\documentclass{article} %% Created with wxMaxima 14.12.1 \setlength{\parskip}{\medskipamount} \setlength{\parindent}{0pt} \usepackage[utf8]{inputenc} \usepackage{graphicx} \usepackage{color} \usepackage{amsmath} \definecolor{labelcolor}{RGB}{100,0,0} \begin{document} \pagebreak{} {\Huge {\sc hydrogen2sb.mac, corrected, from Steeb \& Hardy Ch.26}} \setcounter{section}{0} \setcounter{subsection}{0} \setcounter{figure}{0} \subsection{This script allows to find energy levels of Hydrogen atom from Schroedinger equation, and here we illustrate how to use physical constants in Maxima} \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i1) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} load (physical_constants); \end{verbatim}} \end{minipage} %%% OUTPUT: % \begin{math}\displaystyle {\it Compiling /tmp/gazonk\_4257\_0.lsp.End of Pass 1. End of Pass 2. OPTIMIZE levels: Safety=2, Space=3, Speed=3 Finished compiling /tmp/gazonk\_4257\_0.lsp. Compiling /tmp/gazonk\_4257\_0.lsp.End of Pass 1. End of Pass 2. OPTIMIZE levels: Safety=2, Space=3, Speed=3 Finished compiling /tmp/gazonk\_4257\_0.lsp. Compiling /tmp/gazonk\_4257\_0.lsp.End of Pass 1. End of Pass 2. OPTIMIZE levels: Safety=2, Space=3, Speed=3 Finished compiling /tmp/gazonk\_4257\_0.lsp. Compiling /tmp/gazonk\_4257\_0.lsp.End of Pass 1. End of Pass 2. OPTIMIZE levels: Safety=2, Space=3, Speed=3 Finished compiling /tmp/gazonk\_4257\_0.lsp. Compiling /tmp/gazonk\_4257\_0.lsp.End of Pass 1. End of Pass 2. OPTIMIZE levels: Safety=2, Space=3, Speed=3 Finished compiling /tmp/gazonk\_4257\_0.lsp. Compiling /tmp/gazonk\_4257\_0.lsp.End of Pass 1. End of Pass 2. OPTIMIZE levels: Safety=2, Space=3, Speed=3Finished compiling /tmp/gazonk\_4257\_0.lsp. } % \end{math} \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o1) } /usr/share/maxima/5.32.1/share/ezunits/physical\_constants.mac \end{math} %%%%%%%%%%%%%%% Let us use notation from S.Fluegge ``Practical Quantum Mechanic'', vol.1, problem 61. Kepler's problem for infinetely heavy point-like nucleus of charge Z. For Z=1 we have hydrogen atom. Denote: \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i2) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} 2*m0*E/hb^2 = -%gamma^2; \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o2) } \frac{2\,m0\,E}{{hb}^{2}}=-{\gamma}^{2} \end{math} %%%%%%%%%%%%%%% here hb denotes $\hbar$, i.e. Planck constant divided by $2\pi$. Let \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i3) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} kappa = Z*e0^2*m0/(%gamma*hb^2); \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o3) } \kappa=\frac{{e0}^{2}\,m0\,Z}{\gamma\,{hb}^{2}} \end{math} %%%%%%%%%%%%%%% Now for usual substitution in wavefunction $u(r,\theta,\phi)$ to separate variables: \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i4) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} u = (1/r) * chi(r) * Ylm(theta,phi); \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o4) } u=\frac{\chi\left( r\right) \,\mathrm{Ylm}\left( \theta,\phi\right) }{r} \end{math} %%%%%%%%%%%%%%% we write the Schroedinger equation as (with $y$ in place of $\chi$): \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i5) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} 'diff(y, r, 2) + (-%gamma^2 + 2*%gamma*kappa/r-l*(l+1)/r^2)* y = 0; \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o5) } \frac{{d}^{2}}{d\,{r}^{2}}\,y+\left( \frac{2\,\gamma\,\kappa}{r}-\frac{l\,\left( l+1\right) }{{r}^{2}}-{\gamma}^{2}\right) \,y=0 \end{math} %%%%%%%%%%%%%%% Define orbital number $l$, magnetic $m$, and radial $k$, e.g.: \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i6) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} l: 1; m: 0; k: 1; \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o6) } 1 \end{math} \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o7) } 0 \end{math} \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o8) } 1 \end{math} %%%%%%%%%%%%%%% now the principal quantum number n is \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i9) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} n:k+l+1; \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o9) } 3 \end{math} %%%%%%%%%%%%%%% \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i10) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} alpha: