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The Point Spread Function (PSF) of an imaging system is its impulse response. Therefore $$ I(x,y,z) = \left[O \star PSF\right](x,y,z) $$ where $I$ is the image, $O$ is the object, and $\star$ is the convolution product.

Physically speaking, it is also the power per unit area emitted by a point-like emitter - i.e. $O(x,y,z) = \delta(x,y,z)$. The integral over the image (i.e. the surface of the camera detector) will therefore provide the photon energy collected during the exposure time (assuming no background and a calibrated sensor).

However, if we imagine acquiring a PSF instantaneously (i.e. with a vanishing exposure time) its units are not those of energy, but those of power. Time, however, just describes the propagation of light. Therefore, integrating over time or over space should give the same result, except for a multiplying constant $c$, which is the speed of light. Following this reasoning, I conclude that the 3D space integral of a PSF $$ \int_{\mathbb{R}^3} PSF(x,y,z) dx dy dz$$ is proportional to the fraction of energy emitted from a point-like emitter collected by the imaging system. Therefore, $PSF(x,y,z)$ is proportional to a volumetric density of energy.

Is my interpretation correct?

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  • $\begingroup$ Maybe not related, but why I(x,y,z)? Isn't the image 2D? $\endgroup$ May 20, 2021 at 16:49
  • $\begingroup$ @CharlesTucker3 Not necessarily. Images can be three-dimensional and are just a stack of 2D images. $\endgroup$ May 20, 2021 at 16:56

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I don’t think so. Even if you acquire the PSF very quickly, it still takes a non-zero amount of time for photons to be collected by the detector. It is better to think of the PSF as the probability density function of photons being detected at different coordinates in 3D space over an arbitrary time period.

Also, the emission of a point emitter may not be constant over time. Consider fluorescence life time imaging where emission follows a biexponential distribution.

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