So I am looking at a statistical path integral, meaning that I work with an Euclidean action. The propagator of my (Wiener) path integral is given by: $$ K(x_T,T|x_0,0)=\int\limits_{x(0)=0}^{x(T)=x_T}\mathcal{D}x\exp\left(-\int\limits_0^T\left[\frac{m}{2}\left(\dot{x}\right)^2+fx\right]d{t}\right), $$ which is basically a free particle in a gravity potential. Since the action is quadratic, the WKB formula $$ K(x_T,T|x_0,0)\approx\sqrt{-\frac{1}{2\pi}\frac{\partial^2 S[x_\mathrm{kl}(t)]}{\partial x_0\partial x_T}}\exp\left(-S[x_\mathrm{kl}(t)]\right) $$ should be exact.

The equation of motion gives me that the path of the particle meeting the boundary conditions is given by: $$ x_\mathrm{kl}(t)=\frac{f}{2m}(t-T)t+\frac{x_T- x_0}{T}t + x_0. $$ Using this path, I can calculate the classical action which becomes equal to: $$ S_\mathrm{kl}=-\frac{f^2}{24m}T^3+\frac{f T}{2}(x_T+x_0)+\frac{m}{2}\frac{(x_T-x_0)^2}{T}. $$

Substituting all of my results in the WKB formula then yields that the propagator is now given by: $$ K(x_T,T|x_0,0)=\sqrt{\frac{m}{2\pi T}}\exp\left(-\frac{m}{2}\frac{(x_T-x_0)^2}{T}-\frac{f T}{2}(x_T+x_0)+\frac{f^2T^3}{24m}\right). $$ The problem however with this propagator is that it does not stay normalized. If I demand that the propagator should remain normalized at all times $T$, then my propagator is given by: $$ K(x_T,T|x_0,0)=\sqrt{\frac{m}{2\pi T}}\exp\left(-\frac{m}{2}\frac{(x_T-x_0)^2}{T}-\frac{f T}{2}(x_T+x_0)+\frac{f^2T^3}{24m}\color{blue}-\color{blue}{\frac{fT}{6}}\left[\color{blue}{\frac{fT^2}{m}-6x_0}\right]\right), $$ which yields an extra term compared to the first version.

Extra: Time sliced method

I think that I have found the source of my problem, and it can be seen by looking at the infinitesimal propagator given by $$ K(x_j,t_j|x_{j-1},t_{j-1})=\sqrt{\frac{m}{2\pi\Delta t_j}}\exp\left(x_{j-1} f \Delta t_j + \frac{f^2}{2m}(\Delta t_j)^3\right)\\\times\exp\left(-\frac{m}{2\Delta t_j}\left[x_j-\left(x_{j-1}+\frac{f}{m}(\Delta t)^2\right)\right]^2\right). $$ In the upper part we indeed see that the normalization gets an extra exponential factor which will cause the path integral propagator (in time) to diverge. Also note that also has (more or less) the same form as the needed normalization factor (supporting my claim above)!

Question (new): For computational simplicity I'd like my propagator to still stay normalized. Is it okay to just use the second normalized version for my expectation values, or is that just wrong? Answer to old question of course still welcome as it may become relevant for further exploration of the path integral.

Question (old): Is there some kind of extra theorem that puts limitations on the correctness of the WKB formulas, or did I miss an extra important term here? I have recalculated the result a couple of times and it all seems to be correct at first sight.

About the solution

I also checked my solution in literature (Dittrich & Reuter) but they found the same (divergent) solution without any explanation. So at least I know the found solution is correct. Unfortunately I still have no idea what this means for my physics.

  • $\begingroup$ Yes, there are. The WKB approximation has caustics and the correct WKB wavefunction must be Maslov corrected when crossing one of them. $\endgroup$ Sep 5, 2016 at 14:06
  • $\begingroup$ @QuantumBrick, interesting. But what is the caustic point that my system went through? It seems like a to simple Lagrangian to me to do that kind of special stuff. Of course I can be wrong :s. $\endgroup$
    – Nick
    Sep 5, 2016 at 14:15
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    $\begingroup$ Indeed it is. I was talking in general. Your system is a free particle plus a linear potential, right? Indeed, there are no caustics. I'll try and think about it, hoping that someone wiser comes and answer you first. haha $\endgroup$ Sep 5, 2016 at 14:17
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    $\begingroup$ I think you've hit a higher problem than you (and I) have imagined. In fact, you cannot think of the propagator as normalizable. This is pointed out by QMechanic in this post: physics.stackexchange.com/q/81230 $\endgroup$ Sep 5, 2016 at 19:52
  • $\begingroup$ @QuantumBrick, since I am looking at Wiener path integrals (purely real path integrals) they should be always normalizable. The only way for me to lose/gain normalization is when there is a barrier at which particle get extracted/added, which is not the case. My thought was to try to calculate the path integral using the exact definition (time-slicing), however then I need to manually normalize which makes me end up with the second form. Somehow I have the feeling that this has to do with a freedom of choice in how you choose your action. $\endgroup$
    – Nick
    Sep 6, 2016 at 7:54

