0
$\begingroup$

The WKB approximation gives meaningful approximative solutions to the $1$-dimensional, time-independent Schroedinger equation

$$ -\frac{\hbar^{2}}{2m} \frac{d^{2}}{dx^{2}} \Psi(x) +V(x)\Psi(x)=E\Psi (x)$$

under the assumption that the potential $V(x)$ 'varies slowly' on the scale of the wavelength $\lambda$. I asked here about the precise way to express this vague formulation of 'slowly varying potential' $V(x)$ on wave length scale in mathematical terms.

Now my question is if it's possible to express in general context the vague phrase that 'some quantity $A(x)$ varies slowly on certain given scale $ ds $ (eg say $ds \approx 10^{-16}m$) of $x$' as a concise $ blabla(ds, A(x)) \ll 1 $ statement? To be more precise that blabla term should be an appropiate expression $ B(ds, A(x)) $ we are looking for, which as expected should depend on the scale $ ds $, the quantity $ A(x)$ and possibly it's derivatives and which should formally characterize the statement that $A(x)$ is assumed to vary slowly on the sclae $ds$.

How I think about this: let denote by $ds$ the 'scale' on which $A(x)$ should vary slowly. Then intuitively the statement that '$A(x)$ varies slowly on certain given scale of $x$' could be also roughly rephrased as if $x_0$ is any argument of $A(x)$ and we chose a $x \in [x_0-ds, x_0+ds]$ then the difference $\vert A(x_0)-A(x) \vert$ is 'small'.
By "small" I mean that in all consequent calculations we can set all as "small" considered terms to be zero and would obtain results consistent to experiments / effects performed / detected on the characteristic scale $ \sim ds$.

A naive guess: Can it be formulated like $\vert \frac{A(x_0)-A(x)}{A(x_0)} \vert \ll 1$ for any $x \in [x_0-ds, x_0+ds]$?

Alternatively, can it be expressed in terms of first and second derivatives like

$$\left| \frac{A''(x)}{A'(x)}\right| \cdot ds \ll 1$$

and if yes are these two ways to characterize that $A(x)$ varies slowly on scale $ ds $ of $x$ equivalent or is there a more standard way to formulate this condition in mathematical terms? So it's really a question only about to transform a sloppy statement to a precise one.

$\endgroup$
3
  • $\begingroup$ Neglectible small. So precisely I'm looking for the way to formulate the phrase $A(x)$ varies slowly on certain scale of $x$' as a concise $ blabla \ \ll 1 $ statment. And the meaning of this $ \ll $ would be that in all subsequent calculations it would be allowed to neglect the term with $ \ll 1 $. $\endgroup$
    – user267839
    Commented Nov 17, 2022 at 17:13
  • $\begingroup$ I have modified the question, hope it's a bit clearer now. $\endgroup$
    – user267839
    Commented Nov 17, 2022 at 17:17
  • $\begingroup$ @JBag: ok, I will try to improve some more details you refering to. But I think that mentioning of WKB is important, since essentially this question arose from the considerations on it as "model example" is motivated as attempt to generalize the way the condition of slow varying potential was characterized there. $\endgroup$
    – user267839
    Commented Nov 18, 2022 at 18:08

2 Answers 2

2
$\begingroup$

This is an incomplete answer but hopefully better than nothing. The way to quantify the degree of approximation in general is to give some estimate of the size of the quantity that has been neglected. So rather than saying 'blah $\ll 1$' you would write $$ y = f + \epsilon \simeq f \;\;\;\; \mbox{ where } |\epsilon| \simeq \mbox{blah}. $$ For many calculations it is, however, very difficult to provide an expression for this $\epsilon$ even though you are confident it will tend to zero in some limit. The WKB approximation is like this. It does indeed behave correctly in the appropriate limit, but it's hard to provide a single catch-all expression for how accurate it is. Broadly speaking what the WKB approximation requires, in order to be useful, is $$ \lambda \ll \left|\frac{V'}{V''}\right| $$ where $\lambda$ is the wavelength, but I don't have an expression (even an approximate one, which is all you need) for $\epsilon$ in terms of $\lambda V''/V'$.

