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I have two questions regarding the lineshape of peaks in spectra obtained with detectors (such as germanium detectors) in spectroscopy.

  • What we often read is that the detector's response lineshape for counts is Gaussian (see Wikipedia Gamma spectroscopy: "The peak shape is usually a Gaussian distribution."). I am unsure why, is it from the Central Limit Theorem applied to the sum of the individual distribution functions (Lorentzians) of single photons (events, or counts) energies?

  • But we read as well that the final lineshape we obtain is a convolution of the different effects: I don't understand where this convolution comes from. If the Gaussian came from the CLT then no matter the underlying individual distribution (in this case Lorentzian), as long as they're identical and independent, the final result has to be Gaussian. Why do we have a convolution here? The convolution of Gaussian (detector) and Lorentzian (natural distribution) then gives a Voigt profile.
    See Wikipedia article Spectral line shape: "The observed line shape is a convolution of the intrinsic line shape with the instrument transfer function."

We end up with people sometimes using a Gaussian, sometimes a Voigt profile, but it is not clear to me why and how we get either.

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  • $\begingroup$ For the convolution part of your question I would recommend looking at it like this: Lets say you want to extract the true physical spectra at a specific wavelength $\lambda_{0}$ using an instrument. If your spectra is denoted by $s(\lambda)$ you want $s(\lambda_{0})$. How about this: $s(\lambda_{0})=\int_{\lambda}s(\lambda')\delta(\lambda'-\lambda_{0})d\lambda'$. Looks familiar? $\endgroup$
    – Newbie
    Commented Jan 12, 2022 at 20:08
  • $\begingroup$ You will be able to have a Dirac $\delta$ as the instrument function only if you have a perfect spectrograph. However, for a general instrument function $g(\lambda)$ you'll end up with $s(\lambda_{0})=\int_{\lambda}s(\lambda')g(\lambda'-\lambda_{0})d\lambda'$. Hope you can see the convolution better now. $\endgroup$
    – Newbie
    Commented Jan 12, 2022 at 20:28
  • $\begingroup$ Many real detectors have some probabilistic factors in things like (1) actual charge produced by the incident particle/photon, and (2) getting that charge out of the device to the preamp. $\endgroup$
    – Jon Custer
    Commented Jan 12, 2022 at 21:07
  • $\begingroup$ Thank you very much for your answers @Newbie. Regarding the convolution question, I understand it now. The only question that remains is how we get a Gaussian distribution from the detector? If there is explanation as to why the detector response is commonly said to be Gaussian and not any other shape, I would be curious to read it. Thanks a lot! $\endgroup$
    – Voidt
    Commented Jan 15, 2022 at 10:59
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    $\begingroup$ same wiki article on Gamma spectroscopy has the line "It takes the transient voltage signal and reshapes it into a Gaussian or trapezoidal shape"! $\endgroup$
    – Andrew Steane
    Commented Jan 18, 2022 at 16:53

1 Answer 1

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The shape of spectral lines is indeed normally determined by two factors

  1. the natural line profile associated with the finite decay time of the transition producing the radiation; since the decay is exponential in time, this translates into the Lorentzian line profile via the Fourier theorem

  2. the Doppler shift of these Lorentz profiles due to the velocity of the emitting atom with regard to observer; for a thermal gas the corresponding velocity distribution is a Maxwell distribution i.e. a Gaussian distribution.

The combination (convolution) of these two effects yields then the Voigt function, which is approximately Gaussian in the central region and Lorentzian in the line wings.

The effect of the detector on the line shape can normally be neglected, as you can make the sampling time of the detector in practice sufficiently long so that the instrumental frequency uncertainty becomes negligible.

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  • $\begingroup$ The “normally” is used very elegantly! $\endgroup$
    – Newbie
    Commented Jan 18, 2022 at 19:52
  • $\begingroup$ Thank you very much for your comment @Thomas. The data I am working with comes from a fixed solid target and therefore does not have any Doppler shift. However the model advised to fit transitions in the literature is quite often a Voigt profile due to the convolution with the detector's response (again, assumed to be Gaussian). In my particular case, my goal is to extract the detector's response function properties and see if we can find any patterns with respect to energy (such as sigma as a function of energy). I assume that if we could remove this response/uncertainty, we would! $\endgroup$
    – Voidt
    Commented Jan 19, 2022 at 15:36
  • $\begingroup$ @Voigt The atoms in a solid will also exhibit thermal motion i.e. Doppler broadening. As for the response function of the detector, you should normally know this in order to be able interpret the measured spectrum. $\endgroup$
    – Thomas
    Commented Jan 19, 2022 at 21:49

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