Non-dimensionalization of Schrodinger equation

1. Jul 26, 2014

ShayanJ

I had a course of computational physics in university. When the professor wanted to non-dimensiolize the Schrodinger equation, among other things, he changed the wave function using the relation $|\psi(x)|^2 dx=|\phi(y)|^2 dy$ where y is the non-dimensionalized postion ($y=\frac x a$) and so $\phi(y)=\frac{1}{\sqrt{a}} \psi(x)$. This seems reasonable to me because wave function has dimension of $[L]^{-\frac 1 2}$ in one dimension. But when I search the internet for non-dimensionalization of Schrodinger equation, non of them do this step. Why? What's the point?
Thanks

2. Jul 26, 2014

Jazzdude

The *projective* Hilbert space structure of QT already makes the Schroedinger equation perfectly agnostic of any choice of unit or dimension. Units and dimensions are just linear factors of the amplitude, which are removed by stepping from vectors to rays. They come only back in if you label your measurement outcomes, i.e. they are a choice of how you map your eigenvalues to arbitrary scales.

Cheers,

Jazz

3. Jul 26, 2014

ShayanJ

I can understand that. But its just strange to think that you can multiply a wave function by e.g. $m^\frac 1 2$ to get another wave function on the same ray! That seems like treating units as they are complex numbers.

4. Jul 26, 2014

Jazzdude

Yep, it seems strange. But that's exactly how it works and it is the only way to get a consistent construction. Remember that the the dimensionless probabilities that would motivate a unit or dimension for the wavefunction is defined as a fraction with the wavefunction in the numerator and the denominator. Any choice of unit cancels there and the probability is automatically dimensionless.

Cheers,

Jazz