From Dirac's remarks $$\langle x_2,t_2|x_1,t_1\rangle=\exp\left[ \frac{i\int_{t_1}^{t_2}\mathrm dt\, L_{\text{classical}}{\left(\dot{x},x\right)}}{\hbar}\right].$$ How can I conclude from Huygens principle a space time trajectory is formed by a particle overall contribution of all equal smaller classical path contribution with different phase?

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    $\begingroup$ In the limit $\hbar \rightarrow 0$ you the path $x$ is classical, hence follows the principle of least action (which corresponds to Huygens principle). Notice that $S[x(t)] = \int^{t_2}_{t_1} L(\dot{x},x)dt$ has $\delta S = 0$ if $x$ is classical path by the stationary phase approximation. I can give you a proof for the stationary phase approximation for functions (although here we are dealing with functionals, the proof for that is more difficult) $\endgroup$ Jan 4, 2020 at 15:57
  • $\begingroup$ Reference to Dirac's remark? Which page? $\endgroup$
    – Qmechanic
    Jan 4, 2020 at 17:41
  • $\begingroup$ Related: physics.stackexchange.com/q/351072/2451 $\endgroup$
    – Qmechanic
    Jan 4, 2020 at 17:47
  • $\begingroup$ @Qmechanic actually I get from here google.com/url?sa=t&source=web&rct=j&url=http://… $\endgroup$
    – baponkar
    Jan 4, 2020 at 18:17
  • $\begingroup$ Permalink: doi.org/10.1103/RevModPhys.20.367 p. 19. $\endgroup$
    – Qmechanic
    Jan 4, 2020 at 18:21

1 Answer 1


Let $\lambda \in \mathbb{R}$, $\lambda \gg 1$, let $f,g$ be analytic real/complex functions near $c \in [a,b] \subseteq \mathbb{R}$. Let $g^\prime(c)=0$ for some $c \in (a,b)$ and $g^\prime(t)\neq 0$ for all $t \in [a,b] \setminus \{c\}$. Assume also that $g^{\prime \prime}(c)\neq 0$, $f(c)\neq 0$, let $\mu$ denote the sign of $g^{\prime \prime}(c)$. Then it holds that:

$$I(\lambda) = \int^b_a f(t)e^{i\lambda g(t)}dt \approx f(c)e^{i\pi \mu/4} e^{i\lambda g(c)}\sqrt{\frac{2\pi}{\lambda|g^{\prime \prime}(c)|}}.$$


$$I(\lambda) = \int^b_a f(t)e^{i\lambda g(t)}dt = e^{i\lambda g(c)}\int^b_a f(t)e^{i\lambda(g(t)-g(c))}dt.$$

$\exp(i\lambda(g(t)-g(c)))$ is highly oscillatory for $t\neq c$ and $\lambda \gg 1$, hence for some $\epsilon \ll 1$ we have that:

$$I(\lambda) = e^{i\lambda g(c)}\int^{c+\epsilon}_{c-\epsilon}f(t)e^{i\lambda(g(t)-g(c))}dt$$

$$\approx f(c)e^{i \lambda g(c)}\int^{c+\epsilon}_{c-\epsilon} e^{i\lambda(g(t)-g(c))}.$$

We can Taylor expand $g$ around $t=c$ to second order: $g(t) \approx g(c)+g^{\prime}(c)(t-c)+\frac{1}{2}g^{\prime \prime}(c)(t-c)^2$. $g^{\prime}(c)=0$ by assumption so it follows that:

$$I(\lambda) = f(c)e^{i\lambda g(c)}\int^{c+\epsilon}_{c-\epsilon} e^{i\lambda g^{\prime \prime}(c)(t-c)^2/2}dt \approx f(c)e^{i\lambda g(c)}\int^{\infty}_{-\infty} e^{i\lambda g^{\prime \prime}(c)(t-c)^2/2}dt$$

$$ = f(c)e^{i\lambda g(c)}\int^{\infty}_{-\infty} e^{i\lambda g^{\prime \prime}(c)s^2/2}ds.$$

Using the standard Gaussian integral formula it follows that

$$I(\lambda) = f(c)e^{i\lambda g(c)}\sqrt{\frac{2\pi}{\lambda|g^{\prime \prime}(c)|}}\sqrt{i\mu},$$

where we used that $\mu g^{\prime \prime}(c)=|g^{\prime \prime}(c)|$. Notice that since $\mu \in \{1,-1\}$ and $\sqrt{i}=(e^{i\pi/2})^{1/2} = e^{i\pi/4}$ it holds that $\sqrt{\mu i} = e^{i\pi/4}$. Therefore indeed it follows that

$$I(\lambda) \approx f(c)e^{i\pi \mu/4}e^{i\lambda g(c)}\sqrt{\frac{2\pi}{\lambda|g^{\prime \prime}(c)|}}.$$

Here in our case the function $g$ is replaced by the functional $S[x(t)]$. The idea is similar. $\delta S = 0$ happens for $x=x_{\mathrm{clas}}$ and hence you get that equation. Notice here that $1/\hbar \rightarrow \infty$ and that the stationary phase approximation corresponds to the Huygens principle in this way. The fact that $$\exp(i/\hbar (S[x]-S[x_{\mathrm{clas}}]))$$ is highly oscillatory corresponds to what you want to prove.

I have the idea of this proof from somewhere online and I don't know the reference exactly.

  • $\begingroup$ Thankyou for proof..But I want physics.Please give me some physical way to prove that.@Mathphys meister $\endgroup$
    – baponkar
    Jan 4, 2020 at 18:19
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    $\begingroup$ Did you see my comment above also? There I give more explanation by arguing that $\delta S = 0$ is the principle of least action. I don't understand what you mean with proving something in a physical way. This is just a mathematical result and physics gives intuition and tells us how to interpret the result. $\endgroup$ Jan 4, 2020 at 18:54

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