A Graphical Depiction of Linear Response

Most of the theory of linear response derives from the following equation:

y(t) = \int_{-\infty}^{\infty} \chi(t-t')h(t') dt'

I remember quite vividly first seeing an equation like this in Ashcroft and Mermin in the context of electric fields (i.e. \textbf{D}(\textbf{r}) = \int_{-\infty}^{\infty} \epsilon(\textbf{r}-\textbf{r}')\textbf{E}(\textbf{r}') d\textbf{r}') and wondering what it meant.

One way to think about \chi(t-t') in the equation above is as an impulse response function. What I mean by this is that if I were to apply a Dirac delta-function perturbation to my system, I would expect it to respond in some way, characteristic of the system, like so:

ImpulseResponse

Mathematically, this can be expressed as:

y(t) = \int_{-\infty}^{\infty} \chi(t-t')h(t') dt'= \int_{-\infty}^{\infty} \chi(t-t')\delta(t') dt'=\chi(t)

This seems reasonable enough. Now, though this is going to sound like tautology, what one means by “linear” in linear response theory is that the system responds linearly to the input. Most of us are familiar with the idea of linearity but in this context, it helps to understand it through two properties. First, the strength of the input delta-function determines the strength of the output, and secondly, the response functions combine additively. This means that if we apply a perturbation of the form:

h(t')=k_1\delta(t'-t_1) + k_2\delta(t'-t_2) +k_3\delta(t'-t_3)

We expect a response of the form:

y(t)=k_1\chi(t-t_1) + k_2\chi(t-t_2) +k_3\chi(t-t_3)

This is most easily grasped (at least for me!) graphically in the following way:

MultipleImpulses

One can see here that the response to the three impulses just add to give the total response. Finally, let’s consider what happens when the input is some sort of continuous function. One can imagine a continuous function as being composed of an infinite number of discrete points like so:

continuous

Now, the output of the discretized input function can be expressed as so:

y(t) = \sum_{n=-\infty}^{\infty} [\chi(t-n\epsilon_{t'})][h(n\epsilon_{t'}][\epsilon_{t'}]

This basically amounts to saying that we can treat each point on the function as a delta-function or impulse function. The strength of each “impulse” in this scenario is quantified by the area under the curve. Each one of these areal slivers gives rise to an output function. We then add up the outputs from each of the input points, and this gives us the total response y(t) (which I haven’t plotted here). If we take the limit n \rightarrow \infty, we get back the following equation:

y(t) = \int_{-\infty}^{\infty} \chi(t-t')h(t') dt'

This kind of picture is helpful in thinking about the convolution integral in general, not just in the context of linear response theory.

Many experiments, especially scattering experiments, measure a quantity related to the imaginary part of the  Fourier-transformed response function, \chi''(\omega). One can then use a Kramers-Kronig transform to obtain the real part and reconstruct the temporal response function \chi(t-t'). An analogous procedure can be done to obtain the real-space response function from the momentum-space one.

Note: I will be taking some vacation time for a couple weeks following this post and will not be blogging during that period.

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One response to “A Graphical Depiction of Linear Response

  1. Pingback: Response and Dissipation: Part 1 of the Fluctuation-Dissipation Theorem | This Condensed Life

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