I'm trying to get the eigenfunctions and eigenvalues of the following one-dimensional Schrodinger equation: $$ -\frac{1}{2}\frac{d^2}{dx^2}\psi(x)+V(x)\psi(x)=E\psi(x)\,,$$ where the external potential reads:$$V(x)=2\tanh^2(x/\sqrt{2})\,.$$ Mathematica gives me the following analytical result for the eigenfunctions (I used the method sketched by @Stefen Luttrell in this post):

C[1] LegendreP[1/2 (-1 + Sqrt[17]), Sqrt[2] Sqrt[2 - e], Tanh[x]] + 
C[2] LegendreQ[1/2 (-1 + Sqrt[17]), Sqrt[2] Sqrt[2 - e], Tanh[x]]

and now the problem is to find the correct values for the coefficients $C[1]$ and $C[2]$. I want the wavefunction to vanish for $x\rightarrow \pm \infty$. I tried to use Asymptotic and Series functions, but in all cases Mathematica takes too much time to compute or I cannot get a clue for the coefficients behavior. Any help is greatly appreciated.

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    $\begingroup$ Welcome to Mathematica StackExchange! Have you tried giving it assumptions, e.g. $E<2$? I do get a result, although it looks very complicated: Asymptotic[C[1] LegendreP[1/2 (-1 + Sqrt[17]), Sqrt[2] Sqrt[2 - e], Tanh[x]] + C[2] LegendreQ[1/2 (-1 + Sqrt[17]), Sqrt[2] Sqrt[2 - e], Tanh[x]], x -> ∞, Assumptions -> e < 2 && C[1] ∈ Reals && C[2] ∈ Reals]; Solve[% == 0, {C[1], C[2]}] $\endgroup$
    – Domen
    Commented Aug 15, 2023 at 17:04
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    $\begingroup$ Can we use $\lim_{x \to \pm \infty} \tanh(x) = \pm 1$ to simplify things at all? $\endgroup$
    – ydd
    Commented Aug 15, 2023 at 17:19

1 Answer 1


As pointed out by ydd, instead of looking at $x\to\pm\infty$ it is easier to change variables $y=\tanh(x)$ and look at $y\to\pm1$. So we want to check for which values of $\mu,\nu$ is the expression $c_1 L^\nu_\mu(y)+c_2 Q^\nu_\mu(y)$ finite at $y\to\pm1$, where $\mu=1/2 (-1 + \sqrt{17}), \nu=\sqrt{2(2-E)}$. I'm going to assume that $E<2$ so that $\nu$ is real, I leave it to you to explore other cases.

I use the following code to solve this. I write a and b instead of $\mu,\nu$ to simplify the notation.

First, the divergence at $y\to-1$,

Assuming[y > -1, Series[c1 LegendreP[a, b, y] + c2 LegendreQ[a, b, y], {y, -1, 0}] // Normal // FullSimplify]

I isolate the divergent term and obtain

c1 -> -(1/2) c2 π Cot[a π]

Next, the divergence at $y\to+1$,

Assuming[y < 1, Series[c1 LegendreP[a, b, y] + c2 LegendreQ[a, b, y] /. c1 -> -(1/2) c2 π Cot[a π], {y, 1, 0}] // Normal //   FullSimplify]

Finally, I isolate the divergence and get the condition

Csc[a π] Gamma[b] Sin[(a - b) π] == 0

If you write this in terms of $\mu,\nu$ you get a simple equation for $E$, essentially something like $E_n=1/4 (-1 + \sqrt{17} - 4 n + 4 \sqrt{17} n - 8 n^2)$ where $n$ is an arbitrary integer.

I most likely missed some solutions when performing algebraic manipulations (I was not careful with inverse functions and branch cuts) but you can double check that yourself.

  • $\begingroup$ Thank you very much for you answer! A final question: in the final step (when obtaining an expression for E) one actually gets another solution which reads: $E_n=1/4 (-5 + 3 \sqrt{17} - 12 n + 4 \sqrt{17} n - 8 n^2)$. Why should I discard it? $\endgroup$ Commented Aug 16, 2023 at 8:50
  • $\begingroup$ @Crazydemon92 Perhaps you shouldn't discard it (cf. my last paragraph: I did not pay attention to other solutions; I leave it to you to check whether they are satisfactory solutions or not, which depends on what exactly your goal is). $\endgroup$ Commented Aug 17, 2023 at 14:46
  • $\begingroup$ @Crazydemon92, I am glad to see a clear derivation of the quantum spectrum of bound states. However, this is nothing but the well known spectrum of the Pöschl–Teller potential. Simply note that $\tanh^2(x)=1-\text{sech}^2(x)$. All details needed to find energies and wave functions are already presented in en.wikipedia.org/wiki/Pöschl–Teller_potential $\endgroup$
    – Jam
    Commented Sep 22, 2023 at 1:23

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