# Abstract

I comment on the entropy of waves by using the Shannon Entropy on the density of standing wave solutions to an infinite box and harmonic oscillator potentials, finding closed form results. The standing wave box entropies do not change with excited state, the harmonic oscillator entropies do change for excited states.

# Introduction

The Shannon Entropy is a functional $H[f(x)]=-\int f(x)\mathrm{ln}f(x) \;dx$ which can be written in a similar form $H[f(x)]=\int \mathrm{ln} f^{-f(x)}(x)\;dx$

Consider a wave in a box of length $$\pi$$, from $$0$$ to $$pi$$ which has a spatial wavefunction $\psi_n(x)=\mathrm{sin}(nx)$ Then if we take the density $\rho_n(x)=\mathrm{sin}^2(nx)$ we can evaluate the corresponding entropy of these states.

We find very quickly that $\int_0^\pi \mathrm{ln}\;\mathrm{sin}^{-2\mathrm{sin}^2(nx)}(nx)\;dx=\frac{\pi}{2}(\mathrm{ln}(4)-1)$ for any value of $$n$$! That is the entropy of each state is the same, independant of the level of excitation.

When adjusting for waves of any amplitude $$\alpha$$ we get $s_n=-\int_0^\pi \alpha^2\sin^2(nx)\ln \alpha^2\sin^2(nx) \; dx =\pi\alpha^2\left(\mathrm{ln}\left(\frac{2}{\alpha}\right)-\frac{1}{2}\right)$

# Harmonic Oscillator

We can analyse the entropy of the harmonic oscillator with wavefunctions $\psi_n(x) = \frac{1}{\sqrt{2^nn!}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\exp\left({-\frac{m\omega x^2}{2\hbar}}\right)H_n\left(\sqrt{\frac{m \omega}{\hbar}}\right), \;\;n=0,1,2,\cdots$

we then set $\frac{m\omega}{\hbar}=1$ and analyse the wavefunction integral $\sigma_0=-\int_{-\infty}^\infty \frac{1}{\sqrt[4]{\pi}}e^{-x^2/2}\mathrm{ln}\left(\frac{1}{\sqrt[4]{\pi}}e^{-x^2/2}\right)\;dx = -\frac{\sqrt[4]{\pi}(2+\ln(\pi))}{2\sqrt{2}}$ and the density integral $s_0=-\int_{-\infty}^\infty \frac{1}{\sqrt{\pi}}e^{-x^2}\mathrm{ln}\left(\frac{1}{\sqrt{\pi}}e^{-x^2}\right)\;dx = \frac{1}{2}(\ln(\pi)+1)$ With added constants we find $s_0=-\int_{-\infty}^\infty \frac{a}{\sqrt{\pi}}e^{-bx^2}\mathrm{ln}\left(\frac{a}{\sqrt{\pi}}e^{-bx^2}\right)\;dx = \frac{a}{2\sqrt{b}}(\ln(\frac{\pi}{a^2})+1)$ Then with $a^2=b=\frac{m\omega}{\hbar}$ $s_0= \frac{1}{2}(\ln(\frac{\hbar\pi}{m\omega})+1)$

Then for the next integrals with $$H_1(x)=2x$$ we have $s_1 = \frac{3\sqrt{\pi}}{4} -\int_{-\infty}^\infty e^{-x^2}x^2\ln(x^2) \; dx - \int_{-\infty}^\infty e^{-x^2}x^2\ln(A) \; dx$

Using inverse symbolic methods thanks to RIES and ISC, we find from accurate numerical integration $\sigma_1 = \frac{-\sqrt{2}}{\sqrt[4]{\pi}}\int_{-\infty}^{\infty} xe^{-x^2/2}\mathrm{ln}(\frac{\sqrt{2}}{\sqrt[4]{\pi}}xe^{-x^2/2})\;dx = \sqrt{2}\sqrt[4]{\pi^3}i$ and $s_1 = \frac{-2}{\sqrt{\pi}}\int_{-\infty}^{\infty} x^2e^{-x^2}\mathrm{ln}(\frac{2}{\sqrt{\pi}}x^2e^{-x^2})\;dx = \gamma + \mathrm{ln}(2) + \frac{1}{2}(\mathrm{ln}(\pi)-1)$ where $$\gamma$$ is the Euler-Mascheroni constant.

When solving for the next integrals we have $$H_2(x)=4x^2-2$$, and additional complexity. We find by laking a log expansion some progress can be made to finding a closed form for the total entropy of the density state.

