# Proof for equation of radial probability distribution curve

So my textbook says the equation for this is

$$P(r) = 4\pi r^2\Psi^2$$

It also gives the volume of the shell formula

$$\mathrm{d}V = \frac 43 \pi(r+\mathrm{d}r)^3 - \frac 43 \pi r^3$$

which I understood

But what is the derivation method for the radial probability distribution curve equation mentioned above?

• @Poutnik thank you, can you please elaborate on the proof of the latter?
– G.S.
Jun 29, 2022 at 3:59
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Let take for the question context as the axiom of the quantum atomic theory this:

• $$|\Psi|^2$$ is the probability density an electron occurs at the given point. $$\Psi$$ itself is a wave function as a particular solution of Schroedinger wave equation.

Then the probability differential for the given volume differential is:

$$\mathrm{d}p=|\Psi|^2\mathrm{d}V=|\Psi|^2\mathrm{d}x\mathrm{d}y\mathrm{d}z$$

The wave equation and functions could be formally formulated for Cartesian coordinates x,y,z. But as a general rule, equation solving and the solution results are more simple and elegant, if the symmetry of the problem and the symmetry of the used coordinate system match each other. For that reason, both the equation and functions are formulated for spherical coordinates $$r$$, $$\varphi$$, $$\theta$$, as the electrostatic force with the central charge is spherically symmetric.

$$\mathrm{d}p=|\Psi|^2\mathrm{d}V=|\Psi(x,y,z)|^2\mathrm{d}x\mathrm{d}y\mathrm{d}z=|\Psi(r,\varphi,\theta)|^2r^2\sin{\theta}\mathrm{d}r\mathrm{d}\varphi\mathrm{d}\theta$$

For a spherically symmetric case of s-orbitals, $$\Psi$$ is a function just of $$r$$. The infinitesimal volume of a spherical shell is

$$\mathrm{d}V=4\pi r^2 \mathrm{d}r$$

(where $$4\pi r^2$$ is the surface of a sphere with the radius $$\mathrm{r}$$), therefore:

$$\mathrm{d}p=|\Psi|^2\mathrm{d}V=|\Psi(r)|^2 4\pi r^2 \mathrm{d}r$$

Then, $$P(r)=|\Psi(r)|^2 4\pi r^2$$ is the radial probability distribution function in the equation $$\mathrm{d}p=P(r)\mathrm{d}r$$ for spherically symmetric wave functions of s orbitals.

For general wave function $$\Psi(r,\varphi,\theta)$$: $$P(r)=\int_0^{2\pi}{\left(\int_0^{\pi}{|\Psi(r,\varphi,\theta)|^2r^2\sin{\theta}\mathrm{d}\theta }\right)\mathrm{d}\varphi},$$,

as the radial probability distribution function $$P(r)$$ is $$|\Psi(r,\varphi,\theta)|^2$$ integrated over the spherical surface of the radius $$r$$.

As a check for the symmetrical case above, $$\Psi(r,\varphi,\theta) = \Psi(r)$$

$$P(r)=\int_0^{2\pi}{\left(\int_0^{\pi}{|\Psi(r,\varphi,\theta)|^2r^2\sin{\theta}\mathrm{d}\theta }\right)\mathrm{d}\varphi} = |\Psi(r)|^2r^2 \int_0^{2\pi}{\left(\int_0^{\pi}{\sin{\theta}\mathrm{d}\theta }\right)\mathrm{d}\varphi}= \\|\Psi(r)|^2r^2 \int_0^{2\pi}{\left(2\right)\mathrm{d}\varphi}=|\Psi(r)|^2 4 \pi r^2$$

Just to expand a bit on the detailed answer already given: the problem can be solved in arbitrary coordinate system, and it all comes down to finding the volume element in different coordinate systems, which has a relatively simple general algorithm.

So instead of just stating on how to write a volume integral in spherical coordinates, you can start by writing it in an arbitrary coordinate system as

$$I = \int_{\pmb{V}} f(u_1, u_2, u_3) \left| \frac{\partial (x,y,z)}{\partial(u_1, u_2, u_3)} \right| du_1 du_2 du_3 \, ,$$

with $$\left| \frac{\partial (x,y,z)}{\partial(u_1, u_2, u_3)} \right|$$ is the determinant of the Jacobian matrix, measuring the change of a given volume element when going from Cartesian to the arbitrary $$(u_1,u_2,u_3)$$ system (we don't even need to be restricted to 3D by the way).

Then you write the definition of the spherical coordinates

$$x=\rho\cos\theta\sin\phi\\ y=\rho\sin\theta\sin\phi\\ z=\rho\cos\phi$$

Then you take all the partial derivatives to build the $$3x3$$ matrix, take the determinant, and you end up with the correct formula.

• Good idea and applicable generally for any coordinate transform. But, especially if we keep in mind quite simple way how spherical coordinates are defined, it may be easier to write it directly in them. // Also, integration is easier there. Jul 1, 2022 at 13:21