PX262 - F4 - separation in spherical polar coordinates

the schrödinger equation

considering a spherically symmetric potential: $V (x, y, z) = V (r)$ , where $r$ is the distance from the centre of potential
the schrödinger equation in spherical polar coordinates:

\begin{matrix} - \frac{ℏ^{2}}{2 m} [\frac{1}{r^{2}} \frac{\partial}{\partial r} (r^{2} \frac{\partial ϕ}{\partial r}) + \frac{1}{r^{2} \sin θ} \frac{\partial}{\partial θ} (\sin θ \frac{\partial ϕ}{\partial θ}) \\ (1) & + \frac{1}{r^{2} \sin^{2} θ} \frac{\partial^{2} ϕ}{\partial φ^{2}}] + V ϕ = E ϕ \end{matrix}

where, $ϕ = ϕ (r, θ, ϕ)$

as potential depends on a single variable, using separation of variables in the wavefunction: $$\phi(r,\theta,\varphi) = R(r) , Y(\theta,\varphi)$$
equation $(1)$ becomes:

\begin{matrix} \begin{matrix} - \frac{ℏ^{2}}{2 m} [\frac{Y}{r^{2}} \frac{\partial}{\partial r} (r^{2} \frac{\partial R}{\partial r}) + \frac{R}{r^{2} \sin θ} \frac{\partial}{\partial θ} (\sin θ \frac{\partial Y}{\partial θ}) \\ + \frac{R}{r^{2} \sin^{2} θ} \frac{\partial^{2} Y}{\partial φ^{2}}] + V R Y = E R Y \end{matrix} \\ (2) & \begin{matrix} [- \frac{ℏ^{2}}{2 m} \frac{1}{R} \frac{d}{d r} (r^{2} \frac{d R}{d r}) + r^{2} V - r^{2} E] \\ - \frac{ℏ^{2}}{2 m} [\frac{1}{Y} \frac{1}{\sin θ} \frac{\partial}{\partial θ} (\sin θ \frac{\partial Y}{\partial θ}) + \frac{1}{Y} \frac{1}{\sin^{2} θ} \frac{\partial^{2} Y}{\partial φ^{2}}] = 0 \end{matrix} \end{matrix}

each part in a square bracket must be equal to a constant and the two constants must add up to zero

the angular equation

the second part does not contain $V$ , so solving it for $Y (θ, φ)$ will give the angular parts of the wavefunction:

- \frac{ℏ^{2}}{2 m} [\frac{1}{\sin θ} \frac{\partial}{\partial θ} (\sin θ \frac{\partial Y}{\partial θ}) + \frac{1}{\sin^{2} θ} \frac{\partial^{2} Y}{\partial φ^{2}}] = λ Y

from here, this is the same as the equation for the eigenvalues of the operator ${\hat{L}}^{2} :$

\frac{1}{2 m} {\hat{L}}^{2} Y = λ Y

the solutions are spherical harmonics: $Y_{l m} (θ, ϕ)$ , and the eigenvalues are: $λ = \frac{ℏ^{2}}{2 m} l (l + 1)$
therefore, for every system with a spherically symmetric potential, the angular part will be the same, and it will correspond to the angular momentum
any solution will be an eigenstate of ${\hat{L}}^{2}$ and ${\hat{L}}_{z}$ operators

the radial equation

substituting into equation $(2) :$

\begin{aligned} - \frac{ℏ^{2}}{2 m} \frac{1}{R} \frac{d}{d r} (r^{2} \frac{d R}{d r}) + r^{2} V + \frac{l (l + 1) ℏ^{2}}{2 m} & = r^{2} E \\ (3) & - \frac{ℏ^{2}}{2 m} \frac{1}{r^{2}} \frac{d}{d r} (r^{2} \frac{d R}{d r}) + [V (r) + \frac{l (l + 1) ℏ^{2}}{2 m r^{2}}] R & = E R \end{aligned}

making the substitution $χ (r) = r R (r) :$

\begin{matrix} \frac{d R}{d r} = \frac{d}{d r} (\frac{χ}{r}) = \frac{1}{r} \frac{d χ}{d r} - \frac{χ}{r^{2}} \\ ∴ \frac{d}{d r} (r^{2} \frac{d R}{d r}) = \frac{d}{d r} (r \frac{d χ}{d r} - χ) = r \frac{d^{2} χ}{d r^{2}} ​ \end{matrix}

substituting into equation $(3) :$

- \frac{ℏ^{2}}{2 m} \frac{d^{2} χ}{d r^{2}} + [V (r) + \frac{ℏ^{2}}{2 m r^{2}} l (l + 1)] χ = E χ

apart from the term within the square brackets, it is identical in form to the 1D schrödinger equation
an additional boundary condition applies to avoid divergence: $χ (r = 0) = 0$ , otherwise, $R = r^{- 1} χ$ will be infinite
physical interpretation: $| χ (r) |^{2} d r$ is the probability of finding the electron at a distance between $r$ and $r + d r$ from the origin averaged in all directions