\[\def\rn{\mathbb{R}^n} \def\half{\frac{1}{2}} \def\dd{\ \!\mathrm{d}} \def\dvvp#1#2{\frac{\partial #1}{\partial #2}} \def\dvvpn#1#2#3{\frac{\partial^#3 #1}{\partial #2^#3}} \def\of#1{\tilde{#1}} \def\ten#1{\mathsf{#1}} \def\diag#1{\mathrm{diag}{#1}} \def\abs#1{\left| #1 \right|} \def\pdv#1{\frac{\partial}{\partial #1}} \def\dv#1{\frac{\dd}{\dd #1}} \def\ddv#1#2{\frac{\dd #1}{\dd #2}} \def\vdot{\mathbf{\cdot}}\]

Spherical Harmonics and the Schrodinger Equation¶

In spherical coordinates the time-independent Schrodinger equation is

\[ \begin{align}\begin{aligned}\begin{split} \label{eq:tisespherical} \begin{split} - \frac{\hbar^2}{2m} \frac{1}{r^2 \sin^2 \theta} \bigg( \pdv{r} \qty[ r^2 \sin(\theta) \pdv{\psi}{r}] + \pdv{\theta} \qty[ \sin \theta \pdv{\psi}{\theta}] \\ + \pdv{\phi} \qty[ \frac{1}{\sin \theta} \pdv{\psi}{\phi}] \bigg) + (V-E)\psi = 0 \end{split}\end{split}\\Now, splitting this into the radial part and an angular part,\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned} \label{eq:sepschrod} \psi(r, \theta, \phi) = R(r) Y(\theta, \phi)\\Then, substituting this in, and setting both sides of the equation\end{aligned}\end{align} \]

equal to \(l(l+1)\) we get two independent solutions

\[ \begin{align}\begin{aligned}\begin{split} \begin{aligned} \dv{r} \qty[r^2 \dv[2]{R}{r}] - \frac{2m}{\hbar^2} \qty(V(r)-E) r^2 R - l(l+1) R &= 0 \\\frac{1}{\sin \theta} \pdv{\theta} \qty[ \sin \theta \pdv{Y}{\theta}] + \frac{1}{\sin^2(\theta)} \pdv[2]{Y}{\phi} +l(l+1)Y &= 0\end{aligned}\end{split}\\The angular part is solved by spherical harmonics,\end{aligned}\end{align} \]

\[\label{eq:sphericalharmonics} Y_l^m(\theta, \phi) = N e^{im\phi} P_l^m (\cos \theta)\]

Bessel Functions¶

Bessel functions are the solutions to Bessel’s differential equation,

\[ \begin{align}\begin{aligned} \label{eq:besselde} x^2 \dv[2]{y}{x} + x \dv{y}{x} + (x^2 - \alpha^2)y = 0\\with :math:`p` a constant.\end{aligned}\end{align} \]

Bessel functions from the Generating Function¶

The Bessel functions can be described by a generating function,

\[\label{eq:besselgen} g(x,t) = \exp(\frac{x}{2t}(t^2-1)) = \sum_{\nu=-\infty}^{\infty} J_{\nu}(x) t^{\nu}\]

So, for Bessel functions of integer order we can expand this to form a series expansion,

\[ \begin{align}\begin{aligned} \label{eq:besselseriesexp} J_n(x) = \sum^{\infty}_{s=0} \frac{(-1)^s}{s! (n+s)!} \qty( \frac{x}{2} )^{n+2s} \approx \frac{x^n}{2^n n!}\\for small :math:`x`.\end{aligned}\end{align} \]

Bessel functions with a negative index can be found from the relation

\[\label{eq:negativebessel} J_{-\nu}(x) = (-1)^{\nu} J_{\nu}(x)\]

