Derivation of the thermal noise spectrum The thermal noise spectrum is given by:
$$\mathcal{S}(f) = \frac{\hbar f}{2(e^{\frac{\hbar f}{kT}} - 1)}$$
This equation seems really similar to the Dirac-Fermi distribution but where does it come from?
 A: This is what I've got so far.
The thermal noise is generated by photons, which follow the Bose-Einstein distribution:
$$N_i = \frac{2}{e^{\epsilon_n/(kT)} - 1}$$
Substituting $E = \omega \hbar$ in the above equation and using that the expected energy is the individual energy of a photon multiplied by the number of photons in that energy level gives:
$$\left< E_i \right> = \frac{2\omega \hbar}{e^{\omega \hbar/(kT)} - 1}$$
Consider a 1D case where the photon is described by the usual wavefunction:
$$\Psi_n(x, t) = A_ne^{j(k_n x - \omega_n t)}$$
In a conductor of length $L$ the boundary conditions are:
$$\Psi_n(0, t) = \Psi_n(L,t) \therefore k_n L=2\pi n$$
The wave velocity in the conductor is:
$$v = \frac{dx}{dt} = \frac{dx}{d \Psi_n} \frac{d\Psi_n}{dt} = \frac{\omega_n}{k_n}$$
The density of states is:
$$\frac{1}{L}dn = \frac{1}{L}\frac{dn}{dk_n}\frac{dk_n}{d\omega_n}d\omega = \frac{1}{L}\frac{L}{2\pi}\frac{1}{v}d\omega = \frac{d\omega}{2\pi v}$$
The energy density can now be calculated by multiplying the density of states by the energy of the individual state:
$$U=\frac{d\omega}{2\pi v} \frac{2\omega \hbar}{e^{\omega \hbar/(kT)} - 1}$$
The energy flow is:
$$P(\omega) = vU=\frac{\omega \hbar}{\pi(e^{\omega \hbar/(kT)} - 1)}d\omega$$
And, finally, changing to frequency using $\omega = 2\pi f$ and $d\omega = 2\pi df$:
$$P(f) = \frac{2hf}{(e^{hf/(kT)} - 1)}df$$
The above quantity needs to be halved because it includes both energy flowing out and in the conductor:
$$P(f) = \frac{hf}{e^{hf/(kT)} - 1}df$$
Thoughts?
A: This is one half of the average energy of harmonic oscillator with natural frequency $f$ according to quantum theory when in contact with reservoir with temperature $T$, minus constant $hf/2$.
EDIT: the formula is close the Planck spectrum of thermal fluctuations of EM radiation, and since the latter is everywhere, it may play role in explanation of why electric or other noise has similar spectrum.
