In Sakurai's Problem 1.27 b), we use $\langle \mathbf{r} | \mathbf{p}\rangle = e^{i\mathbf{p}\cdot\mathbf{r}/\hbar}$ to show that

$$ \langle \mathbf{p} | F(\hat{r}) | \mathbf{p}' \rangle = \frac{1}{2 \pi^2 \hbar^2 q} \int_{0}^{\infty}r'\sin(q r'/\hbar) F(r') \,\text{d}r', $$

where $q = |\mathbf{p} - \mathbf{p}'|$. So if I just set $F(r) = 1/r$, then

$$ \langle \mathbf{p} | F(\hat{r}) | \mathbf{p}' \rangle = \frac{1}{2 \pi^2 \hbar^2 q} \int_{0}^{\infty}\sin(q r'/\hbar) \,\text{d}r', $$

but that clearly doesn't converge as it is. So, is that just it?

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    $\begingroup$ Okay, that looks better. What is it you are trying to understand here? Obviously with the divergent integral, $F(r)=1/r$ is not a good operator, but is there something more you're looking for? $\endgroup$
    – Kyle Kanos
    Commented Sep 25, 2014 at 2:24
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    $\begingroup$ @KyleKanos I guess not, it just seemed like a strange question to ask since there's "no answer." $\endgroup$
    – Mr. G
    Commented Sep 25, 2014 at 2:34
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    $\begingroup$ Try finding the expectation value of $e^{-\epsilon r}/r$ instead, then let $\epsilon \to 0$. $\endgroup$ Commented Sep 25, 2014 at 14:50
  • $\begingroup$ @RobinEkman Thanks, that gives a result, but how come you can just do that? I mean, it looks like I could make almost anything converge that way. $\endgroup$
    – Mr. G
    Commented Sep 25, 2014 at 21:59
  • $\begingroup$ @Mr.G I'll post an answer, the explanation is longer than what fits here. $\endgroup$ Commented Sep 25, 2014 at 22:52

1 Answer 1


Consider the general case that we want to calculate $$ \langle p |F(r) |p'\rangle.$$ By inserting the resolution of the identity $\int d^3r\, |r\rangle\langle r|$ we find that we need to compute $$\tilde{F}(q = p-p') = \int d^3 r \, e^{i(p-p')r} F(r). \tag{1}$$ This integral will converge if $\int dr\, |F(r)|$ is finite. Such a function is said to be $L^1$. It will also be the case that $\int dq\, | \tilde F (q)|$ is finite, that is, $\tilde F(q)$ is also $L^1$. The map $F \mapsto \tilde F$ is the Fourier transform, let's call it $\mathcal F$.

Now the problem is that $1/r$ in 3-dimensional space is not $L^1$, but if we can find a way to extend the Fourier transform to be defined for more functions, maybe we can still make sense of $\langle p |\frac{1}{r}|p' \rangle$. The Fourier transform can be extended to functions that satisfy only $$\int d^3 r \, |F(r) g(r)| < \infty \tag{2}$$ for all functions $g(r)$ that decay rapidly enough (more precisely, $g$ needs to decay faster than any power of $r$). However by extending this far, $\tilde F(q)$ isn't necessarily $L^1$, or even a function anymore. It could be a distribution, like a Dirac delta. Maybe you have seen the formula $$\int d^3r\, e^{i (\vec p - \vec p') \cdot \vec r} = \delta(\vec p- \vec p').$$ What is really meant is that $$(\mathcal F[1])(p) = \delta(p).$$

Connecting back to the original question, the function $F(r) = 1/r$ satisfies the condition (2), but we cannot use the integral formula (1) to find $\tilde F(q)$. However, the functions $F_\epsilon(r) = e^{-\epsilon r}/r$, with $\epsilon > 0$, converge to $F$ when $\epsilon \to 0$. Each $F_\epsilon(r)$ is $L^1$, so we can (rather) easily find $\tilde F_\epsilon(q)$. The extension of the Fourier transform is so constructed that if $$F_\epsilon \overset{\epsilon \to 0}{\longrightarrow} F \text{ then } \tilde{F}_\epsilon \overset{\epsilon \to 0}{\longrightarrow} \tilde F.$$ This justifies using the convergence factor $e^{-\epsilon r}$.

(You are right that you can make almost anything converge by damping with $e^{-\epsilon r}$, but that just reflects that since the condition (2) isn't very strong, the Fourier transform is defined for a whole bunch of functions.)

  • $\begingroup$ There's a little mistake in the range of the Fourier transform on $L^1(\Bbb{R}^n)$. By the Riemann-Lebesgue lemma, the range is into $C_\infty(\Bbb{R}^n)$, the continuous functions vanishing at infinity, which is not a subset of $L^1(\Bbb{R}^n)$ $\endgroup$ Commented Sep 29, 2014 at 14:09
  • $\begingroup$ Good catch. I mixed it up with the result for $L^2$ functions. $\endgroup$ Commented Sep 30, 2014 at 12:29

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