quantum mechanics - Quasiclassical QM for central fields

Let's have quasiclassical QM for central field $V(r)$. The Schroedinger equation for radial part of wavefunction $R_{n\ell}$ after substitution $u_{n\ell} = rR_{n\ell}$ takes the form
$$ u_{n\ell}{''} + k_{n\ell}^{2}u_{nl} = 0, \quad k_{n\ell}^{2} = 2m\left(E_{n\ell} - V(r) - \frac{\ell(\ell + 1)}{2mr^{2}} \right). $$ I don't understand the reasoning for the Langer correction

$$\ell(\ell + 1) \to \left(\ell + \frac{1}{2}\right)^{2}.$$

The most of authors usually claim that it is important because we need that the phase of our function on infinity must coincide with phase of exact solution.

What phase is discussed in this statement? In terms of previous question, how to show that this replacement leads to correct phase?

References:

1. L.D. Landau & E.M. Lifshitz, QM, Vol. 3, 3rd ed, 1981; $\S49$.

Answer

I) Let us for simplicity put the physical constants $\hbar=1=m$ to one. OP is considering the usual transcription $u(r)\equiv rR(r)$ of the 3D radial TISE into a 1D TISE,

$$\tag{A} - \frac{1}{2} u^{\prime\prime}(r)+U_{\ell}(r)u(r) ~=~E u(r),$$

where the total potential energy

$$\tag{49.8b} U_{\ell}(r)~:=~ U(r) + \frac{C_{\ell}}{2r^2} $$

is a sum of a central potential energy $U(r)$ and a centrifugal potential energy $\frac{C_{\ell}}{2r^2}$. Here and below the equation numbers refer to Ref. 1. The constant

$$\tag{B} C_{\ell}~:=~\ell (\ell +1)~=~\left(\ell+\frac{1}{2}\right)^2 -\frac{1}{4} $$

in eq. (49.8b) is the eigenvalue of the $\hat{L}^2$ operator.

II) We are investigating a bound state where the angular momentum $\ell>0$ is non-zero.

We are interested in the situation where the centrifugal potential energy numerically completely dominates [and the potential $U(r)$ can be ignored] in a neighborhood $[0,r_0+\epsilon[$ of the classically forbidden interval $[0,r_0[$, where $r_0$ denotes the inner radial turning point. In other words,

$$\tag{C} E~\approx~\frac{C_{\ell}}{2r_0^2}. $$

We also want the semiclassical WKB approximation to be valid in the interval $[0,r_0[$ (away from the turning point). The semiclassical condition

$$\tag{46.6} |\lambda^{\prime}(r)|~\ll~ 1$$ implies that

$$\tag{D} \ell ~\gg~ 1.$$

III) The (absolute value of the) momentum is

$$\tag{46.5} p(r)~:=~\sqrt{2\left|E - U_{\ell}(r)\right|} ~\approx~\sqrt{C_{\ell}\left|r^{ - 2} - r_0^{-2}\right|} \quad\text{for}\quad r\in[0,r_0+\epsilon[,$$

where

$$\tag{E} \sqrt{C_{\ell}} ~\stackrel{(B)}{=}~\ell+\frac{1}{2}-\frac{1}{8\ell}+{\cal O}(\ell^{-2}). $$

The semiclassical connection formulas yield

$$\tag{F} u(r)~\approx~\frac{c}{\sqrt{p(r)}}\exp\left[ \int_{r}^{r_0} \! dr^{\prime}~p(r^{\prime})\right] \quad\text{for}\quad r~<~ r_0, $$

$$\tag{48.1} u(r)~\approx~\frac{c}{\sqrt{p(r)}}\cos\left[ \int_{r_0}^{r} \! dr^{\prime}~p(r^{\prime})-\frac{\pi}{4}\right]$$ $$~=~\frac{c}{\sqrt{p(r)}}\sin\left[ \int_{r_0}^{r} \! dr^{\prime}~p(r^{\prime})+\frac{\pi}{4}\right] \quad\text{for}\quad r~>~ r_0, $$

where $c\in\mathbb{C}$.

IV) The semiclassical approximation (F) behaves as

$$\tag{G} u(r)~\propto~ r^{\sqrt{C_{\ell}}+\frac{1}{2}}\quad\text{for}\quad r~\to~ 0^{+}, $$

while the well-known exact behavior is

$$\tag{32.15} u(r)~\propto~ r^{\ell+1}\quad\text{for}\quad r~\to~ 0^{+}. $$

Hence, the semiclassical approximation would have the correct behavior at the origin $r=0$ if we replace (E) with $\ell+\frac{1}{2}$.

V) Alternatively, in the free case $U(r)=0$, the semiclassical approximation (48.1) behaves as$^1$

$$\tag{H} u(r)~\propto~\sin\left[\sqrt{C_{\ell}}\left(\frac{r}{r_0}-\frac{\pi}{2}\right)+\frac{\pi}{4}\right] \quad\text{for}\quad r~\to~\infty, $$

while the well-known exact behavior is

$$\tag{33.12} u(r)~\propto~\sin\left[\sqrt{C_{\ell}}\frac{r}{r_0}-\ell\frac{\pi}{2}\right] \quad\text{for}\quad r~\to~\infty. $$

This again suggests to replace (E) with $\ell+\frac{1}{2}$.

References:

1. L.D. Landau & E.M. Lifshitz, QM, Vol. 3, 3rd ed, 1981; $\S49$.

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$^1$ Momentum integral becomes

$$ \frac{1}{\sqrt{C_{\ell}}} \int_{r_0}^{r} \! dr^{\prime}~p(r^{\prime}) ~=~\int_1^{\frac{r}{r_0}} \! \frac{dx}{x}\sqrt{x^2-1} ~=~\left[ \sqrt{x^2-1}+\arctan\frac{1}{\sqrt{x^2-1}} \right]_{x=1}^{x=\frac{r}{r_0}}$$ $$\tag{I} ~\approx~\frac{r}{r_0}-\frac{\pi}{2}\quad\text{for}\quad r~\to~\infty. $$