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Suppose we solve the Schrödinger equation, \begin{equation} -\psi''(x) + V(x) \psi(x) = -|E| \psi(x), \end{equation} where two cases are considered, $V(x) = -V_0$ (square well) and $V(x) = -V_0 (1 - |x|)$ (triangular well) in the interval $[-1,1]$, and $V(x) = 0$ outside this interval. The solution is supposed to be zero in zero, $\psi(0) = 0$. In the first case, we have the trigonometric functions of the form $\sin(\sqrt{|E|}x)$, the second one gives the Airy functions of the form $\operatorname{Ai}(V_0^{-2/3}(V_0|x|-(V_0-|E|)))$.

My question is the following, is it possible to consider the square well as a limit of the triangular well by introducing a parameter $\varepsilon$ for the equation inside the interval $x \in [-1,1]$ \begin{equation} -\psi''(x) - V_0 (1 - \varepsilon |x|)\psi(x) = -|E| \psi(x), \end{equation} taking the limit $\varepsilon \rightarrow 0$ in the solution and finishing with the trigonometric functions? The problem is such that the parametrized equation has the solutions of the form $\operatorname{Ai}(\varepsilon^{-2/3}(\varepsilon|x|-(1-|E|)))$, which do not converge to the usual trigonometric function but to the "decaying" one since the Airy functions converge to zero for big negative arguments. Where is the error?

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  • $\begingroup$ How is this going to work out with the boundary conditions (or which BC do you choose)? A triangle does not converge to a square. $\endgroup$ Aug 30, 2019 at 16:54
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    $\begingroup$ You potential for the square well is incomplete as it does not impose any boundary conditions. And when you write a more complete version of the potential your limit procedure will break down. $\endgroup$ Aug 30, 2019 at 16:54
  • $\begingroup$ I'm sorry for forgetting to write down the BC, added them to the question. I use $\psi(0)=0$ and continuity conditions in $x=\pm 1$. $\endgroup$ Aug 30, 2019 at 17:06
  • $\begingroup$ The equation $-\psi''(x) - V_0 (1 - \varepsilon |x|)\psi(x) = -|E| \psi(x)$ implies that $V(x)>0$ for $|x|>1$. If that wasn't what you meant, then you should not write the Schrödinger equation in that form. $\endgroup$ Aug 31, 2019 at 16:56
  • $\begingroup$ This equation is valid, as before, only in the interval $[-1,1]$. Outside this interval the potential is still zero. I've added this remark to the question. $\endgroup$ Sep 2, 2019 at 8:25

3 Answers 3

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In fact, if we calculate the first eigenvalues and eigenfunctions, we will not find a big difference between the eigenfunctions, and the eigenvalues for large $n$ almost coincide. Put $V_0=1$, then the first 4 eigenfunctions for two types of potentials shown in Figure 1 Figure 1 The first 7 eigenvalues for two types of potentials:

1.4674, 8.86974, 21.2081, 38.4868, 60.7166, 87.9195, 120.134

2.16793, 9.37633, 21.7317, 38.9876, 61.2251, 88.4198, 120.638

The difference in eigenfunctions for the two types of potentials is shown in Fig. 2. It can be seen that at large values of $n$ the difference tends to zero, which is already obvious. Figure 2

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  • $\begingroup$ Exactly, at the qualitative level we should expect an oscillation-like behaviour inside the potential, and actually the Airy functions give it for a negative argument. However the problem lies at the quantitative level: the eigenfunctions do not converge to normal trigonometric functions but to $\cos(x^{3/2})/x^{1/4}$, which in turn will converge to zero. This is the thing I do not understand. $\endgroup$ Aug 30, 2019 at 19:21
  • $\begingroup$ This asymptotic behavior is valid for large x for the wave function in a uniform field. The triangular well potential is not a uniform field. And the solution here is different. A numerical solution for the eigenfunctions in two types of potentials demonstrates their proximity. $\endgroup$ Aug 31, 2019 at 3:09
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The problem is such that the parametrized equation has the solutions of the form $\operatorname{Ai}(\varepsilon^{-2/3}(\varepsilon|x|-(1-|E|)))$, which do not converge to the usual trigonometric function but to the "decaying" one since the Airy functions converge to zero for big negative arguments. Where is the error?

