1
$\begingroup$

I am trying to understand the physics behind photon shot noise and have two sources. The first is from "Photon Transfer" by James Janesick which has he following bit on photon shot noise.

Signal shot noise is fundamentally connected to the way photons spatially arrive on a detector. For example, Fig. 3.1 shows a Monte Carlo simulation where $200$ photons are randomly interacting with a $20\times 20$ pixel region. As can be seen, the number of photon interactions varies from zero to four interactions per pixel. The standard deviation (or rms) for the number of interactions per pixel is called photon shot noise. Photon shot noise—a spatially and temporally random phenomenon described by Bose-Einstein statistics—is expressed by \begin{equation} \sigma_{\text{SHOT}}(P_I)^2=P_I\frac{e^{hc/\lambda kT}}{e^{hc/\lambda kT}-1} \end{equation} where $\sigma_{\text{SHOT}}(P_I)^2$ is the interacting photon shot noise variance, $h$ is Planck’s constant, $\lambda$ is the photon wavelength (cm), k is Boltzmann’s constant, $c$ is the speed of light, and $T$ is absolute temperature $(K)$.

Figure 3.2 plots Eq. (3.1) as a function of wavelength $(\mu \mathrm{m})$ and the temperature of the semiconductor. For wavelengths greater than $10\, \mu \mathrm{m}$, photons couple with phonons (i.e., lattice vibrations in a solid) that increase the shot noise. As the operating temperature is reduced, the semiconductor produces less coupling action and variance as seen in the plot.

It would seem that the temperature, $T$, is that of the semiconductor material in the detector. However, in my other source "Optical Radiation Detectors" by Dereniak and Crowe they have the following discussion on shot noise.

Note that Eq. (1.60) can be rewritten \begin{equation} \sigma^2=\bar n\left[\frac{e^{h\nu/kT}}{e^{h\nu/kT}-1}\right] \end{equation} which is the variance for a Poisson distribution $(\sigma^2=\bar n)$ multiplied by the Boson factor. For the majority of optical applications, we are interested in wavelengths from $0.3$ to $30\,\mu$m and $T<500\, K$. These values imply that the energy per photon is much greater than the thermal energy $(h\nu\gg kT)$, and, therefore, \begin{equation} \frac{e^{h\nu/kT}}{e^{h\nu/kT}-1}\approx 1 \end{equation} and we can use the Poisson result for photon noise $(\sigma^2=\bar n)$. When very-high-temperature sources are to be detected, or when working at long wavelengths approaching a millimeter or more, the correction factor should be remembered and applied.

Here is seems to be that $T$ refers to the temperature of the photon source. I am confused by this apparent discrepancy. For my application, I am concerned with understanding the relationship between the variance and mean of a photon stream as observed by a detector. That said, does this mean for my application that the departure of photon shot noise from the Poisson relation, that is $\operatorname{var}(P_I)=E(P_I)$, is dictated by the temperature of the detector? or the source? or both? I am not a physicist but have a background in statistics.

$\endgroup$
  • 2
    $\begingroup$ Hi, welcome to Physics SE! Can you convert the picture of the text into typed-out, formatted text (v1)? It makes the content index-able by search engines, and shows up better on different devices' displays. For formulae, try MathJax instead. $\endgroup$ – user191954 Aug 13 '18 at 15:03
  • $\begingroup$ Sure. It will just be a moment. Thank you for the welcoming. $\endgroup$ – Aaron Hendrickson Aug 13 '18 at 15:04
  • 2
    $\begingroup$ Not an expert in optical radiation detectors and not familiar with the derivation of the shot noise $\sigma^2$, but I'm thinking that the first reference source doesn't make complete sense to me. The B-E statistics of the photons should be determined by the temperature of the photon source and don't see what photon-phonon coupling has to do with it. In fact, there aren't even any parameters in the first equation which describe the photon-phonon coupling or the phonon properties of the detector material. The section from your second reference source makes more sense to me. $\endgroup$ – Samuel Weir Aug 13 '18 at 17:37
  • $\begingroup$ @AaronHendrickson - Well, the first source should have at least provided a reference for where it got the expression for the shot noise variance, but there nothing. The equation is just suddenly popped out with no justification, derivation, or explanation as to where it came from. The second source, on the other hand, derives the variance from an earlier equation (Eqn 1.60) in the same source. So I think that the first source didn't properly introduce its expression for the shot noise variance. It should have had a reference. You shouldn't have had to go searching on your own for a 2nd source. $\endgroup$ – Samuel Weir Aug 13 '18 at 18:46
  • 1
    $\begingroup$ I agree with @SamuelWeir, it makes no sense for the occupancy of photons to depend on the temperature of the detector, and it makes perfect sense for it to depend on the temperature of the emitter. See black body radiation. $\endgroup$ – Bobak Hashemi Aug 14 '18 at 0:20
1
$\begingroup$

Aaron, I spent some time trying to understand CCDs and shot noise. My belief is that shot noise is due to the light source inherent atomic emission process. It is a quantum mechanical probabilistic process and the noise is proportional to the square root of the average numbers of photons measured. For example if on average a pixel receives 100 photons you would tend to measure 90 to 110 photons (1 sigma) on the individual measurements. ( Or if you received 10,000 photons the individual measurements would range from 9,900 to 10,100). There is no such thing as a light source that produces a steady stream of light (even a laser has shot noise) as far as know. The atom emits probabilistically. Hope this helps.

$\endgroup$

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.