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
    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.
    Jul 5, 2014 at 21:02
  • $\begingroup$ LaTeX tip: you can use \Re and \Im to indicate real and imaginary parts respectively. $\endgroup$
    – JamalS
    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
    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
    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
    Jul 7, 2014 at 0:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.