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I tried out using $r = \sqrt{x^2+y^2}$ and $\tan \theta = y/x$, then getting the formula for $\partial/\partial x$ using the chain rule, and then obtaining the second partial derivative operator without differentiating any further but through purely algebraic means, i.e., expanding.

In the fourth line of the below derivation, I could recognise that something was going wrong because the terms consisting $\partial^2/\partial x^2$ should have indeed been five, i.e.,

$$\frac{\partial ^2}{\partial x^2} = \cos^2\theta \frac{\partial ^2}{\partial x^2} - \frac{2 \sin \theta \cos \theta}{r} \frac{\partial ^2}{\partial r \partial \theta} + \frac{\sin ^2 \theta}{r} \frac{\partial }{\partial r} + \frac{2 \sin \theta \cos \theta }{r^2} \frac{\partial }{\partial \theta} + \frac{\sin ^2 \theta}{r^2} \frac{\partial ^2}{\partial \theta ^2}$$, but there are only three terms in my derivation. enter image description here

I heard in quantum mechanics classes that treating linear (differential) operators algebraically is perfectly fine, but this seems to be a counterexample.

If anyone wants to see how I tried out further derivations please see below. Thanks.

enter image description here

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    $\begingroup$ @Allawonder I agree with the fact that $\frac{d^2 u}{dx^2} \neq (\frac{du}{dx})^2$, but doesn't it hold for differential operators themselves? $\endgroup$
    – curious
    Jan 31 '21 at 12:59
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    $\begingroup$ I don't know about quantum mechanics (if your claim is true, then it would definitely have something to do with quantum mechanics), but as far as mathematics is concerned, it is false that the square of a differential is the same as the second differential, provided no additional assumptions are made. $\endgroup$
    – Allawonder
    Jan 31 '21 at 17:27
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    $\begingroup$ @Allawonder Like this: youtu.be/CWK9cW1RwQQ $\endgroup$
    – curious
    Jan 31 '21 at 19:53
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    $\begingroup$ @Allawonder: You seem to be confusing $d/dx$ and $df/dx$. While $f''\ne (f')^2$, it is by definition true that $d^2 f/dx^2 = (d/dx)(d/dx)f$. Thus, $d^2/dx^2 = (d/dx)^2$. $\endgroup$
    – user26872
    Jan 31 '21 at 22:04
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    $\begingroup$ Read my answer here. $\endgroup$
    – K.defaoite
    Feb 1 '21 at 2:05
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$\def\th{\theta}$Manipulation of differential operators in this fashion is perfectly fine and is in fact just shorthand for more cluttered notation. (We leave out the function being operated on.) The rules of manipulation may seem strange, but if one simply considers what is being represented they are straightforward to work out and remember.

We begin with an example from calculus of one variable. Note that $$(f g)' = f'g + f g'.$$ This can be represented in operator notation as \begin{align*} D f = (D f) + f D,\tag{1} \end{align*} where $D=d/dx$. This is product rule in operator notation. It is understood in (1) that $D f$ on the left-hand side is meant to act on another function and that $(D f)$ on the right-hand side is simply $f'$. Think of a differential operator as "acting to the right" in the following way:
\begin{align*} D f g &= f'g+fg' \\ (D f)g &= f'g \\ (f D)g &= f D g = fg'. \end{align*} The uncertainty principle in quantum mechanics essentially follows from the fact that $$xf' - (xf)' = xf' - (f+xf') = -f,$$ or \begin{align*} [x,D] &\equiv x D - D x \\ &= x D - (1 + x D) \\ &= -1. \end{align*}

Regarding the original problem, let $D_x = \partial/\partial x,r_x = \partial r/\partial x$, etc. We can work out $D_x^2$ as follows: \begin{align*} D_x &= r_x D_r + \th_x D_\th \\ D_x^2 &= (r_x D_r + \th_x D_\th)(r_x D_r + \th_x D_\th) \\ &= (r_x D_r)^2 + (r_x D_r)(\th_x D_\th) + \ldots \\ &= r_x r_{xr}D_r + r_x^2 D_r^2 \\ &\quad + r_x \th_{xr}D_\th + r_x \th_x D_{\th r} \\ &\quad + \ldots \end{align*} Using that \begin{align*} r_{xr} &= 0 \\ r_{x\th} &= -\sin\th \\ \th_{xr} &= \frac{\sin\th}{r^2} \\ \th_{x\th} &= -\frac{\cos\th}{r} \end{align*} one will arrive at the correct relation for $D_x^2$.

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