Must a wavefunction always be dimensionless?

[jeebs]

I was just daydreaming for a few minutes about the energy eigenvalue equation $$H\Psi = E\Psi$$. Say H described a particle in zero potential, so that all its energy was kinetic, ie. $$H = 0.5mv^2 = \frac{p^2}{2m} = \frac{-\hbar^2}{2m}\frac{d^2}{dx^2}$$.

Looking at the units of $$\hbar$$ these are Js, so the units of $$\hbar^2 / 2m$$ are $$J^2s^2kg^-^1 = (kgm^2s^-^2)s^2kg^-^1 = kgm^4s^-^2$$, which is (energy)(length)² dimensions.

So, the $$\frac{d^2}{dx^2}$$ part that operates on the $$\Psi$$ must give a factor with units m^-2 to get units of energy overall, which is what you want the energy eigenvalue E to have, right?

So, am I right in thinking that a wavefunction must always be dimensionless overall? I never really considered this before, but I suppose it would make sense given that if you square it you get a position probability, which requires no units.

If this is true I wish I had realized earlier, might have made checking my solutions easier...

 

[grey_earl]

It doesn't matter what dimension the wave function has if you only consider the Schrödinger equation. Since the unit of d²/dx² = 1/m² always, you always get energy, no matter what object H acts on. In fact, from the probability definition (probabaility to find the particle in the volume V)
$$P(V) = \int_V |\psi(x)|² \mathrm{d}^3 x$$
you can infer the units of the wave function: it must be m^(-3/2) so that by squaring and integrating over three spatial dimensions you get a pure number.