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In the lectures I've heard linear response theory was introduced multiple times, however I can't tell how this is any different from solving an inhomogeneous PDE with a greens function and there's nothing linear about that to my knowledge.

Typically it's introduced for a damped harmonic oscillator with an external force $$ \ddot{x} + \gamma \dot{x} + \omega_0^2 x = F(t) $$ The Fourier transform then gives you the susceptibility $\hat{\chi}(\omega)$ so the whole equation corresponds to multiplication in the frequency domain and we can then retrieve the solution by doing the inverse transform or doing a convolution in the time domain $$ x(t) = \int\mathrm{d}t' \chi(t-t') F(t') $$ This is how I would solve the harmonic oscillator using Fourier transforms/Greens functions (since the inverse transformed susceptibility is just the corresponding Greens function) and here I haven't made any assumptions about $F(t)$. But in my 'linear response' lectures they always mention the fact that we neglect 'higher orders of $F$' and they write something like $$ x(t) = \int\mathrm{d}t' \chi(t-t') F(t') + \mathcal{O}(F^2) $$ But the $\mathcal{O}(F^2)$ should simply always be $0$ by the derivation above, so why is there such an emphasis on the 'linear response'?

To clarify what I meant by introducing the susceptibility: take the Fourier transform of the inhomgeneous ODE \begin{align*} \hat{F} &= -\omega^2 \hat{x} -i\gamma\omega \hat{x} + \omega_0^2 \hat{x} = \hat{x}(-\omega^2 - i\gamma\omega + \omega_0^2) \\ \hat{x}(\omega) &= \hat{F}(\omega) \frac{1}{\omega_0^2 - \omega^2 - i\gamma\omega} = \hat{F}(\omega) \hat{\chi}(\omega) \end{align*} To be this procedure seems fairly general, I didn't assume anything about $F(t)$, just that it's some ODE and not a PDE? I'm not familiar with the 'Volterra Expansion', but I can't really see how more terms would get added to this expression that are non linear in $F$ somehow.

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You are largely right. The mathematics you've been presented could indeed be packaged under several other names, including solutions via Fourier transforms or via Green functions. However, that doesn't mean that the "packaging" under the Linear Response Theory is not valuable.

Perhaps the key missing point in the way you have presented things is where the 'linear' in LRT comes from. The linearity in question is that of the differential operator that governs the response to the external force, $$ \mathcal L(x) = \ddot x + \gamma \dot x + \omega^2 x. \tag{$*$} $$ For most systems – and particularly for complicated, highly-coupled systems with a lot of internal dynamics – if the size of the perturbation $F$ (which drives the output $x$ via the dynamics $\mathcal L(x)=F$) is allowed to cover an arbitrarily large range, then there is no a priori requirement that $\mathcal L(x)$ be linear in $x$. However, accounting for a general nonlinear response operator is massively complicated, and it does not gel very well with reality, since we can typically stipulate that $F$ can be attenuated to be low enough that the induced response $x$ falls within the range of linearity of $\mathcal L(x)$, so we can go a very long way for real-world systems by assuming that $\mathcal L(x)$ is linear.

(In addition to this, it is also a good description of real-world devices to assume that they obey second-order ODEs, and that the system is time-invariant in addition to being linear, which then fully confines $\mathcal L(x)$ to the form $(*)$. This explains why the harmonic oscillator is such a common occurrence in LTR texts.)

(As another aside: if you want to understand the role of linearity here (and, similarly, elsewhere), then one useful way to go about things is to flip the keyword, and search for nonlinear response theory. This will turn up valuable resources (though often quite technical) where they do introduce nonlinearity, and the place where they do so shines a bright light at where the original linearity was in the first place.)

That's basically it, really. For a realistic system, it is a good reminder to include that $\cdots + O(F^2)$ to highlight the fact that this is only valid over the range in $x$'s where the "true" response operator is well approximated as a linear one. But, as you have highlighted, this is disingenuous in the way that (probably?) most texts introduce things, since they start off with a linear $\mathcal L(x)$ and neglect to explain how and why this is an approximation to begin with.

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  • $\begingroup$ I see that explains it. The way it has been presented to me so far always seemed to suggest that linear response is about solving the ODE, but if I understood correctly it seems to me that more emphasis should be placed on actually 'finding' that linear differential operator in the first place. $\endgroup$
    – Wihtedeka
    Commented Nov 23, 2021 at 15:18
  • $\begingroup$ @Wihtedeka Indeed. And you will see this happen as you go deeper into the theory (if that's the direction you choose to take). $\endgroup$ Commented Nov 23, 2021 at 16:19

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