You've forgotten that Schrödinger's equation is a second-order equation, and that to satisfy your boundary condition $\psi(0)=0$ you need both kinds of Airy functions: $\operatorname{Ai}$ and $\operatorname{Bi}$. The solution of your equation for the potential $-V_0(1-\varepsilon x)$, satisfying the boundary condition, is then:

$$\psi(x)=\operatorname{Ai}\left(\frac{V_0(\varepsilon |x|-1)+|E|}{(V_0\varepsilon)^{2/3}}\right)-\operatorname{Bi}\left(\frac{V_0(\varepsilon |x|-1)+|E|}{(V_0\varepsilon)^{2/3}}\right)\frac{\operatorname{Ai}\left(\frac{|E|-V_0}{(V_0\varepsilon)^{2/3}}\right)}{\operatorname{Bi}\left(\frac{|E|-V_0}{(V_0\varepsilon)^{2/3}}\right)}.$$

With this expression you'll get

$$\frac{\psi(x)}{\psi'(0)}\to\frac{\sin(\sqrt{V_0-|E|}x)}{\sqrt{V_0-|E|}}\quad\text{as}\quad\varepsilon\to0.$$

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  • $\begingroup$ Thank you, Ruslan! This seems pretty close to what I need. In your result, you use the derivative at zero as the normalization (so that the whole expression would be a kind of inverted transparency of the potential (with $\psi(0)$), which should actually be a constant). Is it possible to perform this limit with the normalization $1$? $\endgroup$ Sep 3, 2019 at 17:34
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    $\begingroup$ @CiruzzBroncio normalization to 1 should be performed on the whole domain of your wavefunction, which includes $|x|>1$. Of course it's possible, but that's not the point of my answer. The point was to show the limiting behavior, and in my case it's not normalization, but rather bringing the function to unit slope at $x=0$, so the limit could be compared with Airy wavefunction. $\endgroup$
    – Ruslan
    Sep 3, 2019 at 17:52
  • $\begingroup$ The problem is resolved, thank you very much for the help! $\endgroup$ Sep 4, 2019 at 14:44
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To see what you've actually done here :

$$-\frac {d^2\psi(x)}{dx^2} - V_0 (1 - \varepsilon |x|)\psi(x) = -|E| \psi(x)$$

Change the coordinate using $x:=ay$ for some constant $a>0$.

Now you get :

$$-\frac{d^2\psi(y)}{dy^2} - a^2V_0 (1 - a\varepsilon |y|)\psi(y) = -a^2|E| \psi(y)$$

So by making $a\varepsilon=1$ (a choice we can make) :

$$-\psi''(y) - a^2V_0 (1 - |y|)\psi(y) = -a^2|E| \psi(y)$$

And you get your triangular problem back with scaled energy and potential values.

It is always a triangular potential problem.

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  • $\begingroup$ I do not get the convention $a\varepsilon=1$, if $a$ is a constant then we the limit $\varepsilon \rightarrow 0$ does not have sense since $\varepsilon$ should be constant as well. $\endgroup$ Aug 30, 2019 at 17:59
  • $\begingroup$ Furthermore, is it possible to see this connection exactly at the level of the solutions, something like $Ai(x) \rightarrow \cos(x)$? $\endgroup$ Aug 30, 2019 at 18:28
  • $\begingroup$ The problem is that Ciruzz is using extremely misleading notation, by neglecting to be crystal clear in specifying that the potential should be truncated when it gets to zero. $\endgroup$ Aug 31, 2019 at 16:55

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