Integral equations for research in quantum mechanics

Question

In my research, I have reached a point where I need to find a nonnegative function which satisfies the following properties:

1. It's an even function:

$$f(-\theta) = f(\theta)$$

2. It's normalised as:

$$\int^\pi_0f(\theta)d\theta = \frac{\pi}{2}$$

3. For any $\phi$ in the interval $[0,\pi]$,the following property is satisfied:

$$\int^\pi_0[f(\theta)+f(\theta - \phi)-2f(\theta)f(\theta - \phi)]d\theta = \pi\sin^2\frac{\phi}{2}$$

I would be very much in your debt if you help me find such a function.

Answer 1

This falls into the problem of computing convolutional square roots, modulo the flaws already outlined by Cameron Williams.

A more sensible problem would be; compute a function $f$ on the circle (that is, a $2\pi$ periodic function) that satisfies $f(\theta)=f(-\theta)$ and such that $$\tag{1} f\ast f (\phi) = \pi \cos^2\frac \phi 2.$$ Here $$ f\ast f(\phi):= \int_{-\pi}^\pi f(\phi-\theta)f(\theta)\,d\theta.$$ The normalisation condition is already encoded in (1); indeed, evaluating (1) at $\phi=0$, $$ f\ast f(0)= 2\int_0^\pi f^2(\theta)\, d\theta = \pi.$$

Convolution square roots have been discussed before on Math.StackExchange. (Here there is an answer of Julián Aguirre (who, unfortunately, recently died). In the comments to that question, I have linked a related problem, answered by Terry Tao. All of these links, however, describe the problem on $\mathbb R$. I doubt they can be useful to the OP).

Answer 2

There are two functions that satisfy the updated conditions.

Let us start by writing $f(\theta)$ as a Fourier series involving only cosines (which are even functions):

\begin{equation} f(\theta)=\sum_{n=0}^{+\infty}f_n\cos(n\theta) \end{equation}

the first condition is immediately satisfied. Let us impose the second condition:

\begin{equation} \int_{0}^{\pi}d\theta \,\,f(\theta)=\sum_{n=0}^{+\infty}f_n\int_{0}^{\pi}\cos(n\theta)=\sum_{n=0}^{+\infty}f_n\frac{\sin{(\pi n)}}{n} \stackrel{!}{=}\frac{\pi}{2} \end{equation}

this expression vanishes except when $n=0$, in that case \begin{equation} \frac{\sin{(\pi n)}}{n}=\pi \end{equation} or, in other words, $f_0=\frac{1}{2}$. Hence,

\begin{equation} f(\theta)=\frac{1}{2}+\sum_{n=1}^{+\infty}f_n\cos(n\theta) \end{equation}

Now we have to look at condition 3. We have
\begin{equation} \int_{0}^{\pi}d\theta\,\,\cos\left(n(\theta-\phi)\right)=\frac{\sin(n\phi )}{n}[1-(-1)^n], \end{equation}

\begin{equation} \begin{split} \int_0^{\pi}d\theta\,\,\left(\frac{1}{2}+\sum_{n= 1}^{+\infty}f_n\cos(n\theta)\right)\left(\frac{1}{2}+\sum_{m=1}^{+\infty}f_m\cos(m(\theta-\phi))\right) \\ \\ =\frac{\pi}{4}+\frac{1}{2}\sum_{n=1}^{+\infty}\frac{\sin(n\phi )}{n}\left[1-(-1)^n\right] +\sum_{n,m\neq0}f_nf_m\int_0^{\pi}d\theta\,\,\cos(n\theta)\cos(m(\theta-\phi)) \end{split} \end{equation}

The last integral is \begin{equation} \int_0^{\pi}d\theta\,\,\cos(n\theta)\cos(m(\theta-\phi))=\frac{m}{m^2-n^2}[\sin(m\phi )-(-1)^n\sin(m(\phi-\pi))] \end{equation}

Let us study the limit $n\to m$ \begin{equation} \lim_{m\to n}\frac{m}{m^2-n^2}[\sin(m\phi )-(-1)^n\sin(m(\phi-\pi))]=\frac{\pi}{2}\cos\left(n\phi\right) \end{equation}

Hence, imposing condition 3 means that

\begin{equation} \frac{\pi}{2}+\frac{\pi}{2}+\sum_{n=1}^{\infty}f_n \frac{\sin(\phi n)}{n}[1-(-1)^n]-2\bigg\{\frac{\pi}{4}+\frac{1}{2}\sum_{n=1}^{+\infty}f_n\frac{\sin(n\phi)}{n}[1-(-1)^n]\\ +\frac{\pi}{2}\sum_{n=1}^{+\infty}f^2_n\cos(n\phi)+\sum_{n\neq m\neq 0}f_nf_m \frac{m}{m^2-n^2}\sin(m\phi )[1-(-1)^{n+m}]\bigg\}\stackrel{!}{=} \pi\sin^2\left(\frac{\phi}{2}\right) \end{equation}

Now the right hand side can be written as

\begin{equation} \pi\sin^2\left(\frac{\phi}{2}\right)=\frac{\pi}{2}(1-\cos\phi) \end{equation}

Therefore we have obtained the equation

\begin{equation} \begin{split} \pi\sum_{n=1}^{+\infty}f^2_n\cos(n\phi)+2\sum_{n\neq m\neq 0}f_nf_m \frac{m}{m^2-n^2}\sin(m\phi )[1-(-1)^{n+m}]=\frac{\pi}{2}\cos(\phi) \end{split} \end{equation}

We can satisfy this equation if $f^2_1\pi=\frac{\pi}{2}$ and if $f_{k}=0$ whenever $k>1$. This means $f_1=\pm \frac{1}{\sqrt{2}}$.

We conclude that the functions that you are looking for are

\begin{equation} f(\theta)=\frac{1}{2}\pm\frac{1}{\sqrt{2}}\cos\theta \end{equation}

If you want $f$ to be non negative, then just choose the plus sign.