Product Wavefunction Wo W1 (Wo=ground state) Approach to the 3D Hypergeometric Equation
| Main Author: | Francesco R. Ruggeri |
|---|---|
| Format: | info publication-preprint Journal |
| Terbitan: |
, 2019
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| Subjects: | |
| Online Access: |
https://zenodo.org/record/3408415 |
Daftar Isi:
- In a previous note, we argued one may write the bound state wavefunction as W(x)=Wo(x)W1(x), where Wo(x) is the bound ground state wavefunction. This was then applied to the one dimensional case, yielding a coupled equation for W1(x) : -1/2m d/dx d/dx W1 -1/m (d/dx Wo/Wo) (d/dx W1) = (E-Eo) W1(x) ((1)) A transformation may bring ((3)) into the hypergeometric form (1): p(x) d/dx d/dx W + q(x) d/dx W + constant W = 0 ((2)) with p(x) at most quadratic in x and q linear. Then, a Rodrigues polynomial solution exists and: Constant = -(E-Eo) = -n/2 ( d/dx d/dx p (n-1) + 2 d/dx q ) ((3)) In a 3D problem with a potential depending on r alone, one may also obtain a “one dimensional” looking Schrodinger equation with an additional term - l(l+1)/(r*r) (l=the angular momentum eigenvalue). As a result, one may try to apply the approach of ((1)) to this radial equation and force the coefficient of d/dr W1 to be ar+b (or ay+b after a transformation y=f(r)). When carrying out the analysis, however, one may find that the imposed ay+b leads to l=0, i.e. eliminates the l(l+1)/(r*r) term. Then, one has a hypergeometric solution, but this does not correspond, it seems, to the original problem, which has an l present. One may try to argue that the original problem with l not zero in general, will have a set of solutions for which l is 0 and that these should obey a DE with the term l(l+1)/(r*r) removed. That may be the case, but solving an l=0 DE does not lead to the same energy levels of the original problem which holds for all l, including l-0. We investigate this in this note. First, we consider the 3D harmonic oscillator with a potential of .5k r*r and transform it using. W=r(l+1) v1. Next, a second transformation s*s=y is used to bring the equation into hypergeometric form for all l values. This then yields correct energy levels. Next, we consider solving the DE with l=0. This is not the original DE and the solutions do not yield the correct energy levels (although the correct levels are included as a subset). Using W=r(l+1) v1 and the coupled equation ((1)) is then investigated. Finally, we consider a potential of: V(r) = [A exp(-2qr) + B + C exp(2qr) ] / [ exp(2qr) (1-exp(-2qr))2] ((4)) with energy levels given by the Nikiforov-Uvarov method given as ((5)) Enl = C - 2q*q/m { [ n(n+1) + l(l+1) +.5(1+ (mB+2mC)/(q*q) + sqrt( (1+21)2 + 2m/(q*q) (C+B-A) (n+.5) ] / [ (1+2n +sqrt( (1+21)2 + 2m/(q*q)(C+B-A)] } We then consider the problem with a DE with l=0. This DE, however, is the one dimensional problem, not the 3D one. We show it leads to very different energy level results, even though it should have the appearance of the 3D equation for l=0 only results. This shows that one cannot simply set l=0 in a 3D equation and obtain information about the l=0 only energy levels. It is possible that a subset of the solutions of the DE with l=0 apply, but one does not know which subset, hence the energy levels of the DE with l set to 0 may not be very meaningful.