Difficulty in understanding step in Deriving WKB approximation

[curious_mind]

TL;DR Summary

I am not able to understand the calculation step in deriving WKB approximation given in Quantum Mechanics book of Zettili. It is more related to Vector Calculus though.

In Zettili book, it is given that $\nabla^2 \psi \left( \vec{r} \right) + \dfrac{1}{\hbar ^2} p^2 \left( \vec{r} \right) \psi ( \vec{r} ) =0$ where $\hbar$ is very small and $p$ is classical momentum.
Now they assumed the ansatz that $\psi ( \vec{r} ) = A ( \vec{r} ) e^{i S( \vec{r} ) / \hbar}$ , where $A(\vec{r})$ is amplitude and $S(\vec{r})$ is phase. Now they write that substituting this ansatz into the above semiclassical Schrodinger equation, it is obtained :
$A \left [ \dfrac{\hbar^2}{A} \nabla^2 A - ( \vec{\nabla} S )^2 + p^2(\vec{r}) \right] + i \hbar \left[ 2 \left( \vec{\nabla} A \right) \cdot \left( \vec{\nabla} S \right) + A \nabla^2 S \right] =0$.

I am not exactly how I this expression can be obtained. I have tried to use laplacian operator property given in https://en.m.wikipedia.org/wiki/Vector_calculus_identities

$$ \nabla^2 (fg) = f \nabla^2 g + g \nabla^2 f + 2 \nabla f ~~ \nabla g $$.

But still I am not able to arrive at that expression given. Can anyone tell me about laplacian operator's chain rule. I think it can be applicable here for getting laplacian of term $e^{i S( \vec{r} ) / \hbar}$ .Or is there any other way around. Using simple formula for laplacian in spherical coordinates is also giving me cumbersome expressions, not equal to this one. Where am I mistaken ?
Any help or hint would be much appreciated.

 

[pasmith]

Chain rules for vector derivative operators are most easily derived using cartesian coordinates:$$\begin{split} \nabla^2g(f) &= \frac{\partial^2}{\partial x_i\,\partial x_i}g(f) = \frac{\partial}{\partial x_i}\left(g'(f)\frac{\partial f}{\partial x_i}\right)\\ &= g''(f) \frac{\partial f}{\partial x_i}\frac{\partial f}{\partial x_i} + g'(f)\frac{\partial^2 f}{\partial x_i\,\partial x_i} \\ &= g''(f)\nabla f \cdot \nabla f + g'(f)\nabla^2 f. \end{split}$$

 

[curious_mind]

Ok thanks, using your expression - $\nabla^2 (e^{i S(\vec{r}) / \hbar})$ simplifies into the form that is derived into the book. But if I apply spherical laplacian on $\nabla^2 (e^{i S(\vec{r}) / \hbar})$, I am getting one additional term involving $\nabla S$ which is not supposed to be obtained. But using the chain rule you obtained I am able to get that's why I want to know what is my exact mistake.

Is the expression I wrote from wikipedia is also for cartesian coordinates ? Because if I use that and then applying spherical laplacian on - $\nabla^2 (e^{i S(\vec{r}) / \hbar})$ I am not getting the expression derived in the book.

These are some obvious mistakes from my side but I am quite weak at math sometimes :) .

 

[vanhees71]

These expressions are written in manifestly covariant form and thus are independent of the choice of coordinates. You only have to be careful when applying $\Delta$ to a vector. There you should use Cartesian components or use the mainfestly covariant definition
$$\Delta \vec{A}=\vec{\nabla} (\vec{\nabla} \cdot \vec{A})-\vec{\nabla} \times (\vec{\nabla} \times \vec{A}),$$
but this special case isn't needed for your calculation, because you deal with the Laplacian applied to a scalar field.

Which formula for the Laplacian in spherical coordinates have you used?

 

[curious_mind]

$\nabla^2 f = \dfrac{1}{r^2} \dfrac{\partial }{\partial r} \left ( r^2 \dfrac{\partial f}{\partial r} \right ) + \dfrac{1}{r \sin \theta} \dfrac{\partial }{\partial \theta} \left ( \sin \theta \dfrac{\partial f}{\partial \theta} \right ) + \dfrac{1}{r^2 \sin^2 \theta} \dfrac{ \partial^2 f} { \partial \phi^2}$

Here, $r^2$ need to be multiplied while taking partial derivative with respect to r. There is no $\theta$ and $\phi$ dependence, so those terms will be 0.

 

[vanhees71]

Why shouldn't there be $\theta$ and $\phi$ dependence in your problem?

 

[curious_mind]

Oh okay, it is written $\vec{r}$, so it would mean $(r, \theta, \phi)$ ? So yes, there will be. I had misunderstood it completely. Thanks for pointing it out, otherwise there would be no point of using Laplacian form of $\nabla^2$.

The mistake I was making also I understood. I used spherical coordinates to differentiate, and then after converting back into vector calculus form I was using cartesian one. That is why confusion arised.

Like, I was taking $\dfrac{1}{r^2} \dfrac{\partial^2 S}{\partial r^2}$ as $\nabla^2 S$ but I can't do that in spherical, I have to write $\dfrac{1}{r^2} \dfrac{ \partial }{\partial r} \left( r^2 S \right )$ as $\nabla^2 S$. So this was main mistake. And it was when I wrongly neglected dependence of $\theta$ and $\phi$. But now confusion is cleared. Thanks.

Actually whole calculation was lengthy that's why I did not post here, but it seems I should have written whole things. :)