(EDIT: Thanks to Nathaniel's comments, I have altered the question to reflect the bits that I am still confused about.)
This is a general conceptual question, but for definiteness' sake, imagine a quantum dot sandwiched between two macroscopic metal leads with different chemical potentials. The chemical potential difference drives a current of electrons that flow through the quantum dot from one lead to another. Conservation of charge leads to the continuity equation for the charge density operator $\hat{\rho}$:
$$ \frac{ \mathrm{d} \hat{\rho}}{\mathrm{d}t} = \mathrm{i}[\hat{H},\hat{\rho}] = – \nabla \hat{j}. $$
Given a Hamiltonian $\hat{H}$, one can in principle use the above formula to calculate the form of the current operator $\hat{j}$, whose expectation value gives the number of electrons that pass from one reservoir to the other per unit time. The expectation value of the current is independent of time in the steady state. The operational procedure to measure this expectation value is clear: sit there and count the number of electrons $n_i$ that pass through the quantum dot in time $t$, then repeat this procedure $M$ times, giving
$$ \langle \hat{j} \rangle \approx \frac{1}{M}\sum\limits_i^M \frac{n_i}{t}. $$
The angle brackets on the left mean quantum mechanical average: $\langle \hat{j} \rangle = \mathrm{Tr}(\hat{\chi} \hat{j})$, where $\hat{\chi}$ is the density operator describing the quantum dot. (I am assuming that conceptual issues relating to quantum measurement are unrelated to this problem — although please tell me if I'm wrong — because a similar question can be posed for a classical stochastic system.)
Now, each measurement $n_i$ will not not be exactly $\langle \hat{j} \rangle t$ due to fluctuations of the current. One can write down the variance of the current
$$(\Delta j)^2 = \langle \hat{j}^2\rangle – \langle \hat{j} \rangle^2. $$
(As Nathaniel pointed out, the calculated variance in the current depends on the choice of time units.) However, the quantity you actually measure is the following:
$$ (\Delta n)^2 = \overline{n^2} – \overline{n}^2, $$
where the overline means the average over the $M$ realisations of a measurement of $n_i$ electrons hopping between the reservoirs in time $t$, i.e. $\overline{n} = \frac{1}{M}\sum_i n_i$.
My confusion relates to the fact that the quantity $\Delta n(t)$ must depend on the measuring time $t$. You can see this easily by considering the limit $t\to\infty$: if you watch and wait for long enough then the fluctuations will average to zero and every measurement $n_i$ that you make will be exactly the expected value. On the other hand, (I think) $\Delta j$ is the expected RMS fluctuation over a single unit of time, i.e. $\Delta j = \Delta n(1)$. Is there a simple relationship between $\Delta j$ and $\Delta n(t)$ measured over arbitrary times?
Best Answer
Current fluctuations are notoriously difficult to calculate and work with. There is no simple relation between the moments of the current and corresponding density. Correlations, autocorrelations, etc. spoil any chance of a simple relation in general.
There is a useful method called full counting statistics (for a review see "Nonequilibrium fluctuations, fluctuation theorems, and counting statistics in quantum systems" by Esposito, Harbola and Mukamel, Rev. Mod. Phys. 81, 1665–1702 (2009), arXiv:0811.3717) which helps us calculate the current distribution.
Sorry I cannot give a better answer, but this is a very broad field of theoretical and experimental research in out-of-equilibrium quantum statistical physics.
Here is a paper which measures the full counting statistics for a quantum dot, which you mentioned: