I have a transmission spectrum of a material which has been fit to a Lorentzian.

According to Wikipedia here and here, FWHM is the spectral width which is wavelength interval over which the magnitude of all spectral components is equal to or greater than a specified fraction of the magnitude of the component having the maximum value.

What I understood from that is the FWHM gives an idea of the frequency components which are greater than or equal to the resonant frequency. I am not sure what 'specified fraction' is.

I am also confused of whether the FWHM tells me something something about the material or the source?


FWHM is Full Width Half Maximum. Take the maximum of your transmission peak, identify the points either side of the peak where the transmission is half the maximum, and the full width is the distance along the wavelength axis between those two points.

It's just a way of measuring the width of a peak. The physical meaning will depend on the system tou're studying.


The Full Width Half Maximum (as defined by John Rennie's answer ) can have the following physical meanings:

  1. For a Lorentzian lineshape spectrum, it is proportional to the square magnitude of the strength of the coupling between electromagnetic field and the atomic transition begetting the spectrum in the first place;
  2. For a Lorentzian lineshape spectrum, it is the reciprocal of the $1/e$ lifetime $\tau$ of the atomic transition, which, in the Lorentzian lineshape case, has the unique memoryless lifetime probability density function, namely $p(t) = \exp(-t/\tau)/\tau$;
  3. For spectrums dominated by Doppler broadening, say in thermalized gas molecules the FWHM is proportional to both the square root of temperature, but it is also proportional to the centre frequency.

For more understanding of the points (1) and (2), see the section "The Shape of the Spectrum Without a Cavity" in my answer to "Why do lasers require mirror at the ends?" and the model of the excited atom coupled to the ground state electromagnetic field there, as well as the alternative and equivalent Wigner-Weisskopf theory - this is referenced in my answer too although by far the best exposition I know of (Weisskopf's / Wigner's own) is in German, there is no direct English language translation and the main English exposition I know of is:

Section 6.3, M. O. Scully and M. S. Zubairy, "Quantum Optics"

which is not the grandest piece of technical writing I know. Basically it works like this: the atom is coupled roughly equally to all modes of the electromagnetic field (at least, the coupling is roughly flat near the transition frequency), so it tries to couple equally to all. However, only those modes that are tuned to the transition can be coupled into successfully; for the others, the beating between the EM field mode and the atomic transition thwarts the radiation into that mode by destructive quantum interference. The stronger the coupling constant near the transition frequency, however, the more this detuning and destructive interference effect can be overcome and thus the broader the spectral line, which, in this theoretical model, is indeed exactly Lorentzian:

$$H(\omega) = \frac{1}{(\omega - \omega_0)^2 \tau^2 + 1}$$

where $2/\tau$ is the angular frequency FWHM,

$$\tau = \frac{1}{2\,\pi\,|\kappa|^2}$$

is the transition lifetime, $\kappa$ is the (complex) coupling co-efficient between each electromagnetic field mode and the transition and $\omega_0$ is the transition frequency. The probability amplitude to find the atom still excited a time $t$ after observing that it is excited is:

$$\psi(t) = \left\{\begin{array}{ll}0;&t<0\\\frac{1}{\sqrt{\tau}} \exp \left(-i\,\omega_0\,t - \frac{t}{2\,\tau}\right); & t\geq0\end{array}\right.$$

A common reason for a nonthermalised line to deviate from Lorentzian lineshape is that it is owing to a sequence of transitions rather than a lone transition. In that case, the spontaneously radiating atom is not memoryless. See my answer to "Are there old aged particles?".

For thermalized gas transitions, the spectrum will be broadened in a way related to the Maxwell-Boltzmann velocity distribution. See the Wikipedia page for "Doppler Broadening".

  • $\begingroup$ Damn, another vastly better answer than mine :-) $\endgroup$ – John Rennie Oct 10 '13 at 9:43
  • $\begingroup$ @JohnRennie Thanks for the compliment. I don't suppose you know the spectrum arising from Doppler shift in thermalized gas, do you? - I don't, so you could put this in your answer! $\endgroup$ – Selene Routley Oct 10 '13 at 9:57
  • $\begingroup$ en.wikipedia.org/wiki/Doppler_broadening possibly? $\endgroup$ – John Rennie Oct 10 '13 at 10:05
  • $\begingroup$ @JohnRennie You still got three upvotes, though! $\endgroup$ – Selene Routley Oct 11 '13 at 4:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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