How noisy are photon detectors? I have a single photon detector and $N$ photons per second arrive at the detector. Then something happens and the number of incoming photons per second changes by the factor of $\alpha$. So now $(1+\alpha) N$ photons per second arrive at the detector.
What are currently the best photon detectors to measure the smallest possible change $\alpha$?
Of course I know that this depends on the noise and also on $N$.
Say I have $N=1000$ photons per second and I want to measure a change of 1%, hence I want to measure $\alpha= 0.01$. Is that possible? If not, is there a limit in principle?
 A: Since your question is purely theoretical I assume you are referring to the intrinsic noise limits. This limit is called shot noise limit. Generally photon detection is complicated by the fact that photons arrive (are generated) in a random fashion. This is because the photon emission/absorption events occur independently and there is no communication between photons. An analogy to photon emission with the same statistics is radioactive decay.
Hence, most sources will generate a photon flow that follows a Poisson distribution (http://en.wikipedia.org/wiki/Poisson_distribution) *. 
If you wait e.g. for 1 sec the detected number of photons is in your case Poisson distributed with $\lambda= 1000$ parameter. Under this conditions this very well corresponds to a normal distribution with mean $\mu = 1000$ and standard deviation of $\sigma = \sqrt{1000}$. 
So the intrinsic 1σ signal to noise ratio is $\sqrt{1000} \approx 31$. This is probably not enough to detect relative changes of 1 % with the reliability you probably would like to have.
If you wait longer the situation improves of course with $\sqrt{T}$. So 10 sec will give you an SNR of around $100$ which now comes close to what you need.
So there is no limit in principle, if you can wait long enough. If the waiting time is fixed there are intrinsic limits which cannot be overcome.
Practically, single photon detectors also generate dark count events, i.e. which do not correspond to incoming photons. For practical experiments this dark noise can (but does not need to) substantially lower your signal to noise ratio (in addition to the more fundamental shot noise). For this you have to consult the detector specifications. Usually one would design the experiment so that the detector is operated above its dark noise limit (i.e. with Photon flow $\lambda > \lambda_{dark}$)
*The above is e.g. not true for squeezed light (where photon events are correlated), which is, however, not relevant to consider for almost all technical cases of light detection.
A: It is certainly possible.  It depends, as others have said, on the detector.  But it also depends on the detection electronics, and the techniques used to do the measurement.   Common sources of noise are shot noise, dark current noise,  statistical fluctuations in the detection mechanism, and  thermal noise in the detection electronics.  Which of these are limiting factors depends on the situation.   Experimenters typically spend a lot of time tracking down and understanding noise sources.
Dark current noise can be made insignificant by cooling the detector.  Thermal noise in the electronics can be reduced by proper choice of amplifiers, and cooling if necessary.  Statistical fluctuations in the detector (e.g. pulse height fluctuations in a photomultiplier tube) can be reduced by proper choice of detector, and perhaps using a photon-counting scheme.  The effects of shot noise can be reduced by using longer integration time.  There are trade-offs to consider.
So the short answer is "yes", but how to do it depends heavily on the particular situation.
A: As a practical matter we generally use these devices in cases where the photon arrivals are random. That is the mean rate may be known, but the actual arrivals are distributed according to a exponential time law rather than being periodic.
In those case, counting statistics usually dominate the uncertainty (in low background detectors shot-noise can generally be made a small effect).
So the question becomes one of how long you can integrate the before and after the change. 

I'm going to use $R$ for the actual rate, and $r$ for the measured rate and use a prime to denote "after the change" while unprimed symbols come before the change.
To put some numbers in it, the expected number of counts in time $t$ is $R t$, and the standard deviation in the number observed is $\sqrt{R t}$. That makes the standard deviation in the measured rate $\Delta r = \sqrt{R/t}$. We have the measurement before the change:
$$ r = R t \pm \sqrt{\frac{R}{t}} \,.$$
Taking $R' = R(1 + \alpha)$ the measurement after the change is
$$ r = R' t \pm \sqrt{\frac{R'}{t'}} \,.$$
Now because we assume that $\alpha \ll 1$ it makes sense to use our time evenly so I set $t' = t$, and form the difference between the measured rates:
$$
\begin{align*}
d &= r' - r\\
&= \left( R' \pm \sqrt{\frac{R'}{t}}\right) -
   \left( R \pm \sqrt{\frac{R}{t}}\right) \\
&= \alpha R \pm \sqrt{\frac{R'}{t} + \frac{R}{t}}\\
&= \alpha R \pm \sqrt{ \frac{R(2+\alpha)} {t} } \,.
\end{align*}
$$
This is only statistically significant when the main terms is several times the error term
$$
\begin{align*}
\alpha R &= \text{(a few)} \times \sqrt{ \frac{R(2+\alpha)} {t} } \\ 
\alpha^2 R^2 &= 10 \times \frac{R(2+\alpha)} {t} \\ 
t &= 10 \left( \frac{2 + \alpha}{\alpha^2} \right)\frac{1}{R} \\
 &\approx \frac{20}{\alpha^2 R}\,.
\end{align*}
$$
In other words, is $\alpha$ is small and $R$ is anything but very large you have a long wait in front of you. Using Andreas' numbers ($R = 1000 \,\mathrm{Bq}$ and $\alpha = 0.01$)--which represent a fairly desirable arrangement--we get $t = 200\,\mathrm{s}$ of counting on either side of the change to get a 3ish sigma measurement of a one percent change in the rate.
This is a specific case of the general rule that measuring small changes is hard.
