I'm just starting to come across path integrals in quantum field theory, and want to get the right intuition for the them from the start. The amplitude for propagation from $x_a$ to $x_b$ is typically written

$$U(x_a,x_b)=\int\mathcal{D}x(t) e^{\frac{i}{\hbar}S[x(t)]}$$

where $S$ is the classical action functional. Some books I read seem to treat $\int \mathcal{D}x(t)$ as a formal sum over all paths connecting $x_a$ and $x_b$. Is this the right way to think about it?

I come from quite a pure maths background so I've been trying to imagine this as some kind of measure on the space of smooth curves $x(t)$. I'm having trouble visualising this though. Does anyone have an intuitive argument as to how to picture this perspective?

Finally, how does one go about evaluating a general path integral in practice? I know that one can get an approximation by breaking the path up into piecewise linear segments. Is this a method that's used? Presumably also one can do the formal sum, and I assume this is the origin of many infinities.

I'd be happy to be told that there is no 'general' method for solution, or 'right' intuition. I'd just be interested to hear a range of ideas on the subject! Many thanks in advance!

  • $\begingroup$ Related: physics.stackexchange.com/q/1894/2451 and links therein. $\endgroup$ – Qmechanic Jan 5 '13 at 16:27
  • $\begingroup$ One way to evaluate a path integral is to use the path integral to derive the schrodinger equation, or something related, and then solve it :) $\endgroup$ – Abhimanyu Pallavi Sudhir Jun 22 '13 at 5:09
  • $\begingroup$ One can analytically solve path integrals that are gaussian in the path object using extensions of typical gaussian integral rules for many fields in a continuum limit. $\endgroup$ – aquirdturtle Nov 27 '16 at 22:30

Your physical intuition is correct, it is indeed sum over all admissible paths.

There is a problem with viewing $Dx(t)$ as a measure on the function space, because it is not well defined (e.g. infinite at some 'points' $x(t)$). This is one of the big (and as far as I know open problems) in the mathematical formulation of path integrals. Often one absorbs the kinetic part of the Hamiltonian in the measure to get an exponential dumping factor, e.g. $D x(t) exp(-\frac{i}{\hbar} \int T[x(t)]$. Salmhofer considers in Renormalization: An Introduction many of the mathematical questions. (If you know other good sources, I would be happy to hear about them!)

Path integrals are a nice way to 'visualize' many calculations (e.g 'I sum xyz over all possible paths), but are hard to compute. Indeed, the only calculations I know are based on breaking the path in linear segments (and even this gets clumsy). Often one performs some kind of Taylor expansion and only considers the first orders. There are then rules how to calculate often recurring terms (see Feymann diagrams).

  • $\begingroup$ FWIW, a prof teaching quantum many-body while I was a Ph.D. student told me that mathematicians have nailed down the $\epsilon$'s and $\delta$'s of functional integration. $\endgroup$ – DanielSank Sep 17 '14 at 0:23

As far as intuitive interpretation of path integrals is concerned (I don't really have experience solving them so I can't tell you how to calculate them), the formal sum over all possible trajectories connecting $x_a$ and $x_b$ is correct. It was, in fact, the original idea of Feynman to consider a particle going through all possible trajectories and adding their amplitudes.

If you have trouble with this idea, consider a double-slit experiment. Here, you have two possible ways how to get from source to detector -- either through the first slit or through the second one. The integration is thus effectively replaced by a sum over two paths but the principle remains the same.

  • $\begingroup$ Well, there aren't just two paths in the double-slit experiment. You could also: (1) go in through one slit, back out, and back in (2) go in through one slit, loop around a few times, go back in (3) ... and so on. $\endgroup$ – Arturo don Juan Nov 2 '17 at 21:42

In the following all integrals are done over the whole real line. The only one I can give an in depth explanation for, in terms of calculation, is the free particle. $$U(x,t;x_0) = <x|U(t)|x_0> = \int dp <x|e^{-ip^2/2m\hbar}|P><P|x_0>$$ where the $P$ is a momentum eigenstate and the Dirac notation indicates a jump from $x_0$ into $P$ which is then acted on by the the time evolution operator $U(t)$ to end up at $x$. The simplest, and easiest to generalize, solution method is with a Dirac Delta function $\delta(x-x_0)$. To see why consider the Fourier transform of it: $$ F(\delta(x-x_0)) =\int \frac{dk}{2\pi} e^{ik(x-x_0)} $$ Which follows from the Dirac Delta's defining property as a distribution. To continue insert a (complete) set of momentum eigenstates as follows: $$ \int dp <x|P>...<P|x_0> $$ $$ = \hbar \int dk <x|k=p/\hbar><k=p/\hbar|x_0> $$ By the deBroglie relation $ p=\hbar k$. So we can conclude that, by inspection of the Dirac Delta identity and our version in position space, that $$ <x|k><k|x_0> = \frac{1}{2\pi\hbar} e^{ik(x-x_0)} $$ $$ <x|k=p/\hbar|x_0> = \frac{1}{\sqrt{2\pi\hbar}} \int e^{ipx/\hbar} = <x|P> $$ So we can therefore reformulate our original integral as a product of Gaussians: $$ U(x,t;x_0) = \frac{1}{2\pi\hbar} \int dp e^{ip(x-x_0)/\hbar} e^{-ip^2/2m\hbar} $$ Completing the square in the exponent we obtain: $$ \frac{1}{2\pi\hbar} \int exp(\frac{-it}{2m\hbar}(p-\frac{m(x-x_0)}{t})^2)exp(im(x-x_0)^2/2\hbar t)$$ (Note that the second exponential is NOT integrated) Which equals, by the standard Gaussian integral: $$ \frac{1}{2\pi\hbar}\sqrt{\frac{( 2m\hbar\pi}{it}} e^{im(x-x_0)^2/2\hbar t}$$ Combining fractions we obtain the textbook result: $$ \sqrt{\frac{m}{2\pi\hbar it}} e^{im(x-x_0)^2/2 \hbar t}$$ More info on the formula for $<x|P>$ can be found in Sakurai Modern Quantum Mechanics 2nd Edition pp. 52-54.

  • $\begingroup$ Comment to the answer (v2): It should probably be stressed that these formulas only hold for a free non-relativistic particle. $\endgroup$ – Qmechanic Sep 17 '14 at 0:44
  • $\begingroup$ Yes it should I forgot that caveat :P $\endgroup$ – Tomas Smith Sep 17 '14 at 0:57

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