[ width=, height=2in, xmin=0, xmax=20, ] gnuplot[raw gnuplot, id=bess, mark=none, muted-blue, ultra thick] set xrange[0:20]; plot besj0(x); ; gnuplot[raw gnuplot, id=bess2, mark=none, muted-green, ultra thick] set xrange[0:20]; plot besj1(x); ; gnuplot[raw gnuplot, id=bess3, mark=none, muted-orange, ultra thick] set xrange[0:20]; fac(n) = (int(n)==0) ? 1.0 : int(n) * fac(int(n)-1.0); besj_eps = 0.1; besj(n,x) = (n==0) ? besj0(x) : (n==1) ? besj1(x) : (abs(x)<besj_eps*(n+1)) ? (x/2.0)**n/fac(n) : 2*(n-1)/x*besj(n-1,x) - besj(n-2,x); plot besj(2,x); ; gnuplot[raw gnuplot, id=bess4, mark=none, accent-purple, ultra thick] set xrange[0:20]; fac(n) = (int(n)==0) ? 1.0 : int(n) * fac(int(n)-1.0); besj_eps = 0.1; besj(n,x) = (n==0) ? besj0(x) : (n==1) ? besj1(x) : (abs(x)<besj_eps*(n+1)) ? (x/2.0)**n/fac(n) : 2*(n-1)/x*besj(n-1,x) - besj(n-2,x); plot besj(3,x); ; gnuplot[raw gnuplot, id=bess5, mark=none, accent-red, ultra thick] set xrange[0:20]; fac(n) = (int(n)==0) ? 1.0 : int(n) * fac(int(n)-1.0); besj_eps = 0.1; besj(n,x) = (n==0) ? besj0(x) : (n==1) ? besj1(x) : (abs(x)<besj_eps*(n+1)) ? (x/2.0)**n/fac(n) : 2*(n-1)/x*besj(n-1,x) - besj(n-2,x); plot besj(4,x); ;

Recurrence Relation for Bessel Functions¶

The Bessel functions can be descried by a pair of recurrence relations, found by differentiating with respect to \(t\),

\[\label{eq:recurrencebessel} J_{\nu-1}(x) + J_{\nu+1}(x) = \frac{2 \nu}{x} J_{\nu}(x)\]

and by differentiating with respect to \(x\),

\[\label{eq:recurrencebessel2} J_{\nu-1}(x) - J_{\nu+1}(x) = 2J_{\nu}^{\prime}(x)\]

A number of other integral relationships also exist.

\[\begin{split}\begin{aligned} \int x^n J_{n-1}(x) \dd{x} &= x^n J_n(x) \\ \int x^{-n} J_{n+1}(x) \dd{x} &= -x^{-n} J_n(x) \\ \int J_1(x) \dd{x} &= -J_0(x)\end{aligned}\end{split}\]

Orthogonality of the Bessel Functions¶

The orthogonality relations for Bessel functions are similar to those of the trigonometric functions, but they include an additional weighting factor, \(r\).

\[\label{eq:orthogbess} \int_0^a r J_p \qty( \frac{\alpha r}{a} ) J_p \qty(\frac{\beta r}{a}) \dd{r} = \delta_{\alpha \beta} \frac{a^2}{2} J_{p+1}^2(\alpha)\]

with

\[\begin{aligned} J_p(\alpha) = J_p(\beta) = 0\end{aligned}\]

Bessel Series¶

The orthogonality relations for Bessel functions allow the definition of Bessel series,

\[ \begin{align}\begin{aligned} \label{eq:besselser} f(x) = \sum_0^{\infty} c_n J_p(k_n x)\\with :math:`J_p(k_na)=0`.\end{aligned}\end{align} \]

Deriving the steady state inside an infinite cyclinder with the curved sides kept at a temperature :math:`T_0`, and the base at :math:`T_1`.

We know \(\nabla^2 T =0\), and we can use seperation of variables to give a solution of the form \(T = R(r)\Theta(\theta)Z(z)\). Then, in cylinderical coordinates,

\[ \begin{align}\begin{aligned}\frac{1}{R}\frac{1}{r} \dv{r} \qty(r \dv{R}{r}) + \frac{1}{\Theta} \frac{1}{r^2} \dv[2]{\Theta}{\theta} + \frac{1}{Z} \dv[2]{Z}{z} = 0\\We now have\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned}\frac{1}{Z} \dv[2]{Z}{z} = k^2\\implying\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned}Z = \exp(\pm kz)\\also,\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned}\frac{1}{R}\frac{1}{r} \dv{r} \qty(r \dv{R}{r}) + \frac{1}{\Theta} \frac{1}{r^2} \dv[2]{\Theta}{\theta} + k^2 = 0\\which we can multiply by :math:`r^2`,\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned}\frac{r}{R} \dv{r} \qty(r \dv{R}{r}) + \frac{1}{\Theta} \dv[2]{\Theta}{\theta} + k^2 r^2 = 0\\from which,\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned}\frac{1}{\Theta} \dv[2]{\Theta}{\theta} = -n^2\\implying that\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned}\Theta = \{ \cos(n \theta), \sin(n \theta) \}\\and the periodicity of :math:`\theta` will force :math:`n` to be a\end{aligned}\end{align} \]

natural number. Then

\[ \begin{align}\begin{aligned}\frac{r}{R} \dv{r} \qty(r \dv{R}{r}) + (k^2 r^2 - n^2) = 0\\and letting :math:`kr = s`,\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned}s \dv{s} \qty( s \dv{R}{s} ) + (s^2 - n^2)R = 0\\which has the form of Bessel’s differential equation, equation\end{aligned}\end{align} \]

