📂Quantum Mechanics

Schrödinger Equation in Spherical Coordinates

Equation

In the spherical coordinate system, the Schrödinger equation is as follows.

$$-\frac{\hbar^{2}}{2M}\left[\frac{1}{r^2}\frac{\partial}{\partial r} \left( r^2\frac{\partial \psi}{\partial r} \right) + \frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\left( \sin\theta \frac{\partial \psi}{\partial \theta} \right) + \frac{1}{r^2\sin^2\theta}\frac{\partial^2 \psi}{\partial^2 \phi} \right]+V\psi=E\psi \tag{1}$$

Explanation

The time-independent Schrödinger equation in three dimensions is as follows.

$$-\frac{\hbar^{2}}{2M}\nabla^{2}\psi+V\psi=E\psi$$

Here, $M$ is the mass of the particle. If the potential depends only on the distance from the origin as in $V = V(r)$, it is preferable to solve the problem in the spherical coordinate system.

Laplacian in the spherical coordinate system

$$\nabla ^{2}f = \frac{1}{r^{2}} \frac{ \partial }{ \partial r }\left(r^{2} \frac{ \partial f}{ \partial r} \right) + \frac{1}{r^{2}\sin\theta}\frac{ \partial }{ \partial \theta }\left( \sin \theta \frac{ \partial f}{ \partial \theta} \right)+\frac{1}{r^{2}\sin^{2}\theta} \frac{\partial ^{2} f}{\partial \phi^{2} }$$

Since the Laplacian in the spherical coordinate system is as shown above, the Schrödinger equation is as follows.

$$-\frac{\hbar^{2}}{2M}\left[\frac{1}{r^2}\frac{\partial}{\partial r} \left( r^2\frac{\partial \psi}{\partial r} \right) + \frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\left( \sin\theta \frac{\partial \psi}{\partial \theta} \right) + \frac{1}{r^2\sin^2\theta}\frac{\partial^2 \psi}{\partial^2 \phi} \right] + V\psi = E\psi \tag{1}$$

Assume that the wave function $\psi$ is separable as follows.

$$\psi (r,\theta,\phi) = R(r) \Theta(\theta) \Phi( \phi)$$

Then the equation $(1)$ is as follows.

$$-\frac{\hbar^{2}}{2M}\left[\frac{\Theta \Phi}{r^2}\frac{d}{d r} \left( r^2\frac{d R}{d r} \right) + \frac{R \Phi}{r^2\sin\theta}\frac{d}{d\theta}\left( \sin\theta \frac{d \Theta}{d \theta} \right) + \frac{R \Theta}{r^2\sin^2\theta}\frac{d^2 \Phi}{d^2 \phi} \right]+VR\Theta \Phi=ER\Theta \Phi$$

By transposing the right-hand side terms to the left-hand side and multiplying both sides by $-\dfrac{2Mr^{2}}{\hbar ^{2}}\dfrac{1}{R\Theta \Phi}$, we get the following.

$$\left[ \frac{1}{R}\frac{d}{d r} \left( r^2\frac{d R}{d r} \right) -\frac{2Mr^{2}}{\hbar ^{2}}(V(r)-E) \right]+\left[ \frac{1}{\Theta \sin\theta}\frac{d}{d\theta}\left( \sin\theta \frac{d \Theta}{d \theta} \right) + \frac{1}{\Phi\sin^2\theta}\frac{d^2 \Phi}{d^2 \phi} \right]=0$$

Although additional potential terms and constants have been added, fundamentally it is the same as solving Laplace’s equation in spherical coordinates. The first term enclosed by curly braces depends only on the variable $r$, and the second term depends only on variables $\theta$ and $\phi$, so each part within the curly braces is a constant. Let the first term be $\ell(\ell + 1)$. Then the second term is $-\ell(\ell + 1)$.

$$\begin{align*} \frac{1}{R}\frac{d}{d r} \left( r^2\frac{d R}{d r} \right) -\frac{2Mr^{2}}{\hbar ^{2}}(V(r)-E) &= \ell(\ell+1) \tag{2} \\ \frac{1}{\Theta \sin\theta}\frac{d}{d\theta}\left( \sin\theta \frac{d \Theta}{d \theta} \right) + \frac{1}{\Phi\sin^2\theta}\frac{d^2 \Phi}{d^2 \phi} &= -\ell(\ell+1) \tag{3} \end{align*}$$

The solution to the angular equation $(3)$ is specifically called the spherical harmonics and is as follows. $$Y_{\ell}^{m}(\theta, \phi) = e^{\i m\phi}P_{\ell}^{m}(\cos \theta)$$

Although $(3)$ doesn’t directly show $m$, during the solving process of variable separation in $\theta$ and $\phi$, they appear as separation constants. $P_{\ell}^{m}(\cos\theta)$ is the associated Legendre polynomial. When normalized, it is as follows.

$$Y_{\ell}^{m} (\theta,\phi)=\sqrt{\frac{2\ell+1}{4\pi}\frac{(\ell-m)!}{(\ell+m)!}}P_{\ell}^{m}(\cos\theta)e^{\i m\phi}$$

Now the radial component equation $(2)$ remains. By multiplying both sides by $R$ and isolating the term related to energy $E$ on the right-hand side,

$$-\frac{ \hbar^{2} }{ 2Mr^{2} }\frac{d}{d r} \left( r^2\frac{d R}{d r} \right) +\left(V- \frac{\hbar ^{2}}{2M}\frac{l(l+1)}{r^{2}} \right)R =ER \tag{4}$$

If we then substitute $rR(r)=u(r)$, the equation simplifies.

$$\begin{align*} R&=\frac{u}{r} \\ \frac{ dR }{ dr }&=\frac{1}{r}\frac{ d u }{ d r}-\frac{1}{r^{2}}u \\ \frac{ d }{ d r }\left(r^{2} \frac{ d R}{ d r } \right)&=\frac{ d }{ dr }\left( r \frac{ d u }{ dr }-u \right)=r\frac{ d ^{2}u}{ d^{2} } \end{align*}$$

Hence, $(4)$ is as follows.

$$\begin{align*} &&-\frac{ \hbar^{2} }{ 2Mr^{2} }r\frac{ d ^{2}u}{ dr^{2} } +\left(V- \frac{\hbar ^{2}}{2M}\frac{\ell(\ell+1)}{r^{2}} \right)\frac{u}{r} &= E\frac{u}{r} \\ \implies &&-\frac{ \hbar^{2} }{ 2M }\frac{ d ^{2}u}{ dr^{2} } +\left(V- \frac{\hbar ^{2}}{2M}\frac{\ell(\ell+1)}{r^{2}} \right)u &= Eu \end{align*}$$

This form closely resembles the one-dimensional Schrödinger equation below.

$$-\frac{\hbar^{2}}{2m}\frac{ d ^{2} \psi}{ d x^{2} }+V\psi = E\psi$$

The difference is that the potential has changed to $V = V- \dfrac{\hbar ^{2}}{2M}\dfrac{\ell(\ell+1)}{r^{2}}$. The second term proportional to $\dfrac{1}{r^{2}}$ is called the centrifugal term. Like $Y_{l}^{m}(\cos\theta)$, $u(r)$ must also satisfy the normalization condition.

$$\int_{0}^{\infty}|R(r)|^{2}r^{2}dr=\int _{0}^{\infty} |u(r)|^{2}dr$$

The general solution for $V(r)$ stops here. Now, if the potential function $V(r)$ is exactly given in the problem, it can be solved accordingly.