0
$\begingroup$

I'm interested in understanding the approximate solution for large values of $\xi$ (as $\xi \rightarrow \infty$) of the following differential equation $$\dfrac{d^2\psi}{d \xi^2} = \xi^2 \psi$$ which is an approximation to the QHO differential equation.

Now Griffiths states that the approximate solution is $$\psi (\xi) = Ae^{-\xi^2/2} + Be^{\xi^2/2}$$

But if I differentiate this function twice I get $$\dfrac{d^2\psi}{d \xi^2} = Be^{\xi^2/2} - Ae^{-\xi^2/2} + \xi^2 \psi$$

Now assuming $\xi$ is large the second term $- Ae^{-\xi^2/2} \rightarrow 0$ but the first term diverges.

So I don't get why that $\psi$ is the approximate solution. If $\psi$ were

$$\psi (\xi) = Ae^{-\xi^2/2}$$

Then

$$\dfrac{d^2\psi}{d \xi^2} = - Ae^{-\xi^2/2} + \xi^2 \psi$$

And for large values of $\xi$ it is approximately $$\dfrac{d^2\psi}{d \xi^2} = \xi^2 \psi$$ which satisfies that equation.

Am I wrong?

I mean Griffiths says that the first term is not physically acceptable but I claim that it's not even mathematically acceptable. The books says that the $\psi$ with the $B$ term is an approximate solution.

Improve this question
New contributor
Jonathan Cellucci is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$
5
  • $\begingroup$ Isnt it just asserting that, in the absence of boundary conditions, the solution with A and B is approximately the solution of the offending ODE, and of course, if you have boundary at infinity, then only A will be tolerable? Why did you not write that the A-only solution gives you $$\psi^{\prime\prime}=(\xi^2-1)\psi$$ and that obviously, as $|\xi|\to\infty$, the -1 can be dropped and thus the solution is indeed approximately correct? $\endgroup$
    – naturallyInconsistent
    Commented Dec 8 at 14:38
  • 1
    $\begingroup$ Hi, welcome to Physics SE. Hint: if you let $A,\,B$ depend on $\xi$ but assume they don't vary too fast when $\xi$ is large, you can use iterative approximations to work out higher-order corrections. The exact result is expressible in terms of the parabolic cylinder functions $D_{-1/2}(\sqrt{2}x),\,D_{-1/2}(i\sqrt{2}x)$, expanded here. See if you can match that. $\endgroup$
    – J.G.
    Commented Dec 8 at 14:39
  • $\begingroup$ Thanks I've never heard of these cylinder functions.... Thanks!!! $\endgroup$
    – Jonathan Cellucci
    Commented Dec 8 at 14:47
  • $\begingroup$ Yes I mean I said that (the first comment) in a slightly different way... But I meant just what you said... I'm just asking why the first solution with the $B$ term is said to be an approximate solution if it diverges. I mean, it is clearly not a solutions. $\endgroup$
    – Jonathan Cellucci
    Commented Dec 8 at 14:49
  • $\begingroup$ If you let $B$ vary, it gives rise to a solution that diverges as $\xi\to\infty$, which is physically problematic, whereas $A$ doesn't have the same problem. $\endgroup$
    – J.G.
    Commented Dec 8 at 15:01

1 Answer 1

Reset to default
2
$\begingroup$

Actually, taking \begin{align} \frac{d^2}{d\xi^2}\left(A e^{-\xi^2/2} + B e^{\xi^2/2}\right)&= A e^{-\xi^2/2}(\xi^2-1) + B e^{\xi^2/2}(\xi^2+1) \\ & = \xi^2 \left(A e^{-\xi^2/2} + B e^{\xi^2/2}\right) - A e^{-\xi^2/2} + B e^{\xi^2/2} \end{align} so clearly the term with $\xi^2 \left(A e^{-\xi^2/2} + B e^{\xi^2/2}\right) $ dominates the other terms because of the extra $\xi^2$ prefactor.

Moreover, since the wavefunction must be $0$ at $\pm \infty$, you should remove the $e^{+\xi^2/2}$ by hand.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Alright!!!! Thanks!!! Yes of course there is the extra factor of $\xi^2$!!! Thank you sooooo much!!! $\endgroup$
    – Jonathan Cellucci
    Commented Dec 8 at 15:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.