Why there can not be a monochromatic pulse? My physics professor told us that we can't generate a monochromatic light pulse and I was wondering what are the physical limitations causing this.


5 Answers 5


The fundamental reason for this is that a truly monochromatic waveform $$ f(t) = f_0e^{-i\omega t} $$ is active for all real times $t$ ─ it doesn't start, and it doesn't stop ─, and this means that you need an infinite time to produce one, and you need an infinite time to detect it. Since the energy density of the wave is constant, the infinite duration also means that you require infinite energy to set this up.

Now, why do we say that you need infinite time to have a truly monochromatic situation? Let's focus first on the detection side, and suppose that you only have a finite time $T$ to measure the waveform, which is centered at a frequency $\omega_0$ such that $\omega_0 T\gg1$, i.e. $T$ fits many cycles of the central frequency. Now, the real question is: can you distinguish between the frequency $\omega=\omega_0$ and some other frequency $\omega=\omega_0+\delta\omega$ which is close to, but not quite, the central frequency $\omega_0$ which you think you have?

Putting feet to the ground, and supposing that the signals start off in sync, the question ultimately asks how well you can distinguish between $e^{i\omega_0T}$ and $$ e^{i\omega T} = e^{i\omega_0T} e^{i\delta\omega \,T}, $$ where the wave has advanced by a phase $\delta\omega\,T$ over the observation window. Now, here's the problem: what happens if $\delta\omega$ is much smaller than $2\pi/T$? In this case, the two waves at $\omega=\omega_0$ and $\omega=\omega_0+\delta\omega$ will have hardly drifted out of step even over your long observation window, and you will find it hard to distinguish between the two.

Note, moreover, that if you could expand your observation window to a longer time $T_\mathrm{longer} = 2\pi/\delta\omega\gg T$, then your observation window would include times where waves at $\omega=\omega_0$ and $\omega=\omega_0+\delta\omega$ would be $\pi$ out of step, and you would be able to distinguish between them. However, so long as your observation window $T_\mathrm{longer}$ is finite, there will always be detunings $\widetilde{\delta\omega}\ll 2\pi/T_\mathrm{longer}$ that are too close for you to be able to resolve between $\omega=\omega_0$ and $\omega=\omega_0+\widetilde{\delta\omega}$.

It's also worth talking about what happens at the edges of the observation window, as well as the production of the wave. Does your wave have a sharp cutoff, going from a finite amplitude to zero instantly? Then at the boundary it is hardly monochromatic. Instead, you might want it to go smoothly from flat to zero over some transition period $\Delta T$, but then that transition period, during which the amplitude is changing, is going to make it harder for you to distinguish between two waves that are only a small fraction of a radian apart in phase.

So, what does this mean? It tells you that a true monochromatic pulse is impossible to realize in real life, because by "true monochromatic pulse" what we mean is the mathematically idealized model that's been on since forever and will remain turned on until eternity. What you can make is waves that are closer to monochromatic than your experiment can resolve, in which case you can just use the monochromatic approximation without worrying ─ but that doesn't make the wave truly monochromatic.

Finally, it's also important to note that the fact that monochromatic waves are unphysical does not make them less useful. Normally, when we consider monochromatic waves, we are considering the dynamics of some oscillatory quantity $u(t)$ which responds to a linear system. In this case, it is often overwhelmingly simpler to view $u(t)$ as a superposition of plane waves $e^{-i\omega t}$ with some weight $\tilde u(\omega)$, i.e. to deconstruct $u(t)$ as its Fourier transform: $$ u(t) = \int_{-\infty}^\infty \tilde u(\omega) e^{-i\omega t}\mathrm d\omega. $$ If the dynamics is linear, then we can just worry independently about how each monochromatic component will react to the dynamics, without needing to pay attention to the fact that they are unphysical, and then put it together later when we Fourier transform back to the time domain. Whenever we deal with the physics of monochromatic waves (as in, say, the Helmholtz equation, or phasor analysis) that's always the underlying mindset.


This follows from classical Fourier analysis. The frequency spread and time duration of a pulse are related by $$ \Delta \omega \Delta t \approx 2 \pi $$ so to make a truly monochromatic pulse where $\Delta \omega$ is basically $0$ implies this pulse is infinite in duration. Thus, any pulse with a finite duration cannot be truly monochromatic.

