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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?

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  • $\begingroup$ 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 $r^2$ $\endgroup$ Jan 17, 2016 at 16:34
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    $\begingroup$ @ChrisCundy in your first example don't you mean $|\psi|^2 \mathrm{d}r$? $\endgroup$
    – Sparkler
    Jan 17, 2016 at 16:40
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    $\begingroup$ "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. $\endgroup$
    – Gert
    Jan 17, 2016 at 16:40
  • $\begingroup$ @Gert, question updated. $\endgroup$
    – Sparkler
    Jan 17, 2016 at 16:42

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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$.

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  • $\begingroup$ are you implying that the interpretation of the wavefunction in 3d is different than in 1d? in other words, how come the 1s wavefunction has the highest amplitude next to r=0? $\endgroup$
    – Sparkler
    Jan 17, 2016 at 16:50
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    $\begingroup$ @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. $\endgroup$
    – ACuriousMind
    Jan 17, 2016 at 16:57
  • $\begingroup$ @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. $\endgroup$ Jan 17, 2016 at 17:14

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