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

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

So, the [tex]\frac{d^2}{dx^2}[/tex] part that operates on the [tex]\Psi[/tex] must give a factor with units m

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 realised earlier, might have made checking my solutions easier...

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

^{2}dimensions.So, the [tex]\frac{d^2}{dx^2}[/tex] part that operates on the [tex]\Psi[/tex] 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 realised earlier, might have made checking my solutions easier...

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