We have $s_2=-\int_{-\infty}^{\infty}\frac{e^{-x^2}(16x^4-16x^2+4)}{8\sqrt{\pi}}\left(\sum_{n=1}^\infty \frac{1}{n}\left(\frac{e^{-x^2}(16x^4-16x^2+4)}{8\sqrt{\pi}} -1 \right)^n \right)\;dx$

Note down each term in the converging sequence $1-\frac{41}{64\sqrt{2}\pi^{1/2}}\\ \frac{1}{2}+\frac{11}{54\sqrt{3}\pi}-\frac{41}{64\sqrt{2}\pi^{1/2}}\\ \frac{1}{3}-\frac{272753}{3145728\sqrt{4}\pi^{3/2}}+2\frac{11}{54\sqrt{3}\pi}-\frac{41}{64\sqrt{2}\pi^{1/2}}\\ \frac{1}{4}+\frac{65209}{1562500\sqrt{5}\pi^2}-3\frac{272753}{3145728\sqrt{4}\pi^{3/2}}+3\frac{11}{54\sqrt{3}\pi}-\frac{41}{64\sqrt{2}\pi^{1/2}}\\ \frac{1}{5}-\frac{61710899}{2866544640\sqrt{6}\pi^{5/2}}+4\frac{65209}{1562500\sqrt{5}\pi^2}-6\frac{272753}{3145728\sqrt{4}\pi^{3/2}}+4\frac{11}{54\sqrt{3}\pi}-\frac{41}{64\sqrt{2}\pi^{1/2}}\\ \vdots\\ \frac{1}{n} + \sum_{k=0}^{n-1} \begin{pmatrix}n-1\\k\end{pmatrix}\frac{(-1)^{n-k}a_k}{b_k\sqrt{k+2}\pi^{(k+1)/2}}\\ s_2=\sum_{n=1}^\infty\frac{1}{n} + \sum_{k=0}^{n-1} \begin{pmatrix}n-1\\k\end{pmatrix}\frac{(-1)^{n-k}a_k}{b_k\sqrt{k+2}\pi^{(k+1)/2}}$

We have some higher terms from Mathematica $\frac{964482097}{83047723206\sqrt{7}\pi^3}\\ \frac{...}{\pi^{7/2}}$

It seems clear that a factor of $$8^{n+1}$$ must be removed from the denominator in the sequence.

$1-\frac{164}{2^2\cdot8^2\sqrt{2}\pi^{1/2}}\\ \frac{1}{2}+\frac{5632}{3^3\cdot8^3\sqrt{3}\pi}-\frac{164}{2^2\cdot8^2\sqrt{2}\pi^{1/2}}\\ \frac{1}{3}-\frac{272753}{3\cdot4^4\cdot8^4\sqrt{4}\pi^{3/2}}+2\frac{5632}{3^3\cdot8^3\sqrt{3}\pi}-\frac{164}{2^2\cdot8^2\sqrt{2}\pi^{1/2}}\\ \frac{1}{4}+\frac{534192128}{5^3\cdot8^5\cdot5^5\sqrt{5}\pi^2}-3\frac{272753}{3\cdot4^4\cdot8^4\sqrt{4}\pi^{3/2}}+3\frac{2816}{3^3\cdot8^3\sqrt{3}\pi}-\frac{164}{2^2\cdot8^2\sqrt{2}\pi^{1/2}}\\ \frac{1}{5}-\frac{3949497536}{15\cdot8^6\cdot6^6\sqrt{6}\pi^{5/2}}+4\frac{65209}{1562500\sqrt{5}\pi^2}-6\frac{272753}{3145728\sqrt{4}\pi^{3/2}}+4\frac{11}{54\sqrt{3}\pi}-\frac{164}{2^2\cdot8^2\sqrt{2}\pi^{1/2}}\\ \vdots\\ s_2=\sum_{n=1}^\infty\frac{1}{n} + \sum_{k=0}^{n-1} \begin{pmatrix}n-1\\k\end{pmatrix}\frac{(-1)^{n-k}a_k}{8^{(k+2)}b_k(k+2)^{(2k+5)}\pi^{(k+1)/2}}$

The series of integers which are peculiar to the above terms are generated in the following manner

$N_n = \frac{n^n\sqrt{n}}{\sqrt{\pi}}\int_{-\infty}^\infty e^{-nx^2}(4x^2-2)^{2n}\;dx$

with \begin{aligned} N_1=8\\ N_2=164\\ N_3=5632\\ N_4=272753\\ N_5=\frac{2136768512}{125}\end{aligned}