# Deriving the Lagrangian form of the Feynman path integral through Gaussian integration

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.

• 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? – Sahand Tabatabaei Feb 7 '18 at 16:04
• 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. – Sahand Tabatabaei Feb 7 '18 at 19:54
• Suggestion: As a check with an arbitrary potential $V$, you might want to consider the diabatic limit, cf. my Phys.SE answer here. – Qmechanic Feb 7 '18 at 20:39