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I was looking at the Schrodinger Equation for the Hydrogen Atom, and saw it in the form $$\left(-\frac{\hbar^2}{2{\mu}r^2}\left(\frac{\partial}{{\partial}r}\left(r^2\frac{\partial}{{\partial}r}\right)+\frac{1}{\sin\theta}\frac{\partial}{{\partial}r}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^2\theta}\frac{\partial^2}{{\partial}\varphi^2}\right)-\frac{e^2}{4\pi\epsilon_0r}\right)\psi(r,\theta,\varphi)=E\psi(r,\theta,\varphi)$$

I understand that $\Large\frac{\partial1}{{\partial}a}=0$, so could the schrodinger equation for the Hydrogen Atom be rewritten as

$$\left(-\frac{\hbar}{2{\mu}r^2}\left(\frac{\partial^2r^2}{{\partial}r^2}+\frac{{\partial}\csc\theta}{\partial\theta}\frac{{\partial}\sin\theta}{\partial\theta}+\frac{\partial^2\csc\theta}{\partial\varphi^2}\right)-\frac{e^2}{4\pi\epsilon_0r}\right)\psi(r,\theta,\varphi)=E\psi(r,\theta,\varphi)$$?

If this way of rewriting the Schrodinger Equation for the Hydrogen Atom is correct, then I am still confused about the $$\frac{\partial^2\csc\theta}{\partial\varphi^2}$$ part.

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closed as unclear what you're asking by Buzz, GiorgioP, Yashas, user191954, Jon Custer Apr 30 at 12:36

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    $\begingroup$ I didn't check the entire thing, but it seems to me like the first $\partial /\partial r$ term isn't being treated properly. If you apply the product rule to that term you get $2r\partial/\partial r + r^2 \partial^2/\partial r^2$. $\endgroup$ – Julius Apr 29 at 18:57
  • $\begingroup$ Having read the title of your question again I'm inclined to say: no, because $f(x)$ contains operators, and they do not (necessarily) commute with differentiation. $\endgroup$ – Julius Apr 29 at 18:59
  • $\begingroup$ There is an error in the first equation. The second is entirely wrong. $\endgroup$ – my2cts Apr 29 at 19:00
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    $\begingroup$ Neither equation illustrates the question in the title. $\endgroup$ – my2cts Apr 29 at 19:09
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No, sadly.

This is an unfortunate consequence we have in our notation, we write $r$ when we mean the function $(r\cdot)$ which multiplies a function by $r$, and we write $A B$ when we mean the composition of functions $A\circ B$. So when we write something like $$r^{-2}\frac{\partial}{\partial r} \left(r^2 \frac{\partial}{\partial r}\right)$$ what we really mean is the function which would more unambiguously be written as $$\psi \mapsto \left\{(r,\theta,\phi)\mapsto \frac{2}{r} \cdot \psi_{(1)}(r, \theta,\phi) + \psi_{(1,1)}(r,\theta,\phi) \right\},$$where $f_{(n)}$ is the partial derivative of $f$ with respect to its $n^\text{th}$ argument holding all its other arguments constant, and $f_{(m,n)}$ is shorthand for $[f_{(m)}]_{(n)},$ the partial derivative with respect to the $n^\text{th}$ argument of the partial derivative with respect to the $m^\text{th}$ argument. So it is a function which takes a scalar field and returns another scalar field.

We typically extend our understanding addition $(+)$ and subtraction $(-)$ to such functions-from-fields-to-fields by saying that for example $$f + g = \psi \mapsto \big\{\mathbf r \mapsto f(\psi)(\mathbf r) + g(\psi)(\mathbf r)\big\},$$to add two field-transformers you construct the field-transformer which provides its argument field to the two constituent transformers and then adds the two fields together pointwise. Indeed this thing in the curly braces could be correctly denoted $f(\psi) + g(\psi)$ when we extend $+$ to fields: we add fields together pointwise, giving the same input to both fields and adding the numbers that they produce; then we add field-transformers together fieldwise, giving the same field input to both transformers and adding the fields that they produce. It's kind of the same idea.

With that idea we could correctly say that due to the product rule of normal calculus, $$\partial_r \circ (r \cdot) = (1 \cdot) + (r\cdot) \circ \partial_r,$$in other words if you multiply a field by $r$ and then take a partial derivative of the result with respect to $r$, that is related to taking the partial derivative of a field and then multiplying the result by $r$, by adding the original field at the end.

But in QM we let ourselves get sloppy with notation and write this as $$\partial_x ~x = 1+ x~\partial_x.$$In some sense this is less sloppy than it sounds because it is correct if we think about $x$ and $\partial_x$ as somehow being like matrices and this as being a matrix equation.

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  • $\begingroup$ As a teacher I would not recommend that a student write that final expression. To my mind $\partial_x x = 1$ and $\partial_x (x f) = f + x \partial_x f$. I advise students to follow the policy "when in doubt, allow your operator to operate on something, and use brackets to make explicit what is being differentiated by any differential operator in your expressions." $\endgroup$ – Andrew Steane Apr 29 at 23:13
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Certainly not. I think your confusion is coming from operator notation.

Let $D$ be a differential operator. Then we can write the action on $xf(x)$ as

$$D (xf) = f(x) + xDf(x)$$

which we could also write, if we wished as the new differential operator

$$\tilde{D}f(x):=(1 + xD)f(x).$$

You have to remember to be careful as to when you've already taken into account the product rule.

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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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