As far as I know, the function:

$$ \vec{E}(\vec{r},t)=\vec{E_0}\cdot e^{i(\vec{k}\cdot \vec{r}-\omega t)} \hspace{2cm}(1) $$

is a mathematical solution of the wave equation:

$$ \nabla^2 \vec{E}=\mu\varepsilon\frac{\partial^2 \vec{E}}{\partial t^2} $$

if and only if $\omega$ satisfies the dispersion relation:

$$\omega(k)=\frac{k}{\sqrt{\mu\varepsilon}}$$ Previously I wrote "mathematical" because the complex fuction $(1)$ has no physical meaning, if we want to have a physically significative function we have to take the real part of (1) $$ \vec{E}(\vec{r},t)=\vec{E_0}\cdot \cos(\vec{k}\cdot \vec{r}-\omega t) $$

Up to now there should be no problem. The problem arises when I consider a 1D wave packet.

Using the complex notation: the initial wave packet is given by: $$ E(x,t=0)=e^{-\frac{(x-x_c)^2}{2\sigma^2}}\cdot e^{i k_{\text{c}}x} $$ where the derm $e^{i k_{\text{c}}x}$ derives from the fact that the initial wave packet is a moving wave packet and not a static one.

Its temporal evolution can be determined doing a Fourier transform.

Let's indicate with $g(k)$ the Fourier transform of the initial wave packet:

$$ g(k)=\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty}E(x,t=0)\cdot e^{-ikx}dx $$ So, using the fact that every component of the spectra evolves according to a specific $\omega(k)$, I can determine the temporal evolution: $$ E(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}g(k)e^{i(kx-\omega(k) t)}dk \hspace{20mm}(2) $$

Finally, according to what I said previouisly, if I am interested in physical quantities like the temporal evoultion of the amplitude, I have to take the real part of (2):

$$ E_{\text{phys}}(x,t)=\Re \left[\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}g(k)e^{i(kx-\omega(k) t)}dk\right]\hspace{20mm}(3) $$

Reasoning on the physical meaning step by step

The initial condition is a real value wave packet: $$ E_r(x,t=0)=\Re\left[e^{-\frac{(x-x_c)^2}{2\sigma^2}}\cdot e^{i k_{\text{c}}x}\right]=e^{-\frac{(x-x_c)^2}{2\sigma^2}}\cdot\cos(k_{\text{c}}x) $$ so, because of I am interested in determining the temporal evolution of the wave packet, I perform a Fourier transform of the initial wave packet: $$ g_r(k)=\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty}E_r(x,t=0)\cdot e^{-ikx}dx $$ where now $g_r(k)$ is could be a complex value function but this does not hurt me because this function lives in the k-space and I am interested in having real function only in the x-space. The problem is thai if I evolve $g_r(k)$ in the function in the Fourier space multiplying it for $e^{-i\omega(k) t}$ and then I come back in the x space antitrasforming: $$ E_r(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}g_r(k)e^{i(kx-\omega(k) t)}dk \hspace{20mm}(4) $$
I obtain that (4) is a function whose amplitude becomes zero in a very rapid time, a very different behaviour from (3). Could anyone give me an explanation of that, or any reference in which the complex notation is well explained?

  • 1
    $\begingroup$ I'm afraid I am unable to discern the difference between (3) and (4). Why do you say that the amplitude of (4) becomes zero? $\endgroup$
    – ACuriousMind
    Commented Jul 5, 2014 at 17:34
  • $\begingroup$ Ok, complex numbers (or any other numbers or vectors) are just a way to model (physical) data when they can be used this way. So in this sense, complex numbers are just as real as real numbers. The problem i think stems form the fact that in using the real part a-priori actually refers to another problem, since by linearity of diff. eq. the solution is a superposition of both $cos$ and $sin$ (which the complex representation maintains). When one use the real part at the end is equivalent to $Re(z) = 1/2(z + z^{*})$ and this makes the physical meaning clear, it selects the retarded waves $\endgroup$
    – Nikos M.
    Commented Jul 5, 2014 at 21:02
  • $\begingroup$ LaTeX tip: you can use \Re and \Im to indicate real and imaginary parts respectively. $\endgroup$
    – JamalS
    Commented Jul 6, 2014 at 8:03

1 Answer 1


Take the wave equation $$\nabla^2\vec{E} = \frac{1}{c^2} \frac{\partial^2 \vec{E}}{\partial t^2},$$ and let $\vec{E}(\vec{r},t)$ be a solution. Indeed taking the real part $\Re(\vec{E}(\vec{r},t))$ yields the physical significant values.

The initial values at $t = 0$ are $\Re(\vec{E}(\vec{r},0))$ and the problem arises here: this does not give you enough information to predict the future evolution of the field!

For example, consider the 1D case with: $$\begin{align} f_1(x,t) &= e^{i(kx-\omega t)}, \\ f_2(x,t) &= \cos(kx - \omega t). \end{align}$$

Note that $\Re(f_1(x,0)) = \Re(f_2(x,0)) = \cos(kx)$. Therefore the two functions have the same initial values at $t = 0$, yet they evolve differently as $t$ progresses.

To fully specify the initial state of a solution, you would need $\Re(\vec{E}(\vec{r},0))$ and one of in addition:

  • $\Im(\vec{E}(\vec{r},0))$,
  • $\Re(\frac{\partial\vec{E}}{\partial t}(\vec{r},0))$,
  • $\Im(\frac{\partial\vec{E}}{\partial t}(\vec{r},0))$. [Added in Edit 2].

In your calculation, the moment you took $\Re(\vec{E}(\vec{r},0))$ and applied Fourier Transform, you've implicitly stipulated $\Im(\vec{E}(\vec{r},0)) = 0$. Therefore you changed the initial state to something else, so of course you'd end up with a different solution.

  • $\begingroup$ why do I need $\Im(\vec{E}(\vec{r},0))$ and $\Im(\frac{\partial\vec{E}}{\partial t}(\vec{r},0))$ to specify the initial state? As far as I know the imaginary part has no physical meaning. $\endgroup$
    – Chaos
    Commented Jul 6, 2014 at 13:32
  • 1
    $\begingroup$ Your wavepacket is complex, so taking the real part altered the result. Meanwhile, yeah the differential equation works fine when everything is real. However, eigenfunctions of the operator $\nabla^2 - \frac{1}{c^2} \frac{\partial^2}{\partial t^2}$ are complex; therefore modal decomposition would would need complex number. So we may as well see solutions in their complex glory, and treat real signals as a special case where imaginary parts are 0. $\endgroup$
    – arccosh
    Commented Jul 7, 2014 at 0:41
  • $\begingroup$ Meanwhile, the wave packet is also seen a lot for for Shrodinger equation of a free particle. In this case, the complex part become essential because time derivative is first-order and there is no freedom to specify initial time derivative. In this case, the imaginary part interacts with the real part, and affects the directions of propagation. $\endgroup$
    – arccosh
    Commented Jul 7, 2014 at 0:46

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