2 Removed ridiculous notation.
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No, this is not a correct way of rewriting the Schr$\ddot{\textrm{o}}$dingerSchrödinger equation for the hydrogen atom. If we take the left-hand side in its current form (and correct the error in your second term) and expand all the terms, we get

\begin{align} &\left(-\frac{\hbar}{2\mu r^2}\left(\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right)+\frac{1}{\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\varphi^2}\right)-\frac{e^2}{4\pi\epsilon_0 r}\right)\psi(r,\theta,\varphi)\\ &=-\frac{\hbar}{2\mu r^2}\left(\frac{\partial}{\partial r}\left(r^2\frac{\partial\psi}{\partial r}\right)+\frac{1}{\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial\psi}{\partial\theta}\right)+\frac{1}{\sin^2\theta}\frac{\partial^2\psi}{\partial\varphi^2}\right)-\frac{e^2\psi}{4\pi\epsilon_0 r}\\ &=-\frac{\hbar}{2\mu r^2}\left(2r\frac{\partial\psi}{\partial r}+r^2\frac{\partial^2\psi}{\partial r^2}+\frac{1}{\sin\theta}\left( \cos\theta\frac{\partial\psi}{\partial\theta}+\sin\theta\frac{\partial^2\psi}{\partial\theta^2}\right)+\frac{1}{\sin^2\theta}\frac{\partial^2\psi}{\partial\varphi^2} \right)-\frac{e^2\psi}{4\pi\epsilon_0 r}\\ &=-\frac{\hbar}{2\mu}\left(\frac{\partial^2\psi}{\partial r^2}+\frac{2}{r}\frac{\partial\psi}{\partial r}+\frac{1}{r^2}\frac{\partial^2\psi}{\partial\theta^2}+\frac{\cot\theta}{r^2}\frac{\partial\psi}{\partial\theta}+\frac{1}{r^2}\frac{\partial\psi}{\partial\varphi^2} \right)-\frac{e^2\psi}{4\pi\epsilon_0 r} \end{align}

which doesn't match up with your expression, since yours doesn't contain any derivatives of $\psi$ anywhere at all. Hopefully this illustrates what we mean in physics when we use this kind of notation.

As a side note, every single line in this answer is equivalent to $\nabla^2\psi$, where $\nabla^2$ is the Laplacian operator (see e.g. https://en.wikipedia.org/wiki/Laplace_operator). The article chose to write out the Laplacian in spherical coordinates; they could have also chosen to write it in Cartesian coordinates, such that $\nabla^2\psi=\frac{\partial^2\psi}{\partial x^2}+\frac{\partial^2 \psi}{\partial y^2}+\frac{\partial^2\psi}{\partial z^2}$. For any choice of coordinates, the underlying physics is the same, which is why we like to write equations in coordinate-free terms by using things like $\nabla^2$. Your choice of coordinates will, however, influence the difficulty of obtaining an elegant solution, and in this case, the hydrogen atom's spherical symmetry makes spherical coordinates the obvious first choice.

No, this is not a correct way of rewriting the Schr$\ddot{\textrm{o}}$dinger equation for the hydrogen atom. If we take the left-hand side in its current form (and correct the error in your second term) and expand all the terms, we get

