I Unveiling the Math Behind Potential Well Boundaries

[DrinkanDerive]

TL;DR Summary

Confused about boundaries

Hello there. I want to understand the mathematical idea behind boundaries that we write for a potential well. Why we use equally greater and smaller than let's say x between -4a and -2a but we only write x is less than -4a ? How to approach this idea with convergence theorem or Hilbert space? Thank you all and please be safe.

 

[BvU]

Hello DD, !

the mathematical idea

No different than when studying functions.
E.g. $\tan(x)$ exists for all $x$ in $-{\pi\over 2} < x < {\pi\over 2}$ but not for all $x$ in $-{\pi\over 2} < x \le {\pi\over 2}$

QM potential well is two discontinuities that have to be treated separately. Usually we require the wave function and its derivative to be continuous, but at the discontinuity the second derivative is discontinuous.

Physicists are notoriously casual with such things, because they are idealizations of continuous phenomena.

We also don't bother to assign a potential value at the discontinuities themselves: the outcome of the analysis doesn't change.

Disclaimer: I'm an experimental physicist -- perhaps a theoretician or a mathematician can improve or even correct my reply.

 

[hilbert2]

The problem with a finite depth potential well has the "particle in box" system as its limit when the well depth increases without bound. So you can approximate it as accurately as you want without discontinuity even in the first derivative of $\psi$, by constructing a finite potential well and setting some large number as the depth.

 

[DrinkanDerive]

A beam comes from - \infty to 0 and we have u(x) = e^(kx) + R e^(-kx) ; x<0. Why we omit the potential barrier when it comes to write the boundary. After the first term reaches in the box or well we write 0\geq x \leq 2a. When we take the limit when x goes to infinity shouldn't we omit the e^(kx) term? If we say x \leq 0 can't we write u_{1} |_{x=0} = u_{2}|_{x=0} ? Thank you for the answers! :)