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I don't understand how to obtain second equation with first part in the equation $$ \nabla \times \vec A_0 e^{-j \vec k\cdot \vec r} = -j\vec k\times \vec A_0 e^{-j \vec k\cdot \vec r}. $$ Can you show me how to derive it?

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  • $\begingroup$ This is easiest done with cartesian tensor notation. Are you familiar with it? $\endgroup$ – DanielC Nov 11 '17 at 22:31
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    $\begingroup$ Why on earth would anyone use $\left({\vec i}, {\vec j}, {\vec k}\right)$ for the basis vectors and $\bf j$ for $\sqrt{-1}$? You're just begging to confuse people. $\endgroup$ – Jerry Schirmer Nov 11 '17 at 23:41
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\begin{equation} \boldsymbol{\nabla}\equiv \begin{bmatrix} \dfrac{\partial\hphantom{ x_{1}}}{\partial x_{1}}\vphantom{\dfrac{\dfrac{}{}}{}}\\ \dfrac{\partial\hphantom{ x_{1}}}{\partial x_{2}}\vphantom{\dfrac{\dfrac{}{}}{}}\\ \dfrac{\partial\hphantom{ x_{1}}}{\partial x_{3}}\vphantom{\dfrac{\dfrac{}{}}{}} \end{bmatrix}\,, \quad \mathbf{A}= \begin{bmatrix} A_{1}\vphantom{\dfrac{\dfrac{}{}}{}}\\ A_{2}\vphantom{\dfrac{\dfrac{}{}}{}}\\ A_{3}\vphantom{\dfrac{\dfrac{}{}}{}}\\ \end{bmatrix}\,, \quad \mathbf{k}\boldsymbol{\cdot}\mathbf{x}=k_{1}x_{1}+k_{2}x_{2}+k_{3}x_{3} \tag{01} \end{equation}

\begin{align} \boldsymbol{\nabla}\boldsymbol{\times}\left(\mathbf{A}\,e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}}\right) & = \begin{bmatrix} \mathbf{e}_{1} & \mathbf{e}_{2} & \mathbf{e}_{3}\vphantom{\dfrac{\dfrac{}{}}{}}\\ \dfrac{\partial\hphantom{ x_{1}}}{\partial x_{1}} & \dfrac{\partial\hphantom{ x_{1}}}{\partial x_{2}} & \dfrac{\partial\hphantom{ x_{1}}}{\partial x_{3}}\vphantom{\dfrac{\dfrac{}{}}{}}\\ A_{1}e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}} & A_{2}e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}} & A_{3}e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}} \vphantom{\dfrac{\dfrac{}{}}{}} \end{bmatrix} = \begin{bmatrix} A_{3}\dfrac{\partial e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}}}{\partial x_{2}}-A_{2}\dfrac{\partial e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}}}{\partial x_{3}}\vphantom{\dfrac{\dfrac{}{}}{}}\\ A_{1}\dfrac{\partial e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}}}{\partial x_{3}}-A_{3}\dfrac{\partial e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}}}{\partial x_{1}}\vphantom{\dfrac{\dfrac{}{}}{}}\\ A_{2}\dfrac{\partial e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}}}{\partial x_{1}}-A_{1}\dfrac{\partial e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}}}{\partial x_{2}}\vphantom{\dfrac{\dfrac{}{}}{}} \end{bmatrix} \nonumber\\ & =\boldsymbol{-}\mathrm{i}\:e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}} \begin{bmatrix} k_{2}A_{3}-k_{3}A_{2}\vphantom{\dfrac{\dfrac{}{}}{}}\\ k_{3}A_{1}-k_{1}A_{3}\vphantom{\dfrac{\dfrac{}{}}{}}\\ k_{1}A_{2}-k_{2}A_{1}\vphantom{\dfrac{\dfrac{}{}}{}} \end{bmatrix} =\boldsymbol{-}\mathrm{i}\:e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}} \underbrace{ \begin{bmatrix} \mathbf{e}_{1} & \mathbf{e}_{2} & \mathbf{e}_{3}\vphantom{\dfrac{\dfrac{}{}}{}}\\ k_{1} & k_{2} & k_{3}\vphantom{\dfrac{\dfrac{}{}}{}}\\ A_{1} & A_{2} & A_{3}\vphantom{\dfrac{\dfrac{}{}}{}}\vphantom{\dfrac{\dfrac{}{}}{}} \end{bmatrix}}_{\mathbf{k}\boldsymbol{\times}\mathbf{A}} \tag{02} \end{align} so \begin{equation} \boxed{\:\boldsymbol{\nabla}\boldsymbol{\times}\left(\mathbf{A}\,e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}}\right) =\boldsymbol{-}\mathrm{i}\:e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}}\left(\mathbf{k}\boldsymbol{\times}\mathbf{A}\right)\:\vphantom{\dfrac12^{\tfrac12}_{\tfrac12}}} \tag{03} \end{equation}


