, for example, in fall under gravity, the distance fallen goes as the square of the time you’ve been falling.</p>
<h3 id="keplers-third-law-k-1">Kepler’s third law <span class="math inline">\(k=-1\)</span></h3>
<p>the Newtonian attraction of two masses, of the Coulomb attraction of two charges, the potential energy is inversely proportional to the distance apart <span class="math inline">\(k=-1\)</span>. Then <span class="math inline">\((t&#39;/t)^2=(l&#39;/l)^3\)</span>. Hence, the square of the time of revolution in the orbit goes as the cube of the size of the orbit.</p>
<p><img src="https://upload.wikimedia.org/wikipedia/commons/5/52/Solar_system_planets_a%28AU%29_vs_period%28terrestrial_years%29.svg"></p>
<p>We can see in this log-log plot from wikipedia that the slope is approximately <span class="math inline">\(2/3\)</span>, as expected.</p>
<h2 id="virial-theorem">Virial theorem</h2>
<p><span class="math display">\[\label{LL1/10.4}\tag*{(10.4)}2T=\sum_a\mathbf{p}_a\mathbf{v}_a=\frac{\mathrm{d}^{}}{\mathrm{d}^{}t}(\sum_a\mathbf{p}_a\mathbf{r}_a)-\sum_a\mathbf{r}_a\dot{\mathbf{p}}_a\]</span></p>

<p><span class="math display">\[\label{LL1/10.5}\tag*{(10.5)}2\overline{T}=\overline{\sum_a\mathbf{r}_a\cdot\partial U/\partial \mathbf{r}_a}\]</span></p>

<p><span class="math display">\[\label{LL1/10.6}\tag*{(10.6)}2\overline{T}=k\overline{U}\]</span></p>

<p><span class="math display">\[\label{LL1/10.7}\tag*{(10.7)}\overline{U}=2E/(k+2),\qquad \overline{T}=kE/(k+2)\]</span></p>

<hr>
<h3>See also</h3>
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<p><span class="math display">\[ L=\sum_a\frac{1}{2}m_av_a^2-U(\mathbf{r}_1,\mathbf{r}_2,\ldots) \]</span> <span class="math display">\[ \frac{\mathrm{d}^{}}{\mathrm{d}^{}t}\frac{\partial L}{\partial \mathbf{v}_a}=\frac{\partial L}{\partial \mathbf{r}_a} \]</span> <span class="math display">\[ m_a\mathrm{d}^{}\mathbf{v}_a/\mathrm{d}^{}t=-\partial U/\partial \mathbf{r}_a \]</span> <span class="math display">\[ \mathbf{F}=-\partial U/\partial \mathbf{r}_a \]</span> <span class="math display">\[ L=\sum_{i,k}a_{ik}(q)\dot{q_i}\dot{q_k}-U(q) \]</span> <span class="math display">\[ L=\frac{1}{2}mv^2-U(\mathbf{r}, t) \]</span> <span class="math display">\[ m\dot{\mathbf{v}}=-\partial U/\partial \mathbf{r} \]</span> <span class="math display">\[ U=-\mathbf{F}\cdot\mathbf{r} \]</span></p>
<h2 id="assumption">Assumption</h2>
<ul>
<li><a href="galileo-relativity.html">Galileo’s Relativity</a></li>
</ul>
<h2 id="model">Model</h2>
<ul>
<li><a href="particle.html">particles</a></li>
</ul>

<hr>
<h3>See also</h3>
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<h1 class="title"; style="text-align:left;float:left">the momentum operator

<h5 style="text-align:right;float:right">theory</h5>

</h1>

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<p>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.</p>
<p><span class="math display">\[ \psi(r_1+\delta r,r_2+\delta r,...)=\psi(r_1,r_2,...)+\delta r\sum_a \nabla_a\psi \]</span></p>
<p>The expresion <span class="math inline">\(\hat{O}\vphantom{O}^{}=1+\delta r\sum_a\nabla_a\)</span> can be regarded as the “infinitely small displacement” operator. This sdisplacement should not change the <a href="hamiltonian-operator.html">Hamiltonian operator</a>, by this we mean that the order of application doesn’t matter</p>
<p><span class="math display">\[ \hat{H}\vphantom{H}^{}\hat{O}\vphantom{O}^{}-\hat{O}\vphantom{O}^{}\hat{H}\vphantom{H}^{}=0 \]</span></p>
<p>Since the unit operator and the multiplication by <span class="math inline">\(\delta r\)</span> both commute with <span class="math inline">\(\hat{H}\vphantom{H}^{}\)</span> we are left with</p>
<p><span class="math display">\[\label{LL3/15.1}\tag*{(15.1)}(\sum_a\nabla_a)\hat{H}\vphantom{H}^{}-\hat{H}\vphantom{H}^{}(\sum_a\nabla_a)=0\]</span></p>

<p>As we know the commutation with <span class="math inline">\(\hat{H}\vphantom{H}^{}\)</span> mean that the corresponding physical quantity is conserved. The conserved quantity that follows from the homogoneity of space is the momentum <span class="math inline">\(p\)</span>.</p>
<p>The operator <span class="math inline">\(\nabla_a\)</span> must then be proportional to the momentum of the <span class="math inline">\(a^{th}\)</span> particle. We find the constatnt of proportionality by passing to the limit of classical mechanics and putting <span class="math inline">\(\hat{p}\vphantom{p}^{}=c\nabla\)</span> and using <span class="math inline">\((6.1)\)</span></p>
<p><span class="math display">\[ \hat{p}\vphantom{p}^{}\Psi=(i/\hbar)cae^{(i/\hbar)S}\nabla S=c(i/\hbar)\Psi\nabla S \]</span></p>
<p>We know from classical mechanics that <span class="math inline">\(p=\nabla S\)</span>. Therefore <span class="math inline">\(c=-i\hbar\)</span>. Thus we have the momentum operator <span class="math inline">\(\hat{p}\vphantom{p}^{}=-i\hbar\nabla\)</span>, or in components</p>
<p><span class="math display">\[\label{LL3/15.2}\tag*{(15.2)}\hat{p}\vphantom{p}^{}_x=i\hbar\partial/\partial x,\qquad
\hat{p}\vphantom{p}^{}_y=i\hbar\partial/\partial y,\qquad
\hat{p}\vphantom{p}^{}_z=i\hbar\partial/\partial z\]</span></p>

<p>We note that these operators are Hermitian, as they should be, as they represent real physical quantities. For arbitrary function <span class="math inline">\(\psi(x)\)</span> and <span class="math inline">\(\phi(x)\)</span> which vanish at inifity, we have</p>
<p><span class="math display">\[ \int\phi\hat{p}\vphantom{p}^{}_x\psi\mathrm{d}^{}x=\int\phi\frac{\partial\psi}{\partial x}\mathrm{d}^{}x=\int\psi\frac{\partial\phi}{\partial x}\mathrm{d}^{}x=\int\psi\hat{p}\vphantom{p}^{*}_x\phi\mathrm{d}^{}x \]</span></p>
<p>and this is the condition that the operator should be Hermitian.</p>
<p>Since differentiating with respect to two different variables is independant of order, it is clear that the operators commute with one another.</p>
<p><span class="math display">\[\label{LL3/15.3}\tag*{(15.3)}\hat{p}\vphantom{p}^{}_x\hat{p}\vphantom{p}^{}y-\hat{p}\vphantom{p}^{}_y\hat{p}\vphantom{p}^{}x=0,\qquad
\hat{p}\vphantom{p}^{}_x\hat{p}\vphantom{p}^{}z-\hat{p}\vphantom{p}^{}_z\hat{p}\vphantom{p}^{}x=0,\qquad
\hat{p}\vphantom{p}^{}_y\hat{p}\vphantom{p}^{}z-\hat{p}\vphantom{p}^{}_z\hat{p}\vphantom{p}^{}y=0\]</span></p>

