Spectral density of fluctuations (white noise/delta-correlated process)

Question

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 $2\pi$.

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.