# What does $(\delta\vec{r}\cdot\nabla)^2$ mean in the derivation of the Lamb shift, and how do you find its expectation?

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}

Then

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

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.

• 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...? Apr 25, 2016 at 8:17
• @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. Apr 25, 2016 at 8:18
• Oh I see. I was confused because the text says "step 3" but that refers to equation (2.2). I'll attempt an edit. Apr 25, 2016 at 8:19
• 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. Apr 25, 2016 at 16:45
• It means "evaluated at". Apr 26, 2016 at 10:43

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? $$\delta x\ \partial_x f = \partial_x(\delta x\ f)$$

• 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. Apr 25, 2016 at 8:40
• Consider this, though: does $\delta x$ depend on $x$? Apr 25, 2016 at 13:00
• 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. Apr 25, 2016 at 14:18
• 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. Apr 26, 2016 at 8:31