0
$\begingroup$

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 $a$.

$\endgroup$
1
  • $\begingroup$ Where is the negative sign for the derivative? $\endgroup$ Commented May 5, 2020 at 17:08

2 Answers 2

4
$\begingroup$

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 $$ i \hbar \lim_{a\to 0}\frac{\psi(r - a) - \psi (r)}{a} = - i \hbar \frac{\partial\psi}{\partial r}(r) $$

$\endgroup$
4
  • $\begingroup$ I forgot that this is the definition of the derivative $\endgroup$ Commented May 5, 2020 at 15:42
  • 2
    $\begingroup$ I think the numerator should be either $f(x+a)-f(x)$ or $f(x)-f(x-a)$. $\endgroup$ Commented May 5, 2020 at 16:59
  • 1
    $\begingroup$ @AccidentalTaylorExpansion Or, more correctly, it brings along a negative sign, which is what we want. $\endgroup$ Commented May 5, 2020 at 17:01
  • 1
    $\begingroup$ @AaronStevens Oops yes the momentum operator has a minus sign. $\endgroup$ Commented May 5, 2020 at 17:02
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.