# Solving free particles with Fourier series

Here's a silly idea : take the action of a free particle,

$$S = \int_{t_1}^{t_2} \dot{x}^2 dt$$

Our configuration space is the space of $$C^1$$ functions over $$[t_1, t_2]$$, which is spanned by the Fourier series

$$x^a(t) = \sum c^a_n e^{i\frac{2\pi n t}{T}}$$

So that every path is represented by some series of vectors $$c^a_n$$. The time derivative is the series

$$\dot{x}^a(t) = \sum_n \frac{2\pi n}{T} c^a_n e^{i\frac{2\pi n t}{T}}$$

By Parseval's theorem, we get

$$S = \int_{t_1}^{t_2} \dot{x}^2 dt = \sum_n \|\frac{2\pi n}{T} c^a_n\|^2$$

Can we show, without using variational methods, that a straight line is the least action? Using the boundary conditions, we can say that (rescaling our time for simplicity)

$$\begin{eqnarray} x^a(0) &=& \sum c^a_n = x^a_0\\ \dot{x}^a(0) &=& \sum n c^a_n = v^a_0 \end{eqnarray}$$

Obviously what we'd like is that our path is a straight line, which is equivalent to the sawtooth function, so that our extremal coefficients should be something similar to

$$\begin{eqnarray} c^a_0 &=& x^a_0\\ c^a_n &=& \Theta(n)(-1)^{n+1}\frac{i}{2}\frac{v^a}{n} \end{eqnarray}$$

Can the least action in this case be proven from this?

## 1 Answer

As it is widely known, Fourier series makes sense on periodic functions. Anyway, let us see what we have in your case. Let us compute the action $$S=\int_{t_1}^{t_2}{\dot x}^2dt=-\sum_k\sum_n\frac{4\pi^2}{T^2}knc^a_kc^a_n\int_{t_1}^{t_2}e^{i\frac{2\pi}{T}(k+n)t}dt.$$ This yields, $$S=-\sum_k\sum_nc^a_kc^a_n\frac{2\pi}{T}kn\frac{e^{i\frac{2\pi}{T}(k+n)t_2}-e^{i\frac{2\pi}{T}(k+n)t_1}}{i(k+n)}.$$ We have no Parseval theorem here of course. You would get it by properly redefining $$x(t)$$ and the action for a complex Fourier series. Instead, you will get, for $$k+n=0$$, $$S=\sum_nc^a_nc^a_{-n}\frac{4\pi^2}{T^2}n^2(t_2-t_1)+(n\ne k\ terms).$$ You get your minimum neglecting the $$n\ne k$$ terms and you are left with the known action for a free particle.