The Wikipedia page on the Lamb shift includes the following first steps:

$$\Delta V = V\bigl(\vec{r}+\delta \vec{r}\bigr)-V(\vec{r})=\delta \vec{r} \cdot \nabla V (\vec{r}) + \frac{1}{2} \bigl(\delta \vec{r} \cdot \nabla\bigr)^2V(\vec{r})+\cdots\tag{1}$$

Because the fluctuations are isotropic:

$$\begin{align} \langle \delta \vec{r} \rangle _\text{vac} &=0 \tag{2.1}\\ \langle (\delta \vec{r} \cdot \nabla )^2 \rangle _\text{vac} &= \frac{1}{3} \langle (\delta \vec{r})^2\rangle _\text{vac} \nabla ^2\tag{2.2} \end{align}$$


$$\langle \Delta V\rangle =\frac{1}{6} \langle (\delta \vec{r})^2\rangle _\text{vac}\left\langle \nabla ^2\left(\frac{-e^2}{4\pi \epsilon _0r}\right)\right\rangle _{at}\tag{3}$$

I understand that step (1) is a Taylor expansion. Step (2.1) also makes intuitive sense, isotropy means that it is the same in every direction so it's no surprise that the expectation of $\delta \vec{r}$ is zero.

I can also understand step (3), which seems to be substituting results from step (2) into step (1), while ignoring higher order terms.

However, I have no clue where step (2.2) comes from. I attempted naively expanding the dot product as you would with $(a \cdot b)^2$, but I don't know if this is allowed.

In the same vein, notation wise, is $\delta \vec{r} \cdot \nabla$ the same as $\nabla \cdot \delta \vec{r}$? I know that the scalar product is commutative, but then $\nabla \cdot (\text{something})$ returns the divergence.

TL;DR: Don't know where step 2 comes from, and getting confused by vector calculus.

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    $\begingroup$ The equation labels seems like something taken verbatim from a textbook, but since we don't have the book in front of us (and the title isn't even mentioned in the post), why not just label the equations 1, 2, 3...? $\endgroup$
    – DanielSank
    Apr 25, 2016 at 8:17
  • $\begingroup$ @DanielSank I made up the labels for convenience, thought it would help anyone explaining. yeah I don't know about the formatting; edits would be welcome. I made a 2.1 and 2.2 because wikipedia suggested that those two results were implications of the isotropy. $\endgroup$ Apr 25, 2016 at 8:18
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    $\begingroup$ Oh I see. I was confused because the text says "step 3" but that refers to equation (2.2). I'll attempt an edit. $\endgroup$
    – DanielSank
    Apr 25, 2016 at 8:19
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    $\begingroup$ Once you go past first order, use index notation... $V(r+\delta r) \approx V(r) + r_i (\partial_i V)|_r + \frac{1}{2}r_i r_j (\partial_i \partial_j V)|_r$. From isotropy $\langle r_i r_j \rangle = \frac{1}{3} \delta_ij $. The usual vector notation with dot and cross products just isn't suited for more complicated expressions, that are unambiguously expressed in index notation. $\endgroup$ Apr 25, 2016 at 16:45
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    $\begingroup$ It means "evaluated at". $\endgroup$ Apr 26, 2016 at 10:43

2 Answers 2


This can be a little subtle the first time you see it, so I'll move through the rationale carefully:

$\langle (\delta \mathbf{r} \cdot \nabla )^2 \rangle _{vac} = \langle (\delta x \ \partial_x + \delta y\ \partial_y + \delta {z}\ \partial_z )^2\rangle _{vac} $

On expanding you get squared and cross terms. The text mentions $\langle \delta \mathbf r \rangle = \langle \delta \mathbf x + \delta \mathbf y + \delta \mathbf z \rangle = 0$ due to isotropy; this will kill all terms linear in $\delta x_i$ (or $\delta x_i \delta x_j$) leaving only the square terms:

= $\langle (\delta x^2 \ \partial_x^2 + \delta y^2\ \partial_y^2 + \delta z^2\ \partial_z^2 )\rangle _{vac} $

Noting again isotropy, the squared quantities (a scalar value) are all equal:

$\delta \mathbf r ^2 = \delta \mathbf x^2 + \delta \mathbf y^2 + \delta \mathbf z^2 = 3 \delta \mathbf x_i^2$

Substitute $\delta \mathbf x_i$ back into the second equation to get:

= $\langle \delta \mathbf x_i^2 ( \partial_x^2 + \partial_y^2 + \partial_z^2 )\rangle _{vac} $

= $\frac{1}{3}\langle \delta \mathbf r^2 \rangle_{vac} \nabla^2 $


The scalar product is just a shortcut notation for multiplication and then addition: $$\vec{a}\cdot\vec{b} = a_x b_x + a_y b_y + a_z b_z\tag{1}$$ It's commutative if the underlying multiplication is commutative, and otherwise it is not.

Notation like $\vec{a}\cdot\nabla$ is not really a scalar product, but it takes the same form of (1) and applies it to operator composition. $$\vec{A}\cdot\vec{B}f = A_x B_x f + A_y B_y f + A_z B_z f$$ where $A$ and $B$ are vectors of operators that act on $f$. (When we can get away with it, we omit $f$, and it should be understood that the results apply to any $f$ the operators might be acting on.) Again, the "dotting" is commutative if the underlying operators commute, and otherwise it's not. $$\vec{A}\cdot\vec{B} = \vec{B}\cdot\vec{A} \quad\Leftrightarrow\quad A_x B_x + A_y B_y + A_z B_z = B_x A_x + B_y A_y + B_z A_z$$ Similarly, a notation like $O^2$ where $O$ is an operator just means you apply $O$ twice. $$O^2 f = O(Of)$$ If $O$ is a scalar "product", $$\bigl(\vec{A}\cdot\vec{B}\bigr)^2f = \vec{A}\cdot\vec{B}\bigl(\vec{A}\cdot\vec{B}f\bigr)$$

In your case, if you want to figure out whether $\delta\vec{r}\cdot\nabla = \nabla\cdot\delta\vec{r}$... take a look at one component, for starters: is this true? $$\delta x\ \partial_x f = \partial_x(\delta x\ f)$$

  • $\begingroup$ Ok thank you for the clarifications on commutativity. And unfortunately I'm a bit unsure as to your last question! I'm leaning towards saying that it's not true. Because $x$ is a variable, $\delta x$ should be a variable too, so it can't be taken out of the partial derivative like that. $\endgroup$ Apr 25, 2016 at 8:40
  • $\begingroup$ Consider this, though: does $\delta x$ depend on $x$? $\endgroup$
    – David Z
    Apr 25, 2016 at 13:00
  • $\begingroup$ I would have thought so. But on the other hand, since it means an (infinitesimally?) small change in $x$, I guess it could be argued that it's a very small constant. I really don't know either way. $\endgroup$ Apr 25, 2016 at 14:18
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    $\begingroup$ Think of doing a limiting process, where you have some finite shift $\delta x$ and at each step you make it smaller and smaller. Imagine doing this at different values of $x$. Would you have to use different shifts for different values of $x$ at the same step of the limiting process? You probably can't come up with any reason to do so, and you'd be right: there is no reason to change $\delta x$ depending on $x$. Thus it's effectively a constant. In other, more complicated cases, there could be some reason to do so. $\endgroup$
    – David Z
    Apr 26, 2016 at 8:31

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