([eq:besselde]), and thus the solutions are Bessel functions, \(J_n(s)\), the complete solution is thus

\[ \begin{align}\begin{aligned}J_n (kr) \qty( A \sin(n\theta) + B \cos(n \theta) ) e^{-kz}\\We can ignore the Bessel functions which are infinite at :math:`r=0`,\end{aligned}\end{align} \]

as we need a finite solution there, so the first-order functions are the appropriate solutions. We know that \(T_1 > T_0\), so \(T>T_0\) everywhere, and so \(T_0\) can be taken as a constant. The boundary condition of the curved surface at \(r=a\) is where \(J_n(ka) = 0\). We now need to know the zeros of the Bessel functions, and our solution becomes

\[ \begin{align}\begin{aligned}T = T_0 + \sum_{m=0}^{\infty} c_m J_0 \qty(\alpha_{0m}\frac{r}{a}) \exp(-\qty(\frac{\alpha_{0m}z}{a}))\\The boundary condition at :math:`z=0` is that :math:`T=T_0`, so\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned}T_1 - T_0 = \sum_m c_m J_0 \qty( \alpha_{0m} \frac{r}{a})\\and using the orthogonality condition,\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned}\int_0^a (T_1 - T_0) J_0 \qty( \alpha_{0m} \frac{r}{a} ) r \dd{r} = c_m \frac{a^2}{2} J_1^2(\alpha_{0m})\\and then, from the indefinite integral relationship\end{aligned}\end{align} \]

\(\int x J_0(x) \dd{x} = x J_1(x)\),

\[ \begin{align}\begin{aligned}\begin{split} \begin{aligned} (T_1-T_0) \frac{a}{\alpha_{0m}} \qty[ r J_1 \qty( \alpha_{0m} \frac{r}{a})]_0^a &= (T_1 - T_0) \frac{a^2}{\alpha_{0m}} J_1 (\alpha_{0m})\\ &= c_m \frac{a^2}{2} J_1^2 (\alpha_{0m})\end{aligned}\end{split}\\with\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned}c_m = \frac{2}{\alpha_{0m}} \frac{1}{J_1(\alpha_{0m}} (T_1-T_0)\\and the overarching solution is thus\end{aligned}\end{align} \]

\[T = T_0 + \sum_m \frac{2 (T_1-T_0)}{\alpha_{0m}J_1(\alpha_{0m})} J_0 \qty( \alpha_{0m} \frac{r}{a}) \exp( - \qty(\frac{\alpha_{0m}z}{a}) )\]

Spherical Bessel Functions¶

The spherical Bessel functions are a class of Bessel function related to the half-integer order order Bessel functions by

\[\label{eq:sphericalbess} j_n(x) = \sqrt{\frac{\pi}{2x}} J_{n+\frac{1}{2}}(x) = x^n \qty(- \frac{1}{x} \dv{x})^n \frac{\sin(x)}{x}\]

Finding energy levels of particles inside a spherical box using Schrodinger’s equation.

Starting at

\[ \begin{align}\begin{aligned}- \frac{\hbar^2}{2m} \nabla^2 \Psi = E \Psi\\after seperating variables\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned}\pdv{r} \qty(r^2 \pdv{R}{r}) + \qty( \frac{2mEr^2}{\hbar^2} - l(l+1) )R=0\\letting\end{aligned}\end{align} \]

\[k^2 = \frac{2mE}{\hbar^2} \quad \text{and} \quad s=kr\]

\[ \begin{align}\begin{aligned}s^2 \pdv[2]{R}{s} + 2s \pdv{R}{s} + \qty(s^2 - l(l+1))R = 0\\and letting\end{aligned}\end{align} \]

\[R = \frac{Z}{s^{\frac{1}{2}}}\]

\[ \begin{align}\begin{aligned}s^2Z^{\prime \prime} + s Z^{\prime} + (s^2 - \qty(l + \frac{1}{2})^2 ) Z = 0\\Which is Bessel’s equation of order :math:`l+\half`, so\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned}R = j_l \qty( \frac{\sqrt{2mE}}{\hbar} r)\\which is a finite solution as :math:`r \to 0`. The lowest energy\end{aligned}\end{align} \]

state will have \(l=0\) (so no angular variation), and to satisy the boundary condition of \(R=0\) when \(r=a\), we need

\[ \begin{align}\begin{aligned}j_0 \qty( \frac{\sqrt{2mE}}{\hbar}a)=0\\the zeros of :math:`j_0` are the same as those of :math:`\sin(x)`,\end{aligned}\end{align} \]

since

\[ \begin{align}\begin{aligned}j_0(x) = \frac{\sin(x)}{x}\\so\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned}\frac{a \sqrt{2mE_{\rm min}}}{\hbar} = \pi\\thus\end{aligned}\end{align} \]

\[E_{\rm min} = \frac{\pi^2 \hbar^2}{2ma^2}\]