$\endgroup$
4
  • $\begingroup$ thank you, so in general it's too optimistic to expect existence of an expression like $ bla \ll 1$ to characterize that $A(x)$ vary slowly on some prescribed scale? but let come back to WKB as 'model example'. Is there an intuitive reason why the statement that the potential $V(x)$ varies slowly as long as we vary $ x $ on typical wavelength scale can be expressed equivalently via $ \lambda \ll \left|\frac{V'}{V''}\right| $? What I'm getting at can't we try to literaly mimic it in order obtain a general characterization of what it means that $A(x)$ vary slowly on scale $ds $ via $\endgroup$
    – user267839
    Commented Nov 17, 2022 at 19:14
  • $\begingroup$ condition like $ ds \ll \left|\frac{A'}{A''}\right| $? Or is it really only a WKB specific estimation? $\endgroup$
    – user267839
    Commented Nov 17, 2022 at 19:15
  • $\begingroup$ Or let go a step back, how do you explicitly derive for the specific case of WKB approximation that the potential $ V $ varies slowly on the wavelength scale $ \lambda $ is equivalent to condition $ \lambda \ll \left|\frac{V'}{V''}\right| $ ? Do you use exacly the same derivation / trick as in Hossein's answer here: physics.stackexchange.com/a/731182/347719 ? The point I want to understand in the derivation if to obtain this expression you have to exploit some 'additional WKB specific' informations like that the wave function has roughly the shape $ \sim e^{ikx} $. $\endgroup$
    – user267839
    Commented Nov 17, 2022 at 19:37
  • $\begingroup$ I can't add much. The concept of wavelength implies the shape $\sim e^{ikx}$. Also: $V'/V''$ is a distance, and it gives the distance over which $V'$ changes appreciably in comparison to $V'$. $\endgroup$ Commented Nov 18, 2022 at 9:52
1
$\begingroup$

First of all, a small but important correction: WKB approximation is not limited to 1-dimensional problems to give meaningful approximative solutions.

Second, the condition for the validity of WKB approximation should be provided in any decent QM textbook. In particular, you can check chapter 7 in Landau&Lifshits, vol.3, equation (46.6) (in the second (revised) edition from 1965): $$\left|\frac{d \lambda(x)}{d x} \right|\ll 2\pi$$ where $$\lambda(x)=\frac{h}{\sqrt{2m(E-V(x))}}$$

$\endgroup$
7
  • $\begingroup$ ok, so that's the neccessary condition when WKB approach works. But question was if there is a way (like you did in case of WKB) to express the formulation that some quantity $A(x)$ varies slowly on certain predetermined scale $ds$ of the parameter $x$. For WKB one says in sloppy way that this approach is legal if 'the potential $V(x)$ varies slowly on the wavelength scale'. So that's a special case of above with $A(x)=V(x)$ and $ds$ the wavelength scale. And you gave a concise condition in $blabla \ll 1$ form when WBK works. Can this procedure be generalized to expressing the sloppy $\endgroup$
    – user267839
    Commented Nov 18, 2022 at 14:51
  • $\begingroup$ statement above that $A(x)$ varies slowly on certain predetermined scale $ds$ of $x$ in concrete mathematical form like you did for WKB? $\endgroup$
    – user267839
    Commented Nov 18, 2022 at 14:52
  • $\begingroup$ Sorry, I missed that. When one is asking is a quantity A(x) varies significantly over distance ds, the standard criterion is |A(x)'|ds<<|A(x)|. If your original question "does A(x) vary $\textit{slowly}$?" can be recast as "does the $\textit{rate}$ of change of A(x) (which is (A(x))') vary $\textit{significantly}?$", then my answer would be the same as that of @Andrew Steane, i.e. |(A(x)')'|ds<<|(A(x))'|. $\endgroup$
    – John
    Commented Nov 18, 2022 at 15:25
  • $\begingroup$ That's another important point in specific case of WKB I not understand. In a plently of sources (eg: theory.physics.manchester.ac.uk/~judith/AQMI/PHYS30201se8.xhtml ; but the net is full of them, just try to google 'wkb approximation vary' ) it is stated that the WKB approach is legal for the situation where "$V(x)$ vary slowly on the wavelength scale", but on the other hand the usual formal condition is $\vert V'' \vert ds \ll \vert V' \vert $ as @Andrew noticed, which in turn as you said is equivalent $\endgroup$
    – user267839
    Commented Nov 18, 2022 at 18:01
  • $\begingroup$ *plenty not *plently $\endgroup$
    – user267839
    Commented Nov 18, 2022 at 18:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.