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