Let I be the current flowing across some junction as a result of N charge carriers of charge q. And let $\langle I (t) \rangle$ be its average.

Assume a particle number distribution such that its fluctuation is given by $\langle (\Delta N)^2 \rangle=\langle N\rangle$.

So $\langle I\rangle = q \langle N \rangle$ and by definition of the correlation function $K_I(\tau)=\langle (\Delta I)^2 \rangle$ (where $\tau$ be the time difference $t' - t$) we have

$$ K_f(\tau)= q \langle I\rangle $$

Simplifying this comes from the fact that the charge carriers flow randomly and independently. So we use the following:

Let the spectral density of fluctuations be defined as the Fourier transform of the correlation function $K_f(\tau)$

$$ S_I(\omega) = \frac{1}{2\pi} \int_{- \infty}^{\infty} K_f(\tau) e^{i \omega \tau} d\tau $$

The random and independent nature of the system means that this is a delta-correlated process, where we have $S_f(\omega)= constant = S_f(0)$,

so that via an inverse fourier transform we have

\begin{eqnarray} K_I(\tau) &=& 2\pi S_I(0) \delta(\tau) \\ &=& 2 \pi \bigg(\frac{1}{2\pi} \int_{- \infty}^{\infty} K_f(\tau) e^{i 0 \tau} d\tau \bigg) \delta(\tau) \\ &=& \int_{- \infty}^{\infty} K_f(\tau) d\tau \delta(\tau) \\ &=& K_f(0) \delta(\tau) \\ \end{eqnarray}

I am trying to understand if this next step I take is legitimate:

Could I not rearrange the first line of the above equation to say

\begin{eqnarray} K_I(\tau) &=& 2\pi S_I(0) \delta(\tau) \\ q \langle I\rangle &=& 2\pi S_I(0) \delta(\tau) \\ \frac{q \langle I\rangle}{2\pi \delta(\tau)} &=& S_I(0) \end{eqnarray}

I hope this is more clear now.

My motivation behind the original question

if , for an average quantity $\langle I \rangle$, does

$$ \frac{\langle I \rangle}{\delta(\tau)} = I $$

Would $$S_I(\omega)=S_I(0)= \frac{q \langle I \rangle}{2 \pi \delta(\tau)}$$ or $$S_I(\omega)=S_I(0)= \frac{q \langle I \rangle}{\delta(\tau)}$$

is that I know the answer to be

$$ S_I = q I $$

with no average $\langle I \rangle$ or $2\pi$.

  • 3
    $\begingroup$ usually it is bad enough to multiply with $\delta(\cdot)$ but dividing with it is utter nonsense $\endgroup$
    – hyportnex
    Nov 14, 2020 at 21:41
  • $\begingroup$ Also note that the units are not consistent in the equation $\langle I \rangle / \delta(\tau) = I$, since $\delta(\tau)$ has units of 1/time. I think we need a bit more detail on what you want to calculate. $\endgroup$
    – Andrew
    Nov 14, 2020 at 22:31
  • $\begingroup$ How does $\delta(\tau)$ have units of time? I didn't know a distribution like the Dirac delta could have units @Andrew $\endgroup$
    – Lopey Tall
    Nov 14, 2020 at 23:43
  • 4
    $\begingroup$ $\int {\rm d} \tau \delta(\tau)=1$, and ${\rm d} \tau$ has units of time, so $\delta(\tau)$ must have units of 1/time. You can also see this from the scaling rule $\delta(a x)=1/a\delta(x)$. $\endgroup$
    – Andrew
    Nov 15, 2020 at 6:52
  • 1
    $\begingroup$ When you integrate, the delta function automatically disappears. That's how it works. It is meant to be used in an integral to define the value of some function no matter what the integral (assuming the value is within the bounds of integration). So I don't see how the delta function survived the integration of $K_{f}$. $\endgroup$ Nov 17, 2020 at 15:22

1 Answer 1


I believe that your definition of the correlation function is incorrect, and much confusion follows from there. If $I(t)$ is a random process (i.e. a variable randomly changing in time), then we can define:

  • Average/mean: $\langle I(t)\rangle$
  • Fluctuation (which is also a random process, but with zero mean): $\Delta I(t) = I(t) - \langle I(t)\rangle$
  • Variance : $Var(I(t)) = \langle (\Delta I(t))^2\rangle$
  • Correlation function: $K(t,t_1)=\langle \Delta I(t)\Delta I(t_1)\rangle$

In many situations the random process can be argued to be stationary, i.e. its moments, such as mean and variance, do not depend on time, whereas the correlation function depends only on the difference of times: $$ K(\tau) = \langle \Delta I(t+\tau)\Delta I(t)\rangle $$ Obviously variance is the value of correlation function at equal times or, for a stationary process: $$ Var(I) = K(0) = \langle(\Delta I)^2\rangle, $$ which is what the first equation in the original question should have been.

  • $\begingroup$ Hi, the definition you give are what I use. $\endgroup$
    – Lopey Tall
    Nov 20, 2020 at 13:45
  • $\begingroup$ @LopeyTall You equate the variance with the correlation function: $K_I(\tau)=\langle(\Delta I)^2\rangle$. One is a constant, the other is a function of time. $\endgroup$ Nov 20, 2020 at 13:48

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