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The widely recognized theory for the detected image in microscope (I believe) is calculated by taking the convolution of the object and a point spread function (PSF). Deconvolution tries to reverse this operation to get back the original object. This explanation sounds easy and intuitive but after some digging, I found that I don't understand what deconvolution is doing anymore.

So this PSF is actually made up of two components: detection (PSF_det) and illumination PSF (PSF_ill). If we talk about widefield microscopy, illumination is constant so PSF_ill=1 which means the PSF in the original theory is just PSF_det. This is still ok but a quick question will be why doesn't this operation return the image of the object (image with infinite resolution)?

If we talk about confocal microscopy, PSF_ill is not 1 and we can basically approximate to PSF_det giving PSF=PSF_det^2. So we can see that the convolution with this PSF actually gives higher resolution. My main question with this is what happens if I deconvolve this image? Am I removing the effect of PSF_det? If so, do I end up with PSF=PSF_det (without the square) which has lower resolution then I started with?

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    $\begingroup$ Can you explain what PSF_ill and PSF_det are? They are not generally known terms, and may be specific to the kind of microscopy you are doing. $\endgroup$
    – garyp
    Commented Jun 14, 2023 at 10:58
  • $\begingroup$ @garyp Are you questioning the abbreviations, which literally follow the full text phrase? Or do you mean the actual PSFs used for each? $\endgroup$
    – Kyle Kanos
    Commented Jun 14, 2023 at 21:40
  • $\begingroup$ The PSFs. Why are two needed, and what do they represent. I suspect my issue is ignorance of the application. I also suspect that I'm not the only ignoramous. $\endgroup$
    – garyp
    Commented Jun 15, 2023 at 12:56
  • $\begingroup$ Usually only 1 PSF is used in image formation theory for widefield microscopy since the illumination is constant. However in confocal microscope, the illumination is localized so it is represented by a PSF (which is what almost every single text I read talks about: resolution is not infinite because of diffraction limit). If it helps, we can use the idea used in Image Scanning Microscopy (PRL 104, 198101 (2010)) to think about the PSF. Illumination PSF describes the probability portion of the sample is illuminated while detection PSF is detection probability by detector from a point source. $\endgroup$
    – CoffeeBiscuit
    Commented Jun 15, 2023 at 13:37

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This is still ok but a quick question will be why doesn't this operation return the image of the object (image with infinite resolution)?

  1. The detection process introduces noise. Deconvolution is much more likely to enhance this noise (at least for some spatial frequency bands) than to reduce it. It is particularly likely to enhance the noise for high spatial frequencies where the imaging system has most attenuated the signal.

  2. It's possible that the solution to a deconvolution is not unique. Then the deconvolution process has to choose one particular solution. The usual choice is to choose the minimum norm (if I recall the terminology correctly) solution. But there's no guarantee this was the actual object image you started with.

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  • $\begingroup$ I assumed it was noise but I thought that less than 2 times improvement in resolution is so much off from infinite. I am not sure about the second point. Why is the inverse operation not unique? Do you mean because some points are 0 in the k-space? Or do you mean we can't find unique solution? $\endgroup$
    – CoffeeBiscuit
    Commented Jun 15, 2023 at 1:20
  • $\begingroup$ @CoffeeBiscuit, Sorry I don't know the terminology of confocal microscopy, I'm just talking about deconvolutions in general. But you probably have it right. If the object is convolved with a PSF that doesn't have full coverage of frequency space, then information will be lost in forming the image. No deconvolution can recover it. Even if the PSF has full coverage, if it is very weak in some frequency band (like high frequencies), then practically noise will dominate those frequencies in the image, and again we won't be able to recover those frequencies by deconvolution. ... $\endgroup$
    – The Photon
    Commented Jun 15, 2023 at 15:46
  • $\begingroup$ And if we lose the high-frequency information, then that is very clearly going to limit the resolution of the recovered object. $\endgroup$
    – The Photon
    Commented Jun 15, 2023 at 15:47
  • $\begingroup$ I see. I guess you are right, this makes sense $\endgroup$
    – CoffeeBiscuit
    Commented Jun 16, 2023 at 1:20

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