<p>This means that all 3 components can have definite values simultaneously, which means the momentum <span class="math inline">\(\mathbf{p}\)</span> can have a definite direction and amplitude.</p>
<p>To find the eigenfunctions and eigenvalues of the momentum operators we have to solve</p>
<p><span class="math display">\[\label{LL3/15.4}\tag*{(15.4)}-i\hbar\partial\psi/\partial x=p_x\psi,\qquad
-i\hbar\partial\psi/\partial y=p_y\psi,\qquad
-i\hbar\partial\psi/\partial z=p_z\psi\]</span></p>

<p>The solution to the first equation is</p>
<p><span class="math display">\[ \psi=f(y,z)e^{(i/\hbar)p_xx} \]</span></p>
<p>Thus the eigenvalues form a continuous spectrum from <span class="math inline">\(-\infty\)</span> to <span class="math inline">\(+\infty\)</span>.</p>
<p>There is a common solution to all three equations, which correspond to a state where the momentum <span class="math inline">\(\mathbf{p}\)</span> is entirely defined.</p>
<p><span class="math display">\[\label{LL3/15.5}\tag*{(15.5)}\psi=Ce^{(i/\hbar)\mathbf{p}\cdot\mathbf{r}}\]</span></p>

<p>This is a completely determined wave function, therefore the momentum <span class="math inline">\(\mathbf{p}=(p_x,p_y,p_z)\)</span> forms a complete basis. Let us now find the normalization factor <span class="math inline">\(C\)</span>. The rule for normalization is</p>
<p><span class="math display">\[\label{LL3/15.6}\tag*{(15.6)}\int\Psi_{p&#39;}^*\Psi_p\mathrm{d}^{}v=\delta(\mathbf{p&#39;}-\mathbf{p})\]</span></p>

<p>The integration can be effected with the formula</p>
<p><span class="math display">\[\label{LL3/15.7}\tag*{(15.7)}\textstyle\frac{1}{\tau}\displaystyle\int_{-\infty}^{+\infty}e^{i\alpha x}\mathrm{d}^{}x=\delta(\alpha)\]</span></p>

<p>We have</p>
<p><span class="math display">\[\begin{align}
    \int\Psi_{p&#39;}^*\Psi_p\mathrm{d}^{}V&amp;=C^2\int e^{(i/\hbar)(\mathbf{p&#39;}-\mathbf{p})\mathbf{r}}\mathrm{d}^{}V\\
    &amp;=C^2h^3\delta(\mathbf{p&#39;}-\mathbf{p})
\end{align}\]</span></p>
<p>Therefore we mut have <span class="math inline">\(C^2h^3=1\)</span>. Thus the normalized eigenfunction <span class="math inline">\(\psi_p\)</span> is</p>
<p><span class="math display">\[\label{LL3/15.8}\tag*{(15.8)}\psi=h^{-3/2}e^{i\tau(\mathbf{p}\cdot\mathbf{r}/h)}\]</span></p>

<p>We can expand any wave function <span class="math inline">\(\psi(\mathbf{r})\)</span> in terms of the eigenfunctions <span class="math inline">\(\psi_\mathbf{p}\)</span>, as a Fourier integral</p>
<p><span class="math display">\[\label{LL3/15.9}\tag*{(15.9)}\psi(\mathbf{r})=
\int a(\mathbf{p})\psi_\mathbf{p}(\mathbf{r})\mathrm{d}^{3}p=
\int a(\mathbf{p})e^{i\tau(\mathbf{p}\cdot\mathbf{r}/h)}\mathrm{d}^{3}p\]</span></p>

<p>(where <span class="math inline">\(\mathrm{d}^{3}p=\mathrm{d}^{}p_x\mathrm{d}^{}p_y\mathrm{d}^{}p_z\)</span>). The expansion coefficients <span class="math inline">\(a(\mathbf{p})\)</span> are, according to formula <span class="math inline">\((5.3)\)</span></p>
<p><span class="math display">\[\label{LL3/15.10}\tag*{(15.10)}a(\mathbf{p})=
\int \psi(\mathbf{r})\psi_\mathbf{p}^*(\mathbf{r})\mathrm{d}^{}V=
h^{-3/2}\int\psi(\mathbf{r})e^{i\tau(\mathbf{p}\cdot\mathbf{r}/h)}\mathrm{d}^{}V\]</span></p>

<p>The function <span class="math inline">\(a(\mathbf{p})\)</span> can be regarded as the wave function in the <span class="math inline">\(\mathbf{p}\)</span> representation. <span class="math inline">\(|a(\mathbf{p})|^2\mathrm{d}^{3}p\)</span> is the probability that the momentum <span class="math inline">\(\mathbf{p}\)</span> has a value in the interval <span class="math inline">\(\mathrm{d}^{3}p\)</span>.</p>
<h2 id="the-position-operator-hatmathbfrvphantommathbfr">The position operator <span class="math inline">\(\hat{\mathbf{r}}\vphantom{\mathbf{r}}^{}\)</span></h2>
<p>TODO 15.11 and 15.12</p>
<h2 id="uncertainty-relations">Uncertainty relations</h2>
<p>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</p>
<p><span class="math display">\[\label{LL3/16.1}\tag*{(16.1)}\hat{p}\vphantom{p}^{}_xy-y\hat{p}\vphantom{p}^{}_x=0,\qquad
\hat{p}\vphantom{p}^{}_xz-z\hat{p}\vphantom{p}^{}_x=0,\qquad
\hat{p}\vphantom{p}^{}_yz-z\hat{p}\vphantom{p}^{}_y=0\]</span></p>

<p>For the commutation of <span class="math inline">\(\hat{p}\vphantom{p}^{}_x\)</span> and <span class="math inline">\(x\)</span> we write</p>
<span class="math display">\[\begin{aligned}
(\hat{p}\vphantom{p}^{}_xx-x\hat{p}\vphantom{p}^{}_x)\psi&amp;=-i\hbar(\partial(x\psi)/\partial x)+i\hbar x\partial\psi/\partial x\\
&amp;=-i\hbar\psi
\end{aligned}\]</span>
<p>We find that the commutator reduces to a multiplication by <span class="math inline">\(-i\hbar\)</span>, the same is true for <span class="math inline">\(y\)</span> and <span class="math inline">\(z\)</span></p>
<p><span class="math display">\[\label{LL3/16.2}\tag*{(16.2)}\hat{p}\vphantom{p}^{}_xx-x\hat{p}\vphantom{p}^{}_x=
\hat{p}\vphantom{p}^{}_yy-y\hat{p}\vphantom{p}^{}_y=
\hat{p}\vphantom{p}^{}_zz-z\hat{p}\vphantom{p}^{}_z=
-i\hbar\]</span></p>

