# Noise spectral density for waveguide: I don't understand this calculation

In Introduction to Quantum Noise, Measurement and Amplification, on page 64 is computed the power spectral density of noise on a classical waveguide. I am really struggling to understand a step of the calculation.

We consider a classical 1D waveguide. We study the physics by looking at the fluctuations of voltage waves propagating on it. Calling $$c$$ the velocity of the waves, we apply periodic boundary condition, on a period $$L$$ and we can then write down the expression of the left & right movers. We have:

$$V_{\rightarrow}=\sqrt{\frac{1}{2 Lc}}\sum_{k>0} A_k(0)e^{i(kx-\omega_k t)} + A_k^*(0)e^{-i(kx-\omega_k t)}$$

We are interested in the spectral density of the voltage. I remind that it is defined as (because of Wiener-Khintchine theorem we can take the following as its definition here):

$$S_{V_{\rightarrow}V_{\rightarrow}}(t)=\int_{-\infty}^{+\infty} d \omega ~ e^{i \omega t} \langle V_{\rightarrow}(t) V_{\rightarrow}(0) \rangle$$

In the paper, he finds:

$$S_{V_{\rightarrow}V_{\rightarrow}}(t)=\frac{2 \pi}{2 Lc} \sum_{k>0} \langle A_k A_k^* \rangle \delta(\omega-\omega_k) + \langle A_k^* A_k \rangle \delta(\omega+\omega_k)$$

I don't understand how he finds it.

Indeed, for example, how does his two other (non crossed) term disappear ? How does the sum on $$k'$$ disappear. For instance, if I do the calculation, I would have:

$$S_{V_{\rightarrow}V_{\rightarrow}}(t)=\frac{1}{2Lc} \sum_{k,k'} \langle A_k(0) A_{k'}(0) \rangle e^{i(k+k')x} \int_{-\infty}^{+\infty} d\omega e^{i (\omega-\omega_k) t}+\text{other terms}$$ $$S_{V_{\rightarrow}V_{\rightarrow}}(t)=\frac{1}{2Lc} \sum_{k,k'} \langle A_k(0) A_{k'}(0) \rangle e^{i(k+k')x} (2 \pi \delta(\omega-\omega_k))+\text{other terms}$$

As you can see, my "non cross term" does not vanish, and I have the two sum still.

How to make sense of all this ? Are there some implicit assumption that are made ?

• In your equations, did you lose an $\omega_{k'}$? – Daniel Jul 13 '20 at 22:40
• @Daniel I dont think. One of the voltage must be evaluated at time $0$. There is not $\omega_{k'}$ because of that ! – StarBucK Jul 13 '20 at 22:42
• Ah, right. In the paper, it looks like the quantity computed is not quite $S_{V_{\rightarrow}V_{\rightarrow}}$. It is $S_{V V_{\rightarrow}}$. If this is right, one of the sums includeds negative $k$. Together with the fact that $A_k^* = A_{-k}$, this might solve your problem. – Daniel Jul 14 '20 at 16:17
• @Daniel no I don't think. It wouldn't make any sense physically to compute $S_{V V^{\rightarrow}}$. Furthermore he explicitly says that $S^{\rightarrow}_{VV}$ is the voltage spectral density for the right moving waves which is then $S_{V^{\rightarrow} V^{\rightarrow}}$ and has the definition that I wrote. There must have a physical reason why he doesn't have the same expression that I have. – StarBucK Jul 14 '20 at 16:21

In order to use Wiener-Khintchine, we assumed that the process was stationary. This implies that $$\langle A_k(0)A_{k'}(0) \rangle = \langle A_k(t)A_{k'}(t) \rangle$$ - our choice of initial time was arbitrary. But $$\langle A_k(t)A_{k'}(t) \rangle = e^{i(\omega_k + \omega_k')t}\langle A_k(0)A_{k'}(0) \rangle$$ and this equation can be satisfied only if either $$\omega_k' = -\omega_k$$ or $$\langle A_k(0)A_{k'}(0) \rangle = 0$$. All terms except $$\langle A_k(0)A_k^*(0) \rangle$$ and $$\langle A_k^*(0)A_k(0) \rangle$$ thus vanish. You can also get this result by computing the autocorrelation function directly, without using Wiener-Khintchine.