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    $\begingroup$ i have this formula in my course but no proof or derivation , am curious to the similarity with the Heisenberg uncertainty principle is there a link ? $\endgroup$
    – Mac Sat
    Commented Feb 12, 2017 at 18:48
  • $\begingroup$ @Mac Sat: Yes and no. The Heisenberg uncertainty relations are quantum in nature, and depend on non-commutativity of operators. There is nothing quantum about the Fourier relations, which are entirely classical. The version $\Delta E\Delta t\approx h$ is actually tricky to handle because time is not an observable in quantum mechanics, so the usual interpretation of this inequality as quantum must be done rather carefully. The relations certainly look the same (much to the relief of Heisenberg, apparently) but are not strictly linked. $\endgroup$ Commented Feb 12, 2017 at 18:55
  • $\begingroup$ Well they are linked. Actually you can prove momentum-position uncertainty relation from the fact that momentum-space wavefunction is a Fourier image of the position-space one. As for time-energy — yes, this one is a bit more tricky. $\endgroup$
    – Ruslan
    Commented May 18, 2017 at 20:46

To create pulses you start with an pure monochromatic wave and add a pulse shape. If you now apply a Fourier transform to this pulse shaped wave, you will see that your delta spike got broader. So the Fourier transform is your link between the space and time domain. Basically the sharper you want your pulses to be, the more high frequency components you need to archive the desired pulse form.

Here is a visualization of the Fourier transform of a pulsed signal. You can think of the low frequency blue line as your starting monochromatic frequency and the higher order ones are those you need to get your pulse shape

enter image description here


As a real example google this paper (Experimental realization of Wheeler’s delayed-choice GedankenExperiment, V. Jacques1, E Wu1,2, F. Grosshans1, F. Treussart1, P. Grangier3, A. Aspect3, and J.-F. Roch3") where they use a small diamond crystal to generate single photons of light.

The pulse rate is set at 4Mhz but the photons are well separated. The color drifts a few nanometers but of course each pulse has its own color.



An electromagnetic radiation is monochromatic if all constituents of this radiation will have the same wavelength respectively the same frequency respectively the same energy content. An examples for EM radiation is a light bulb. This light is hardly monochromatic because the electrons which emit the photons are excited on different levels.

Somehow an exception are sodium-vapor lamps, for which Wikipedia says:

These lamps produce a virtually monochromatic light averaging a 589.3 nm wavelength (actually two dominant spectral lines very close together at 589.0 and 589.6 nm).

What about a laser? Here a picture from Wikipedia about Laser: enter image description here

Spectrum of a helium neon laser illustrating its very high spectral purity (limited by the measuring apparatus). The 0.002 nm bandwidth of the lasing medium is well over 10,000 times narrower than the spectral width of a light-emitting diode.

Taking in account that the engineering possibilities to create pure crystals and constant temperatures and constant currents and so on are limited then an bandwidth of 2 picometre are nearly monochromatic.

An electromagnetic pulse is an electromagnetic radiation of a certain duration. To say it with Wikipedia:

A rapid, transient change in the amplitude of a signal from a baseline value to a higher or lower value, followed by a rapid return to the baseline value.

So switching on and off a light bulb we create a pulse. But if this light is not monochromatic, the pulse then more is not monochromatic. The emitting photons electrons will emit at different temperatures and by this emit photons of more different energy values than in the continuous mode.

The same dilemma for a laser pulse. Switching on and off the power source the lasers crystal heats up and cools down and the bandwith will be broader.


Which possibilties do we have to create a monochromatic pulse? @PhysicsDave mentiones in his answer a

.. small diamond crystal to generate single photons of light.

  • Suppose that the pulses for the stimulation of the photon emissions are following one by one at a constant frequency (this is important because each pulse increases the thermal energy of the crystal) and suppose that the heat dissipation is equilibration with the heating from the pulses. So suppose we are able to stabilize the thermal energy of the crystal at a constant level.
  • Suppose the phonon oscillations of the crystal - stimulated from the external pulses - are a whole multiplicity of this pulses.
  • Suppose futhermore that the stimulating pulses are hitting always the same atoms respectively the electrons on the same places in the crystal structure.

Even if all supposes would be solvable, the last one is not solvable under room temperature conditions. The electrons are dislocated and this dislocation has a random distribution. Never under such conditions theoretical one will get monochromatic photons. Would it be possible to create photons which are "more" monochromatic as the detection methods I don't know.

Cooling a material nearly to zero and using material, in which the atoms have integer spins (like helium-4) this atoms wil behave as an particle and a excitision should lead to a photons emission at one wavelength. Unfortunately - if i remember right - only some part of the helium-4 atoms are in superposition, so even for superfluid helium-4 one will not get only one wavelength.

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    $\begingroup$ "Nearly" and "virtually" monochromatic are not the same as monochromatic. $\endgroup$ Commented May 18, 2017 at 20:59

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