e0^2/(hb*c); \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o10) } \frac{{e0}^{2}}{c\,hb} \end{math} %%%%%%%%%%%%%%% alpha is fine structure constant. Now substitute a polynomial $v1$ and $\exp$ as an ansatz for $y$, where \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i11) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} v1: (sum (a [j] *r^j, j, 1 ,k)); \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o11) } {a}_{1}\,r \end{math} %%%%%%%%%%%%%%% \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i12) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} v1: v1+1; \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o12) } {a}_{1}\,r+1 \end{math} %%%%%%%%%%%%%%% \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i13) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} y1: r^(l+1)*exp(-sqrt(g2)*r)*v1; \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o13) } {r}^{2}\,\left( {a}_{1}\,r+1\right) \,{e}^{-\sqrt{g2}\,r} \end{math} %%%%%%%%%%%%%%% $y1$ is the trial function of the radial wave function. Insert the ansatz into Schroedinger differential equation \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i14) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} left: diff(y1,r,2)+(-g2+2*m0*c*alpha/(hb*r)-l*(l+l)/(r*r))*y1; left: left*exp(sqrt(g2)*r); left: left/r; left: ratsimp(left); left: num(left); left: expand(left); left: subst(x,sqrt(g2),left); \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o14) } -2\,{a}_{1}\,\sqrt{g2}\,{r}^{2}\,{e}^{-\sqrt{g2}\,r}+\left( \frac{2\,{e0}^{2}\,m0}{{hb}^{2}\,r}-\frac{2}{{r}^{2}}-g2\right) \,{r}^{2}\,\left( {a}_{1}\,r+1\right) \,{e}^{-\sqrt{g2}\,r}+g2\,{r}^{2}\,\left( {a}_{1}\,r+1\right) \,{e}^{-\sqrt{g2}\,r}-4\,\sqrt{g2}\,r\,\left( {a}_{1}\,r+1\right) \,{e}^{-\sqrt{g2}\,r}+2\,\left( {a}_{1}\,r+1\right) \,{e}^{-\sqrt{g2}\,r}+4\,{a}_{1}\,r\,{e}^{-\sqrt{g2}\,r} \end{math} \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o15) } \left( -2\,{a}_{1}\,\sqrt{g2}\,{r}^{2}\,{e}^{-\sqrt{g2}\,r}+\left( \frac{2\,{e0}^{2}\,m0}{{hb}^{2}\,r} \\ -\frac{2}{{r}^{2}}-g2\right) \,{r}^{2}\,\left( {a}_{1}\,r+1\right) \,{e}^{-\sqrt{g2}\,r}+g2\,{r}^{2}\,\left( {a}_{1}\,r+1\right) \,{e}^{-\sqrt{g2}\,r}-4\,\sqrt{g2}\,r\,\left( {a}_{1}\,r+1\right) \,{e}^{-\sqrt{g2}\,r}+2\,\left( {a}_{1}\,r+1\right) \,{e}^{-\sqrt{g2}\,r}+4\,{a}_{1}\,r\,{e}^{-\sqrt{g2}\,r}\right) \,{e}^{\sqrt{g2}\,r} \end{math} \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o16) } \frac{\left( -2\,{a}_{1}\,\sqrt{g2}\,{r}^{2}\,{e}^{-\sqrt{g2}\,r}+\left( \frac{2\,{e0}^{2}\,m0}{{hb}^{2}\,r}-\frac{2}{{r}^{2}}-g2\right) \,{r}^{2}\,\left( {a}_{1}\,r+1\right) \,{e}^{-\sqrt{g2}\,r}+g2\,{r}^{2}\,\left( {a}_{1}\,r+1\right) \,{e}^{-\sqrt{g2}\,r} \\ -4\,\sqrt{g2}\,r\,\left( {a}_{1}\,r+1\right) \,{e}^{-\sqrt{g2}\,r}+2\,\left( {a}_{1}\,r+1\right) \,{e}^{-\sqrt{g2}\,r}+4\,{a}_{1}\,r\,{e}^{-\sqrt{g2}\,r}\right) \,{e}^{\sqrt{g2}\,r}}{r} \end{math} \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o17) } -\frac{\sqrt{g2}\,\left( 6\,{a}_{1}\,{hb}^{2}\,r+4\,{hb}^{2}\right) -2\,{a}_{1}\,{e0}^{2}\,m0\,r-2\,{e0}^{2}\,m0-4\,{a}_{1}\,{hb}^{2}}{{hb}^{2}} \end{math} \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o18) } -\sqrt{g2}\,\left( 6\,{a}_{1}\,{hb}^{2}\,r+4\,{hb}^{2}\right) +2\,{a}_{1}\,{e0}^{2}\,m0\,r+2\,{e0}^{2}\,m0+4\,{a}_{1}\,{hb}^{2} \end{math} \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o19) } 2\,{a}_{1}\,{e0}^{2}\,m0\,r-6\,{a}_{1}\,\sqrt{g2}\,{hb}^{2}\,r+2\,{e0}^{2}\,m0-4\,\sqrt{g2}\,{hb}^{2}+4\,{a}_{1}\,{hb}^{2} \end{math} \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o20) } -6\,{a}_{1}\,{hb}^{2}\,r\,x-4\,{hb}^{2}\,x+2\,{a}_{1}\,{e0}^{2}\,m0\,r+2\,{e0}^{2}\,m0+4\,{a}_{1}\,{hb}^{2} \end{math} %%%%%%%%%%%%%%% \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i21) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} h[0]: coeff(left,r,0); h[1]: coeff(left,r,1); h[2]: coeff(left,r,2); h[3]: coeff(left,r,3); h[4]: coeff(left,r,4); \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o21) } -4\,{hb}^{2}\,x+2\,{e0}^{2}\,m0+4\,{a}_{1}\,{hb}^{2} \end{math} \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o22) } 2\,{a}_{1}\,{e0}^{2}\,m0-6\,{a}_{1}\,{hb}^{2}\,x \end{math} \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o23) } 0 \end{math} \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o24) } 0 \end{math} \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o25) } 0 \end{math} %%%%%%%%%%%%%%% \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i26) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} s: solve ([h[0]=0, h[1]=0, h[2]=0, h[3] =0, h[4]=0], [x,a[1],a[2],a[3],a[4]]); \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o26) } [[x=\frac{{e0}^{2}\,m0}{2\,{hb}^{2}},{a}_{1}=0,{a}_{2}=\%r1,{a}_{3}=\%r2,{a}_{4}=\%r3],[x=\frac{{e0}^{2}\,m0}{3\,{hb}^{2}},{a}_{1}=-\frac{{e0}^{2}\,m0}{6\,{hb}^{2}},{a}_{2}=\%r4,{a}_{3}=\%r5,{a}_{4}=\%r6]] \end{math} %%%%%%%%%%%%%%% \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i27) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} sol: last(s); \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o27) } [x=\frac{{e0}^{2}\,m0}{3\,{hb}^{2}},{a}_{1}=-\frac{{e0}^{2}\,m0}{6\,{hb}^{2}},{a}_{2}=\%r4,{a}_{3}=\%r5,{a}_{4}=\%r6] \end{math} %%%%%%%%%%%%%%% \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i28) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} sol: first(sol); \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o28) } x=\frac{{e0}^{2}\,m0}{3\,{hb}^{2}} \end{math} %%%%%%%%%%%%%%% \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i29) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} sol: rhs(sol); \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o29) } \frac{{e0}^{2}\,m0}{3\,{hb}^{2}} \end{math} %%%%%%%%%%%%%%% \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i30) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} E_kl: -hb^2*sol*sol/2/m0; \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o30) } -\frac{{e0}^{4}\,m0}{18\,{hb}^{2}} \end{math} %%%%%%%%%%%%%%% energy eigenvalue for k, l \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i31) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} print ("E_kl=" ,E_kl); \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle E\_kl=-\frac{{e0}^{4}\,m0}{18\,{hb}^{2}} \end{math} \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o31) } -\frac{{e0}^{4}\,m0}{18\,{hb}^{2}} \end{math} %%%%%%%%%%%%%%% The output is the energy eigenvalue E\_kl. Now we may use maxima package "physical\_constant" which has a set of many dimensional constants. It uses package ezunits which by default employs SI. One can try demo(ezunits); display\_known\_unit\_conversions; We can list the constants: \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i32) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} propvars (physical_constant); \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o32) } [\%c,\%mu\_0,\%e\_0,\%Z\_0,\%G,\%h,\%h\_bar,\%m\_P,\%\%k,\%T\_P,\%l\_P,\%t\_P,\\ \%\%e,\%Phi\_0,\%G\_0,\%K\_J, \\ \%R\_K,\%mu\_B,\%mu\_N,\alpha,\%R\_inf,\%a\_0,\%E\_h,\%ratio\_h\_me,\%m\_e,\%N\_A,\%m\_u,\%F,\%R,\%V\_m,\\ \%n\_0,\%ratio\_S0\_R,\sigma,\%c\_1,\%c\_1L,\%c\_2,\%b,\%b\_prime] \end{math} %%%%%%%%%%%%%%% And print some of them, e.g. \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i33) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} constvalue(%alpha); \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o33) } 0.0072973525376 \end{math} %%%%%%%%%%%%%%% \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i34) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} constvalue (%c); \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o34) } 299792458 ` \frac{m}{s} \end{math} %%%%%%%%%%%%%%% we may copy paste this SI value into CGS value for speed of light (adding two orders): \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i35) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} c0:29979245800; \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o35) } 29979245800 \end{math} %%%%%%%%%%%%%%% \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i36) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} constvalue (%h_bar); \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o36) } \frac{41412931}{125000000000000000000000000000000000000000\,\pi } ` s\,J \end{math} %%%%%%%%%%%%%%% \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i37) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} float(%); \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o37) } 1.054571628251774\,{10}^{-34} ` s\,J \end{math} %%%%%%%%%%%%%%% we may copy paste this SI value into CGS value (7 orders difference in J and erg): \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i38) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} hb:1.054571628251774e-27; \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o38) } 