I do remember learning with a shock all the extremely confusing physicist talk about feynmann diagrams and connected correlators was just about cumulants of multivariate gaussians. One wonders how much faster and deeper one could learn theoretical physics if somebody could write a sensible exposition shorn from vague terms like energy, temperature, connected correlators, propagators and particles...
Anyway.
I don’t know about these low temperature perturbative expansions.
In SLT one is interested in a tempered Boltzmann distribution… do you see a way in which this perturbative expansion story might come into play or is a no go because of singularities ? (Hence failure of gaussianity)
Yes, I actually thought about this a bit. It is definitely the case that the LC (or RLCT) in the SLT context is also exactly a (singular) stationary phase expansion. Unfortunately, the Fourier transform of a random variable, including a higher-dimensional one, really does have an isolated nondegenerate maximum at 0 (unless the support of your random variable is contained in a union of linear subspaces, which is kinda boring/ reducible to simpler contexts). Maybe if you think about some kind of small perturbation of a lower-dimensional system, you can get some components of the singular free energy expansion, but the expansion relevant here is really nonsingular. This is also the type signature of the expansion you see in most physical QFT systems, at least if they have a perturbative form (in which case, the free theory will in general be nondegenerate).
So the oscillating phase formula is about approximately integrating the function exp(−f(x)/ℏ) against various “priors” p(x) (or more generally any fixed function g), where f is a Lagrangian (think energy) and (\hbar) is a small parameter. It gives an asymptotic series in powers of ℏ. The key point is that (more or less) the kth perturbative term only depends on the kth-order power series expansion of f around the “stationary points” (i.e., saddlepoints, Jac(f) = 0) when f is imaginary, on the maxima of f when f is real, and there is a mixed form that depends on stationary points of the imaginary part which are also maxima of the real part (if these exist); the formulae are all exactly the same, with the only difference between real and imaginary f (i.e. statistical vs. quantum mechanics) being whether you only keep maxima or all saddle points.
Now in SLT, you’re exactly applying the “real” stationary phase formula, i.e., looking at maxima of the (negative) loss function -L(w). The key thing that can happen is that there can be infinitely many maxima, and these might be singular (both in the sense of having higher degree of stationarity, and in the sense of forming a singular manifold). In this case the stationary phase formula is more complicated and AFAIK isn’t completely worked out; Watanabe was the first person who contributed to finding expressions for the general case here beyond the leading correction.
In the case of maxima which are nondegenerate, i.e., have positive-definite Hessian, the full perturbative expansion is known; in fact, at least in one very useful frame on it, terms are indexed by Feynman diagrams.
Now the energy function f that appears in this context is the log of the Fourier transform ^p(θ) of a probability distribution p(x). Notice that p(x) satisfies p(x)≥0 and ∫p(x)dx=1. This means that (\hat{p}(0) = \int f(x) dx) is 1 and its log is 0. You can check that all other values of the Fourier transform are ≤1 in absolute value (this follows from the fact that ∫|f(x)|≥|∫f(x)|). In fact, the Hessian is equal (up to scale) to the variance of p. Now the point is that the only way that variance can be zero is if your pd is concentrated on a lower-dimensional affine subspace, in which case you can simply reduce your problem to a lower-dimensional one with nonsingular Hessian. When this doesn’t happen, the function you’re applying stationary phase to has only nondegenerate maxima, and so the “standard” Feynman-diagram formula applies instead of the more sophisticated Watanabe one that’s used in SLT.
Wonderful.
I do remember learning with a shock all the extremely confusing physicist talk about feynmann diagrams and connected correlators was just about cumulants of multivariate gaussians. One wonders how much faster and deeper one could learn theoretical physics if somebody could write a sensible exposition shorn from vague terms like energy, temperature, connected correlators, propagators and particles...
Anyway.
I don’t know about these low temperature perturbative expansions. In SLT one is interested in a tempered Boltzmann distribution… do you see a way in which this perturbative expansion story might come into play or is a no go because of singularities ? (Hence failure of gaussianity)
Yes, I actually thought about this a bit. It is definitely the case that the LC (or RLCT) in the SLT context is also exactly a (singular) stationary phase expansion. Unfortunately, the Fourier transform of a random variable, including a higher-dimensional one, really does have an isolated nondegenerate maximum at 0 (unless the support of your random variable is contained in a union of linear subspaces, which is kinda boring/ reducible to simpler contexts). Maybe if you think about some kind of small perturbation of a lower-dimensional system, you can get some components of the singular free energy expansion, but the expansion relevant here is really nonsingular. This is also the type signature of the expansion you see in most physical QFT systems, at least if they have a perturbative form (in which case, the free theory will in general be nondegenerate).
Sorry these words are not super meaningful to me. Would you be able to translate this from physics speak ?
So the oscillating phase formula is about approximately integrating the function exp(−f(x)/ℏ) against various “priors” p(x) (or more generally any fixed function g), where f is a Lagrangian (think energy) and (\hbar) is a small parameter. It gives an asymptotic series in powers of ℏ. The key point is that (more or less) the kth perturbative term only depends on the kth-order power series expansion of f around the “stationary points” (i.e., saddlepoints, Jac(f) = 0) when f is imaginary, on the maxima of f when f is real, and there is a mixed form that depends on stationary points of the imaginary part which are also maxima of the real part (if these exist); the formulae are all exactly the same, with the only difference between real and imaginary f (i.e. statistical vs. quantum mechanics) being whether you only keep maxima or all saddle points.
Now in SLT, you’re exactly applying the “real” stationary phase formula, i.e., looking at maxima of the (negative) loss function -L(w). The key thing that can happen is that there can be infinitely many maxima, and these might be singular (both in the sense of having higher degree of stationarity, and in the sense of forming a singular manifold). In this case the stationary phase formula is more complicated and AFAIK isn’t completely worked out; Watanabe was the first person who contributed to finding expressions for the general case here beyond the leading correction.
In the case of maxima which are nondegenerate, i.e., have positive-definite Hessian, the full perturbative expansion is known; in fact, at least in one very useful frame on it, terms are indexed by Feynman diagrams.
Now the energy function f that appears in this context is the log of the Fourier transform ^p(θ) of a probability distribution p(x). Notice that p(x) satisfies p(x)≥0 and ∫p(x)dx=1. This means that (\hat{p}(0) = \int f(x) dx) is 1 and its log is 0. You can check that all other values of the Fourier transform are ≤1 in absolute value (this follows from the fact that ∫|f(x)|≥|∫f(x)|). In fact, the Hessian is equal (up to scale) to the variance of p. Now the point is that the only way that variance can be zero is if your pd is concentrated on a lower-dimensional affine subspace, in which case you can simply reduce your problem to a lower-dimensional one with nonsingular Hessian. When this doesn’t happen, the function you’re applying stationary phase to has only nondegenerate maxima, and so the “standard” Feynman-diagram formula applies instead of the more sophisticated Watanabe one that’s used in SLT.