Calculating the Gradient of a Complex Exponential Function

[LagrangeEuler]

Homework Statement

Calculate \nabla e^{i\vec{k}\cdot \vec{r}}

Homework Equations

\nabla f(r)=\frac{df}{dr}\nabla r=\frac{df}{dr}\frac{\vec{r}}{r}

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}}

 

[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 $\vec r = x \hat x + y \hat y + z \hat z$

 

[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

 

[RUber]

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

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.

 

[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.