I've been looking at the path integral + saddle point method derivation of the propagator of the harmonic oscillator recently, and I came to conclusion that the measure bit of the path integral confuses me a bit for this problem. I would like to note that yes, there exist methods to reason what the measure should be. This only really gives me the feeling that I in fact don't understand the mathematics. Hence this question.

So deriving the general formulas wasn't much of a problem (see my work below*). The problem arises in that I cannot use zeta function regularization to regularize the infinite product that remains. Since this is really the only technique that I know of that could possibly lead to the correct prefactors, I'm a bit stumped now.

So thinking out loud: one possibility is that the Fourier sine series may introduce a constant change in the measure, but this feels rather counter intuitive since I always treated Fourier expansions as unitary transformations (why should this not be the case?). Another option is that my original equation for the propagator is incorrect (although I don't see how this could be the case...). In the derivation by Feynman himself, it appears that he also gets these weird prefactors (see the problem 3.13 directly below the derivation of the propagator "by keeping track of constants...). His measure doesn't seem to add up either if I only consider zeta functions to regularize this expression.

Regardless, how would I go about calculating the measure explicitly?

My derivation


  • First and second order variations of the action are vanishing at the boundaries. This yields respectively the equations of motion and semi-explicit form $\det^{\frac12}\left(A\right)$ where the matrix $A$ is the usual matrix to be evaluated in multi-dimensional Gaussian integrals.
  • First order variation is vanishing completely by the saddle-point condition / formula.
  • The calculation of the "classical" part of the propagator will omitted here for brevity, since it is not relevant to the constant prefactors of the propagator.

Note: in the below the Fourier sine decomposition is already missing a factor $\sqrt{2}$, leading now to an overall extra factor $1/\sqrt{2}$ (rather than $8^{-1/4}$) missing from the propagator. Thanks for Wakabaloola for his useful link/comment.

The second order variation of the measure is \begin{equation} \frac{\delta^2 S_\text{HO}[x]}{\delta x_i\delta x_f} = \frac M2 \int_0^{\Delta t} \delta x(t')\left(-\partial_{t'}^2 - \omega^2\right)\delta x(t'), \end{equation}

Using the Fourier sine series representation \begin{equation} \delta x(t') = \sqrt{\frac2{\Delta t}} \sum_{n=1}^\infty x_n \sin\left(\frac{\pi n t'}{\Delta t}\right) \end{equation} the diagonal form of the above second order variation becomes \begin{align} \frac{\delta^2 S_\text{HO}[x]}{\delta x_i\delta x_f} &= \frac M{2}\sum_{n=1}^\infty \left[\left(\frac{\pi n}{\Delta t}\right)^2 - \omega^2\right] \lvert x_n \rvert^2 \end{align} where I used the sine orthogonality condition, integrated over half a period.

Inserting the above into the definition of the propagator, we get \begin{align} i\mathcal G\left(x_f,\Delta t;x_i,0\right) &= \lim_{n\rightarrow\infty} \left(\sqrt{\frac{M}{2\pi i\hbar \Delta t}}\right)^{n+1} \left(\prod_{j=1}^n \int_{-\infty}^\infty dx_j\right) \exp\left(\frac i\hbar S_\text{HO}[x_j]\right)\\ % 20 &= \lim_{n\rightarrow\infty} \left(\sqrt{\frac{M}{2\pi i\hbar \Delta t}}\right)^{n+1} \exp\left(\frac i\hbar S_\text{HO}[x_\text c]\right)\left(\prod_{m=1}^n \int_{-\infty}^\infty dx_m\right) \exp\left(\frac i{\hbar} \delta_x^2 S_\text{HO}[x_m(x_j)]\Big|_{x=x_\text c}\right)\\ % 21 &= \lim_{n\rightarrow\infty} \left(\sqrt{\frac{M}{2\pi i\hbar \Delta t}}\right)^{n+1} \exp\left(\frac i\hbar S_\text{HO}[x_\text c]\right)\\ &\phantom=\times\left(\prod_{m=1}^n \int_{-\infty}^\infty dx_m\right) \exp\left(\frac {Mi}{2\hbar}\sum_{k=1}^\infty \left[\left(\frac{\pi k}{\Delta t}\right)^2 - \omega^2\right] \lvert x_n\rvert^2\right)\\ % 22 &= \lim_{n\rightarrow\infty} \left(\sqrt{\frac{M}{2\pi i\hbar \Delta t}}\right)^{n+1} \exp\left(\frac i\hbar S_\text{HO}[x_\text c]\right)\prod_{k=1}^n\left[ \int_{-\infty}^\infty dx \exp\left(\frac {Mi}{\hbar}\left[\left(\frac{\pi k}{\Delta t}\right)^2 - \omega^2\right] \frac{x^2}{2}\right)\right]\\ % 23 &= \lim_{n\rightarrow\infty} \left(\sqrt{\frac{M}{2\pi i\hbar \Delta t}}\right)^{n+1} \exp\left(\frac i\hbar S_\text{HO}[x_\text c]\right) \prod_{k=1}^n\sqrt{\frac{2\pi}{-\frac {Mi}{\hbar}\left[\left(\frac{\pi k}{\Delta t}\right)^2 - \omega^2\right]}}\\ % 24 &= \lim_{n\rightarrow\infty} \left(\sqrt{\frac{M}{2\pi i\hbar \Delta t}}\right)^{n+1} \exp\left(\frac i\hbar S_\text{HO}[x_\text c]\right)\left(\sqrt{\frac{2\pi i\hbar}{M}}\right)^n\frac{1}{\sqrt{\prod_{k=1}^n \left[\left(\frac{\pi k}{\Delta t}\right)^2 - \omega^2\right]}}\\ % 25 &= \lim_{n\rightarrow\infty} \left(\sqrt{\frac{M}{2\pi i\hbar \Delta t}}\right)^{n+1} \exp\left(\frac i\hbar S_\text{HO}[x_\text c]\right) \left(\sqrt{\frac{2\pi i\hbar\Delta t}{M}}\right)^n \\ &\phantom=\times\frac{1}{\sqrt{\prod_{k=1}^n \left(\pi k\right)^2} \sqrt{\prod_{\ell=1}^n\left[1 - \frac{\omega^2\Delta t^2}{\pi^2 \ell^2 }\right]}}\\ % 26 &= \sqrt{\frac{M}{2\pi i\hbar \Delta t}} \exp\left(\frac i\hbar S_\text{HO}[x_\text c]\right) \frac{1}{\sqrt{\prod_{k=1}^\infty \left(\pi k\right)^2} \sqrt{\prod_{\ell=1}^\infty\left[1 - \frac{\omega^2\Delta t^2}{\pi^2 \ell^2 }\right]}} \end{align}