More generally :

If $\:\psi\left(x_{1},x_{2},x_{3}\right)\:$ and $\:\mathbf{A}\left(x_{1},x_{2},x_{3}\right)\:$ are scalar and vector functions respectively of the coordinates in $\:\mathbb{R}^{3}\:$ then \begin{equation} \boxed{\:\boldsymbol{\nabla}\boldsymbol{\times}\left(\psi\mathbf{A}\right)=\boldsymbol{\nabla}\psi\boldsymbol{\times}\mathbf{A}+\psi\boldsymbol{\nabla}\boldsymbol{\times}\mathbf{A}\:\vphantom{\dfrac12^{\tfrac12}_{\tfrac12}}} \tag{04} \end{equation} Equation (03) is a special case of (04) with \begin{equation} \psi\left(\mathbf{x}\right)\equiv e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}}\,,\quad \mathbf{A}\left(\mathbf{x}\right)=\text{constant} \tag{05} \end{equation} and the fact that \begin{equation} \boldsymbol{\nabla}\psi=\boldsymbol{\nabla} e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}}=\boldsymbol{-}\mathrm{i}\:e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}}\mathbf{k} \tag{06} \end{equation}


Proof of identity (04): \begin{align} \boldsymbol{\nabla}\boldsymbol{\times}\left(\psi\mathbf{A}\right) & = \begin{bmatrix} \mathbf{e}_{1} & \mathbf{e}_{2} & \mathbf{e}_{3}\vphantom{\dfrac{\dfrac{}{}}{}}\\ \dfrac{\partial\hphantom{ x_{1}}}{\partial x_{1}} & \dfrac{\partial\hphantom{ x_{1}}}{\partial x_{2}} & \dfrac{\partial\hphantom{ x_{1}}}{\partial x_{3}}\vphantom{\dfrac{\dfrac{}{}}{}}\\ \psi A_{1} & \psi A_{2} & \psi A_{3} \vphantom{\dfrac{\dfrac{}{}}{}} \end{bmatrix} = \begin{bmatrix} \dfrac{\partial \left(\psi A_{3}\right)}{\partial x_{2}}-\dfrac{\partial \left(\psi A_{2}\right)}{\partial x_{3}}\vphantom{\dfrac{\dfrac{}{}}{}}\\ \dfrac{\partial \left(\psi A_{1}\right)}{\partial x_{3}}-\dfrac{\partial \left(\psi A_{3}\right)}{\partial x_{1}}\vphantom{\dfrac{\dfrac{}{}}{}}\\ \dfrac{\partial \left(\psi A_{2}\right)}{\partial x_{1}}-\dfrac{\partial \left(\psi A_{1}\right)}{\partial x_{2}}\vphantom{\dfrac{\dfrac{}{}}{}} \end{bmatrix} = \begin{bmatrix} A_{3}\dfrac{\partial \psi}{\partial x_{2}}+ \psi\dfrac{\partial A_{3}}{\partial x_{2}}-A_{2}\dfrac{\partial \psi}{\partial x_{3}}-\psi\dfrac{\partial A_{2}}{\partial x_{3}}\vphantom{\dfrac{\dfrac{}{}}{}}\\ A_{1}\dfrac{\partial \psi}{\partial x_{3}}+ \psi\dfrac{\partial A_{1}}{\partial x_{3}}-A_{3}\dfrac{\partial \psi}{\partial x_{1}}-\psi\dfrac{\partial A_{3}}{\partial x_{1}}\vphantom{\dfrac{\dfrac{}{}}{}}\\ A_{2}\dfrac{\partial \psi}{\partial x_{1}}+ \psi\dfrac{\partial A_{2}}{\partial x_{1}}-A_{1}\dfrac{\partial \psi}{\partial x_{2}}-\psi\dfrac{\partial A_{1}}{\partial x_{2}}\vphantom{\dfrac{\dfrac{}{}}{}} \end{bmatrix} \nonumber\\ & = \begin{bmatrix} A_{3}\dfrac{\partial \psi}{\partial x_{2}}-A_{2}\dfrac{\partial \psi}{\partial x_{3}}\vphantom{\dfrac{\dfrac{}{}}{}}\\ A_{1}\dfrac{\partial \psi}{\partial x_{3}}-A_{3}\dfrac{\partial \psi}{\partial x_{1}}\vphantom{\dfrac{\dfrac{}{}}{}}\\ A_{2}\dfrac{\partial \psi}{\partial x_{1}}-A_{1}\dfrac{\partial \psi}{\partial x_{2}}\vphantom{\dfrac{\dfrac{}{}}{}} \end{bmatrix} + \begin{bmatrix} \psi\dfrac{\partial A_{3}}{\partial x_{2}}-\psi\dfrac{\partial A_{2}}{\partial x_{3}}\vphantom{\dfrac{\dfrac{}{}}{}}\\ \psi\dfrac{\partial A_{1}}{\partial x_{3}}-\psi\dfrac{\partial A_{3}}{\partial x_{1}}\vphantom{\dfrac{\dfrac{}{}}{}}\\ \psi\dfrac{\partial A_{2}}{\partial x_{1}}-\psi\dfrac{\partial A_{1}}{\partial x_{2}}\vphantom{\dfrac{\dfrac{}{}}{}} \end{bmatrix} = \underbrace{ \begin{bmatrix} \mathbf{e}_{1} & \mathbf{e}_{2} & \mathbf{e}_{3}\vphantom{\dfrac{\dfrac{}{}}{}}\\ \dfrac{\partial \psi}{\partial x_{1}} & \dfrac{\partial\psi}{\partial x_{2}} & \dfrac{\partial\psi}{\partial x_{3}}\vphantom{\dfrac{\dfrac{}{}}{}}\\ A_{1} & A_{2} & A_{3} \vphantom{\dfrac{\dfrac{}{}}{}} \end{bmatrix}}_{\boldsymbol{\nabla}\psi\boldsymbol{\times}\mathbf{A}} + \psi\, \underbrace{ \begin{bmatrix} \mathbf{e}_{1} & \mathbf{e}_{2} & \mathbf{e}_{3}\vphantom{\dfrac{\dfrac{}{}}{}}\\ \dfrac{\partial\hphantom{ x_{1}}}{\partial x_{1}} & \dfrac{\partial\hphantom{ x_{1}}}{\partial x_{2}} & \dfrac{\partial\hphantom{ x_{1}}}{\partial x_{3}}\vphantom{\dfrac{\dfrac{}{}}{}}\\ A_{1} & A_{2} & A_{3} \vphantom{\dfrac{\dfrac{}{}}{}} \end{bmatrix}}_{\boldsymbol{\nabla}\boldsymbol{\times}\mathbf{A}} \nonumber\\ & =\boldsymbol{\nabla}\psi\boldsymbol{\times}\mathbf{A}+\psi\boldsymbol{\nabla}\boldsymbol{\times}\mathbf{A} \tag{04} \end{align}

