.. include:: macros.rst


Spherical Harmonics and the Schrodinger Equation
------------------------------------------------

In spherical coordinates the time-independent Schrodinger equation is

.. math::

   \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}

 Now, splitting this into the radial part and an angular part,

.. math::

   \label{eq:sepschrod}
     \psi(r, \theta, \phi) = R(r) Y(\theta, \phi)

 Then, substituting this in, and setting both sides of the equation
equal to :math:`l(l+1)` we get two independent solutions

.. math::

   \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}

 The angular part is solved by spherical harmonics,

.. math::

   \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,

.. math::

   \label{eq:besselde}
     x^2 \dv[2]{y}{x} + x \dv{y}{x} + (x^2 - \alpha^2)y = 0

 with :math:`p` a constant.

Bessel functions from the Generating Function
---------------------------------------------

The Bessel functions can be described by a generating function,

.. math::

   \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,

.. math::

   \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`.

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

.. math::

   \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 :math:`t`,

.. math::

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

and by differentiating with respect to :math:`x`,

.. math::

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

A number of other integral relationships also exist.

.. math::

   \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}

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, :math:`r`.

.. math::

   \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

.. math::

   \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,

.. math::

   \label{eq:besselser}
     f(x) = \sum_0^{\infty} c_n J_p(k_n x)

 with :math:`J_p(k_na)=0`.

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

  .. math:: \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

  .. math:: \frac{1}{Z} \dv[2]{Z}{z} = k^2

   implying

  .. math:: Z = \exp(\pm kz)

   also,

  .. math:: \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`,

  .. math:: \frac{r}{R} \dv{r} \qty(r \dv{R}{r}) + \frac{1}{\Theta} \dv[2]{\Theta}{\theta} + k^2 r^2 = 0

   from which,

  .. math:: \frac{1}{\Theta} \dv[2]{\Theta}{\theta} = -n^2

   implying that

  .. math:: \Theta = \{ \cos(n \theta), \sin(n \theta) \}

   and the periodicity of :math:`\theta` will force :math:`n` to be a
  natural number. Then

  .. math:: \frac{r}{R} \dv{r} \qty(r \dv{R}{r}) + (k^2 r^2 - n^2) = 0

   and letting :math:`kr = s`,

  .. math:: s \dv{s} \qty( s \dv{R}{s} ) + (s^2 - n^2)R = 0

   which has the form of Bessel’s differential equation, equation
  ([eq:besselde]), and thus the solutions are Bessel functions,
  :math:`J_n(s)`, the complete solution is thus

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

  .. math:: 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

  .. math:: T_1 - T_0 = \sum_m c_m J_0 \qty( \alpha_{0m} \frac{r}{a})

   and using the orthogonality condition,

  .. math:: \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
  :math:`\int x J_0(x) \dd{x} = x J_1(x)`,

  .. math::

     \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}

   with

  .. math:: c_m = \frac{2}{\alpha_{0m}} \frac{1}{J_1(\alpha_{0m}} (T_1-T_0)

   and the overarching solution is thus

  .. math:: 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

.. math::

   \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

  .. math:: - \frac{\hbar^2}{2m} \nabla^2 \Psi = E \Psi

   after seperating variables

  .. math:: \pdv{r} \qty(r^2 \pdv{R}{r}) + \qty( \frac{2mEr^2}{\hbar^2} - l(l+1) )R=0

   letting

  .. math:: k^2 = \frac{2mE}{\hbar^2} \quad \text{and} \quad s=kr

  .. math:: s^2 \pdv[2]{R}{s} + 2s \pdv{R}{s} + \qty(s^2 - l(l+1))R = 0

   and letting

  .. math:: R = \frac{Z}{s^{\frac{1}{2}}}

  .. math:: 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

  .. math:: R = j_l \qty( \frac{\sqrt{2mE}}{\hbar} r)

   which is a finite solution as :math:`r \to 0`. The lowest energy
  state will have :math:`l=0` (so no angular variation), and to satisy
  the boundary condition of :math:`R=0` when :math:`r=a`, we need

  .. math:: j_0 \qty( \frac{\sqrt{2mE}}{\hbar}a)=0

   the zeros of :math:`j_0` are the same as those of :math:`\sin(x)`,
  since

  .. math:: j_0(x) = \frac{\sin(x)}{x}

   so

  .. math:: \frac{a \sqrt{2mE_{\rm min}}}{\hbar} = \pi

   thus

  .. math:: 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,

.. math::

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

.. math::

   \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 :math:`t`,

.. math::

   \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,

.. math::