Hermite Polynomials¶

[ width=, height=2in, xmin=-3, xmax=3, ymin=-60, ymax=50, ] gnuplot[raw gnuplot, id=bess, mark=none, muted-blue, ultra thick] set xrange[-3:3]; set yrange[-40:50]; herm(n,x) = (n==0) ? 1 : (n==1) ? 2*x : 2*x*herm(n-1,x)-2*n*herm(n-2,x); plot herm(0,x); ; gnuplot[raw gnuplot, id=bess2, mark=none, muted-green, ultra thick] set xrange[-3:3]; set yrange[-40:50]; herm(n,x) = (n==0) ? 1 : (n==1) ? 2*x : 2*x*herm(n-1,x)-2*n*herm(n-2,x); plot herm(1,x); ; gnuplot[raw gnuplot, id=bess3, mark=none, muted-orange, ultra thick] set xrange[-3:3]; set yrange[-40:50]; herm(n,x) = (n==0) ? 1 : (n==1) ? 2*x : 2*x*herm(n-1,x)-2*n*herm(n-2,x); plot herm(2,x); ; gnuplot[raw gnuplot, id=bess4, mark=none, accent-purple, ultra thick] set xrange[-3:3]; set yrange[-40:50]; herm(n,x) = (n==0) ? 1 : (n==1) ? 2*x : 2*x*herm(n-1,x)-2*n*herm(n-2,x); plot herm(3,x); ; gnuplot[raw gnuplot, id=bess5, mark=none, accent-red, ultra thick] set xrange[-3:3]; set yrange[-40:50]; herm(n,x) = (n==0) ? 1 : (n==1) ? 2*x : 2*x*herm(n-1,x)-2*n*herm(n-2,x); plot herm(4,x); ;

Hermite polynomials are the solutions to the hermite equation,

\[ \begin{align}\begin{aligned} \label{eq:hermitede} \dv[2]{y}{x} - 2x \dv{y}{x} + 2n y = 0\\Hermite polynomials are solutions to the radial part of the Schrodinger\end{aligned}\end{align} \]

equation for the simple harmonic oscillator. Just like Legendre polynomials and Bessel functions we can define Hermite polynomials, \(H_n (x)\) via a generating function:

\[\label{eq:hermite} g(x,t) = e^{-t^2 + 2tx} = \sum^\infty_{n=0} H_n(x) \frac{t^n}{n!}\]

Recurrence Relations for Hermite polynomials¶

First we diferentiate with respect to \(t\),

\[\frac{\partial}{\partial t} g(x,t) = (-2t+2x) e^{-t^2+2tx} = \sum^{\infty}_{n=1} H_n(x) \frac{t^{n-1}}{n!}\]

Expanding, and putting into the generating function again,

\[ \begin{align}\begin{aligned} -2 \sum^{\infty}_{n=0} H_n(x) \frac{t^{n+1}}{n!} + 2x \sum^{\infty}_{n=0} H_n(x) \frac{t^n}{n!} = \sum^{\infty}_{n=1} H_n(x)\frac{t^{n-1}}{(n-1)!}\\Relabelling the indices,\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned} -2 \sum^{\infty}_{n=1} nH_{n-1}(x) \frac{t^{n}}{n!} + 2x \sum^{\infty}_{n=0} H_n(x) \frac{t^n}{n!} = \sum^{\infty}_{n=1} H_{n+1}(x)\frac{t^n}{n!}\\and finally equating coefficients of :math:`t^n`,\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned} \label{eq:recurrencehermite} H_{n+1}(x) = 2x H_n(x) - 2n H_{n-1}(x) \qquad (n \ge 1)\\If we instead differentiate with respect to :math:`x`,\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned}\pdv{x}g(x,t) = 2t e^{-t^2+2tx} = \sum_{n=0}^{\infty} H^{\prime}_n(x) \frac{t^n}{n!}\\and substitute in :math:`g`,\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned}2 \sum_{n=0}^{\infty} H_n(x) \frac{t^{n+1}}{n!} = \sum_{n=1}^{\infty} H^{\prime}_n(x) \frac{t^n}{n!}\\and relabelling,\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned}2 \sum_{n=1}^{\infty} H_{n-1}(x) \frac{t^{n}}{(n-1)!} = \sum_{n=1}^{\infty} H^{\prime}_n(x) \frac{t^n}{n!}\\and then equating coeffients of :math:`t^n`,\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned} \label{eq:recurrencehermite2} H_n^{\prime}(x) = 2n H_{n-1}(x)\\These can be used to derive the ordinary differential equation which\end{aligned}\end{align} \]