\begin{align} &\left(-\frac{\hbar}{2\mu r^2}\left(\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right)+\frac{1}{\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\varphi^2}\right)-\frac{e^2}{4\pi\epsilon_0 r}\right)\psi(r,\theta,\varphi)\\ &=-\frac{\hbar}{2\mu r^2}\left(\frac{\partial}{\partial r}\left(r^2\frac{\partial\psi}{\partial r}\right)+\frac{1}{\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial\psi}{\partial\theta}\right)+\frac{1}{\sin^2\theta}\frac{\partial^2\psi}{\partial\varphi^2}\right)-\frac{e^2\psi}{4\pi\epsilon_0 r}\\ &=-\frac{\hbar}{2\mu r^2}\left(2r\frac{\partial\psi}{\partial r}+r^2\frac{\partial^2\psi}{\partial r^2}+\frac{1}{\sin\theta}\left( \cos\theta\frac{\partial\psi}{\partial\theta}+\sin\theta\frac{\partial^2\psi}{\partial\theta^2}\right)+\frac{1}{\sin^2\theta}\frac{\partial^2\psi}{\partial\varphi^2} \right)-\frac{e^2\psi}{4\pi\epsilon_0 r}\\ &=-\frac{\hbar}{2\mu}\left(\frac{\partial^2\psi}{\partial r^2}+\frac{2}{r}\frac{\partial\psi}{\partial r}+\frac{1}{r^2}\frac{\partial^2\psi}{\partial\theta^2}+\frac{\cot\theta}{r^2}\frac{\partial\psi}{\partial\theta}+\frac{1}{r^2}\frac{\partial\psi}{\partial\varphi^2} \right)-\frac{e^2\psi}{4\pi\epsilon_0 r} \end{align}

which doesn't match up with your expression, since yours doesn't contain any derivatives of $\psi$ anywhere at all. Hopefully this illustrates what we mean in physics when we use this kind of notation.

As a side note, every single line in this answer is equivalent to $\nabla^2\psi$, where $\nabla^2$ is the Laplacian operator (see e.g. https://en.wikipedia.org/wiki/Laplace_operator). The article chose to write out the Laplacian in spherical coordinates; they could have also chosen to write it in Cartesian coordinates, such that $\nabla^2\psi=\frac{\partial^2\psi}{\partial x^2}+\frac{\partial^2 \psi}{\partial y^2}+\frac{\partial^2\psi}{\partial z^2}$. For any choice of coordinates, the underlying physics is the same, which is why we like to write equations in coordinate-free terms by using things like $\nabla^2$. Your choice of coordinates will, however, influence the difficulty of obtaining an elegant solution, and in this case, the hydrogen atom's spherical symmetry makes spherical coordinates the obvious first choice.

No, this is not a correct way of rewriting the Schrödinger equation for the hydrogen atom. If we take the left-hand side in its current form (and correct the error in your second term) and expand all the terms, we get

\begin{align} &\left(-\frac{\hbar}{2\mu r^2}\left(\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right)+\frac{1}{\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\varphi^2}\right)-\frac{e^2}{4\pi\epsilon_0 r}\right)\psi(r,\theta,\varphi)\\ &=-\frac{\hbar}{2\mu r^2}\left(\frac{\partial}{\partial r}\left(r^2\frac{\partial\psi}{\partial r}\right)+\frac{1}{\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial\psi}{\partial\theta}\right)+\frac{1}{\sin^2\theta}\frac{\partial^2\psi}{\partial\varphi^2}\right)-\frac{e^2\psi}{4\pi\epsilon_0 r}\\ &=-\frac{\hbar}{2\mu r^2}\left(2r\frac{\partial\psi}{\partial r}+r^2\frac{\partial^2\psi}{\partial r^2}+\frac{1}{\sin\theta}\left( \cos\theta\frac{\partial\psi}{\partial\theta}+\sin\theta\frac{\partial^2\psi}{\partial\theta^2}\right)+\frac{1}{\sin^2\theta}\frac{\partial^2\psi}{\partial\varphi^2} \right)-\frac{e^2\psi}{4\pi\epsilon_0 r}\\ &=-\frac{\hbar}{2\mu}\left(\frac{\partial^2\psi}{\partial r^2}+\frac{2}{r}\frac{\partial\psi}{\partial r}+\frac{1}{r^2}\frac{\partial^2\psi}{\partial\theta^2}+\frac{\cot\theta}{r^2}\frac{\partial\psi}{\partial\theta}+\frac{1}{r^2}\frac{\partial\psi}{\partial\varphi^2} \right)-\frac{e^2\psi}{4\pi\epsilon_0 r} \end{align}

which doesn't match up with your expression, since yours doesn't contain any derivatives of $\psi$ anywhere at all. Hopefully this illustrates what we mean in physics when we use this kind of notation.