<p>or they can be rewritten in the form</p>
<p><span class="math display">\[\label{LL3/16.3}\tag*{(16.3)}\hat{p}\vphantom{p}^{}_ik-k\hat{p}\vphantom{p}^{}_i=-i\hbar\delta_{ki}\qquad(i,k=x,y,z)\]</span></p>

<p>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.</p>
<p>Let us suppose a particle has a definite momentum <span class="math inline">\(\mathbf{p}_0\)</span>, mathematically, the wave function has the form <span class="math inline">\(\psi=u(\mathbf{r})e^{(i/\hbar)\mathbf{r}\cdot\mathbf{p}_0}\)</span>, where $u(). Let us suppose that the particle is in a finite region of space, whose dimensions is of the order of magnitude <span class="math inline">\(\Delta x, \Delta y, \Delta z\)</span>, so that <span class="math inline">\(u(\mathbf{r})\)</span> only differs significantly from 0 inside this volume.</p>
<p>TODO 16.6 -&gt; 16.11</p>

<hr>
<h3>See also</h3>
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<p><span class="math display">\[ \sum_a\partial L/\partial \mathbf{r}_a=0 \]</span> <span class="math display">\[ \mathbf{P}\equiv\sum_a\partial L/\partial \mathbf{v}_a \]</span> <span class="math display">\[ \mathbf{P}=\sum_am_a\mathbf{v}_a \]</span> <span class="math display">\[ \sum_a\mathbf{F}_a=0 \]</span> <span class="math display">\[ p_i=\partial L/\partial \dot{q}_i \]</span> <span class="math display">\[ F_i=\partial L/\partial q_i \]</span> <span class="math display">\[ \dot{p}_i=F_i \]</span></p>
<h2 id="model">Model</h2>
<ul>
<li><a href="mechanical-system.html">Classical mechanics</a></li>
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<h3>See also</h3>
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</h1>

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<hr style="clear:both;"/>


<ul>
<li>A system of <span class="math inline">\(N\)</span> particles has <span class="math inline">\(3N\)</span> degrees of freedom.</li>
<li>If all the co-ordinates and velocities are specified, the mechanical state of the system is determined.</li>
</ul>
<h3 id="for-a-free-particle">For a free particle</h3>
<p><span class="math display">\[\label{LL1/4.1}\tag*{(4.1)}L=\frac{1}{2}mv^2\]</span></p>

<p><span class="math display">\[\label{LL1/4.2}\tag*{(4.2)}L=\sum\frac{1}{2}m_av_a^2\]</span></p>

<p><span class="math display">\[\label{LL1/4.3}\tag*{(4.3)}v=(\mathrm{d}^{}l/\mathrm{d}^{}t)^2=(\mathrm{d}^{}l)^2/(\mathrm{d}^{}t)^2\]</span></p>

<p><span class="math display">\[\label{LL1/4.4}\tag*{(4.4)}\mathrm{d}^{}l^2=\frac{1}{2}(\dot{x}^2+\dot{y}^2+\dot{z}^2)\]</span></p>

<p><span class="math display">\[\label{LL1/4.5}\tag*{(4.5)}\mathrm{d}^{}l^2=\frac{1}{2}(\dot{r}^2+r^2\dot{\phi}^2+\dot{z}^2)\]</span></p>

<p><span class="math display">\[\label{LL1/4.6}\tag*{(4.6)}\mathrm{d}^{}l^2=\frac{1}{2}(\dot{r}^2+r^2\dot{\theta}^2+r^2\dot{\phi}^2\sin^2\theta)\]</span></p>

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<h3>See also</h3>
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<p>Real physical quantities <span class="math inline">\(f\)</span> are contained in the <a href="wave-function.html">wave function</a> <span class="math inline">\(\Psi\)</span>, measurement is done by the application of an operator <span class="math inline">\((\hat{f}\vphantom{f}^{}\Psi)\)</span>. The values that are taken by a physical quantities are eigenvalues <span class="math inline">\(f_n\)</span> 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 <span class="math inline">\(\Psi_n\)</span> which is also normalized.</p>
<p><span class="math display">\[\label{LL3/3.1}\tag*{(3.1)}\int|\Psi_n|^2\mathrm{d}^{}q=1\]</span></p>

<p>If a measurement <span class="math inline">\(\hat{f}\vphantom{f}^{}\)</span> is carried out on a system <span class="math inline">\(\Psi\)</span>, the result will be one of <span class="math inline">\(f_n\)</span>. The wavefunction must be a linear combination of the eigen-wavefunctions. (see principle of superposition).</p>
<p><span class="math display">\[\label{LL3/3.2}\tag*{(3.2)}\Psi=\sum a_n\Psi_n\]</span></p>

<p>A wavefunction can be expanded in terms of eigenfunction of any physical quantity. These <span class="math inline">\(a_n\)</span> represent the probability of a given eigenfunction which are <span class="math inline">\(|a_n|^2\)</span>. We also must have unit probability over the eigenfunction.</p>
<p><span class="math display">\[\label{LL3/3.3}\tag*{(3.3)}\sum_n|a_n|^2=1\]</span></p>

<p>This relation doesn’t hold if <span class="math inline">\(\Psi\)</span> is not normalized. The sum <span class="math inline">\(\sum|a_n|^2\)</span> must be bilinear in <span class="math inline">\(\Psi, \Psi^*\)</span>, and become unit when <span class="math inline">\(\Psi\)</span> is normalized, thus we have</p>
<p><span class="math display">\[\label{LL3/3.4}\tag*{(3.4)}\sum_n a_na_n^*=\int\Psi\Psi^*\mathrm{d}^{}q\]</span></p>

<p>we can find a simple form for <span class="math inline">\(a_n\)</span></p>
<p><span class="math display">\[\begin{align}
\Psi^*&amp;=\sum a_n^*\Psi_n^* &amp;&amp;\text{conjugate of $(3.2)$}\\ 
\int\Psi\Psi^*\mathrm{d}^{}q&amp;=\sum a_n^*\int\Psi_n^*\Psi\mathrm{d}^{}q &amp;&amp;\text{by substituting $\Psi^*$}\\
\sum_n a_na_n^*&amp;=\sum a_n^*\int\Psi_n^*\Psi\mathrm{d}^{}q &amp;&amp;\text{replacing with $(3.4)$}
\end{align}\]</span></p>
<p><span class="math display">\[\label{LL3/3.5}\tag*{(3.5)}a_n=\int\Psi\Psi_n^*\mathrm{d}^{}q\]</span></p>

