1
$\begingroup$

I'm a bit stuck on a problem relating to statistics in photon counting. I'm measuring a spectra with a spectrometer and can set a measurement time and number of times to repeat the measurement in order to get an average. I know that a stream of photons exhibits shot noise and can be modeled by a Poisson distribution, yielding a standard deviation of $\sqrt{N}$.

So if I plot the results of a large number of measurements as a histogram, the standard deviation would be $\sqrt{N}$. If this is true then what advantage does taking the average of a number of photon counting measurements have? The values clearly gets closer to its 'true' average, but the standard deviation always remains $\sqrt{N}$.

In my head I have this idea that taking a larger number of measurements should reduce the standard deviation. Is this something inherently different in 'randomly' occurring events as in a Poisson distribution?

EDIT: Okay I just got a bit confused. I think it's just the standard error of the mean. The square root of the average number of photons would be the standard deviation of the distribution. And then the SE of the mean is equal to that divided by the square root of the amount of measurements.

$s_{mean}=\frac{\sqrt{N_{photons,avg}}}{\sqrt{N_{measurements}}}$

$\endgroup$
1
$\begingroup$

What you say is true. If you are dealing with pure counting statistics and a constant source then the uncertainty in the countrate determined from the average of 10 measurements ought to be the same as the uncertainty derived from a measurement that is 10 times as long.

So what could be gained? Firstly you get to test whether the statistics are Poissonian and that there are no additional sources of uncertainty. Secondly, you get to test (within the sensitivity allowed by your experiment), whether the source really is constant. Of course the latter can be accomplished in a single measurement if events are time-tagged. Thirdly, if something goes wrong with your detector, you get some data, rather than none!

$\endgroup$
9
  • $\begingroup$ Thank you for the reply, although I'm not quite sure yet how to handle my values. Say I average a measurement 10 times and it results in the number 100, can I divide the standard deviation of the distribution $\sqrt{100}$ by the square root of the amount of measurements $\sqrt{10}$? This is the standard error of the mean right? $\endgroup$ – Rubenknex Oct 7 '15 at 19:05
  • $\begingroup$ When I apply $s=\sigma/\sqrt{N_{meas}}$ to your example, I get something like $s=\sqrt{N}/\sqrt{10}$ for an average of 10 measurements, and $s=\sqrt{10N}$ for a single measurement 10 times as long. I think I am still not getting something. $\endgroup$ – Rubenknex Oct 7 '15 at 19:19
  • $\begingroup$ Yes. You reduced the signal to noise ratio (SNR) by $\sqrt(N)$ If you instead measured for ten times longer, you would have gotten an average of 1000, with a std dev of the square root of that. The (SNR) is $\sqrt(1000)$. We can compare this to the SNR obtained by averaging: $\sqrt(100)\sqrt(N) = \sqrt(100)\sqrt(10) = \sqrt(1000)$. Exactly the same. The advantages of averaging over longer collection are pointed out nicely in the answer. $\endgroup$ – garyp Oct 7 '15 at 19:21
  • $\begingroup$ Perhaps what you are missing is that the important quantity is signal-to-noise ratio. $\endgroup$ – garyp Oct 7 '15 at 19:23
  • $\begingroup$ @Rubenknex You are assuming the s.d. would be $\sqrt{100}$? The essence of my answer is that a repeat measure analysis allows you to check that it is. $\endgroup$ – ProfRob Oct 7 '15 at 19:43

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.