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If we have a thermal noise generated by Brownian stochastic force $\xi (t)$, it has zero mean value. And its correlation function at temperature T is : \begin{equation} \langle\xi(t) \xi(t^{\prime})\rangle=\frac{\gamma_m}{\omega_m}\int\frac{d\omega}{2\pi}e^{-i\omega(t-t^{\prime})}\omega\left[\coth\left(\frac{\hbar \omega}{2K_BT}\right )+1\right] \end{equation} And it said that the thermal noise spectrum is: \begin{equation} S_T(\omega)=(\gamma_m/\omega_m)\omega \coth(\hbar \omega/ 2k_BT) \end{equation}

So how we derive the noise spectrum?

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The noise spectrum is the Fourier transform of the correlation function, as long as the correlation function is invariant under global time translation. This is nice, because the Fourier transform of the correlation function you have there produces a delta function in frequency, which then makes the frequency integral totally trivial.

\begin{align} S_T(\Omega) &\equiv \int \langle \xi(0)\xi(t) \rangle e^{i \Omega t} dt\\ &= \frac{\gamma_m}{\omega_m} \int \int \frac{d\omega}{2\pi} \, dt \, e^{- i \omega t} e^{i \Omega t} \omega \left[ \coth\left(\frac{\hbar \omega}{2 k_b T} \right) + 1 \right] \\ &= \frac{\gamma_m}{\omega_m} \int \frac{d \omega}{2\pi} (2\pi) \delta(\Omega - \omega) \, \omega \left[ \coth \left( \frac{\hbar \omega}{2 k_b T} \right) + 1 \right] \\ &= \frac{\gamma_m}{\omega_m} \Omega \left[ \coth \left( \frac{\hbar \Omega}{2 k_b T} \right) + 1 \right] \end{align}

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  • $\begingroup$ I seem to have an extra term in the square brackets which OP does not have in their formula. I'll have to go look at my notes to see whether OP's or my formula has an error. In any case, this is the general idea. $\endgroup$ – DanielSank May 20 '16 at 16:26
  • $\begingroup$ It's also possible that I goofed a minus sign in the exponential, but again that doesn't change the basic form of the equation. $\endgroup$ – DanielSank May 20 '16 at 16:30
  • $\begingroup$ @Mr.an 1) Did you figure out where either of us missed the +1 term? 2) Double check the signs in my equation. I don't remember which sign convention is used in the Fourier transform between the correlation function and the noise spectral density. 3) Don't forget to mark an answer as accepted once there is one which does in fact answer your question. $\endgroup$ – DanielSank May 20 '16 at 20:38
  • $\begingroup$ Hi, 1. About the +1 term, at the moment I am not sure. Maybe there is some kind of assumption, the process is Markovian or something like that. 2. The choice of sign for the Fourier transform is correct. $\endgroup$ – Mr. an May 20 '16 at 21:58
  • $\begingroup$ @Mr.an I have a strange feeling that I've encountered this mysterious +1 before. I think we just missed something simple. I need to check my notes though. $\endgroup$ – DanielSank May 21 '16 at 0:45

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