1 Answer 1


We are given the action$^1$

$$ S[x]~=~\int \! dt ~L, \qquad L~=~\frac{m}{2}\dot{x}^2-V~=~\frac{m}{2}\dot{x}^2+Fx, \qquad V~:=~ -Fx ,\tag{A}$$

where $F$ is a constant external force. The Dirichlet boundary conditions reads

$$ x(t_i)~=~x_i \qquad \text{and}\qquad x(t_f)~=~x_f. \tag{B}$$

OP correctly calculates the on-shell action

$$ S_{\rm cl}(x_f,t_f;x_i,t_i) ~=~\frac{m}{2} \frac{(\Delta x)^2}{\Delta t} + F \bar{x} \Delta t - \frac{F^2}{24m}(\Delta t)^3 ,\tag{C} $$


$$\Delta t~ :=~t_f-t_i, \qquad \Delta x~ :=~x_f-x_i, \qquad \bar{x}~ :=~ \frac{x_f+x_i}{2}. \tag{D} $$

Let us consider the quantum mechanical system in Minkowski space. (The statistical Euclidean formulation can be found via analytic continuation/Wick rotation $\tau_E=it_M$.) Before the first caustic, the kernel/path integral is given by the exact quantum mechanical formula

$$ K(x_f,t_f;x_i,t_i) ~=~\sqrt{\frac{m}{2\pi i\hbar} \frac{1}{\Delta t}} \exp\left[ \frac{i}{\hbar} S_{\rm cl}(x_f,t_f;x_i,t_i)\right], \tag{E}$$

where the on-shell action $S_{\rm cl}(x_f,t_f;x_i,t_i)$ is given by the expression (C). One may check via Gaussian integration that this formula (E) precisely satisfies the (semi)group property $$ K(x_f,t_f;x_i,t_i) ~=~ \int_{\mathbb{R}}\!\mathrm{d}x_m ~ K(x_f,t_f;x_m,t_m) K(x_m,t_m;x_i,t_i),\tag{F}$$

which is vital for the path integral formulation. We stress that the third & last term on the rhs. of eq. (C) plays a crucial role to the validity of eq. (F). Any of OP's proposed modifications would destroy the (semi)group property (F).

We emphasize that the normalization property $$ \left| \int_{\mathbb{R}}\! \mathrm{d}x_f~K(x_f,t_f;x_i,t_i) \right| ~\stackrel{?}{=}~1 \qquad(\leftarrow\text{Does not hold for a generic propagator!})\tag{G}$$ cannot be maintained for a generic potential, cf. this and this Phys.SE posts.

However, the normalization property (G) happens to hold for the propagator (E), so there is no problem with normalization in Minkowski space! It seems that OP's normalization trouble is spurred by a somewhat counter-intuitive normalization condition in Euclidean space dictated by the analytic continuation/Wick rotation.


$^1$ Note that the interpretation of $F$ as force and $V$ as a potential in the action (A) changes sign under Wick rotation. In other words, the sign of the potential term in the Euclidean and the Minkowski action are interpreted oppositely. This is of course a well-known effect, cf. e.g. my Phys.SE answer here.

  • $\begingroup$ ,thanks for the answer. I indeed remember my doubts about the normalization of the path integral in one of the linked questions. If I understood correctly, there are only two true properties for the propagator: the delta normalization and the (semi)group property. So indeed I can't expect that if I'm normalized in Minkowski space, that I still stay remain normalized in imaginary time. If I understand you correctly, imaginary time path integrals are not normalized (which is why we define this as the partition function). So for my expectation values I should divide by this partition function? $\endgroup$
    – Nick
    Sep 15, 2016 at 7:58
  • $\begingroup$ @Dominique: Yes. $\endgroup$
    – Qmechanic
    Sep 15, 2016 at 8:53
  • $\begingroup$ sounds fair enough :-). Thanks again for helping me out with path integrals! $\endgroup$
    – Nick
    Sep 15, 2016 at 9:04
  • $\begingroup$ Maybe one last side-question: Doesn't the addition of the terms (that normalize my path intgral) have the same effect as gauging my action, making it correct? $\endgroup$
    – Nick
    Sep 15, 2016 at 9:32
  • $\begingroup$ Gauging what symmetry? $\endgroup$
    – Qmechanic
    Sep 15, 2016 at 10:12

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