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I studying the Schrödinger equation applied to the hydrogen atom, so i need to change from retangular to spheric coordinates. So:

$$\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} rcos(\theta )sin(\varphi ) \\ rsin(\theta )sin(\varphi ) \\ rcos(\varphi ) \end{matrix} \right] $$ $$dx=\frac { \partial x }{ \partial r } dr+\frac { \partial x }{ \partial \theta } d\theta +\frac { \partial x }{ \partial \varphi } d\varphi $$ Analogously for dy and dz, then: $$\left[ \begin{matrix} dx \\ dy \\ dz \end{matrix} \right] =\left[ \begin{matrix} cos(\theta )sin(\varphi )dr-rsin(\theta )sin(\varphi )d\theta \quad +rcos(\theta )cos(\varphi )d\varphi \\ sin(\theta )sin(\varphi )dr+rsin(\varphi )cos(\theta )d\theta \quad +rsin(\theta )cos(\varphi )d\varphi \quad \\ cos(\varphi )dr\quad -rsin(\varphi )d\varphi \end{matrix} \right] $$ In the Schrodinger equation have a laplacian: $$\nabla ²$$ So i need to know $$\nabla =\left( \frac { \partial }{ \partial x } ,\frac { \partial }{ \partial y } ,\frac { \partial }{ \partial z } \right) $$ How can I found or discover this from what i have done? In other words, how can i found: $$\frac { \partial }{ \partial x } ,\frac { \partial }{ \partial y } and \frac { \partial }{ \partial z }? $$

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  • $\begingroup$ The Laplacian of $\psi$, $\nabla^2 \psi$ means the sum of the second derivatives of $\psi$ taken in Cartesian coordinates. It really means $\nabla \cdot (\nabla \psi)$, so $\nabla^2$, despite being common in physics, is slightly misleading notation. So you need to get the second derivatives now. $\endgroup$ – Ian Mar 21 '17 at 14:51
  • $\begingroup$ See also en.wikipedia.org/wiki/Laplace_operator#Coordinate_expressions $\endgroup$ – reuns Mar 21 '17 at 15:07
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As noted in the comments the Laplace operator is defined using the second derivatives: $$ \nabla^2 f=\frac{\partial^2 f}{\partial x ^2}+\frac{\partial^2 f}{\partial y ^2}+\frac{\partial^2 f}{\partial z ^2} $$

there is a lot of work to derive it in spherical coordinates (that you can see here), but the final result is surprisingly simple:

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

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