Speaking in general about plane waves propagating along $z$ (electro-magnetic waves, for example; not necessarily particles represented as waves), a wave packet can be defined as

$$A(z,t) = \int_{\omega_1}^{\omega_2} A ( \omega ) e^{j (\omega t - kz) } d\omega$$

In particular, this expression is used when dealing with group velocity. But a single plane wave is usually expressed as

$$B(z, t) = B_0 \cos ( \omega t - kz )$$

So, why is the complex exponential used above? Or is the actual $A(z,t)$ of the wave packet just the real part of the first expression?

Observation: I did not consider the form $B_0 \cos ( \omega t - kz )$ because I necessarily want real functions, but because this is the standard form a plane wave is presented and written.


2 Answers 2


A real-valued wavepacket solution of the dispersionless 1D wave equation can always be defined as $$A(z,t) = \int_{\omega_1}^{\omega_2} A ( \omega ) e^{j (\omega t - kz) } d\omega, $$ where $\omega_1=-\omega_2$ and the frequency-domain amplitude satisfies $A(-\omega)=A(\omega)^*$; if this is not the case then $A(z,t)$ will have some complex values.

This form is consistent with the plane-wave function you wrote, $$B(z, t) = B_0 \cos ( \omega_0 t - k_0z )= \frac{B_0}{2} \left(e^{j (\omega_0 t - k_0z) }+e^{j (-\omega_0 t + k_0z) }\right),$$ with a frequency-domain amplitude $B(\omega)=\frac12 B_0 \left[ \delta(\omega-\omega_0) + \delta(\omega+\omega_0)\right]$. As such, there is no contradiction.

  • $\begingroup$ You're correct about the conjugate, I think, that was hasty writing. Your objection about the $kz$ term is unfounded, however (e.g. it can simply be folded into the time; keep in mind that $k=\omega/c$). And either way it's not crucial to the core of the question as posed. $\endgroup$ Jan 26, 2017 at 11:49
  • $\begingroup$ Ok and thank you also for the precisation about the complex conjugate. The term $ikz$ gives little confusion in fact. But, first of all: are you basically stating that $A(\omega)$ is the Fourier Transform of the time-domain signal corresponding to the wave packet? $\endgroup$
    – BowPark
    Jan 26, 2017 at 12:08

The complex form $$A(z,t) = \int_{\omega_1}^{\omega_2} A ( \omega ) e^{j (\omega t - kz) } d\omega$$ is the most common because it is compact (easy to write and read). As usual in physics, one writes complex expressions and when one wants the "real" thing, one takes the real part of the expression.

However, if you really want to have an expression without complex numbers, you can write your wavepacket under the form $$B(z,t) = \int_{\omega_1}^{\omega_2} \alpha( \omega ) \cos(\omega t - kz) + \beta( \omega ) \sin(\omega t - kz)\,d\omega,$$ both $\alpha( \omega ) $ and $\beta( \omega )$ being real numbers (and possibly $0 \leq \omega_1 \leq \omega_2$ if you like to consider only positive frequencies, which is all you need in the real world). This is for a packet travelling towards positive $z$. Change $\omega t - kz$ into $\omega t + kz$ for a packet travelling towards negative $z$.

Note that for $$\alpha( \omega ) = \Re(A( \omega )),\ \beta( \omega )=-\Im(A( \omega )),$$ you have exactly $$B(z,t)=\Re(A(z,t)).$$

Indeed, using this definition of $\alpha$ and $\beta$, that is: $$A(\omega)=\alpha(\omega)-j\beta(\omega),$$ and the identity $\Re((\alpha-j\beta)(C+jS))=\Re(\alpha C+\beta S+j(\alpha S-\beta C))=\alpha C+\beta S$ where $C$ and $S$ stand for $\cos$ and $\sin$, you have: $$\Re(A(z,t))=\Re \int_{\omega_1}^{\omega_2} (\alpha( \omega ) - j \beta( \omega ))\; e^{j (\omega t - kz) } d\omega=\\ \int_{\omega_1}^{\omega_2} \Re ((\alpha( \omega ) - j \beta( \omega )) (\cos (\omega t - kz) +j\sin (\omega t - kz))) \;d\omega=\\ \int_{\omega_1}^{\omega_2} \alpha( \omega ) \cos (\omega t - kz) + \beta( \omega )\sin (\omega t - kz) \;d\omega=B(z,t).\\$$

  • 1
    $\begingroup$ Thank you for your answer. I can't understand the relation between $\alpha(\omega)$, $\beta(\omega)$ and $A(\omega)$. Can you rephrase it? $\endgroup$
    – BowPark
    Jan 26, 2017 at 12:26
  • $\begingroup$ @BowPark I edited my answer, hope it helps. $\endgroup$
    – user130529
    Jan 26, 2017 at 12:45
  • $\begingroup$ Do we in fact add negative frequencies to the wave packet in order to receive a more compact form? $\endgroup$
    – Mercury
    Dec 17, 2021 at 17:30

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