<p>if we substitute 
        <span class="tooltip" class="math inline">
        \((3.2)\)
        <span class="tooltiptext"><span><span class="math inline">\(\Psi=\sum a_n\Psi_n\)</span></span>
</span>
        </span>
         we find <span class="math inline">\(a_n=\sum a_m\int\Psi_m\Psi_n^*\mathrm{d}^{}q\)</span>, from which it is evident the eigenfunctions must satisfy be an orthogonal set</p>
<p><span class="math display">\[\label{LL3/3.6}\tag*{(3.6)}\int\Psi_m\Psi_n^*\mathrm{d}^{}q=\delta_{nm}\]</span></p>

<p>Thus the spectrum of eigenfunctions is orthonormal.</p>
<p>The mean value <span class="math inline">\(\overline{f}\)</span> of a physical quantity <span class="math inline">\(f\)</span>, with the usual definition using wighted probabilities we get</p>
<p><span class="math display">\[\label{LL3/3.7}\tag*{(3.7)}\overline{f}=\sum_nf_n|a_n|^2\]</span></p>

<p>Let <span class="math inline">\((\hat{f}\vphantom{f}^{}\Psi)\)</span> be the result of the operator acting on the function <span class="math inline">\(\Psi\)</span>. We define <span class="math inline">\(\hat{f}\vphantom{f}^{}\)</span> in such a way that</p>
<p><span class="math display">\[\label{LL3/3.8}\tag*{(3.8)}\overline{f}=\int \Psi^*(\hat{f}\vphantom{f}^{}\Psi)\mathrm{d}^{}q\]</span></p>

<p>In general <span class="math inline">\(\hat{f}\vphantom{f}^{}\)</span> is linear, if we plug 
        <span class="tooltip" class="math inline">
        \((3.5)\)
        <span class="tooltiptext"><span><span class="math inline">\(a_n=\int\Psi\Psi_n^*\mathrm{d}^{}q\)</span></span>
</span>
        </span>
         in 
        <span class="tooltip" class="math inline">
        \((3.7)\)
        <span class="tooltiptext"><span><span class="math inline">\(\overline{f}=\sum_nf_n|a_n|^2\)</span></span>
</span>
        </span>
         we find <span class="math inline">\(\overline{f}=\sum_n f_na_na_n^*=\int\Psi(\sum_n a_nf_n\Psi_n)\mathrm{d}^{}q\)</span></p>
<p>Comparing with 
        <span class="tooltip" class="math inline">
        \((3.8)\)
        <span class="tooltiptext"><span><span class="math inline">\(\overline{f}=\int \Psi^*(\hat{f}\vphantom{f}^{}\Psi)\mathrm{d}^{}q\)</span></span>
</span>
        </span>
         we get that the effect of <span class="math inline">\(\hat{f}\vphantom{f}^{}\)</span> on <span class="math inline">\(\Psi\)</span> is</p>
<p><span class="math display">\[\label{LL3/3.9}\tag*{(3.9)}(\hat{f}\vphantom{f}^{}\Psi)=\sum_nf_na_n\Psi_n\]</span></p>

<p>If we substitute in 
        <span class="tooltip" class="math inline">
        \((3.5)\)
        <span class="tooltiptext"><span><span class="math inline">\(a_n=\int\Psi\Psi_n^*\mathrm{d}^{}q\)</span></span>
</span>
        </span>
         for <span class="math inline">\(a_n\)</span>, we find that <span class="math inline">\(\hat{f}\vphantom{f}^{}\)</span> is an integral operator of the form</p>
<p><span class="math display">\[\label{LL3/3.10}\tag*{(3.10)}(\hat{f}\vphantom{f}^{}\Psi)=\int K(q,q&#39;)\Psi(q&#39;)\mathrm{d}^{}q\]</span></p>

<p>where the kernel function <span class="math inline">\(K(q,q&#39;)\)</span> is</p>
<p><span class="math display">\[\label{LL3/3.11}\tag*{(3.11)}K(q,q&#39;)=\sum_n f_n\Psi_n^*(q&#39;)\Psi_n(q)\]</span></p>

<p>Thus, for every physical quantity there is a definite linear integral operator. From <span class="math inline">\((3.9)\)</span> we see that if <span class="math inline">\(\Psi\)</span> is one the the eigenfunction <span class="math inline">\(\Psi_n\)</span>, then</p>
<p><span class="math display">\[\label{LL3/3.12}\tag*{(3.12)}\hat{f}\vphantom{f}^{}\Psi=f_n\Psi_n\]</span></p>

<p>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 <span class="math inline">\(\Psi_n\)</span>. Because the mean is real, we can equate <span class="math inline">\((3.8)\)</span> to it’s conjugate form</p>
<p><span class="math display">\[\label{LL3/3.13}\tag*{(3.13)}\int\Psi^*(\hat{f}\vphantom{f}^{}\Psi)\mathrm{d}^{}q=\int\Psi(\hat{f}\vphantom{f}^{*}\Psi^*)\mathrm{d}^{}q\]</span></p>

<p>This doens’t hold in general for any linear operator <span class="math inline">\(\hat{f}\vphantom{f}^{}\)</span>, it is a restriction for physical quantities.</p>
<p>For an arbitrary <span class="math inline">\(\hat{f}\vphantom{f}^{}\)</span> we can find the transposed operator <span class="math inline">\(\hat{f}\vphantom{f}^{}\hat{f}\vphantom{f}^{t}\)</span> such that</p>
<p><span class="math display">\[\label{LL3/3.14}\tag*{(3.14)}\int\Psi(\hat{f}\vphantom{f}^{}\Phi)\mathrm{d}^{}q\equiv\int\Phi(\hat{f}\vphantom{f}^{t}\Psi)\mathrm{d}^{}q\]</span></p>

<p>where <span class="math inline">\(\Psi\)</span> and <span class="math inline">\(\Phi\)</span> are different, if we take <span class="math inline">\(\Phi=\Psi^*\)</span> then from <span class="math inline">\((3.13)\)</span> we have</p>
<p><span class="math display">\[\label{LL3/3.15}\tag*{(3.15)}\hat{f}\vphantom{f}^{t}=\hat{f}\vphantom{f}^{*}\]</span></p>

<p>Thus, the operators that represent physical quantities must be Hermitian.</p>
<p>If we consider a complex physical quantity <span class="math inline">\(f\)</span> (need example here), it’s complex conjugate is <span class="math inline">\(f^*\)</span>, whose eigenvalues are the complex conjugate of those of <span class="math inline">\(f\)</span>. We denote <span class="math inline">\(\hat{f}\vphantom{f}^{+}\)</span> the operator corresponding to the physical quantity <span class="math inline">\(f^*\)</span>. It is the Hermitian conjugate of <span class="math inline">\(\hat{f}\vphantom{f}^{}\)</span> and in general is different from <span class="math inline">\(\hat{f}\vphantom{f}^{*}\)</span>: from the condition <span class="math inline">\(\overline{f^*}=\overline{f}^*\)</span> we find that</p>
<p><span class="math display">\[\label{LL3/3.16}\tag*{(3.16)}\hat{f}\vphantom{f}^{+}=\hat{f}\vphantom{f}^{*t}\]</span></p>

