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I'm trying to show the permitted energies of the 3D simple harmonic oscillator (which is spherically symmetrical) are: $E_n = \hbar \omega(N + \dfrac{3}{2})$

In particular, $V(x) = \dfrac{1}{2} m \omega x^2 + \dfrac{\hbar^2l(l+1)}{2mr^2}$

The steps I'm using are:

  1. Rearrange schrodinger's time independent equation and solve for an approximation to $u$ for large $r$. This gave me $u(\rho) \approx Ae^{\rho^2/2}$. Where $\rho = \sqrt{\dfrac{m\omega}{\hbar}} r$

  2. Assume $u(\rho) = P(\rho) e^{\rho^2/2}$, I plug this into the original differential equation to get:

$$\dfrac{-\hbar\omega}{2}(P'' - 2\rho P' + \rho^2P)e^{\rho^2/2} = (\rho^2 +\dfrac{m\omega}{\hbar\rho^2}l(l+1))Pe^{\rho^2/2}$$

  1. I understand that the next step is to assume a taylor expansion for $P(\rho)$ and thence derive a recurrence relationship which will ultimately yield the energies

My problem is that whenever I plug in taylor expansions for $P$ and compare coefficients I end up with a 3 term recurrence relationship. I have no idea where I could have gone wrong, any help would be greatly appreciated.

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  1. Rearrange Schrödinger's time independent equation and solve for an approximation to $u$ for large $r$. This gave me $u(\rho) \approx Ae^{\rho^2/2}$.

Your step 1 is basically OK. But you need to choose the decreasing solution $u(\rho) \approx Ae^{-\rho^2/2}$ instead of the increasing one.

In addition to the above step you also need to solve for an approximation to $u$ for small $r$. This will give you something like $u(\rho) \approx A \rho^{l+1}$.

  1. Then, using the two approximations of the steps above, you can then assume $u(\rho) = P(\rho) \rho^{l+1} e^{-\rho^2/2}$, plug this into the original differential equation, and get a differential equation for $P(\rho)$.
  1. I understand that the next step is to assume a Taylor expansion for $P(\rho)$ and thence derive a recurrence relationship which will ultimately yield the energies.

Your step 3 is fine.

Now, when you plug in the Taylor expansion for $P(\rho)$ and compare coefficients, you should end up with a 2 term recurrence relationship (instead of your 3 term recurrence relationship).

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  • $\begingroup$ So if we analyse the behaviour as $\rho \rightarrow 0$ we get $\dfrac{d^2u}{d\rho^2} \approx \dfrac{l(l+1)}{\rho^2}u $ but I'm not quite sure how that gives the desired $u(\rho) \approx A\rho^l$ $\endgroup$ – Rzmwood May 14 at 15:24
  • $\begingroup$ You are right. It gives the solution $u(\rho) \approx A\rho^{l+1}$. I'll update my answer. $\endgroup$ – Thomas Fritsch May 14 at 15:42
  • $\begingroup$ This might be a stupid question, but is there a way derive the approximate solution $u(\rho) \approx A\rho^{l+1}$ other than by inspection. I see how to justify it as an approximation but not how to reach the approximation itself. $\endgroup$ – Rzmwood May 14 at 17:55
  • $\begingroup$ @Rzmwood Good question. You have the differential equation $\dfrac{d^2u}{d\rho^2} \approx \dfrac{l(l+1)}{\rho^2}u$. With a certain amount of intuition you try the approach $u(\rho)=A\rho^{s}$ with a still unknown $s$. When you plug this into the differential equation, you get $s(s-1)=l(l+1)$. This has two solutions: $s=l+1$ and $s=-l$. So you have two candidate solutions: $u(\rho)\approx\rho^{l+1}$ and $u(\rho)\approx\rho^{-l}$. The first one behaves well at $\rho=0$ and is therefore acceptable. The second one is $\infty$ at $\rho=0$ and is therefore to be ruled out. $\endgroup$ – Thomas Fritsch May 14 at 18:11

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