I'm trying to recreate some work that a professor explained to me in his office, specifically deriving the free particle propagator going from $(y,0)$ to $(x,T)$ using the Feynman Path Integral. I'm trying to reproduce $$K(x,T;y,0) = \sqrt{\frac{m}{2\pi i\hbar T}}\mathrm{exp}[\frac{im(x-y)^2}{2\hbar T}]$$Here's what I've done so far:
$K(x,T;y,0) = \int_y^x\mathscr{D}[x(t)]e^{iS[x(t)]/\hbar}.$
So first I compute the action:
$S[x(t)] = \int_0^T \frac{1}{2}m\dot{x}^2 \mathrm{d}t$
We can always split the path $x(t)$ in the following way:
$x(t) = x_{cl} (t) +q(t)$, where $x_{cl}(t)$ given by $$x_{cl}(t) = \frac{(x-y)t}{T} + y$$ is the classical path and $q(t)$ is a "quantum fluctuation".
Because the endpoints of $x(t)$ and $x_{cl}(t)$ are the same, we get that $q(0)=q(T)=0$, and because any path should be piecewise differentiable, we can represent $q(t)$ in a Fourier Series:
$$q(t) = \sum_{n=1}^{\infty} a_n sin(\frac{n\pi t}{T})$$.
The action is then $$S[x(t)] = \frac{1}{2}m\int_0^T (\frac{x-y}{T})^2 + 2\frac{x-y}{T} \dot{q} + \dot{q}^2 \mathrm{d}t$$
The first term is trivial, the second term vanishes due to the fundamental theorem of calculus and the fact that $q(t)$ vanishes at the endpoints. Now for the last term we get $$\int_0^T \dot{q}^2 \mathrm{d}t = \int_0^T \sum_{n=1}^{\infty}\sum_{m=1}^{\infty} a_n a_m (\frac{n\pi}{T}) (\frac{m\pi}{T})cos(\frac{n\pi t}{T})cos(\frac{m\pi t}{T})\mathrm{d}t$$
but due to orthogonality only the $n=m$ terms survive so we get $$ = \sum_{n=1}^{\infty}(\frac{n\pi }{T})^2\int_0^T a_{n}^2cos^2(\frac{n\pi t}{T})\mathrm{d}t = \sum_{n=1}^{\infty}\frac{(n\pi)^2}{2T}a_{n}^2$$.
Now to do the actual path integral, "all possible paths" would correspond to "all possible $q(t)$'s" which would mean all possible $a_n$'s. Thus our path integral becomes:
$$K(x,T;y) = \lim_{N\to\infty}\int_{-\infty}^{\infty}\mathrm{d}a_1\dotsi\int_{-\infty}^{\infty}\mathrm{d}a_N \mathrm{exp}\{\frac{im}{2\hbar}[\frac{(x-y)^2}{T} + \sum_{n=1}^{\infty}\frac{(n\pi)^2}{2T}a_{n}^2]\}$$
Now the first term in the exponential is clearly the same as the one in the original propagator however for the other integrals, i get an infinite amount of integrals which are infinite! Where does my reasoning or algebra go wrong?
PS I know there's probably a simpler way to do it, but since we started out this way I wanna know how it can be done with this method.
Best Answer
Further to Jonathan's answer, it seems to me that the integrals you're worried about are not actually infinite: $$\int_{-\infty}^\infty da_1\cdots \int_{-\infty}^\infty da_N e^{\frac{im}{2\hbar}\sum_{n=1}^N\frac{(n\pi)^2}{2T}a_n^2} =\prod_{n=1}^N\int_{-\infty}^\infty da_n e^{i\frac{m \pi^2}{4\hbar T}n^2 a_n^2}$$ and each of the individual integrals is a Fresnel integral with a finite result, including a nontrivial phase. However, the $a_n$ are lengths and therefore carry dimensional information, so that your final result (proportional to $(\hbar T/m)^{N/2}$ from dimensional analysis) is wrong by some $N$-dependent constant that comes from the measure normalization. Fixing that should let you get on with the fun.