One-particle, time-independent Schrodinger equation in spherical coordinates

The Laplacian $\nabla^{2}$ in spherical coordinates can be derived from eq87, eq88 and eq89. Substituting eq78 through eq86 in eq87, eq88 and eq89, adding the resulting three equations and simplifying, we have

$\nabla^{2}=\frac{\partial^{2}}{\partial r^{2}}+\frac{2}{r} \frac{\partial}{\partial r}+\frac{1}{r^{2}}\frac{\partial^{2}}{\partial \theta^{2}}+\frac{cos\theta}{r^{2}sin\theta}\frac{\partial}{\partial\theta}+\frac{1}{r^{2}sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}\; \; \; \; \; \; \; \; 46$

$\nabla^{2}=\frac{1}{r^{2}} \frac{\partial}{\partial r}\left ( r^{2}\frac{\partial}{\partial r} \right )+\frac{1}{r^{2}}\left [ \frac{1}{sin^{2}\theta}\frac{\partial^{2}}{\partial \phi^{2}}+\frac{1}{sin\theta}\frac{\partial}{\partial\theta}\left ( sin\theta\frac{\partial}{\partial\theta}\right )\right ]\; \; \; \; \; \; \; \; 47$

It can be easily shown that eq47 is equivalent to eq46 by computing the derivatives in eq47. Hence, eq45 in spherical coordinates is

$\small \hat{H}=-\frac{\hbar^{2}}{2m}\left \{ \frac{1}{r^{2}}\frac{\partial}{\partial r}\left ( r^{2}\frac{\partial}{\partial r}\right ) +\frac{1}{r^{2}}\left [ \frac{1}{sin^{2}\theta}\frac{\partial^{2}}{\partial \phi^{2}}+\frac{1}{sin\theta}\frac{\partial}{\partial\theta}\left ( sin\theta\frac{\partial}{\partial\theta} \right )\right ]\right \}+V(r)\; \; \; \; \; \; \; \; 48$
For the case of a particle confined to a spherical surface of zero relative potential, $\small r$ is constant and $\small V(r)=0$ . The first term of the Laplacian operating on a wavefunction $\small \psi(r,\theta,\phi)$ will return a result of zero. We can therefore discard it and eq48 becomes:

$\hat{H}_{angular}=-\frac{\hbar^{2}}{2mr^{2}}\left [ \frac{1}{sin^{2}\theta}\frac{\partial^{2}}{\partial \phi^{2}}+\frac{1}{sin\theta}\frac{\partial}{\partial\theta}\left ( sin\theta\frac{\partial}{\partial\theta} \right )\right ]\; \; \; \; \; \; \; \; 49$

Substitute eq96 in the above equation

$\hat{H}_{angular}=\frac{\hat{L}^{2}}{2I}\; \; \; \; \; \; \; \; 50$

where $I=mr^{2}$ .

Hence, an eigenvalue of $\hat{H}_{angular}$ is the energy associated with the angular motion of a particle. This implies that the linear (or radial) part of the Hamiltonian is:

$\hat{H}_{radial}=-\frac{\hbar^{2}}{2m} \frac{1}{r^{2}}\frac{\partial}{\partial r}\left ( r^{2}\frac{\partial}{\partial r}\right ) +V(r)=-\frac{\hbar^{2}}{2m} \left ( \frac{2}{r}\frac{\partial}{\partial r}+\frac{\partial^{2}}{\partial r^{2}}\right ) +V(r)\; \; \; \; \; \; \; \; 51$

Similarly, an eigenvalue of $\hat{H}_{radial}$ is the energy associated with the linear motion of a particle. We can, therefore, rewrite eq48 as:

$\hat{H}=\hat{H}_{radial}+\hat{H}_{angular}$

One-particle, time-independent Schrodinger equation in spherical coordinates

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