Recall that a Gibbs measure gives a probability distribution on states $x$ of the form

$$ p_X(x) = \frac{1}{Z(\beta)}\exp(-\beta E(x)) $$

As I understand, the function $E$ is interpreted as the energy of the state. I'm wondering if there is a physical interpretation or significance to the characteristic function of $p$:

$$ \phi_X(k) = \mathbb{E}[\exp(ik^tx)] $$

Obviously, if one has $\phi_X$, one can compute moments and other interesting things about the ensemble, but I'm wondering (as a math guy) if $\phi_X$ has a physical significance.

To provide a little more context for my particular problem, I'm really thinking of $x$ as a digital image, and hence $p_X(x)$ is a probability distribution on possible images (see e.g. texture synthesis). In many applications, it's more convenient to derive/work with $\phi_X(k)$ instead of $p_X(x)$, and I'm trying to think of a way to do MCMC using $\phi_X$ instead of $p_X$. A physical intuition for $\phi_X$ might help, if there is one.

  • $\begingroup$ With an appropriate representation of the state space, the characteristic function can have a number of useful physical interpretations. (As written, the physical significance is 'profound but extremely vague', because you haven't given coordinates to the states). $\endgroup$ – TotallyRhombus Apr 11 '16 at 16:57
  • $\begingroup$ @fs137 Interesting! I'd be happy to see an example, if you happen to have one (or just a reference is fine). $\endgroup$ – icurays1 Apr 11 '16 at 17:01
  • $\begingroup$ One example that I had in mind involves applications of the Hubbard Stratonovich transformation: en.wikipedia.org/wiki/…. Let me know if this looks relevant to your initial question. $\endgroup$ – TotallyRhombus Apr 11 '16 at 17:25
  • $\begingroup$ @fs137 It might, but I'm a bit unclear on where the Fourier transform arises. It seems like that transformation (which is just completing the square?) is just used to evaluate partition functions, but maybe I'm missing something. In my application I'm thinking of $x$ as a "classical" state (in fact, for me, $x$ is really just a digital image...) $\endgroup$ – icurays1 Apr 11 '16 at 18:03
  • $\begingroup$ @fs137 I edited the question to add a little more context. $\endgroup$ – icurays1 Apr 11 '16 at 18:08

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