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Must a wavefunction always be dimensionless?

  • Thread starter jeebs
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  • #1
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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)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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Answers and Replies

  • #2
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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)
[tex]P(V) = \int_V |\psi(x)|² \mathrm{d}^3 x[/tex]
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.
 

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