motivates these polynomials, from the previous results we can find

\[ \begin{align}\begin{aligned}H_{n+1}(x) = 2x H_n(x) - H^{\prime}_n(x)\\and then differentiate with respect to :math:`x`,\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned}\begin{split} \begin{aligned} H^{\prime}_{n+1}(x) &= 2 H_n(x) + 2x H^{\prime}_n(x) - H^{\prime \prime}_n(x) \\ 2(n+1)H_{n}(x) &= 2 H_n(x) + 2x H^{\prime}_n(x) - H^{\prime \prime}_n(x) \end{aligned}\end{split}\\and so\end{aligned}\end{align} \]

\[\dv[2]{H_n(x)}{x} - 2x \dv{H_n(x)} + 2n H_n(x) = 0\]

It is possible to use the recurrence relations to find the Hermite polynomials, so

\[\begin{split}\begin{aligned} H_0(x) &= 1 \\ H_1(x) &= 2x \\ H_2(x) &= 4x^2 - 2 \\ H_3(x) &= 8x^3 - 12x \\ H_4(x) &= 16x^4 - 48x^2 + 12\end{aligned}\end{split}\]

Properties of the Hermite Polynomials¶

The Hermite polynomials are symmetric about \(x=0\), so

\[\label{eq:parityhermite} H_n(-x) = (-1)^n H_n(x)\]

The Hermite polynomials can be described by a specific series of the form

\[\label{eq:hermiteseriesspef} H_n(x) = \sum_{m=0}^{\frac{n}{2}}(-1)^m (2x)^{n-2m} \frac{n!}{(n-2m)!m!}\]

And Rodrigues’s equation for Hermite polynomials also exists proof is an exercise

\[\label{eq:rodrigueshermite} H_n(x) = (-1)^n e^{x^2} \dv[n]{x} \qty(e^{-x^2})\]

Orthogonality of Hermite Polynomials¶

It is possible to show the orthogonality of the Hermite polynomials. Starting at Hermite’s equation,

\[ \begin{align}\begin{aligned}\begin{split} \begin{aligned} H_n^{\prime \prime}(x) - 2x H^\prime_n (x) + 2n H_n (x) &= 0 \\ \dv{x} \qty( e^{-x^2} \dv{x} H_n (x) ) + 2n e^{-x^2} H_n(x) &=0 \end{aligned}\end{split}\\then, proceeding in much the same way as with Legendre polynomials in\end{aligned}\end{align} \]

section [sec:orthogonallegendre],

\[\begin{split}\begin{aligned} \begin{split} H_m(x) \dv{x} \qty[ e^{-x^2} \dv{x} H_n(x) ] - H_n(x) \dv{x} \qty[ e^{-x^2} \dv{x} H_m(x)] \\= -H_m(x) \cdot 2 n e^{-x^2} H_n(x) + H_n(x) \cdot 2 m e^{-x^2} H_m(x) \end{split}\end{aligned}\end{split}\]

\[\begin{split}\begin{aligned} \begin{split} \int_{-\infty}^{\infty} H_m(x) \dv{x} \qty[ e^{-x^2} \dv{x} H_n(x)] \dd{x} \\= \qty[ H_m(x) e^{-x^2} \dv{x} H_n(x)]_{-\infty}^{\infty} - \int_{-\infty}^{\infty} \qty[ \dv{x} H_m(x)] e^{-x^2} \dv{x} H_n(x) \dd{x} \end{split}\end{aligned}\end{split}\]

\[ \begin{align}\begin{aligned}\begin{split} \begin{aligned} 2(m-n) \int_{-\infty}^{\infty} H_n(x) H_m(x) e^{-x^2} \dd{x} &= 0 \\ \therefore \int_{-\infty}^{\infty} H_n(x) H_m(x) e^{-x^2} \dd{x} &= 0 \text{ iff } n \neq m\end{aligned}\end{split}\\Hermite polynomials are orthogonal on the interval\end{aligned}\end{align} \]

