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I googled for the above question, and I got the answer to be $$[\Psi]~=~L^{-\frac{3}{2}}.$$

Can anyone give an easy explanation for this?

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    $\begingroup$ What happens if you square the wave function and integrate over some volume? $\endgroup$
    – DanielSank
    Aug 3, 2017 at 16:08

4 Answers 4

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The physical interpretation of the wavefunction is that $|\psi(\vec r)|^2dV$ gives the probability of finding the electron in a region of volume $dV$ around the position $\vec r$. Probability is a dimensionless quantity. Hence $|\psi(\vec r)|^2$ must have dimension of inverse volume and $\psi$ has dimension $L^{-3/2}$.

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Think about it, its square integrated over a volume (that is multiplied for infinitesimal volumes and summed over all those volumes) is a pure number ("probability of finding the particle in that volume") therefore $(\text{wave function})^2 \cdot (\text{Length})^3 = (\text{adimensional quantity})$

Thus the square of the wavefunction has the dimension of a $(\text{Length})^{-3}$. So the wavefunction is, dimensionally, a $(\text{Length}) ^{-3/2}$

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The three-dimensional integral of the norm square of the wave function is a probability, so it should be dimensionless. Therefore $\text{length}^3[\psi]^2=1$, so $[\psi]=\text{length}^{-3/2}$.

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The wave function of an electron does not signify anything in itself. The only useful thing we can get from it is the probability density (probability per unit volume), which is the square of its amplitude. In terms of SI units, probability has no unit, and volume has (meter)^3. So, unit of the wave function (√probability/√volume) will be (meter^-3/2).

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