   -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,

.. math::

   -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`,

.. math::

   \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`,

.. math:: \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`,

.. math:: 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,

.. math:: 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`,

.. math::

   \label{eq:recurrencehermite2}
     H_n^{\prime}(x) = 2n H_{n-1}(x)

 These can be used to derive the ordinary differential equation which
motivates these polynomials, from the previous results we can find

.. math:: H_{n+1}(x) = 2x H_n(x) - H^{\prime}_n(x)

 and then differentiate with respect to :math:`x`,

.. math::

   \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}

 and so

.. math:: \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

.. math::

   \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}

Properties of the Hermite Polynomials
-------------------------------------

The Hermite polynomials are symmetric about :math:`x=0`, so

.. math::

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

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

.. math::

   \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*

.. math::

   \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,

.. math::

   \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}

 then, proceeding in much the same way as with Legendre polynomials in
section [sec:orthogonallegendre],

.. math::

   \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}

.. math::

   \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}

.. math::

   \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}

 Hermite polynomials are orthogonal on the interval
:math:`[-\infty, \infty]` with a weighting of :math:`\exp(-x^2)`.

.. math::

   \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}

 Finally, equating powers of :math:`t^{2n}` gives

.. math:: \int_{-\infty}^{\infty} \qty[ H_n(x)]^2 \exp(-x^2) = 2^n \sqrt{\pi} n!

 so,

.. math::

   \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
polynomial as

.. math:: \phi_n(x) := \exp(-x^2) H_n(x)

 in this case

.. math::

   \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
slightly variant equation,

.. math::

   \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,

.. math::

   \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
the simple harmonic oscillator,

.. math:: V(x) \half m \omega^2 x^2

 so

.. math::

   \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
previous section, and these describe the quantum harmonic oscillator.

Let :math:`y = ax` with :math:`a = \sqrt{\frac{m \omega}{\hbar}}`, so

.. math::

   \dv[2]{\psi}{y} + \qty( -y^2 + \frac{2mE}{\hbar^2 a^2} ) \psi =
   0

 Comparing the two equations, we get the solutions

.. math::

   \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
the equation

.. math::

   \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}

 but why does :math:`n` need to be an integer? The oscillator must have
:math:`\Psi
\to 0` as :math:`x \to \infty`. Taking solutions of the form

.. math:: \Psi \approx \exp( - \frac{x^2}{2} ) H_n(x)

 only guarantees this if :math:`n` is an integer; this can be
demonstrated by returning to Hermite’s equation, equation
([eq:hermitede]), and letting :math:`y =
\sum_{k=0}^{\infty} c_k x^k`, so that

.. math:: \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

.. math:: 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
behaviour

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

.. math:: \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

.. math::

   \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}

 This is a lowering operator, it is also possible, using either
recurrence relations or the Rodrigues’ formula, that

.. math:: \frac{1}{\sqrt{2}} \qty( x - \dv{x} )

 is a raising operator.

Laguerre Polynomials
====================

The Laguerre polynomials are the solutions to the Laguerre equation,

.. math::

   \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

.. math::

   \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 
---------------------

.. math::

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

.. math::

   \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,

.. math::

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

Orthogonality
-------------

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

.. math::

   \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

.. math::

   \label{eq:rodrigueslag}
     L_n(x) = \frac{e^x}{n!} \dv[n]{x} \qty(x^n e^{-x} )

 and a series expansion

.. math::

   \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,

.. math::

   \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

.. math::

   \label{eq:assoclagfromlag}
     L_n^k(x) = (-1)^n \dv[k]{x} L_{n+k}(x)

 They are also orthogonal, with

.. math::

   \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

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

.. math:: E = \qty( n+ \frac{3}{2}) \hbar \omega

 for :math:`n = n_x + n_y +
  n_z`. Separating Schrödinger into spherical coordinates allows the
solutions to take the form

.. math::

   \Psi = N r^l \exp( - \half \alpha r^2) L_{\half(n-l)}^{l+\half}
     (\alpha^2 r^2) Y_{lm}(\theta, \phi)

 for

.. math::

   a = \sqrt{\frac{m
         \omega}{\hbar}}

 With :math:`n` a normalisation factor, and :math:`l` taking the values
:math:`n, n-2, \dots, 0`. Associated Laguerre polynomials with
non-integer :math:`p` are obtained.

in 0,1,2,3,4

[yshift=-1.8in\*]

[ title=\ :math:`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); ;

.. |image| image:: figures/spharmonics.png
.. |image| image:: figures/spharmonics2.png