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For context; In quantum mechanics with continuous eigenvalues lets suppose we have a probabiltity density like this:

$$\left\lvert a(p) \right\rvert^2=\frac{2a \lvert A \rvert^2\sin^2(ap/\hbar)}{(ap/\hbar)^2}\tag{1}$$ and you wish to show that the relative probability of measuring the particle with momentum $p=\dfrac{\pi \hbar}{2a}$ and with $p=0$ is $\dfrac{4}{\pi^2}$.

The solution simply states that

the relative probability for $p=\dfrac{\pi \hbar}{2a}$ to $p=0$ is $$\frac{1}{(\pi/2)^2}:\frac{\sin(0)}{0}$$


Is this answer really mathematically correct? I wasn't aware that you could invoke the $\dfrac{\sin x}{x}\to 1$ without having limit notation: $\lim \limits_{x \to 0} \dfrac{\sin x}{x}=1$. It seems to me the author has simply evaluated $(1)$ at $p=0$. I thought that this was wrong and we can only say that the fraction tends to $1$.

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    $\begingroup$ No, obviously it is not correct. $\endgroup$ – Andrés E. Caicedo Dec 18 '17 at 23:17
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    $\begingroup$ As an example where it would get you into trouble: $\frac{\sin(2x)}{x} \to 2$ as $x \to 0$, whereas a "substitution" could incorrectly tell you it goes to $\frac{\sin 0}{0} = 1$. $\endgroup$ – Daniel Schepler Dec 18 '17 at 23:23
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The solution uses somewhat sloppy notation; what they're actually using is $$ \lvert a(p)\rvert^2=2a \lvert A \rvert^2(f(ap/\hbar))^2 $$ where $$ f(x)=\begin{cases} \dfrac{\sin x}{x} & x\ne 0 \\[6px] 1 & x=0 \end{cases} $$ In other words, they're using the unique extension of $\frac{\sin x}{x}$ by continuity at $0$.

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$\dfrac {\sin x}x$ is just a number, so in this sense $\dfrac{\sin 0}0$ is undetermined.

However you can define a function $f(x)=\dfrac{\sin x}x$ everywhere $x\neq 0$ and since $\lim\limits_{x\to 0}f(x)$ exists we can extend $f$ by continuity to $\tilde f$.

$\tilde f:\begin{cases}\tilde f(0)=1\\\tilde f(x)=\dfrac{\sin x}x & \forall x\neq 0\end{cases}\quad$ this function is defined over whole $\mathbb R$.

$f$ is not the same as $\tilde f$, because one is defined in zero, while the other is not, but modulo continuity they are the same.

What about I'm now interested in $g(x)=\dfrac{\sin(x^2)}x$ ?

I'm conducting the same methodology and extend $g$ by continuity to $\tilde g$ where $\tilde g(0)=0$.


Now the reason I cannot talk about $\dfrac{\sin 0}0$, is because I do not know which one between $\tilde f(0)=1$ and $\tilde g(0)=0$, I have to consider.

Both are possible interpretations of this ratio, but they give a different value, this is why we call this form an indeterminate.


So when you are conducting calculations, you always have to refer to the function ($f$ in this case) you consider. And the value in $0$ whether you obtain it via a limit or via the continuously extended $\tilde f$, it is just a matter of presentation.

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No you can’t since the expression $\frac{\sin(0)}{0}$ is indeterminate.

https://en.m.wikipedia.org/wiki/Indeterminate_form

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    $\begingroup$ No. That is not quite what an indeterminate form means: the link you give discusses limits. The quotient is simply not defined. $\endgroup$ – Andrés E. Caicedo Dec 18 '17 at 23:35
  • $\begingroup$ Yes it’s indeterminate in the sense that it could have an infinite number of answers. Indeed if we set: $$\frac{0}{0}=a \implies 0=a\cdot 0=0 \quad \forall a$$ The link is aimed to give an example to how handle this expressions. $\endgroup$ – user Dec 18 '17 at 23:43
  • $\begingroup$ You are not applying the definitions correctly. $\endgroup$ – Andrés E. Caicedo Dec 18 '17 at 23:46
  • $\begingroup$ The example posted by egreg show clearly that it’s indeterminate in the sense that you can arbitrarily give at the expression any value. Maybe it’s only a matter of language or definition but I hope it’s clear to you what I mean with “indeterminate”. Thanks. $\endgroup$ – user Dec 18 '17 at 23:50
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    $\begingroup$ No. In mathematics much more so than in regular speech, words have a precise meaning. You are using technical words incorrectly. It is not an issue of whether one sort of maybe kinda gets what you mean. $\endgroup$ – Andrés E. Caicedo Dec 18 '17 at 23:54

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