The second square root is manifestly a sine function, up to prefactors, the first product I have chosen to regularize with zeta functions. Define \begin{equation} \hat\zeta(s) = \sum_{n=1}^\infty \frac1{\left(an\right)^s} \qquad \rightarrow \qquad \partial_s \hat\zeta(s) = -\sum_{n=1}^\infty \frac1{\left(an\right)^s} \log an \qquad \rightarrow \qquad -\partial_s \hat \zeta(s) \big|_{s=0} = \sum_{n=1}^\infty \log an \end{equation}

Another way of expressing the derivative with respect to $s$ is: \begin{equation} \partial_s \hat\zeta(s) = -a^{-s}\log a \sum_{n=1}^\infty n^{-s} + a^{-s} \partial_s\sum_{n=1}^\infty n^{-s} = a^{-s}\left[-\log a \zeta(s) + \partial_s\zeta(s)\right] \end{equation} where $\zeta(s)$ is the Riemann zeta function. It then follows \begin{equation} \partial_s \hat\zeta(s)\big|_{s=0} = -\log a \zeta(0) + \partial_s \zeta(s)\big|_{s=0} = \frac12\log a-\frac12 \log2\pi = \frac12 \log\frac{a}{2\pi} \end{equation} and \begin{equation} \prod_{n=1}^\infty \left(an\right)^2 = \exp\log \prod_{n=1}^\infty \left(an\right)^2 = \exp\left(2\sum_{n=1}^\infty\log an\right) = \exp\left(-2\partial_s \hat\zeta(s) \big|_{s=0}\right) = \exp\left(\log\frac{2\pi}{a}\right) = \frac{2\pi}a \end{equation} Plugging in $a = \pi $, we find $\prod_{k=1}^\infty \left(\pi k\right)^2 = 2$. Also using the identity $\sin x / x = \prod_{j=1}^\infty \left( 1 - \frac{x}{j^2\pi^2}\right)$ for $x= \omega \Delta t$, we get \begin{align} i\mathcal G\left(x_f,\Delta t;x_i,0\right) &= \sqrt{\frac{M}{2\pi i\hbar \Delta t}} \exp\left(\frac i\hbar S_\text{HO}[x_\text c]\right) \sqrt{\frac{\omega\Delta t}{2\sin\left(\omega \Delta t\right)}} \end{align} which is a factor $\sqrt{2}$ off.

  • $\begingroup$ don’t have time for a detailed explanation but check the Details section of this post: physics.stackexchange.com/a/384919/83405 where two equivalent definitions of the path integral measure are given along with all the glory detail $\endgroup$ Mar 4, 2018 at 12:23
  • $\begingroup$ @Wakabaloola Thank you for your comment. This fixes part of it, but in that post again some external constraint is used (i.e. normalization through consistent factorization). But this is a good start. $\endgroup$
    – user55789
    Mar 4, 2018 at 13:28
  • $\begingroup$ the path integral measure always has an ambiguity associated to an overall constant, so the normalisation must be fixed by an independent method (such as unitarity, factorisation, consistency with the Schrodinger equation, or operator method matching) $\endgroup$ Mar 4, 2018 at 13:59
  • $\begingroup$ the crucial point is that this constant is independent of any of the parameters that appears in the theory of interest $\endgroup$ Mar 4, 2018 at 14:00
  • $\begingroup$ @Wakabaloola My question is then, why is this constant not a priori known? The only part where I can see the normalization changing is either during the insertion of the action, or at the point of choosing the sine expansion.... $\endgroup$
    – user55789
    Mar 4, 2018 at 14:14


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