How to find the roots of an equation which is almost singular everywhere

Question

I'm trying to find the roots of the equation as a variable of k:

a = 0.81; b = 1.94;
LegendreP[(-a + Sqrt[8 b + a^2])/(2 a),  k/a, 1] == 0

The problem is that this function is almost divergent everywhere except only a few points, and FindRoot and Reduce seems to fail on this

FindRoot[LegendreP[(-a + Sqrt[8 b + a^2])/(2 a),  k/a, 1] == 0, {k, 0}]

FindRoot::nlnum: The function value {ComplexInfinity} is not a list of numbers with dimensions {1} at {k} = {1.49012*10^-8}. >>

Reduce[LegendreP[(-a + Sqrt[8 b + a^2])/(2 a),  k/a, 1] == 0, k]

Reduce::inex: Reduce was unable to solve the system with inexact coefficients or the system obtained by direct rationalization of inexact numbers present in the system. Since many of the methods used by Reduce require exact input, providing Reduce with an exact version of the system may help. >>

With[{a = 81/100, b = 194/100}, 
 Reduce[LegendreP[(-a + Sqrt[8 b + a^2])/(2 a),  k/a, 1] == 0, k]]

Reduce::nsmet: This system cannot be solved with the methods available to Reduce. >>

How would I solve this equation? The roots are k = n*a, where n is integer.

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Edit

The motivation of this problem is to determine the bound states of a one-dimensional potential well

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

where

$$ V(x)=-\frac{\hbar^2b}{m}sech^2(a x) $$

,$k$ is a parameter.

This equation has solution

ψ[x_]:=A*LegendreP[(-a + Sqrt[8 b + a^2])/(2 a),  k/a, Tanh[a x]] + 
 B*LegendreQ[(-a + Sqrt[8 b + a^2])/(2 a),  k/a, Tanh[a x]]

V[x_] := -((ħ^2 b)/m) Sech[a x]^2
-(ħ^2/(2 m)) ψ''[x] + V[x] ψ[x] == -((ħ^2 k^2)/(2 m)) ψ[x] // FullSimplify
(*True*)

Quantum mechanics requires the wave function is finite, which gives a constrain on parameter k and thus the eigen-energies($\sim k^2$) are some discretized values. And I'm trying to find these eigen-energies. Mathematically, this is equivalent to finding the $k's$ that keep associate Legendre polynomials finite in region 0 <= x <= 1

A*LegendreP[(-a + Sqrt[8 b + a^2])/(2 a),  k/a, x]+B*LegendreQ[(-a + Sqrt[8 b + a^2])/(2 a),  k/a, x]

.

Edit2

For the wave function must be normalizable and square-integratable, here is the explanation in the Griffiths's Introduction to Quantum Mechanics:

For some solutions to the Schrödinger equation the integral is infinite; in that case no multiplicative factor is going to make it 1. The same goes for the trivial solution $\Psi = 0$. Such non-normalizable solutions cannot represent particles, and must be rejected. Physically realizable states correspond to the square-integrable solutions to Schrödinger's equation.

For the case of free particle, he also explained why it's not normalizable:

In the case of the free particle, then, the separable solutions do not represent physically realizable states. A free particle cannot exist in a stationary state; or, to put it another way, there is no such thing as a free particle with a definite energy.

Answer 1

Quantum mechanics requires the wave function is finite, which gives a constrain on parameter k and thus the eigen-energies(∼k2) are some discretized values.

Not true! The wave function does not need to be finite. It does not even have to be an $L^2$ function. If your teacher insists on it, bring the free-particle wave function as an example.

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For the sake of a short argument, let us assume that there is no such a case where the divergence of LegendreP exactly cancels the divergence of LegendreQ. Also, let us assume that $k>0$.

With the replacements {(-a + Sqrt[a^2 + 8 b])/(2 a) -> Δ, k/a -> κ, a x -> 1/ξ} you get a simpler function to analyze:

LegendreP[Δ, κ, Tanh[1/ξ]]

The reasoning behind the following calculation is quite obvious: find the behavior of ψ at infinity.

( Series[LegendreP[Δ, κ, Tanh[1/ξ]], {ξ, 0, 0}] // FullSimplify) /. 
   (1 - Tanh[1/ξ])^(-κ/2) (1 + Tanh[1/ξ])^(κ/2) -> E^(κ/ξ) /. 
   ξ -> 1/(a x)

Now, $k>0$ implies $κ>0$ which in turns causes a divergent exponential. That is bad unless the other factors go to zero faster. Thus, either $κ=1$ (from the numerator) or $κ=2,3,4,…$ (from the denominator).

Similarly, the behavior at minus infinity

gives you another constraint (because $\lim_{x\to-∞} sech(x)=0$ and $-κ$ is a negative integer): $Δ=2n$ with $n$ an integer (positive, negative, or zero).

Now, from the physical point of view, this is a rather interesting constraint. If the calculations above are correct, it says that bound states do not happen unless b = 1/2 a^2 n (1 + n).

Lather, rinse, and repeat for LegendreQ. Check that linear combinations of LegendreP and LegendreQ do not cancel each other for special cases.

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Update

(1) The procedure above shows that the behavior at +∞ (computed using Series) imply $κ$ must be an integer. Since $κ=k/a$, the original question has been answered.

(2) From the analysis of the behavior at -∞, a relationship between $a$ and $b$ was found. That relationship is necessary for the solutions posted by OP. The following plot shows how using the values provided by the OP, it is impossible to get a linear combination of LegendrePand LegendreQ that does not diverge at -∞. The numerical calculation quoted by OP imply solutions more general than those posted by OP.

On the other hand, modifying $b$ slightly to comply with $Δ=2$, we obtain the following plot where LegendreP is a function that converges at both +∞ and -∞.

(3) Instead of changing the variable from $x$ to $ξ$, you could have used

Series[LegendreP[Δ, κ, Tanh[x]], {x, ∞, 0}]