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The momentum operator $\hat{p}$ can be written as $$ \hat{p} = i\hbar \lim_{a \rightarrow 0} \frac{\hat{T}(a\hat{r})- \hat{\mathbb{I}}}{a} $$ where $\mathbb{\hat{I}}$ is the identity operator and $\hat{r}$ being the unit vector in a given direction. For the case of a single pasrticle with wavefunction $\psi(r)$ it can be written as $$ ( \hat{p} \psi )(r) = i\hbar \lim_{a \rightarrow 0} \frac{(\hat{T}\psi)(r)- \psi(r)}{a} = i\hbar \lim_{a \rightarrow 0} \frac{\psi(r-a)- \psi(r)}{a} \, . $$

Now I am not sure how to proceed. My thought was to use L'Hôpital's rule: $$ = i\hbar \lim_{a \rightarrow 0} \frac{\frac{d}{da}[\psi(r-a) - \psi(r)]}{\frac{d}{da}a} = i\hbar \lim_{a \rightarrow 0 } \frac{\partial \psi(r-a)}{\partial a} = i \hbar \frac{\partial \psi(r)}{\partial a} = 0 \, . $$ However, it is stated on Wikipedia that the result should be $$ (\hat{p}\psi )(r) = -i \hbar \frac{\partial \psi(r)}{\partial r} \, . $$ I am not sure as to why the wavefuction is being derived with respect to $r$ and not $a$.

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  • $\begingroup$ Where is the negative sign for the derivative? $\endgroup$ May 5 '20 at 17:08
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As you already mentioned $$ (\hat p \psi)(r) = i \hbar \lim_{a\to 0}\frac{\psi(r - a) - \psi (r)}{a} $$ but this is already the derivative of $\psi(r)$ with respect to $r$ so $$ i \hbar \lim_{a\to 0}\frac{\psi(r - a) - \psi (r)}{a} = - i \hbar \frac{\partial\psi}{\partial r}(r) $$

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  • $\begingroup$ I forgot that this is the definition of the derivative $\endgroup$ May 5 '20 at 15:42
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    $\begingroup$ I think the numerator should be either $f(x+a)-f(x)$ or $f(x)-f(x-a)$. $\endgroup$ May 5 '20 at 16:59
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    $\begingroup$ @AccidentalTaylorExpansion Or, more correctly, it brings along a negative sign, which is what we want. $\endgroup$ May 5 '20 at 17:01
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    $\begingroup$ @AaronStevens Oops yes the momentum operator has a minus sign. $\endgroup$ May 5 '20 at 17:02
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You already have an answer but I'll show you why your method is actually consistent. When you take the derivative with respect to something in the argument you have to be careful. I define $y=r-a$ as a helping tool. That means that \begin{align}\frac d{da}\psi(x-a)&=\frac d{da}\psi(y(a))\\ &=\frac {dy}{da}\frac {d\psi(y)}{dy}\\ &=-\psi'(y) \end{align} After taking the limit this becomes $$-\lim_{a\rightarrow 0}\psi'(r-a)=-\psi'(r)=-\frac{\partial\psi(r)}{\partial r}$$ You have to do this because a derivative with respect to $a$, $\lim_{a\rightarrow 0}\frac{\partial}{\partial a}$, makes no sense when you take the limit to zero. You might be wondering about the minus sign in my answer. The derivative is defined as $$\lim_{a\rightarrow 0}\frac{f(x+a)-f(x)}{a}$$ so with a plus sign. The momentum operator is $\hat p=-i\hbar\frac \partial{\partial x}$ so this checks out.

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