\([-\infty, \infty]\) with a weighting of \(\exp(-x^2)\).

\[ \begin{align}\begin{aligned}\begin{split} \begin{aligned} \int_{-\infty}^{\infty} g^2(x,t) e^{(-x^2)} \dd{x} &= \int_{-\infty}^{\infty} \exp(-2t^2+4tx-x^2) \dd{x} \\ &= \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \frac{t^{n+m}}{n! m!} \int_{-\infty}^{\infty} H_n(x) H_m(x) e^{(-x^2)} \dd{x} \\ &= e^{2t^2}\int_{-\infty}^{\infty} e^{-x^2} \dd{x}\\ &= e^{2t^2} \sqrt{\pi} \\ &= \sqrt{\pi} \sum_{n=0}^{\infty} \frac{2^n}{n!} t^{2n}\end{aligned}\end{split}\\Finally, equating powers of :math:`t^{2n}` gives\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned}\int_{-\infty}^{\infty} \qty[ H_n(x)]^2 \exp(-x^2) = 2^n \sqrt{\pi} n!\\so,\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned} \label{eq:hermiteorthoweight} \int_{-\infty}^{\infty} H_n(x) H_m(x) \exp(-x^2) \dd{x} = 2^n \sqrt{\pi} n! \delta_{nm}\\it is also possible to remove the weighting by redefining the\end{aligned}\end{align} \]

polynomial as

\[ \begin{align}\begin{aligned}\phi_n(x) := \exp(-x^2) H_n(x)\\in this case\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned} \label{eq:hermiteorthonoweight} \int_{-\infty}^{\infty} \phi_n(x) \phi_m(x) \dd{x} = 2^n \sqrt{\pi} n! \delta_{nm}\\these, however, are solutions not of Hermite’s equation, but of a\end{aligned}\end{align} \]

slightly variant equation,

\[\label{eq:modhermiteequation} \phi^{\prime \prime}_n(x) + (1-x^2+2n) \phi_n(x) = 0\]

The Quantum Harmonic Oscillator¶

Returning to the one-dimensional time-independent Schrödinger equation,

\[ \begin{align}\begin{aligned} \label{eq:1} - \frac{\hbar^2}{2m} \dv[2]{x} \psi(x) + V(x) \psi(x) = E \psi(x)\\with :math:`m` the mass of the particle, and :math:`E` its energy. For\end{aligned}\end{align} \]

the simple harmonic oscillator,

\[ \begin{align}\begin{aligned}V(x) \half m \omega^2 x^2\\so\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned} \psi^{\prime \prime} (x) + \qty( - \frac{m^2 \omega^2}{\hbar^2} x^2 + \frac{2m E}{\hbar^2} ) \psi(x) = 0\\which has a form very similar to the modified Hermite equation of the\end{aligned}\end{align} \]

previous section, and these describe the quantum harmonic oscillator.

Let \(y = ax\) with \(a = \sqrt{\frac{m \omega}{\hbar}}\), so

\[ \begin{align}\begin{aligned} \dv[2]{\psi}{y} + \qty( -y^2 + \frac{2mE}{\hbar^2 a^2} ) \psi = 0\\Comparing the two equations, we get the solutions\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned} \label{eq:2} \psi_n (x) = \sqrt{\frac{a}{2^n \sqrt{\pi} n!}} \exp( - \frac{a^2 x^2}{2} ) H_n(ax)\\which includes a normalisation constant. The energy is then given by\end{aligned}\end{align} \]

the equation

\[ \begin{align}\begin{aligned}\begin{split} \begin{aligned} \frac{2 m E}{\hbar^2 a^2} &= 1 + 2n \nonumber\\ \frac{2E}{\hbar \omega} &= 1 + 2n \nonumber\\ E &= \hbar \omega \qty(n + \half)\end{aligned}\end{split}\\but why does :math:`n` need to be an integer? The oscillator must have\end{aligned}\end{align} \]

