simple harmonic oscillator - scaled variables

[randybryan]

The TISE can be written as

-\frac{\hbar^{2}}{2m}\frac{d^{2}u}{dx^{2}} + \frac{1}{2}m\omega_{0}^{2}x^{2}u = Eu

Now my lecture notes say that it is convenient to define scaled variables

y = \sqrt{\frac{m\omega_{0}}{\hbar} x}

and \alpha = \frac{2E}{\hbar\omega_{0}}

Hence

\frac{d}{dx} = \sqrt{\frac{\hbar}{m\omega_{0}} x} \frac{d}{dy}

so \frac{2}{\hbar\omega_{0}} times the TISE can be written as

\frac{d^{2}u}{dy^{2}} + (\alpha - y^{2})u= 0

now, it is not at all obvious to me where these scaled variables came from? I know this leads on to deriving hermite polynomials, but I'm just wondering if there is some triviality behind the scaled variables. It really helps me to know the order in which things were derived/formulated. It might have been the case that these scaled variables were used after much painstaking trial and error, or some mathematician just noticed that it would be easier to write the equation this way.

Does anyone have any background information on this? Is it usual to scale the variables in this way?

Thanks

[Avodyne]

The scaled variables are dimensionless.

[RedX]

The TISE can be written as

-\frac{\hbar^{2}}{2m}\frac{d^{2}u}{dx^{2}} + \frac{1}{2}m\omega_{0}^{2}x^{2}u = Eu

Now my lecture notes say that it is convenient to define scaled variables

y = \sqrt{\frac{m\omega_{0}}{\hbar} x}

and \alpha = \frac{2E}{\hbar\omega_{0}}


Well the way it works is that you substitute x=\beta y , where y is dimensionless, and then choose \beta to make as many coefficients equal to 1.

Maybe it's possible to see

y = \sqrt{\frac{m\omega_{0}}{\hbar} x}

directly, but you usually you set x=\beta y and then do the math (determine beta).

[randybryan]

Thanks very much guys, but I'm still slightly confused.

what difference does it make if the variables are dimensionless?

and what do you mean by choosing beta to make as many coefficients equal to 1?

I apologise if I appear stupid, I'm just getting so confused by quantum mechanics in general.

[RedX]

and what do you mean by choosing beta to make as many coefficients equal to 1?

I apologise if I appear stupid, I'm just getting so confused by quantum mechanics in general.

This is not really quantum mechanics. Quantum mechanics tells you what the equation is, but not what to do after that.

\begin{equation*}\begin{split}

a\frac{d^{2}u}{dx^{2}}+bx^2u=Eu \\
\frac{a}{\beta^2}\frac{d^{2}u}{dy^{2}}+b(\beta y)^2u=Eu \\

\frac{d^{2}u}{dy^{2}}+\frac{b}{a}\beta^4 y^2u=\frac{\beta^2}{a}Eu

\end{split}\end{equation*}

so you set beta such that \frac{b}{a}\beta^4=1

Then you combine everything on the RHS into one dimensionless constant alpha.