# 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?

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.