# What is the unit (dimension) of the 3-dimensional position space wavefunction $\Psi$ of an electron?

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?

• What happens if you square the wave function and integrate over some volume? Commented Aug 3, 2017 at 16:08

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 $$L^{-3/2}$$.
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 $(\text{Length}) ^{-3/2}$
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 $[\psi]=\text{length}^{-3/2}$.