Why the action is taking phase in considering Huygens principle in matter waves?

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?

• 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) Jan 4, 2020 at 15:57
• Reference to Dirac's remark? Which page? Jan 4, 2020 at 17:41
• Jan 4, 2020 at 17:47
• @Qmechanic actually I get from here google.com/url?sa=t&source=web&rct=j&url=http://… Jan 4, 2020 at 18:17
• Permalink: doi.org/10.1103/RevModPhys.20.367 p. 19. Jan 4, 2020 at 18:21

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:

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

Proof:

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

$$\exp(i\lambda(g(t)-g(c)))$$ 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.

• Thankyou for proof..But I want physics.Please give me some physical way to prove that.@Mathphys meister Jan 4, 2020 at 18:19
• 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. Jan 4, 2020 at 18:54