# Free Particle Propagator Using Path Integrals

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.

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.
• Since the limits are $-\infty \to \infty$, each factor is an (analytically continued) Gaussian, which is explicitly computable with no special functions needed. This is exactly what happens in the piecewise-linear regularization, so it's a good sign that the right answer is sure to pop out (once the $N$-dependent normalization is determined). Aug 17, 2012 at 2:12
• You don't actually need to analytically continue a gaussian. The integral converges as it is (albeit more slowly, to be sure) and it is a Fresnel integral of the type seen in optics. To make it clearer, change to $u=v^2$ in $\int_0^\infty \cos(v^2) dv=\int_0^\infty \cos(u)\frac{du}{2\sqrt{u}}$, which converges conditionally. Aug 17, 2012 at 2:28
I have not checked the details of your method, but the usual way of computing the path integral in QM is to approximate the trajectory $x(t)$ as a piecewise linear function, with $N$ "pieces", and then taking the limit $N \to \infty$. Now, the absolutely key part of this procedure, is that for each $N$ the integral appears with a certain weight $C_N$ (which is explictly computable), and it is the limit of $C_N \int \prod_{i=1}^N dx_i \cdots$ that exists (and is equal to the propagator), not the naive limit of $\int \prod_{i=1}^N dx_i$.
In your setup, you should consider, for each $N$, the space of trigonometric polynomials of degree at most $N$ (i.e., paths $x(t) = \sum_{n=-N}^N a_n \sin(n\pi t/T$). By comparison with the Schrodinger equation, it should be possible to work out the appropriate constant $C_N$. Then it is the limit of $C_N \int \prod_{n=-N} da_n \cdots$ that will tend to the propagator.
To be more precise, the more correct version of the path integral is with respect to the first order action, i.e. $$\int \mathcal{D} x \mathcal{D} p\ e^{\frac{i}{\hbar} \int p \dot{q} - H dt}$$ Because $x(t)$ and $p(t)$ are canonically conjugate, the "measure" $\mathcal{D}x \mathcal{D}p$ is natural and does not require a regularization-dependent constant (the $C_N$ mentioned above). For most theories, the $p$ dependence above is Gaussian, and so we can integrate it out. However, while this is often convenient, the resulting measure is of the form $C \mathcal{D}x$, where the constant $C$ is regularization dependent.