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.


2 Answers 2


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.

  • $\begingroup$ 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). $\endgroup$
    – Jonathan
    Aug 17, 2012 at 2:12
  • 2
    $\begingroup$ 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. $\endgroup$ 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.

Hopefully, this has been clear enough to give you an idea of how to finish the calculation, but vague enough not to spoil the (fun!) details for you. If you are still stuck then please let me know in a comment and I will supply you with more details.


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