<h3 id="addition-of-operators">Addition of operators</h3>
<p>Let <span class="math inline">\(f\)</span> and <span class="math inline">\(g\)</span> be two physical quantities. The eigenvalues of the sum <span class="math inline">\(f+g\)</span> are equal to the sums of the eigenvalues of <span class="math inline">\(f\)</span> and <span class="math inline">\(g\)</span>. The quantity <span class="math inline">\(f+g\)</span> is represented by the operator <span class="math inline">\(\hat{f}\vphantom{f}^{}+\hat{g}\vphantom{g}^{}\)</span>. Sometimes <span class="math inline">\(f\)</span> and <span class="math inline">\(g\)</span> can’t take definite values at the same time, in this case we define the mean of the sum</p>
<p><span class="math display">\[\label{LL3/4.1}\tag*{(4.1)}\overline{f+g}=\overline{f}+\overline{g}\]</span></p>

<p>In this case, the eigenvalues of the new operator <span class="math inline">\(\hat{f}\vphantom{f}^{}+\hat{g}\vphantom{g}^{}\)</span> are real valued, but they don’t bear any more relation to those of the quantities <span class="math inline">\(f\)</span> and <span class="math inline">\(g\)</span> separately.</p>
<p>Let <span class="math inline">\(f_0\)</span> and <span class="math inline">\(g_0\)</span> be the smallest eigenvalues of the quantities <span class="math inline">\(f\)</span> and <span class="math inline">\(g\)</span>, and <span class="math inline">\((f+g)_0\)</span> of the quantity <span class="math inline">\(f+g\)</span>, then</p>
<p><span class="math display">\[\label{LL3/4.2}\tag*{(4.2)}(f+g)_0\geq f_0+g_0\]</span></p>

<p>(need Hilbert algebra for proof of this)</p>
<h3 id="product-of-operators">Product of operators</h3>
<p>Let <span class="math inline">\(f\)</span> and <span class="math inline">\(g\)</span> be quantities that can be measured simultaneously. The product <span class="math inline">\(\hat{f}\vphantom{f}^{}\hat{g}\vphantom{g}^{}\)</span>, it is the successive application of <span class="math inline">\(\hat{g}\vphantom{g}^{}\)</span>, then <span class="math inline">\(\hat{f}\vphantom{f}^{}\)</span>. If <span class="math inline">\(\Psi_n\)</span> are eigenfunctions common to <span class="math inline">\(\hat{f}\vphantom{f}^{}\)</span> and <span class="math inline">\(\hat{g}\vphantom{g}^{}\)</span> we hav <span class="math inline">\(\hat{f}\vphantom{f}^{}\hat{g}\vphantom{g}^{}\Psi=\hat{f}\vphantom{f}^{}g_n\Psi_n=g_n\hat{f}\vphantom{f}^{}\Psi_n=g_nf_n\Psi_n\)</span>, we could have equally taken the operator <span class="math inline">\(\hat{g}\vphantom{g}^{}\hat{f}\vphantom{f}^{}\)</span>. Sinec <span class="math inline">\(\Psi\)</span> can always be written as a linear combination of <span class="math inline">\(\Psi_n\)</span>, it follows that <span class="math inline">\(\hat{f}\vphantom{f}^{}\hat{g}\vphantom{g}^{}\)</span> is the same as <span class="math inline">\(\hat{g}\vphantom{g}^{}\hat{f}\vphantom{f}^{}\)</span>.</p>
<p><span class="math display">\[\label{LL3/4.3}\tag*{(4.3)}\hat{f}\vphantom{f}^{}\hat{g}\vphantom{g}^{}-\hat{g}\vphantom{g}^{}\hat{f}\vphantom{f}^{}=0\]</span></p>

<p>Thus, we arrive at this important result: if two quantities <span class="math inline">\(f\)</span> and <span class="math inline">\(g\)</span> can simultaneously take definite values, then their operators <span class="math inline">\(\hat{f}\vphantom{f}^{}\)</span> and <span class="math inline">\(\hat{g}\vphantom{g}^{}\)</span> commute. (there is also proof of the converse in par11).</p>
<p>Let us consider an operator raised to a power <span class="math inline">\(p\)</span>. From the above we deduce that the eigenvalues of <span class="math inline">\(\hat{f}\vphantom{f}^{p}\)</span> are <span class="math inline">\(f_n^p\)</span>. Any function <span class="math inline">\(\Phi(\hat{f}\vphantom{f}^{})\)</span> of an operator can be defined as an operator whose eigenvalues are equal to the same function <span class="math inline">\(\Phi(f)\)</span> of the eigenvalues of the operator <span class="math inline">\(\hat{f}\vphantom{f}^{}\)</span>.</p>
<p>The inverse operator <span class="math inline">\(\hat{f}\vphantom{f}^{-1}\)</span> is such that <span class="math inline">\(\hat{f}\vphantom{f}^{}\hat{f}\vphantom{f}^{-1}=\hat{f}\vphantom{f}^{-1}\hat{f}\vphantom{f}^{}=1\)</span>.</p>
<p>If the quantities <span class="math inline">\(f\)</span> and <span class="math inline">\(g\)</span> cannot simultaneously take definite values, the product cannot be defined as above. This appears in the fact that <span class="math inline">\(\hat{f}\vphantom{f}^{}\hat{g}\vphantom{g}^{}\)</span> is not self-conjugate, so it cannot represent a physical quantity. By definition of the transpose we have</p>
<p><span class="math display">\[
\int\Psi\hat{f}\vphantom{f}^{}\hat{g}\vphantom{g}^{}\Phi\mathrm{d}^{}q=
\int\Psi\hat{f}\vphantom{f}^{}(\hat{g}\vphantom{g}^{}\Phi)\mathrm{d}^{}q=
\int(\hat{g}\vphantom{g}^{}\Phi)(\hat{f}\vphantom{f}^{t}\Psi)\mathrm{d}^{}q=
\int\Phi\hat{g}\vphantom{g}^{t}\hat{f}\vphantom{f}^{t}\Psi\mathrm{d}^{}q
\]</span></p>
<p>from which we have directly</p>
<p><span class="math display">\[\label{LL3/4.4}\tag*{(4.4)}(\hat{f}\vphantom{f}^{}\hat{g}\vphantom{g}^{})^t=\hat{g}\vphantom{g}^{t}\hat{f}\vphantom{f}^{t}\]</span></p>

<p>Taking the complex conjugate on both sides we have</p>
<p><span class="math display">\[\label{LL3/4.5}\tag*{(4.5)}(\hat{f}\vphantom{f}^{}\hat{g}\vphantom{g}^{})^+=\hat{g}\vphantom{g}^{+}\hat{f}\vphantom{f}^{+}\]</span></p>

