Answer to Question #154872 in Quantum Mechanics for ali

Question

. Derive Schrodinger wave equation from Cartesian coordinates into Spherical polar coordinates. 1 𝑟 2 𝜕 𝜕𝑟 (𝑟 2 𝜕𝜓 𝜕𝑟 ) + 1 𝑟 2𝑠𝑖𝑛𝜃 𝜕 𝜕𝜃 (𝑠𝑖𝑛𝜃 𝜕𝜓 𝜕𝜃) + 1 𝑟 2𝑠𝑖𝑛2𝜃 𝜕 2𝜓 𝜕𝜙2 + 2𝑚 ℏ 2 (𝐸 − 𝑈)𝜓 = 0

Answer 1

Let's first write (stationary) Schrodinger equation in cartesian coordinates :

$-\frac{\hbar^2}{2m} (\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2})\psi +V\psi = E\psi$

$-\frac{\hbar^2}{2m} \Delta\psi + (V-E)\psi=0$ , where $\Delta$ is Laplace operator

$\Delta\psi + \frac{2m}{\hbar^2} (E-V)\psi=0$

Now it is just enough to use the expression of $\Delta$ in spherical coordinates :

$\frac{1}{r^2} \frac{\partial}{\partial r}(r^2\frac{\partial \psi}{\partial r}) + \frac{1}{r^2 \sin\theta} \frac{\partial}{\partial\theta} (\sin\theta \frac{\partial\psi}{\partial\theta}) + \frac{1}{r^2\sin^2\theta} \frac{\partial^2\psi}{\partial\phi^2}+\frac{2m}{\hbar^2}\psi =0$

The expression of $\Delta$ in spherical coordinates can be found, for example, in Wikipedia : Laplace operator - Wikipedia . We can also, of course, calculate it directly :

$\frac{\partial}{\partial x} = \frac{\partial r}{\partial x}\frac{\partial}{\partial r} + \frac{\partial \theta}{\partial x}\frac{\partial}{\partial \theta} + \frac{\partial \phi}{\partial x}\frac{\partial}{\partial \phi}$ (chain rule)

$\frac{\partial}{\partial x} = \sin\theta \cos\phi \frac{\partial}{\partial r} -\frac{\cos\theta \cos\phi}{r}\frac{\partial}{\partial\theta}-\frac{\sin\phi}{r\sin\theta} \frac{\partial}{\partial\phi}$ (using the expressions of spherical coordinates in cartesian coordinates)

The same calculation for $y, z$ gives :

$\frac{\partial}{\partial y} = \sin\theta \sin\phi \frac{\partial}{\partial r} -\frac{\cos\theta \sin\phi}{r}\frac{\partial}{\partial\theta} + \frac{\cos\phi}{r\sin\theta} \frac{\partial}{\partial\phi}$

$\frac{\partial}{\partial z} = \cos\theta \frac{\partial}{\partial r} -\frac{\sin\theta}{r}\frac{\partial}{\partial\theta}$

And now we find by direct calculation (as $\frac{\partial^2}{\partial x^2} = (\frac{\partial}{\partial x})^2$ and same for other coordinates) :

$\Delta = \frac{1}{r^2} \frac{\partial}{\partial r}(r^2\frac{\partial}{\partial r}) + \frac{1}{r^2 \sin\theta} \frac{\partial}{\partial\theta} (\sin\theta \frac{\partial}{\partial\theta}) + \frac{1}{r^2\sin^2\theta} \frac{\partial^2}{\partial\phi^2}$