Why wavefunction is sometimes multiplied by the radius to get probability density?

When solving 1d particle in a box, the probability density is said to be proportional to $|\psi|$, but when solving 3d orbitals, the probability density is said to be proportional to $|\psi|^2 r^2$. Why this difference?

• The probability of finding the particle in an infinitesimal box is proportional to $|\psi|^2$ but the probability of finding the particle in an shell between $r$ and $r + dr$ is proportional to $|\psi|^2 r^2$ because the volume of the shell is proportional to $r^2$ Jan 17, 2016 at 16:34
• @ChrisCundy in your first example don't you mean $|\psi|^2 \mathrm{d}r$? Jan 17, 2016 at 16:40
• "probability is said to be proportional to $|\psi|^2$". Not really true: $|\psi|^2$ is the probability density: it needs to be integrated over part of the domain to get actual probability.
– Gert
Jan 17, 2016 at 16:40
• @Gert, question updated. Jan 17, 2016 at 16:42

It's not "multiplied by $r^2$ to get the probility density". The issue is that the volume element in spherical coordinates is $$\mathrm{d}V = r^2\sin(\theta)\mathrm{d}r\mathrm{d}\theta\mathrm{d}\phi$$ and since the probability to find a particle in a subspace $X\subset \mathbb{R}^3$ is $$P(X) = \int_X\lvert \psi(r)\rvert^2\mathrm{d}V$$ by definition of a probability density, the quantity $r^2\lvert \psi(r)\rvert^2$ is what behaves like the "normal" probability density in flat coordinates: The probability to find the particle between $r_1$ and $r_2$ is proportional to $\int_{r_1}^{r_2} r^2\lvert \psi \rvert^2\mathrm{d}r$.
• @Sparkler: No. The interpretation is the same. The wavefunction is a probability density, and the probability to be in a subspace $X$ is $\int_X\lvert \psi \rvert^2\mathrm{d}V$. For different coordinate systems, the expression for $\mathrm{d}V$ is different (but also, correspondingly, the expression for $\psi$ is different). One often draws $r^2\lvert\psi\rvert^2$ because our intiuition is used to probability densities in Cartesian coordinates where $\mathrm{d}V$ is just the product of $\mathrm{d}x_i$s without any prefactors. Jan 17, 2016 at 16:57
• @Sparkler The interpretation is the same, but at the same probability per unit volume you're more likely to find a particle between $2r$ and $2r+\Delta r$ than between $r$ and $r+\delta r$, because the volume of the spherical shell is bigger. That's all that's happening here. Jan 17, 2016 at 17:14