\(\Psi \to 0\) as \(x \to \infty\). Taking solutions of the form

\[ \begin{align}\begin{aligned}\Psi \approx \exp( - \frac{x^2}{2} ) H_n(x)\\only guarantees this if :math:`n` is an integer; this can be\end{aligned}\end{align} \]

demonstrated by returning to Hermite’s equation, equation ([eq:hermitede]), and letting \(y = \sum_{k=0}^{\infty} c_k x^k\), so that

\[ \begin{align}\begin{aligned}\sum_k c_k \qty( k(k-1) x^{k-2} - 2kx^k + 2nx^k ) = 0\\This must be true for each power of :math:`x` individually, so\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned}c_{k+2} (k+2) (k+1) - c_k(2k-2n)=0\\and if the series in :math:`k` goes on *ad infinitum*, we have the\end{aligned}\end{align} \]

behaviour

\[ \begin{align}\begin{aligned}\frac{c_{k+2}}{c_k} = \frac{2k - 2n}{(k+1)(k+2)} \to \frac{2}{k} \quad \text{as} \quad k \to \infty\\This has the power series behaviour of :math:`\exp(x^2)`, which would\end{aligned}\end{align} \]

imply that \(\Psi \approx e^{x^2} e^{-\frac{x^2}{2}} \approx e^{\frac{x^2}{2}}\), giving “bad” behaviour as \(x \to \infty\). If the series truncates this behaviour will not occur. This happens if \(2n=2k\) for some \(k\), that is, for \(n \in \mathbb{Z}\). The solution of Hermite’s equation is a finite polynomial, and the solution for \(\Psi\) must be physical, so this forces \(n\) to be an integer.

The harmonic oscillator can also be solved using ladder operators, these work due to the recurrence relation in equation ([eq:recurrencehermite2]). Writing

\[ \begin{align}\begin{aligned}\psi_n(x) = \sqrt{\frac{1}{2^n \sqrt{\pi} n!}} \exp( - \frac{x^2}{2} ) H_n(x)\\and, for simplicity, letting :math:`a=1`, then\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned}\begin{split} \begin{aligned} \frac{1}{\sqrt{2}} \qty(x + \dv{x}) \psi_n(x) &= \sqrt{\frac{1}{2^{n+1} \sqrt{\pi} n!}} \qty( x+ \dv{x}) \exp(- \frac{x^2}{2}) H_n(x) \\ %&= \sqrt{\frac{1}{2^{n+1} \sqrt{\pi} n!}} \qty( x \exp( - \frac{x^2}{2} ) H_n(x) - x \exp( - \frac{x^2}{2} ) H_n(x) + \exp( - \frac{x^2}{2}) H^{\prime}_n(x) ) \\ & = \sqrt{n} \psi_{n-1}(x)\end{aligned}\end{split}\\This is a lowering operator, it is also possible, using either\end{aligned}\end{align} \]

recurrence relations or the Rodrigues’ formula, that

\[ \begin{align}\begin{aligned}\frac{1}{\sqrt{2}} \qty( x - \dv{x} )\\is a raising operator.\end{aligned}\end{align} \]

Laguerre Polynomials¶

The Laguerre polynomials are the solutions to the Laguerre equation,

\[ \begin{align}\begin{aligned} \label{eq:laguerrede} x L_n^{\prime \prime} (x) + (1-x) L_n^{\prime}(x) + n L_n(x) = 0\\The Laguerre polynomials are generated by the function\end{aligned}\end{align} \]

\[\label{eq:3} g(x,t) = \frac{\exp( - \frac{xt}{(1-t)})}{1-t} = \sum_{n=0}^{\infty} L_n(x) t^n\]

Recurrence Relations¶

\[\label{eq:4} (n+1) L_{n+1}(x) = (2n +1 -x) L_n(x) - nL_{n-1}(x)\]

\[\label{eq:5} xL^{\prime}_n(x) = nL_n(x) - nL_{n-1}(x)\]

It is possible to use these recurrence relations to find the first few Laguerre polynomials,

\[\begin{split}\begin{aligned} L_0(x) &= 1 \\ L_1(x) &= 1-x \\ L_2(x) &= \half \qty(x^2 - 4x + 2)\end{aligned}\end{split}\]

Orthogonality¶

Using similar techniques as for other special functions, it can be demonstrated that

\[\label{eq:orthoglag} \int_0^{\infty} L_n(x) L_m(x) \exp(-x) \dd{x} = \delta_{nm}\]

Properties¶

The Laguerre polynomials have a Rodrigues’ formula

\[ \begin{align}\begin{aligned} \label{eq:rodrigueslag} L_n(x) = \frac{e^x}{n!} \dv[n]{x} \qty(x^n e^{-x} )\\and a series expansion\end{aligned}\end{align} \]

\[\label{eq:serieslag} L_n(x) = \sum_{s=0}^n (-1)^{n-s} \frac{n! x^{n-s}}{(n-s)!(n-s)!s!}\]

