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Born-Oppenheimer approximation leads to this equation for the eigenfunctions and eigenvalues describing molecular vibrations: $$\left[-\frac{\hbar^2}{2\mu}\frac{1}{R^2}\frac{\partial}{\partial R}\left(R^2\frac{\partial}{\partial R}\right)+\frac{\langle\Phi_s|\mathbf{N^2}|\Phi_s\rangle}{2\mu R^2}+E_s(R)-E\right]F_s(R)=0$$ (faithfully copied from page 484, Physics of Atoms and Molecules, Bransden & Joachain 2003). Assuming that we are dealing with a molecule which is not rotating (i.e. $\mathbf{N}=0)$, and approximating the curve $E_s(R)$ with the harmonic potential $\frac{1}{2}k(R-R_0)^2$ in the vicinity of its point of minimum, the B.O. equation above is totally equal to that of a 3D isotropic harmonic oscillator with null angular momentum. Therefore, its solutions should be given by associated Laguerre polynomials times some stuff. Turns out that Hermite polinomials are used instead. Why? The idea that a 3d harmonic oscillator which does not rotate should boil down to a linear harmonic oscillator makes total sense to me. But the equations give these two different polynomials and there's no way to say that $L_n^{1/2}(x^2)\propto H_n(x)$ (namely that associated Laguerre polynomials for null angular momentum case are proportional to Hermite p. of the same degree to make sure that eigenfunctions for 3d and 1d are the same). What am I doing wrong?

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  • $\begingroup$ A 3D isotropic harmonic oscillator with null angular momentum is just a 1 DOF harmonic oscillator. $\endgroup$ Commented Aug 27, 2021 at 9:30
  • $\begingroup$ @ArnoldNeumaier not quite: the kinetic energy operator is very different. $\endgroup$
    – Ruslan
    Commented Aug 27, 2021 at 11:23

2 Answers 2

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For the radial part of the Schrödinger equation for 3D harmonic oscillator with $l=0$,

$$-\frac1{r^2}\frac{\mathrm d}{\mathrm d r}\left(r^2f'(r)\right)+Ar^2 f(r)=Ef(r),$$

one can use the substitution

$$f(r)\to \frac{g(r)}r,$$

which will lead to the equation

$$-g''(r)+Ar^2g(r)=Eg(r),$$

which is isomorphic to the Schrödinger equation of 1D harmonic oscillator. Thus, the solutions of Schrödinger equations for both 1D and (the radial part of) 3D harmonic oscillators are expressible in terms of the same kind of functions.

Interestingly, this works only in number of dimensions $n=1$ or $n=3$ (with generalized substitution $f(r)\to g(r)r^{\frac{1-n}2}$). For any other $n$, you'll get additional terms in the ODE for $g$ that will require Laguerre polynomials instead of Hermite ones.

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  • $\begingroup$ Thanks! Now I understand why the book uses Hermite polynomials for calculating the expectation values of $\textit{R}$, the internuclear distance, on vibrational states with null angular momentum. The expval integral would read $$\int \frac{g_n(R)}{R}R\frac{g_n(R)}{R} Y_0^0(\Theta,\Phi)R^2sin\Theta dRd\Theta d\Phi$$ where the $g_n(R)$ are the 1D harmonic oscillator eigenfunctions. This amounts to calculating an expectation value on 1D oscillator states: $$\int g_n(R)R g_n(R) dR$$. I obtained different results with Laguerre polinomials of grade $n$, probably I goofed or used wrong formulae $\endgroup$ Commented Aug 28, 2021 at 18:06
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I think Ruslan gave an excellent answer, and I thought I could add a bit of intuition from another perspective. Now, maybe I'm completely wrong about this because currently I don't deal with molecules, but in the field of optics, one encounters the equations that deal with diffraction of a laser beam (Fresnel diffraction) which lead to eigen modes who obey the 2D quantum harmonic oscillator equation.

These can be described with Hermite-Gauss modes when working in cartesian coordinates: $$I(x,y,z)\propto H_n \left(\frac{x}{\frac{w}{\sqrt{2}}}\right ) H_m \left(\frac{y}{\frac{w}{\sqrt{2}}}\right ) \exp(-\frac{x^2+y^2}{2q^2}) $$ where q,w are parameters related to the size and divergence of the beam and $n,m$ are the quantum numbers of the mode (and z is the direction the beam is propagating in). See image taken from Wikipedia: enter image description here

and in cylindrical coordinates the modes can be described with Laguerre-Gauss modes, where instead of Hermite polynomials as functions of x,y one encounters a generalized Laguerre polynomial as function of r and another $\exp (i m \varphi)$ term giving the beam "orbital angular momentum". (Image taken from this site) enter image description here

As you can see, due to "degeneracy", the modes with $l>0$ can be represented both by LG modes and both by HG modes (their both valid bases to describe the propagation of the beam). However, $l=0$ is the same at both situations, because Hermite and Laguerre 0's polynomial are both constants.

What I'm trying to say is: you're doing nothing wrong, just working in a different, equivalent base, which for $l=0$ also give exactly the same modes.

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  • $\begingroup$ Well one can indeed rewrite any polynomial as a finite sum of either Hermite or Laguerre or even Chebyshev polynomials, but is the expansion in LG basis "natural" for the QHO? Particularly, are the general expressions in terms of Laguerre polynomials similarly simple to those in terms of Hermite polynomials? $\endgroup$
    – Ruslan
    Commented Aug 27, 2021 at 14:35
  • $\begingroup$ @Ruslan , I think so, because the Laguerre polynomials are not only solving the equation of a QHO in polar coordinates, they also commute with the angular momentum operator! (I didn't check that but from my experience in Optics and from what I understand when I think about it, I think I'm correct), meaning that in a way, they are a "better" base for a 2D QHO than multiplying hermite gauss in each cartesian coordinate. Again I want to express that because I didn't do any calculation or dealt with the material in this context, I may be wrong. It would be nice if someone could verify :) $\endgroup$ Commented Aug 27, 2021 at 14:57
  • $\begingroup$ Actually, the Hermite polynomials appear to only work for 1D and 3D cases. In 2D and 4D+ one needs Laguerre polynomials. I've edited my answer with this. $\endgroup$
    – Ruslan
    Commented Aug 27, 2021 at 15:18
  • $\begingroup$ @Ruslan what do you mean by saying the Hermite polynomials work only for 1D and 3D? Here in 2D I just showed you it works as well, but just not as good :) $\endgroup$ Commented Aug 27, 2021 at 15:53
  • $\begingroup$ I mean that the natural solution of the radial equation is in terms of Laguerre polynomials in 2D and 4D+, and in terms of Hermite ones in 1D and 3D. Of course, the separable solutions along Cartesian axes will be different, since the ODEs will be different. $\endgroup$
    – Ruslan
    Commented Aug 27, 2021 at 16:32

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