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The $i$th component of the left-hand side is $\epsilon_{ijk}\partial_j(A_{0k}e^{-j\vec{k}\cdot\vec{r}})$, where $\sum_{jk}$ is implicit. (Don't confuse dummy indices with the square root of $-1$.) The $i$th component of the right-hand side is $-j\epsilon_{ijk}k_jA_{0k}e^{-j\vec{k}\cdot\vec{r}}$. The rest is the product rule.

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J.G.'s answer is perfectly correct. However, rather than getting lost in symbols, a good thing to remember here is the key intuition behind solving this problem:

Exponentials are the eigenfunctions of the derivative operator

The whole problem is simply a souped up application of the fundamental equation $\mathrm{d}_x \exp(x) = \exp(x)$; now simply use this together with a comparison of the the mnemonics for the curl and cross product:

$$\mathbf{u\times v} = \begin{vmatrix} \mathbf{e}_x&\mathbf{e}_y&\mathbf{e}_z\\ u_1&u_2&u_3\\ v_1&v_2&v_3\\ \end{vmatrix};\quad \mathbf{\nabla\times v} = \begin{vmatrix} \mathbf{e}_x&\mathbf{e}_y&\mathbf{e}_z\\ \partial_1&\partial_2&\partial_3\\ v_1&v_2&v_3\\ \end{vmatrix}$$

and your formula is immediately true by inspection, because $-i\,\vec{k}_j$ replaces $\partial_j$, by dint of the exponential's defining property cited above


Note my notation for the basis vectors and Jerry Schirmer's comment:

Why on earth would anyone use $\left({\vec i}, {\vec j}, {\vec k}\right)$ for the basis vectors and $j$ for $\sqrt{-1}$? You're just begging to confuse people.

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