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I have been attempting to solve the following central potential using schrodinger equation.$$V(r)=a_{1}r^{2}+\frac{a_{2}}{r^{2}}+a_{3}$$

The radial equation becomes, $$\Bigg[\frac{-\hbar^{2}}{2m}\dfrac{d^{2}}{dr^{2}}+V(r)+\frac{l(l+1)}{r^{2}}\Bigg]u(r)=Eu(r)$$ I have used laplace transform to solve this problem to get the eigen values. Defining $$u(y)=y^{-\frac{\nu}{2}}\phi(y)$$ I get the following differential equation,$$\Bigg[y\dfrac{d^{2}}{dy^{2}}-\bigg(\nu-\dfrac{1}{2}\bigg)\dfrac{d}{dy}-\dfrac{1}{4}\bigg(\mu^{2}y-\epsilon^{2}\bigg)\Bigg]\phi(y)=0$$ where, $$\nu(\nu+1)=\dfrac{2ma_{2}}{\hbar^{2}}+l(l+1)$$ $$\mu^{2}=\dfrac{2ma_{1}}{\hbar^{2}}$$ $$\epsilon^{2}=\dfrac{2mE}{\hbar^{2}}$$ Now after doing laplace transform I have the following result $$\bigg(t^{2}-\dfrac{\mu^{2}}{4}\bigg)\dfrac{df(t)}{dt}+\bigg[\bigg(\nu+\dfrac{3}{2}\bigg)t-\dfrac{\epsilon^{2}}{4}\bigg]f(t)$$ The soluion in laplace space is, $$f(t)=N\bigg(t+\dfrac{\mu}{2}\bigg)^{-\dfrac{\epsilon^{2}}{4\mu} -\dfrac{1}{2}\bigg(\nu+\dfrac{3}{2}\bigg)}\bigg(t-\frac{\mu}{2}\bigg)^{\dfrac{\epsilon^{2}}{4\mu}+\dfrac{1}{2}\bigg(\nu+\dfrac{3}{2}\Bigg)}$$ To get a finite wave functions the solutions have to be a finite polynomial $$\dfrac{\epsilon^{2}}{4\mu}-\dfrac{1}{2}\bigg(\nu + \dfrac{3}{2}\bigg)=n$$ where, $$n=0,1,2,3...$$ The energy eighen values are ,$$E_{nl}=\sqrt{\dfrac{8\hbar^{2}a_{1}}{m}}\bigg(\dfrac{1}{2}+\dfrac{1}{4}\sqrt{1+4l(l+1)+\dfrac{8ma_{2}}{\hbar^{2}}}\bigg)$$

I have understood everything except the condition to be imposed on the angular quantum number $l$ for a given value of radial quantum number $n$. For instance, suppose $n$=3 the minimum value of $l$ = 0, but the what is the maximum value of $l$ ? What is the condition on $l$ for a given $n$ ? I have followed the following research article

*Exact solutions of the Schrödinger equation via Laplace transform approach: pseudoharmonic potential and Mie-type potentialshttps://link.springer.com/journal/10910 *

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  • $\begingroup$ Another $1/r^2$ question by OP: physics.stackexchange.com/q/370901/2451 $\endgroup$
    – Qmechanic
    Jan 11 '18 at 12:36
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    $\begingroup$ Link to preprint: arxiv.org/abs/1202.4268 $\endgroup$
    – Qmechanic
    Jan 11 '18 at 12:41
  • $\begingroup$ The question is similar yet there is a difference, the potential that you mentioned is about attractive potential and what I am looking for is positive potential. $\endgroup$
    – user135580
    Jan 11 '18 at 16:42
  • $\begingroup$ wiki says: "The angular momentum quantum number, ℓ, governs the number of planar nodes going through the nucleus. In an s orbital, no nodes go through the nucleus, therefore the corresponding azimuthal quantum number ℓ takes the value of 0. In a p orbital, one node traverses the nucleus and therefore ℓ has the value of 1." en.wikipedia.org/wiki/Azimuthal_quantum_number $\endgroup$ Jan 16 '18 at 12:27
  • $\begingroup$ one may also look at this presentation there is one sentence here says: "differential equation plus a boundary condition gives a quantisation" nat.vu.nl/~wimu/EDUC/MNW-lect-2.pdf $\endgroup$ Jan 16 '18 at 12:34

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