I was studying Linear Response Theory from 'A modern course in statistical physics' by Reichl, and some doubts came up.

The response function is defined as

$$<\alpha(t)>_{F} = \int_{-\infty}^{+\infty}dt'\bar{K}(t-t')\cdot F(t') = \int_{-\infty}^{+\infty}d\tau\bar{K}(\tau)\cdot F(t-\tau)$$

where $\bar{K}(t-t')$ is real and is called the response matrix.

Since this equation is linear in the force and using

$$<\alpha(t)>_{F} = \frac{1}{2\pi} \int_{-\infty}^{+\infty}d\omega<\alpha(\omega)>_{F}e^{-i\omega t}$$

we should arrive to this expression

$$<\alpha(\omega)>_{F} = \bar{\chi}(\omega) \cdot \tilde{F}(\omega) $$


$$\bar{\chi}(\omega) = \int_{-\infty}^{+\infty}\bar{K}(t)e^{i\omega t}dt$$

However, I cannot derive this result.

My attempt

I tried pluging in the definition of Fourier tranform of each variable in the first equation, and using

$$\delta (t) = \frac{1}{2\pi} \int_{-\infty}^{+\infty} e^{-i\omega t}dt$$

I obtained

$$\int_{-\infty}^{+\infty} d\tau \int_{-\infty}^{+\infty} \frac{1}{2\pi} \bar{K}(\omega) \cdot F(\omega) e^{-i\omega \tau} d\omega$$

which is a bit far from the desired result.

  • 2
    $\begingroup$ Hint: convolution theorem. $\endgroup$
    – user110971
    Jan 3, 2021 at 12:29
  • $\begingroup$ @user110971 So $<\alpha(t)>$ is basically a function AND a convolution at the same time. Applying the fourier tranform, one gets $F(\alpha) = F(K(t)) \cdot F(F(t)) \Leftrightarrow <\alpha (\omega)> = \bar{\chi}(\omega) \cdot \tilde{F}(\omega)$ which is our final answer. Am I right? $\endgroup$
    – miniplanck
    Jan 3, 2021 at 13:14

1 Answer 1


Thanks to user110971 in the comments, I think I've managed to find the solution.

According to Wikipedia, the Convolution Theorem states that if $f$ and $g$ are two functions, then $f \ast g$ denotes their convolution and

$$ \mathfrak{F}[f \ast g] = \mathfrak{F}[f] \cdot \mathfrak{F}[g]$$

where $\mathfrak{F}$ is the Fourier Transform and $\cdot$ is the pointwise multiplication.

In our case,

$$ \langle\alpha(t)\rangle_{F} = \int_{-\infty}^{+\infty} dt'\bar{K}(t-t')F(t')$$

which is the definition of convolution, according to Wikipedia. Then,

$$ \mathfrak{F}[\langle\alpha(t)\rangle_{F}] = \langle\alpha(\omega)\rangle_{F} = \mathfrak{F}[\bar{K}(t)] \cdot \mathfrak{F}[F(t)] $$

$$\Leftrightarrow \langle\alpha(\omega)\rangle_{F} = \bar{\chi}(\omega) \cdot \tilde{F}(\omega)$$

as requested.


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