# Relating the variance of the current operator to measurements

(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?

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Is it not just $(\Delta j)^2 = (\Delta n/t)^2 = \frac{\overline{n^2} - \overline{n}^2}{t^2}$? The fluctuations encoded by $j$ are, operationally, the variance you'd measure if you averaged over a large number $M$ of trials of duration one time unit, and if you change the units of time you'll change the relationship between $j$ and its variance. (Let me know if this makes sense and I'll make it into an answer - I didn't want to do that already because I'm not 100% sure I've understood the question correctly.) – Nathaniel Nov 27 '12 at 9:58
@Nathaniel yes I think you are right, $\Delta j$ must be the variance of measurements of duration 1 time unit. Indeed this must be the case since $\Delta j$ is dimensionful; $[\Delta j] = T^{-1}$. However, this does not fully answer my question, since the time duration $t$ is not necessarily the unit of time. If the measurement takes a longer time then the observed fluctuations in $n_i$ must be smaller. The quantity $\Delta j$, however, has a fixed numerical value that depends only on the choice of time unit and the dynamics. What is the relation between $\Delta j$ and $\Delta n (t)$? – Mark Mitchison Nov 27 '12 at 11:32
Is it not $\Delta j$ = $\Delta n(t)/t$? That seems intuitively right and has the right units... – Nathaniel Nov 27 '12 at 11:43
I quite agree that your expression looks like it should be the right answer. But $\Delta j$ is a constant number that depends only on the steady-state density operator $\chi$. Both $\Delta n(t)$ and $t^{-1}$ should be decreasing functions of $t$, which is just a variable parametrising the measurement(s), so I don't see how $\Delta j = \Delta n /t$ can be right. – Mark Mitchison Nov 27 '12 at 11:49
@Nathaniel Possibly the conceptual issue is to do with the two types of averaging (over a time period $t$ or a number of realisations $M$). Intuitively, a single measurement with $t\to \infty$ should give the same result as the average of many measurements $M\to\infty$ over a finite time period. This is what is known as ergodicity, I believe? Perhaps this is an unwarranted assumption in the non-equilibrium case... – Mark Mitchison Nov 27 '12 at 11:56