Analytic solution of the SHO for 1D Schrodinger

[weak_phys]

Homework Statement

This is not a homework Q, only in the sense that I am revisiting my degree and wondering about this part of the problem (because it always bothered me): but for reference the leap is made in Griffiths 2.3.2 (2nd edition equation 2.7.4 and 2.7.5) and Liboff 7.20 (4th edition)

Relevant Equations

$$\frac{d^{2}\psi}{du^2} \simeq u^2 \psi$$

At the point where we 'guess' a solution to this 2nd order ODE that cannot be done analytically, I was wondering why Griff and others choose $$e^{-x^2 / 2}$$ rather than just $$e^{-x^2}$$ I've plotted both here and am left wondering what's so different? If we guessed instead the unpopular $$e^{-x^2}$$ surely we still have the same recursion formula and quantum number when we press on?

 

[weak_phys]

Tried to upload an image of my plots but can't keep getting 'uploaded file was not an image as expected' - I've tried .png .jpg with the same error?

 

[MathematicalPhysicist]

Probably for normalization issues. (Can't say I remember why, it's a guess).

 

[weak_phys]

Probably for normalization issues. (Can't say I remember why, it's a guess).

Fair enough, I'm being a bit lazy by asking really - I might try and go on with the non standard choice and see what happens, but as it's an approximation anyway I can't for the life of me see it affecting the 2nd half of the derivation, but I shall see, cheers!

 

[MathematicalPhysicist]

Fair enough, I'm being a bit lazy by asking really - I might try and go on with the non standard choice and see what happens, but as it's an approximation anyway I can't for the life of me see it affecting the 2nd half of the derivation, but I shall see, cheers!

Well, physics is for the masochists no doubt about it...

 

[vela]

The factor of 1/2 cancels the factor of 2 you get from the chain rule. If it wasn't there, you'd get $\psi'' \cong 4 u^2 \psi$.

 

[weak_phys]

The factor of 1/2 cancels the factor of 2 you get from the chain rule. If it wasn't there, you'd get $\psi'' \cong 4 u^2 \psi$.

Thanks for the reply. So yes, I did realize this, but neither this or the 1/2 factor delivers an ODE that =0, they are just approximations (I decided to do both, and the plots show that $\psi$ has great boundary conditions as u approaches +- infinity, but needs some correction around the origin of the wavefunction.
I don't know why I can't attach a photo of the plot (the attach file tab gives me an error) but take both approx solutions and sub them back into the ODE, you get...
$$\psi = (-1-3u^2)e^{\frac{-u^2}{2}}$$
or
$$\psi = (-2-5u^2)e^{-u^2}$$

Both are nice to plot and it's obvious to see why either cannot be a homogenous solution, just an approximation. So, obviously we look for another $f(u)$ to bolt onto $\psi$ to fix this and find our recursion relation.
I haven't had time to push on with this yet, but I will before the week is done!