1. Nov 3, 2014

LagrangeEuler

1. The problem statement, all variables and given/known data
Calculate $$\nabla e^{i\vec{k}\cdot \vec{r}}$$

2. Relevant equations
$$\nabla f(r)=\frac{df}{dr}\nabla r=\frac{df}{dr}\frac{\vec{r}}{r}$$

3. The attempt at a solution
I have a problem. I know result
$$=\nabla e^{i\vec{k}\cdot \vec{r}}=i\vec{k} e^{i\vec{k}\cdot \vec{r}}$$

Last edited: Nov 3, 2014
2. Nov 3, 2014

RUber

This problem is much clearer when broken up componentwise:
$e^{i \vec k \cdot \vec r } = e^{i (k_x x + k_y y + k_z z) }$
Where $\vec k = k_x \hat x + k_y \hat y + k_z \hat z$ and $\vec r = x \hat x + y \hat y + z \hat z$

3. Nov 4, 2014

LagrangeEuler

Yes but for that way we need a lot of time. Perhaps
$$\nabla=\sum_w\vec{e}_w\frac{\partial}{\partial x_w}$$
$$e^{i \vec{k}\cdot \vec{r}}=e^{i\sum_q k_q x_q}$$
but I get a problem with this sums. Maybe
$$\sum_w\vec{e}_w\frac{\partial}{\partial x_w}e^{i\sum_q k_q x_q}=$$
$$=\sum_w\vec{e}_we^{i\sum_q k_q x_q}(\frac{\partial}{\partial x_w}i\sum_q k_q x_q)=$$
$$=e^{i\vec{k}\cdot \vec{r}}i \sum_w \vec{e}_w k_{w}=$$
$$=i\vec{k}e^{i\vec{k}\cdot \vec{r}}$$

but again I did not use theorem
$$\nabla f(r)=\frac{df}{dr}\nabla r$$

4. Nov 4, 2014

RUber

let $f(\vec r) = e^{ik_x \hat x +ik_y \hat y+ik_z \hat z}$ and $\vec r = ik_x \hat x +ik_y \hat y+ik_z \hat z = i \vec k \cdot \vec r$
then use the theorem.

5. Nov 4, 2014

dslowik

I think that "relevant equation",
$\nabla f(r)=\frac{df}{dr}\nabla r=\frac{df}{dr}\frac{\vec{r}}{r}$,
is not that relevant here:
$e^{i\vec{k}\cdot \vec{r}}\neq f(r)$ since $r = (x^2 + y^2 + z^2)^{1/2}$.

Rather,
$\nabla f(g(\vec r))=\frac{df}{dg}\nabla g(\vec r)$
with
$f(g) = e^g$ and $g(\vec r) = i \vec k \cdot \vec r$.