<p>If both <span class="math inline">\(\hat{f}\vphantom{f}^{}\)</span> and <span class="math inline">\(\hat{g}\vphantom{g}^{}\)</span> are Hermitian, then <span class="math inline">\((\hat{f}\vphantom{f}^{}\hat{g}\vphantom{g}^{})^+=\hat{g}\vphantom{g}^{}\hat{f}\vphantom{f}^{}\)</span>. It follows that the operator <span class="math inline">\(\hat{f}\vphantom{f}^{}\hat{g}\vphantom{g}^{}\)</span> is Hermitian if and only if the factors <span class="math inline">\(\hat{f}\vphantom{f}^{}\)</span> and <span class="math inline">\(\hat{g}\vphantom{g}^{}\)</span> commute.</p>
<p>We note that for non commuting operators, we can form an Hermitian operator by taking the symmetrical combination</p>
<p><span class="math display">\[\label{LL3/4.6}\tag*{(4.6)}\textstyle\frac{1}{2}\displaystyle(\hat{f}\vphantom{f}^{}\hat{g}\vphantom{g}^{}+\hat{g}\vphantom{g}^{}\hat{f}\vphantom{f}^{})\]</span></p>

<p>The difference <span class="math inline">\(\hat{f}\vphantom{f}^{}\hat{g}\vphantom{g}^{}-\hat{g}\vphantom{g}^{}\hat{f}\vphantom{f}^{}\)</span> is an anti-Hermitian operator. It can be made Hermitian by mutlplying by <span class="math inline">\(i\)</span></p>
<p><span class="math display">\[ i(\hat{f}\vphantom{f}^{}\hat{g}\vphantom{g}^{}-\hat{g}\vphantom{g}^{}\hat{f}\vphantom{f}^{}) \]</span></p>
<p>For brevity we note the commutator</p>
<p><span class="math display">\[\label{LL3/4.7}\tag*{(4.7)}\{\hat{f}\vphantom{f}^{},\hat{g}\vphantom{g}^{}\}=\hat{f}\vphantom{f}^{}\hat{g}\vphantom{g}^{}-\hat{g}\vphantom{g}^{}\hat{f}\vphantom{f}^{}\]</span></p>

<p>It is easily seen that</p>
<p><span class="math display">\[\label{LL3/4.8}\tag*{(4.8)}\{\hat{f}\vphantom{f}^{}\hat{g}\vphantom{g}^{},\hat{h}\vphantom{h}^{}\}=\{\hat{f}\vphantom{f}^{},\hat{h}\vphantom{h}^{}\}\hat{g}\vphantom{g}^{}-\hat{f}\vphantom{f}^{}\{\hat{g}\vphantom{g}^{},\hat{h}\vphantom{h}^{}\}\]</span></p>

<p>And we notice that if <span class="math inline">\(\{\hat{f}\vphantom{f}^{},\hat{h}\vphantom{h}^{}\}=0\)</span> and <span class="math inline">\(\{\hat{g}\vphantom{g}^{},\hat{h}\vphantom{h}^{}\}=0\)</span>, it does not in general follow that <span class="math inline">\(\hat{f}\vphantom{f}^{}\)</span> and <span class="math inline">\(\hat{g}\vphantom{g}^{}\)</span> commute, i.e. commutation is not transitive. (math example is easy to find, need an example here with a physical meaning).</p>
<h3 id="the-continuous-spectrum">The continuous spectrum</h3>
<p>We can generalize the above results for operator an operator <span class="math inline">\(\hat{f}\vphantom{f}^{}\)</span> with a continuous spectrum. We shall denote it’s eigenvalues <span class="math inline">\(f\)</span>, without suffix, because they take a continuous range of values. We note <span class="math inline">\(\Psi_f\)</span> the eigenfunction corresponding to the eigenvalue <span class="math inline">\(f\)</span>. Just as <span class="math inline">\(\Psi\)</span> can be exanded in a series <span class="math inline">\((3.2)\)</span> 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.</p>
<p><span class="math display">\[\label{LL3/5.1}\tag*{(5.1)}\Psi(q)=\int a_f\Psi_f(q)\mathrm{d}^{}f\]</span></p>

<p>where the integration is taken over the whole range of values that can be taken by the quantity <span class="math inline">\(f\)</span>.</p>
<p>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 <span class="math inline">\(|a_f|^2\mathrm{d}^{}f\)</span> is the probability that the physical quantity has a value between <span class="math inline">\(f\)</span> and <span class="math inline">\(f+\mathrm{d}^{}f\)</span>. Since the sum of all probabilities must equal unity, we have</p>
<p><span class="math display">\[\label{LL3/5.2}\tag*{(5.2)}\int|a_f|^2\mathrm{d}^{}f=1\]</span></p>

<p>In the same way we found 
        <span class="tooltip" class="math inline">
        \((3.5)\)
        <span class="tooltiptext"><span><span class="math inline">\(a_n=\int\Psi\Psi_n^*\mathrm{d}^{}q\)</span></span>
</span>
        </span>
        , we can write</p>
<p><span class="math display">\[\begin{align}
\int\Psi\Psi^*\mathrm{d}^{}q&amp;=\int|a_f|^2\mathrm{d}^{}f\\
\int\Psi\Psi^*\mathrm{d}^{}q&amp;=\int\int a_f^*\Psi_f^*\Psi\mathrm{d}^{}f\mathrm{d}^{}q &amp;&amp;\text{by substituting $\Psi^*$ from $(5.1)$}
\end{align}\]</span></p>
<p>By comparing these two we find the expression for the expansion coefficients</p>
<p><span class="math display">\[\label{LL3/5.3}\tag*{(5.3)}a_f=\int\Psi(q)\Psi_f^*(q)\mathrm{d}^{}q\]</span></p>

<p>in exact analogy to 
        <span class="tooltip" class="math inline">
        \((3.5)\)
        <span class="tooltiptext"><span><span class="math inline">\(a_n=\int\Psi\Psi_n^*\mathrm{d}^{}q\)</span></span>
</span>
        </span>
        .</p>
<p>To derive the normalisation condition, we substitute 
        <span class="tooltip" class="math inline">
        \((5.1)\)
        <span class="tooltiptext"><span><span class="math inline">\(\Psi(q)=\int a_f\Psi_f(q)\mathrm{d}^{}f\)</span></span>
</span>
        </span>
         in 
        <span class="tooltip" class="math inline">
        \((5.3)\)
        <span class="tooltiptext"><span><span class="math inline">\(a_f=\int\Psi(q)\Psi_f^*(q)\mathrm{d}^{}q\)</span></span>
</span>
        </span>
        </p>
<p><span class="math display">\[ a_f=\int a_{f&#39;}(\Psi_{f&#39;}\Psi_f^*\mathrm{d}^{}q)\mathrm{d}^{}f&#39; \]</span></p>
<p>This relation must hold for any <span class="math inline">\(a_f\)</span>. The only solution is</p>
<p><span class="math display">\[\label{LL3/5.4}\tag*{(5.4)}\int\Psi_{f&#39;}\Psi_f^*\mathrm{d}^{}q=\delta(f&#39;-f)\]</span></p>

