The hamiltonian form of the path integral for the time evolution of a single particle in one dimension (in non-relativistic quantum mechanics) is:

$$\langle x|\hat U(t_2,t_1)|x'\rangle=\int \mathcal Dx \mathcal Dp \ e^{\frac i {\hbar} \int_{t_1}^{t_2} dt \ \big(p(t) \dot{x}(t) - \mathcal H(x(t),p(t))\big)}$$ Now consider the single particle to have a standard quadratic hamiltonian $\mathcal H(x,p)=\frac{p^2}{2m} +V(x)$. Plugging this into the above equation, we can use the definition of the path integral to get (equation 3.5 of Condensed matter field theories by Atland and Simons): $$\langle x|\hat U(t_2,t_1)|x'\rangle=\lim_{N\to \infty} \int_{-\infty}^{\infty}...\int_{-\infty}^{\infty} \prod_{n=1}^{N-1}dx_n \prod_{n=1}^{N}\frac {dp_n}{2\pi} \ e^{\frac {i}{\hbar} \sum_{n=1}^{N} \delta t\big(p_n \dot x_n -\frac{p_n^2}{2m}-V(x) \big)} $$ Where $\delta t = \frac{t_2-t_1}{N}$ and $\dot x_n \equiv \frac{x_n-x_{n-1}} {\delta t} $. Now we can see that all integrals over momenta are gaussian, and can be evaluated easily. By writing the sum in the exponent as a product of exponentials $e^{\frac {i}{\hbar} \sum_{n=1}^{N} \delta t\big(p_n \dot x_n -\frac{p_n^2}{2m}-V(x) \big)} = \prod_{n=1}^{N}e^{\frac {i}{\hbar} \delta t\big(p_n \dot x_n -\frac{p_n^2}{2m}-V(x) \big)}$ , and merging the two products, we can easily identify products of Gaussian integrals over the momenta, which can be evaluated. The final result is (equation 3.8 of Atland and Simons): $$\langle x|\hat U(t_2,t_1)|x'\rangle=\int \mathfrak Dx \ e^{\frac i {\hbar} \int_{t_1}^{t_2} dt \ \big(\frac {1}{2m} \dot{x}^2(t) - V(x(t))\big)}=\int \mathfrak Dx \ e^{\frac i {\hbar} \int_{t_1}^{t_2} dt \ \mathcal L(x,\dot x)}$$ Where $\mathfrak Dx$ is the redefined "measure" : $$\mathfrak Dx \equiv \lim_{N \to \infty} \big(\frac {mN}{i2 \pi \hbar (t_2-t_1)}\big)^{N/2} \prod_{n=1}^{N-1}dx_n$$

My question is:

The redefined measure has a constant coefficient that diverges as $N^{N/2} \to \infty$. I don't understand what this means physically. I assume that the integrals without this factor all go to zero, so that multiplied by this divergent factor gives a finite plausible answer for $\langle x|\hat U|x'\rangle$. Another way I look at this is that at least in the context of statistical mechanics, all physical quantities such as expectation values are given as a ratio of path integrals so that the divergent terms cancel. However, I can't find a way to show this in the context of quantum mechanics, with the path integral playing the role of the propagator.

  1. That path integral measure factor is famously known as Feynman's fudge factor. It ensures that the Gaussian (=free) Lagrangian path integral (i) has a finite value, and (ii) is form-invariant wrt. to extra intermediate Gaussian integrations, cf. e.g. my Phys.SE answer here.

  2. As OP already noted, Feynman's fudge factor can be derived from the Hamiltonian phase space path integral by integrating out the momentum variables. See also this related Phys.SE post.

  • $\begingroup$ Thank you for your answer. I see. But what about the divergence of the factor? How can one make sense of that? Either all the integrals without the factor should go to zero so that $0 \times \infty \to finite$, or one should be able to show that similar to the statistical mechanics case, all physically relevant quantities can be written as ratios of path integrals. I want to see which one is true, and why? $\endgroup$ – Sahand Tabatabaei Feb 7 '18 at 16:04
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Feb 7 '18 at 17:27
  • $\begingroup$ I understand, so my first guess was correct. Is there a way to show why the integral goes to zero without that factor in general for any potential $V(x)$? Even a hand-waving argument would be enough for me. $\endgroup$ – Sahand Tabatabaei Feb 7 '18 at 19:54
  • $\begingroup$ Suggestion: As a check with an arbitrary potential $V$, you might want to consider the diabatic limit, cf. my Phys.SE answer here. $\endgroup$ – Qmechanic Feb 7 '18 at 20:39

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