# Infinite square well: wall with infinitesimal thickness

Given an infinite square well, it doesn't matter how thick the wall is, the particle is trapped inside the two walls. If we make the wall of arbitrarily small but finite thickness, the particle is still trapped inside the wall, i.e. it is not possible to find the particle outside of the potential well:

However, if we take a limit of the thickness of the wall to zero, the potential effectively becomes a double Dirac delta distribution. And for this scenario, the derivative of the wavefunction will be discontinuous at the two points of infinite potential:

What is the qualitative difference between finite and infinitesimal thickness of the wall that results in whether the particle is trapped within the walls or leak outside of the walls?

• Interesting question. My guess is that infinite well with finite thickness is made up of infinitely many Dirac delta potential, each of them is separated by an infinitesimally small distance. (Not sure if this can help) – K_inverse Jan 11 at 8:46

Well, one has to be precise when discussing an infinitely tall & infinitesimal thick wall. The order of limits matters. E.g.

1. If the wall is modelled as $$V(x)~=~\left\{\begin{array}{ccc} \infty &{\rm for}& x~=~0\cr\cr 0 &{\rm for}& x~\neq~0\end{array}\right\}~=~\lim_{\varepsilon\to 0^{+}}V_{\varepsilon}(x),\tag{1a}$$ where $$V_{\varepsilon}(x)~=~\frac{1}{\varepsilon}\delta_{x,0},\tag{1b}$$ and where $$\delta_{x,0}$$ is the Kronecker delta, then the potential is zero almost everywhere: $$V(x)~=~0\text{ a.e.}\tag{1c}$$ Therefore the Lebesgue measure (and the particle) can not detect the wall in the first place. In other words, there effectively is no wall.

2. If the wall is modelled by the Dirac delta distribution $$V(x)~=~A \delta(x),\qquad 0 which is a distribution/generalized function, then $$V(x)~=~\lim_{\varepsilon\to 0^{+}}V_{\varepsilon}(x)\tag{2b}$$ is a limit of a finite rectangular wall $$V_{\varepsilon}(x)~=~\frac{A}{\varepsilon} \theta(|x|\!-\!\varepsilon/2),\tag{2c}$$ with height $$\frac{A}{\varepsilon}$$ and thickness $$\varepsilon$$, i.e. with a fixed area $$A$$. Here $$\theta$$ is the Heaviside step function. Then quantum tunnelling through the wall is possible.

3. If the infinitesimal thick wall $$V(x)~=~\lim_{\varepsilon\to 0^{+}}V_{\varepsilon}(x)\tag{3a}$$ is a limit of an infinitely tall wall $$V_{\varepsilon}(x)~=~\left\{\begin{array}{ccc} \infty &{\rm for}& |x|~<~\varepsilon/2,\cr\cr 0 &{\rm for}& |x|~>~\varepsilon/2,\end{array}\right. \tag{3b}$$ then there is no quantum tunnelling.

• Nice nice nice nice nice ............ - ∞ – debo.stackoverflow Jan 11 at 13:10
• Graphically, is the first example you gave equivalent to $V(0)=\infty$ and $V(x)=0 \:\forall x\neq 0$? – Leo L. Jan 11 at 15:52
• $\uparrow$ Yes. – Qmechanic Jan 11 at 17:43
• Then, I am still confused as to the difference between the potential of the 1st scenario and of the 2nd scenario. Are they different kinds of "infinity" at $x=0$? Also, does a particle in the first potential behaves as a free particle? – Leo L. Jan 11 at 22:22
• I updated the answer. – Qmechanic Jan 11 at 23:05