manipulating the TISE for a SHM

Exulus

Hi guys, hoping someone can help with this manipulation. I need to transform this:
[tex] \frac{-\hbar^2}{2m}\frac{d^2}{dx^2}u(x) + \frac{1}{2}m\omega^2 x^2 u(x) = Eu(x)[/tex]
Into its dimensionless form:
[tex]\frac{d^2}{dy^2}u(y) + (2\epsilon - y^2)u(y) = 0[/tex]
I have the following info:
[tex]E = \epsilon\hbar\omega[/tex]
[tex]x = y\sqrt{\frac{\hbar}{m\omega}}[/tex]
Heres what ive done so far:
[tex] \frac{-\hbar^2}{2m}\frac{d^2}{dx^2}u(x) + \frac{1}{2}m\omega^2 y^2 \frac{\hbar}{m\omega} u(x) = \epsilon\hbar\omega u(x)[/tex]
[tex] \frac{-\hbar}{m}\frac{d^2}{dx^2}u(x) + \omega^2 y^2 u(x) = 2\epsilon\omega u(x)[/tex]
[tex] \frac{\hbar}{m}\frac{d^2}{dx^2}u(x) + 2\epsilon\omega u(x) - \omega^2 y^2 u(x) = 0[/tex]
[tex] \frac{\hbar}{m}\frac{d^2}{dx^2}u(x) + (2\epsilon - y^2)\omega u(x) = 0[/tex]
But i cant see where to go next..i know i must be close to the end though..surely!

marlon

at first sight, there is one thing that remains to be done : the transformation of the derivatives.

you know that u(x) = u(y)
Hint : [tex]\frac {du(x)}{dx}= \frac{du(y)}{dx} = \frac {du(y)}{dy} \cdot \frac{dy}{dx}[/tex] . More specifically, can you calculate [tex]\frac {dy}{dx} = constant[/tex] ? What is that constant ? Then, do the same to get the second derivatives.

marlon

ps i did not check the calculations you have done so far but let us first look at the derivatives

Exulus

hmm, the only relationships i have in my notes are:

[tex]u(x) = Ce^{-\gamma x}[/tex]

So [tex] \frac{d^2}{dx^2}u(x) = \gamma^2 u(x)[/tex]

Where [tex]\gamma^2 = \frac{2m}{\hbar^2}(V_0 - E)[/tex]

edit - just saw your hint..will inspect further!


edit again - would i be correct to assume that [tex]u(y) = Ce^{-\gamma y \sqrt{\frac{\hbar}{m\omega}}}[/tex] ?

marlon

This is all true but you do not need this right now. The problem is quite simple, trust me. Once you get the second derivatives right, the problem is very elementary. just try it, i will keep an eye on it.

marlon

marlon

meanwhile, your calculations are ok. Once you get my hint (ie the constant), you are done !!!

marlon

Exulus

hmm..no im totally lost...sorry

I dont see how you can differentiate the function when you say i dont need to know what it is. I can make a wild guess that dy/dx = 2m/hbar^2..but that would just be guessing and not actually understanding.

Could you give me a further hint? To be honest functions have always confused me slightly.

marlon

your guess is wrong.

Well it is very easy. if you know that [tex] x = y \sqrt {\frac{\hbar}{m\omega}}[/tex] then [tex] dx = dy \sqrt{\frac{\hbar}{m \omega}}[/tex] and you know the constant.

But, we need to have second derivatives, this is also easy :
[tex]\frac {d^2u(x)}{dx^2}= \frac{d^2u(y)}{dx^2} = \frac {d^2u(y)}{dy^2} \cdot ( \frac{dy}{dx} )^2 [/tex]

So you need to replace the second derivative of u to x by [tex]\frac {d^2u(y)}{dy^2} \frac {m \omega}{\hbar}[/tex]

Just substitute this into your last equation and all is done
marlon

Exulus

I cannot believe i didnt see that. Whoops! Simple things like that...pretty easy to overlook i guess. Thanks for the help..gonna try it out now! :)

marlon

no, problem, this happens to all of us

marlon