<p>This gives the normalisation rule for the eigenfunctions; replacing condition 
        <span class="tooltip" class="math inline">
        \((3.6)\)
        <span class="tooltiptext"><span><span class="math inline">\(\int\Psi_m\Psi_n^*\mathrm{d}^{}q=\delta_{nm}\)</span></span>
</span>
        </span>
        . Similarly, we have <span class="math inline">\(\Psi_f\)</span> and <span class="math inline">\(\Psi_{f&#39;}\)</span> orthogonal for <span class="math inline">\(f\neq f&#39;\)</span>. However the integrals of <span class="math inline">\(|\Psi_f|^2\)</span> diverge for a continuous system. The eigenfunctions satisfy another relation by subsituting the other way around</p>
<p><span class="math display">\[ \Psi(q)=\int\Psi(q&#39;)\left(\int\Psi_f^*(q&#39;)\Psi_f(q)\mathrm{d}^{}f\right)\mathrm{d}^{}q&#39; \]</span></p>
<p>From which we deduce immediately</p>
<p><span class="math display">\[\label{LL3/5.11}\tag*{(5.11)}\int\Psi_f^*(q&#39;)\Psi_f(q)\mathrm{d}^{}f=\delta(q-q&#39;)\]</span></p>

<p>The analogous for a discrete spectrum is</p>
<p><span class="math display">\[\label{LL3/5.12}\tag*{(5.12)}\sum_n\Psi_n^*(q&#39;)\Psi_n(q)\mathrm{d}^{}f=\delta(q-q&#39;)\]</span></p>

<p>
        <span class="tooltip" class="math inline">
        \((5.1)\)
        <span class="tooltiptext"><span><span class="math inline">\(\Psi(q)=\int a_f\Psi_f(q)\mathrm{d}^{}f\)</span></span>
</span>
        </span>
         and 
        <span class="tooltip" class="math inline">
        \((5.3)\)
        <span class="tooltiptext"><span><span class="math inline">\(a_f=\int\Psi(q)\Psi_f^*(q)\mathrm{d}^{}q\)</span></span>
</span>
        </span>
         are analogous: <span class="math inline">\(\Psi(q)\)</span> can be expanded in terms of the functions <span class="math inline">\(\Psi_f(q)\)</span> with expansion coefficients <span class="math inline">\(a_f\)</span>, or otherwise we can expand <span class="math inline">\(a_f\equiv a(f)\)</span> in terms of the functions <span class="math inline">\(\Psi_f^*(q)\)</span> while <span class="math inline">\(\Psi(q)\)</span> play the expansion coefficients. The function <span class="math inline">\(a(f)\)</span>, like <span class="math inline">\(\Psi(q)\)</span> completely determines the state of the system. Just as <span class="math inline">\(|\Psi(q)|^2\)</span> determines the probability for the system to have coordinates lying in an interval <span class="math inline">\(\mathrm{d}^{}q\)</span>, so <span class="math inline">\(|a(f)|^2\)</span> determines the probability for the values of the quantity <span class="math inline">\(f\)</span> to lie in a given interval <span class="math inline">\(\mathrm{d}^{}f\)</span>.</p>
<p>(TODO: 5.13 -&gt; 5.17)</p>
<h3 id="differentiating-operators-with-respect-to-time">Differentiating operators with respect to time</h3>
<p>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 <span class="math inline">\(\dot{f}\)</span> of a quantity <span class="math inline">\(f\)</span> as the quantity whose mean value is equal to the derivative of the mean value <span class="math inline">\(\bar{f}\)</span></p>
<p><span class="math display">\[\label{LL3/9.1}\tag*{(9.1)}\overline{\dot{f}}=\dot{\overline{f}}\]</span></p>

<p>by definition we have the mean as <span class="math inline">\(\overline{f}=\int\Psi^*\hat{f}\vphantom{f}^{}\Psi\mathrm{d}^{}q\)</span>,</p>
<p><span class="math display">\[ \overline{\dot{f}}=\dot{\overline{f}}=\frac{\mathrm{d}^{}}{\mathrm{d}^{}t}\int\Psi^*\hat{f}\vphantom{f}^{}\Psi\mathrm{d}^{}q=
\int\Psi^*\frac{\partial\hat{f}\vphantom{f}^{}}{\partial t}\Psi\mathrm{d}^{}q+
\int\frac{\partial\Psi^*}{\partial t}\hat{f}\vphantom{f}^{}\Psi\mathrm{d}^{}q+
\int\Psi\hat{f}\vphantom{f}^{}\frac{\partial\Psi}{\partial t}\mathrm{d}^{}q
\]</span></p>
<p>Substituting in 
        <span class="tooltip" class="math inline">
        \((8.1)\)
        <span class="tooltiptext"><span><span class="math inline">\(i\hbar\frac{\partial\Psi}{\partial t}=\hat{H}\vphantom{H}^{}\Psi\)</span></span>
</span>
        </span>
        , the time derivative is simply the <a href="hamiltonian-operator.html">Hamiltonian operator</a> <span class="math display">\[ \overline{\dot{f}}=
\int\Psi^*\frac{\partial\hat{f}\vphantom{f}^{}}{\partial t}\Psi\mathrm{d}^{}q+
\frac{i}{\hbar}\int(\hat{H}\vphantom{H}^{*}\Psi^*)\hat{f}\vphantom{f}^{}\Psi\mathrm{d}^{}q-
\frac{i}{\hbar}\int\Psi^*\hat{f}\vphantom{f}^{}(\hat{H}\vphantom{H}^{}\Psi)\mathrm{d}^{}q
\]</span></p>
<p>since <span class="math inline">\(\hat{H}\vphantom{H}^{}\)</span> is Hermitian we have <span class="math display">\[ \overline{\dot{f}}=
\int\Psi^*
\left(\frac{\partial\hat{f}\vphantom{f}^{}}{\partial t}+
\frac{i}{\hbar}\hat{H}\vphantom{H}^{}\hat{f}\vphantom{f}^{}-
\frac{i}{\hbar}\hat{f}\vphantom{f}^{}\hat{H}\vphantom{H}^{}\right)
\Psi\mathrm{d}^{}q
\]</span></p>
<p>Since, by definition of the mean value, the operator inside the parenthesis must be the derivative operator of the quantity <span class="math inline">\(f\)</span></p>
<p><span class="math display">\[\label{LL3/9.2}\tag*{(9.2)}\hat{\dot{f}}\vphantom{\dot{f}}^{}=\frac{\partial\hat{f}\vphantom{f}^{}}{\partial t}+\frac{i}{\hbar}\left(\hat{H}\vphantom{H}^{}\hat{f}\vphantom{f}^{}-\hat{f}\vphantom{f}^{}\hat{H}\vphantom{H}^{}\right)\]</span></p>

<p>We notice that if <span class="math inline">\(f\)</span> does not depend explicitly on time, then the derivative <span class="math inline">\(\hat{\dot{f}}\vphantom{\dot{f}}^{}\)</span> is, apart from a constant factor, just a commutation of <span class="math inline">\(f\)</span> with the Hamiltonian <span class="math inline">\(\hat{H}\vphantom{H}^{}\)</span>.</p>

<hr>
<h3>See also</h3>
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<h5 style="text-align:right;float:right">model</h5>

</h1>

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


<p>the study of motion of a rigid body</p>
<p><span class="math display">\[\label{LL1/31.1}\tag*{(31.1)}\mathrm{d}^{}\mathfrak{r}/\mathrm{d}^{}t=\mathbf{v},\quad
\mathrm{d}^{}\mathbf{R}/\mathrm{d}^{}t=\mathbf{V},\quad
\mathrm{d}^{}\mathbf{\phi}/\mathrm{d}^{}t=\mathbf{\Omega}\]</span></p>

