5

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.

lagrange103
  • 457

2 Answers2

3

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

David Z
  • 77,804
3

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 $

anon01
  • 1,631