Associate Laguerre Polynomials¶

The associate Laguerre polynomials are solutions to the associate Laguerre equation,

\[ \begin{align}\begin{aligned} \label{eq:assoclag} x y^{\prime \prime} (x) + (k+1-x) L_n^{k \prime}(x) + nL_n^k(x) = 0\\and are derived from the Laguerre polynomials by the expression\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned} \label{eq:assoclagfromlag} L_n^k(x) = (-1)^n \dv[k]{x} L_{n+k}(x)\\They are also orthogonal, with\end{aligned}\end{align} \]

\[\label{eq:assoclagortho} \int_0^{\infty} L_n^k(x) L_m^k(x) x^k \exp(-x) \dd{x} = \frac{(n+k)!}{n!} \delta_{nm}\]

3D Quantum Harmonic Oscillator Consider a quantum harmonic oscillator with a potential

\[ \begin{align}\begin{aligned}V = \half m \omega^2 \qty(x^2+y^2+z^2) = \half m \omega^2 r^2\\First separating Schrödinger’s equation into cartesian coordinates and\end{aligned}\end{align} \]

then deriving the form of the wavefunction leads to the conclusion that the energies are quantised as

\[ \begin{align}\begin{aligned}E = \qty( n+ \frac{3}{2}) \hbar \omega\\for :math:`n = n_x + n_y + n_z`. Separating Schrödinger into spherical coordinates allows the\end{aligned}\end{align} \]

solutions to take the form

\[ \begin{align}\begin{aligned} \Psi = N r^l \exp( - \half \alpha r^2) L_{\half(n-l)}^{l+\half} (\alpha^2 r^2) Y_{lm}(\theta, \phi)\\for\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned} a = \sqrt{\frac{m \omega}{\hbar}}\\With :math:`n` a normalisation factor, and :math:`l` taking the values\end{aligned}\end{align} \]

\(n, n-2, \dots, 0\). Associated Laguerre polynomials with non-integer \(p\) are obtained.

in 0,1,2,3,4

[yshift=-1.8in*]

[ title=\(k=\k\), width=, height=2in, xmin=0, xmax=5, ymin=-3, ymax=5 ] gnuplot[raw gnuplot, id=alag, mark=none, muted-blue, ultra thick] set xrange[-1:5]; lag(n,x) = (n==0) ? 1 : (n==1) ? -x+1 : ( (2*(n-1)+1-x) * lag(n-1,x) - (n-1) * lag(n-2, x) ) / (n); alag(n,k,x) = (k==0) ? lag(n,x) : ( (n+k)*alag(n, k-1, x) - (n+1)* alag(n+1, k-1, x) ) /x; plot alag(0,,x); ; gnuplot[raw gnuplot, id=alag, mark=none, muted-green, ultra thick] set xrange[-1:5]; lag(n,x) = (n==0) ? 1 : (n==1) ? -x+1 : ( (2*(n-1)+1-x) * lag(n-1,x) - (n-1) * lag(n-2, x) ) / (n); alag(n,k,x) = (k==0) ? lag(n,x) : ( (n+k)*alag(n, k-1, x) - (n+1)* alag(n+1, k-1, x) ) /x; plot alag(+1,,x); ; gnuplot[raw gnuplot, id=alag, mark=none, muted-orange, ultra thick] set xrange[-1:5]; lag(n,x) = (n==0) ? 1 : (n==1) ? -x+1 : ( (2*(n-1)+1-x) * lag(n-1,x) - (n-1) * lag(n-2, x) ) / (n); alag(n,k,x) = (k==0) ? lag(n,x) : ( (n+k)*alag(n, k-1, x) - (n+1)* alag(n+1, k-1, x) ) /x; plot alag(+2,,x); ; gnuplot[raw gnuplot, id=alag, mark=none, accent-purple, ultra thick] set xrange[-1:5]; lag(n,x) = (n==0) ? 1 : (n==1) ? -x+1 : ( (2*(n-1)+1-x) * lag(n-1,x) - (n-1) * lag(n-2, x) ) / (n); alag(n,k,x) = (k==0) ? lag(n,x) : ( (n+k)*alag(n, k-1, x) - (n+1)* alag(n+1, k-1, x) ) /x; plot alag(+3,,x); ; gnuplot[raw gnuplot, id=alag, mark=none, accent-red, ultra thick] set xrange[-1:5]; lag(n,x) = (n==0) ? 1 : (n==1) ? -x+1 : ( (2*(n-1)+1-x) * lag(n-1,x) - (n-1) * lag(n-2, x) ) / (n); alag(n,k,x) = (k==0) ? lag(n,x) : ( (n+k)*alag(n, k-1, x) - (n+1)* alag(n+1, k-1, x) ) /x; plot alag(+4,,x); ;