1.054571628251774\,{10}^{-27} \end{math} %%%%%%%%%%%%%%% \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i39) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} constvalue(%%e); \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o39) } \frac{1602176487}{10000000000000000000000000000} ` C \end{math} %%%%%%%%%%%%%%% \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i40) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} float(%); \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o40) } 1.602176487\,{10}^{-19} ` C \end{math} %%%%%%%%%%%%%%% \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i41) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} e0:1.602176487e-19*c0/10; \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o41) } 4.80320427187535\,{10}^{-10} \end{math} %%%%%%%%%%%%%%% \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i42) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} constvalue(%m_e); \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o42) } 9.10938215\,{10}^{-31} ` kg \end{math} %%%%%%%%%%%%%%% In the same way \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i43) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} m0:9.10938215e-28; \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o43) } 9.10938215\,{10}^{-28} \end{math} %%%%%%%%%%%%%%% Now we get the energy of the level in ergs \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i44) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} E: ev(E_kl); \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o44) } -2.4220799671188296\,{10}^{-12} \end{math} %%%%%%%%%%%%%%% and transform to electron-Volts (do not confuse evaluation operator ev above with eV) \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i45) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} load("physconst.mac"); \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o45) } /usr/share/maxima/5.32.1/share/physics/physconst.mac \end{math} %%%%%%%%%%%%%%% \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i46) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} float(%Ry); \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o46) } \frac{1.0973731568549\,{10}^{7}}{m} \end{math} %%%%%%%%%%%%%%% Here we get Rydberg in inverse meters. Next line gives its uncertainty: \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i47) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} get(%Ry, RSU); \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o47) } 7.6\,{10}^{-12} \end{math} %%%%%%%%%%%%%%% We better get Ry in eV. We must multiply electron charge e0 by one Volt, expressed in CGS. The unit of voltage in CGS is 1 statvolt = 299.792458 volts. (The conversion factor 299.792458 is simply the numerical value of the speed of light in cm/s divided by 10\verb|^|8). Hence, \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i48) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} eV0:e0/(c0*1e-8); \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o48) } 1.602176487\,{10}^{-12} \end{math} %%%%%%%%%%%%%%% This is the value of eV in ergs, now we have the energy of our level in eV: \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i49) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} E_eV:E/eV0; \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o49) } -1.511743548086928 \end{math} %%%%%%%%%%%%%%% Since Rydberg is \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i50) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} Ry0:m0*e0^4/hb^2; \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o50) } 4.3597439408138926\,{10}^{-11} \end{math} %%%%%%%%%%%%%%% in ergs, and in eV \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i51) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} Ry0eV:Ry0/eV0; \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o51) } 27.2113838655647 \end{math} %%%%%%%%%%%%%%% We obtain the energy of our level in Rydbergs: \noindent %%%%%%%%%%%%%%% %%% INPUT: \begin{minipage}[t]{8ex}{\color{red}\bf \begin{verbatim} (%i52) \end{verbatim}} \end{minipage} \begin{minipage}[t]{\textwidth}{\color{blue} \begin{verbatim} E/Ry0; \end{verbatim}} \end{minipage} %%% OUTPUT: \begin{math}\displaystyle \parbox{8ex}{\color{labelcolor}(\%o52) } -0.055555555555555 \end{math} %%%%%%%%%%%%%%% It is easy to check that we get $-1/(2n^2)$ when the principal quantum number of the level is $n$. \end{document}