<p><span class="math display">\[\label{LL1/31.2}\tag*{(31.2)}\mathbf{v}=\mathbf{V}+\mathbf{\Omega}\times\mathbf{r}\]</span></p>

<h2 id="assumptions">Assumptions</h2>
<ul>
<li><a href="particle.html">Particles</a> within solids are attached firmly, no deformation.</li>
<li><a href="particle.html">Particles</a> within solids move at the same instant, no delay.</li>
</ul>
<h2 id="model">Model</h2>
<ul>
<li><a href="particle.html">Particles</a></li>
</ul>

<hr>
<h3>See also</h3>
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<h1 class="title"; style="text-align:left;float:left">scattering

<h5 style="text-align:right;float:right">model</h5>

</h1>

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


<p><span class="math display">\[\begin{equation} \chi=|\textstyle\frac{\tau}{2}\displaystyle-2\phi_0| \end{equation}\]</span></p>
<p><span class="math display">\[\begin{equation} \phi_0=\int\limits_{r_\text{min}}^\infty\frac{(M/r^2)\mathrm{d}^{}r}{\sqrt{2m[E-U(r)]-M^2/r^2}} \end{equation}\]</span> <span class="math display">\[\begin{equation} E=\textstyle\frac{1}{2}\displaystyle mv_\infty^2,\qquad M=m\rho v_\infty \end{equation}\]</span> <span class="math display">\[\begin{equation} \phi_0=\int\limits_{r_\text{min}}^\infty\frac{(\rho/r^2)\mathrm{d}^{}r}{\sqrt{1-(\rho^2/r^2)-(2U/mv_\infty^2)}} \end{equation}\]</span> <span class="math display">\[\begin{equation} \mathrm{d}^{}\sigma=\mathrm{d}^{}N/n \end{equation}\]</span> <span class="math display">\[\begin{equation} \mathrm{d}^{}\sigma=\tau\rho\mathrm{d}^{}\rho \end{equation}\]</span> <span class="math display">\[\begin{equation} \mathrm{d}^{}\sigma=\tau\rho\left|\frac{\mathrm{d}^{}\rho}{\mathrm{d}^{}\chi}\right|\mathrm{d}^{}\chi \end{equation}\]</span> <span class="math display">\[\begin{equation} \mathrm{d}^{}\sigma=\frac{\rho}{\sin\chi}\left|\frac{\mathrm{d}^{}\rho}{\mathrm{d}^{}\chi}\right|\mathrm{d}^{}o \end{equation}\]</span></p>
<h2 id="assumptions">Assumptions</h2>
<ul>
<li><a href="two-body-problem.html">Two body problem</a></li>
</ul>

<hr>
<h3>See also</h3>
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<h1 class="title"; style="text-align:left;float:left">small oscillations

<h5 style="text-align:right;float:right">model</h5>

</h1>

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


<p>We consider a <a href="mechanical-system.html">mechanical system</a> near a stable equilibrium.</p>
<h2 id="in-one-dimension">In one dimension</h2>
<p>A stable equilibrium at a position <span class="math inline">\(q_0\)</span> is where the potential <span class="math inline">\(U(q)\)</span> is a local minimum at <span class="math inline">\(U(q_0)\)</span>. A movement away from this position leads to setting up a force <span class="math inline">\(-\mathrm{d}^{}U/\mathrm{d}^{}q\)</span> which tends to return the system to equilibrium. We choose a co-ordinate system where the equilibrium corresponds to <span class="math inline">\(x=0\)</span></p>
<p><span class="math display">\[\label{LL1/21.1}\tag*{(21.1)}x=q-q_0\]</span></p>

<p>We put <span class="math inline">\(U(q_0)=0\)</span> as a base energy<br />
We put <span class="math inline">\(U&#39;(q_0)=0\)</span> because we don’t consider asymetrical potentials<br />
We note <span class="math inline">\(U&#39;&#39;(q_0)=k\neq 0\)</span> because we don’t consider potential of higher order.<br />
Then consider the series expansion of <span class="math inline">\(U(q-q_0)\)</span>, and keeping the lowest terms for small deviations of equilibrium we have</p>
<p><span class="math display">\[\label{LL1/21.2}\tag*{(21.2)}U(x)=\textstyle\frac{1}{2}\displaystyle kx^2\]</span></p>

<p>The kinetic energy, with one degree of freedom is of the form <span class="math inline">\(\textstyle\frac{1}{2}\displaystyle a(q)\dot{q}^2=\textstyle\frac{1}{2}\displaystyle a(q)\dot{x}^2\)</span>. In the same approximation as above, <span class="math inline">\(a(q)=a(q_0)\)</span>.</p>
<p>We note <span class="math inline">\(a(q_0)=m\)</span>, this is the mass only if <span class="math inline">\(x\)</span> is the Cartesian co-ordinate.</p>
<p><span class="math display">\[\label{LL1/21.3}\tag*{(21.3)}L=\textstyle\frac{1}{2}\displaystyle m\dot{x}^2-\textstyle\frac{1}{2}\displaystyle kx^2\]</span></p>

<p>Using 
        <span class="tooltip" class="math inline">
        \((2.6)\)
        <span class="tooltiptext"><span><span class="math inline">\(\frac{\mathrm{d}^{}}{\mathrm{d}^{}t}\frac{\partial L}{\partial \dot{q_i}}-\frac{\partial L}{\partial q_i}=0\qquad(i=1,2,...,s)\)</span></span>
</span>
        </span>
         we derive the equation of motion, which is called the harmonic oscillator</p>
<p><span class="math display">\[\label{LL1/21.5}\tag*{(21.5)}\ddot{x}+\omega^2 x=0\]</span></p>

<p>where</p>
<p><span class="math display">\[\label{LL1/21.6}\tag*{(21.6)}w=\sqrt{k/m}\]</span></p>

<p>The general solution the harmonic oscillator is</p>
<p><span class="math display">\[\label{LL1/21.11}\tag*{(21.11)}x=\text{re}\left[Ae^{i\omega t}\right]\]</span></p>

<p>Where <span class="math inline">\(A=ae^{i\alpha}\)</span> is the complex amplitude, composed of the real amplitude <span class="math inline">\(a\)</span> and the phase <span class="math inline">\(\alpha\)</span> which depend on the initial condition of the system <span class="math inline">\(x(0)\)</span> and <span class="math inline">\(\dot{x}(0)\)</span>.</p>
<p>We note that the frequency <span class="math inline">\(\omega\)</span> doesn’t depend on the inital condition, but only on the parameters of the system <span class="math inline">\(k\)</span> and <span class="math inline">\(m\)</span>.</p>
<p>We also note, that the frequency of the motion is independant on the amplitude, which we have already predicted with 
        <span class="tooltip" class="math inline">
        \((10.2)\)
        <span class="tooltiptext"><span><span class="math inline">\(t&#39;/t=(l&#39;/l)^{1-k/2}\)</span></span>
</span>
        </span>
         for a quadratic potential <span class="math inline">\(k=2\)</span>.</p>

<hr>
<h3>See also</h3>
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