As a side note, every single line in this answer is equivalent to $\nabla^2\psi$, where $\nabla^2$ is the Laplacian operator (see e.g. https://en.wikipedia.org/wiki/Laplace_operator). The article chose to write out the Laplacian in spherical coordinates; they could have also chosen to write it in Cartesian coordinates, such that $\nabla^2\psi=\frac{\partial^2\psi}{\partial x^2}+\frac{\partial^2 \psi}{\partial y^2}+\frac{\partial^2\psi}{\partial z^2}$. For any choice of coordinates, the underlying physics is the same, which is why we like to write equations in coordinate-free terms by using things like $\nabla^2$. Your choice of coordinates will, however, influence the difficulty of obtaining an elegant solution, and in this case, the hydrogen atom's spherical symmetry makes spherical coordinates the obvious first choice.

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No, this is not a correct way of rewriting the Schr$\ddot{\textrm{o}}$dinger equation for the hydrogen atom. If we take the left-hand side in its current form (and correct the error in your second term) and expand all the terms, we get

\begin{align} &\left(-\frac{\hbar}{2\mu r^2}\left(\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right)+\frac{1}{\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\varphi^2}\right)-\frac{e^2}{4\pi\epsilon_0 r}\right)\psi(r,\theta,\varphi)\\ &=-\frac{\hbar}{2\mu r^2}\left(\frac{\partial}{\partial r}\left(r^2\frac{\partial\psi}{\partial r}\right)+\frac{1}{\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial\psi}{\partial\theta}\right)+\frac{1}{\sin^2\theta}\frac{\partial^2\psi}{\partial\varphi^2}\right)-\frac{e^2\psi}{4\pi\epsilon_0 r}\\ &=-\frac{\hbar}{2\mu r^2}\left(2r\frac{\partial\psi}{\partial r}+r^2\frac{\partial^2\psi}{\partial r^2}+\frac{1}{\sin\theta}\left( \cos\theta\frac{\partial\psi}{\partial\theta}+\sin\theta\frac{\partial^2\psi}{\partial\theta^2}\right)+\frac{1}{\sin^2\theta}\frac{\partial^2\psi}{\partial\varphi^2} \right)-\frac{e^2\psi}{4\pi\epsilon_0 r}\\ &=-\frac{\hbar}{2\mu}\left(\frac{\partial^2\psi}{\partial r^2}+\frac{2}{r}\frac{\partial\psi}{\partial r}+\frac{1}{r^2}\frac{\partial^2\psi}{\partial\theta^2}+\frac{\cot\theta}{r^2}\frac{\partial\psi}{\partial\theta}+\frac{1}{r^2}\frac{\partial\psi}{\partial\varphi^2} \right)-\frac{e^2\psi}{4\pi\epsilon_0 r} \end{align}

which doesn't match up with your expression, since yours doesn't contain any derivatives of $\psi$ anywhere at all. Hopefully this illustrates what we mean in physics when we use this kind of notation.

As a side note, every single line in this answer is equivalent to $\nabla^2\psi$, where $\nabla^2$ is the Laplacian operator (see e.g. https://en.wikipedia.org/wiki/Laplace_operator). The article chose to write out the Laplacian in spherical coordinates; they could have also chosen to write it in Cartesian coordinates, such that $\nabla^2\psi=\frac{\partial^2\psi}{\partial x^2}+\frac{\partial^2 \psi}{\partial y^2}+\frac{\partial^2\psi}{\partial z^2}$. For any choice of coordinates, the underlying physics is the same, which is why we like to write equations in coordinate-free terms by using things like $\nabla^2$. Your choice of coordinates will, however, influence the difficulty of obtaining an elegant solution, and in this case, the hydrogen atom's spherical symmetry makes spherical coordinates the obvious first choice.