# Maxwell's Equations: Taking the real part of the product $\mathbf{E}\exp(j \omega t)$ in order to get the "complete field"

My textbook, Laser Electronics, Third Edition, by Verdeyen, says the following in a section on Maxwell's equations:

To describe an electromagnetic wave, we need two field-intensity vectors, $$\mathbf{e}$$ and $$\mathbf{h}$$, which are related to each other by

$$\nabla \times \mathbf{h} = \mathbf{j} + \epsilon_0 \dfrac{\partial{\mathbf{e}}}{\partial{t}} + \dfrac{\partial{\mathbf{p}}}{\partial{t}} \tag{1.2.1a}$$

$$\nabla \times \mathbf{e} = -\mu_0 \dfrac{\partial{\mathbf{h}}}{\partial{t}} \tag{1.2.1b}$$

where $$\mathbf{p}$$ is the polarization current induced by the electric field. (A term of the form $$\partial{\mathbf{m}}/\partial{t}$$ can be added to (1.2.1b) but will be ignored for now.) We use lowercase letters to represent vectors that are explicit functions of time $$t$$ and the three spatial coordinates $$x$$, $$y$$, and $$z$$. Most of the time we will be talking about sinusoidal variations of the field and use the phasor representation

$$\begin{matrix} \mathbf{e}(\mathbf{r}, t) = {\rm Re} \left[ \mathbf{E}(\mathbf{r})e^{j \omega t} \right]& \mathbf{h}(\mathbf{r}, t) = {\rm Re} \left[ \mathbf{H}(\mathbf{r})e^{j \omega t} \right] \\ \mathbf{j}(\mathbf{r}, t) = {\rm Re} \left[ \mathbf{J}(\mathbf{r})e^{j \omega t} \right]& \mathbf{p}(\mathbf{r}, t) = {\rm Re} \left[ \mathbf{P}(\mathbf{r})e^{j \omega t} \right] \tag{1.2.2} \end{matrix}$$

where $$Re$$ is real part, $$\mathbf{r} = x\mathbf{a}_x, + y\mathbf{a}_y + z\mathbf{a}_z$$, $$\mathbf{a}_i$$, is the unit vector in the $$i$$th direction, and the capital letters $$\mathbf{E}$$ and $$\mathbf{H}$$ are complex vector quantities depending on space coordinates but not on time. We recognize that if we want the complete field, we must take the real part of the product $$\mathbf{E}\exp(j \omega t)$$.

What does the author mean by "complete field" here, and why must we take the real part of the product $$\mathbf{E}\exp(j \omega t)$$ in order to get it? I would greatly appreciate it if people would please take the time to explain this.

One of the advantages of the complex representation of the fields—as this passage is trying to emphasize—is that the normal modes of the solution are simple products of a function of the spatial position, times a function of time. However, this only true of the complexified solutions. The true, real-valued solutions are not just products—e.g. the "complete" electric field strength is $${\bf e}({\bf r},t)=\frac{1}{2}\left[{\bf E}({\bf r})e^{j\omega t}+{\bf E}^{*}({\bf r})e^{-j\omega t}\right],$$ which does not have the product form.
• Thanks for the answer. Hmm, but $\frac{1}{2}\left[{\bf E}({\bf r})e^{j\omega t}+{\bf E}^{*}({\bf r})e^{-j\omega t}\right]$ is still complex, right? I find this confusing, because I'm interpreting your answer as saying that ${\bf e}({\bf r},t)=\frac{1}{2}\left[{\bf E}({\bf r})e^{j\omega t}+{\bf E}^{*}({\bf r})e^{-j\omega t}\right]$ is the true, real-valued solution. What am I misunderstanding here? Shouldn't the complete electric field strength be real valued (the real part of $\mathbf{E}\exp(j \omega t)$), as the author says? Dec 31, 2019 at 1:51
• No, $\frac{1}{2} \left( a + a^\ast \right) = \mathrm{Re} \left( a \right)$. So the complete field strength given in @Buzz's answer is indeed real. Dec 31, 2019 at 2:32
• @ClaraDiazSanchez Oh, $\mathbf{E}^*$ is the complex conjugate! Ok, I understand. Thank you for the clarification. Dec 31, 2019 at 3:02