Thermodynamic limit "vs" the method of steepest descent

Question

Let me use this lecture note as the reference.

• I would like to know how in the above the expression (14) was obtained from expression (12).

In some sense it makes intuitive sense but I would like to know of the details of what happened in between those two equations. The point being that if there were no overall factor of "$\sqrt{N}$" in equation (12) then it would be a "textbook" case of doing the "method of steepest descent" in the asymptotic limit of "N".

• I am wondering if in between there is an unwritten argument that in the "large N" limit one is absorbing the $\sqrt{N}$ into a re(un?)defined measure and then doing the steepest descent on only the exponential part of the integrand.

  I don't know how is the "method of steepest descent" supposed to be done on the entire integrand if the measure were not to be redefined.

• But again if something like that is being done then why is there an approximation symbol in equation (14)?

After taking the thermodynamic limit and doing the steepest descent shouldn't the equation (14) become an equality with a sum over all the $\mu_s$ which solve equation (15)?

Though naively to me the expression (12) looks more amenable to a Dirac delta function interpretation since in the "large N" limit it seems to be looking like the standard representation of the Dirac delta function, $\frac{n}{\sqrt{\pi}} e^{-n^2x}$

• I would like to know of some comments/explanation about this general philosophy/proof by which one wants to say that the "method of steepest descent" is exact in the "thermodynamic limit".

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Answer 1

This is a straightforward application of the steepest descent method. So, we have the integral (taken from the lectures you cite)

$$Q_N=\sqrt{\frac{N\beta J}{2\pi}}\int_{-\infty}^\infty d\mu e^{[Nq(\beta J,\beta b,\mu)]}$$

being

$$q(\beta J,\beta b,\mu)=\ln\{2\cosh[\beta(J\mu+b)]\}-\frac{\beta J\mu^2}{2}.$$

Now, you will have, by applying steepest descent method to the integral,

$$\frac{\partial q}{\partial\mu}=\beta J\tanh[\beta(J\mu+b)]-\beta J\mu=0.$$

Let us call $\mu_s$ the solution of this equation and expand the argument of the exponential around this value. You will get

$$q(\beta J,\beta b,\mu)=q(\beta J,\beta b,\mu_s)-\frac{J\beta}{2}[1-J\beta(1-\mu_s^2)](\mu-\mu_s)^2.$$

This shows that

$$Q_N\approx e^{Nq(\beta J,\beta b,\mu_s)}\sqrt{\frac{N\beta J}{2\pi}}\int_{-\infty}^\infty d\mu e^{-\frac{NJ\beta}{2}[1-J\beta(1-\mu_s^2)](\mu-\mu_s)^2}$$

i.e.

$$Q_N\approx e^{Nq(\beta J,\beta b,\mu_s)}\sqrt{\frac{1}{1-J\beta(1-\mu_s^2)}}$$

Note that the $N$ factor is completely removed after integration and remains just into the argument of the exponential. Eq. (16-18) follow straightforwardly.

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Comments

Thanks for the reply but I guess my question was not framed clearly. I CAN see what you have done BUT this is NOT what I would have thought the "method of steepest descent" to do! You seem to only do a Taylor expansion of the exponent. Point being that the only kind of steepest descent analysis that I have seen works on integrals of the form $F(s) = \int f(z)e^{sg(z)} dz$ in the large $s$ limit. This is not in that standard form. May be there is some other version of "method of steepest descent" that you have in mind and it would be great if you can give a reference for that.

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Also I want to know how this (or any valid steepest descent method) on this particular example will effectively manage to take the "large N" (thermodynamic) limit.

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@user6818 : this is standard textbook material. You should read, e.g., chapter VIII in [this nice book](http://algo.inria.fr/flajolet/Publications/book.pdf).

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But THIS IS the steepest descent (it would be better to call it Laplace) method and the one that was applied in that lectures. Firstly, I find the extremum and then I expand around it and the integral becomes a Gaussian one and so computable. Just check http://en.wikipedia.org/wiki/Laplace%27s_method.

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@Jon Okay let me put it this way - this particular version of "steepest descent method" that you are using is not obviously equivalent to the one familiar to me from say the exposition in the book by Arfken and Weber. Is that and the one you did equivalent? (..a priori they look far away from each other..) Also in what you are doing is there any sense in which you are taking the thermodynamic "large N" limit? (..In the Arfken-Weber way of thinking there is a notion of an asymptotics in the "steepest descent" method though in this Laplace thinking I don't see any notion of asymptotics..)

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@Jon Another point that I was making in my original question is that this analysis as in my linked file - goes only as far as to produce an approximation for the integral (same as what you seem to do) but I guess the statement that is true and is also implied in my linked article is that in the thermodynamic limit this method of steepest descent is infact exact! Thats the non-trivial idea for which I am looking for an explanation.

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@Jon In the Wikipedia article they do take a $M \rightarrow \infty$ limit - but why? This is where much of the confusion lies - (1) Why is there an approximation symbol if there is an asymptotics? In whatever asymptotics they consider, Arfen-Weber's version doesn't have an approximation at the end. (2) Could they have done this on an integral of the form $\int \sqrt{N}e^{Nf(x)} dx$ ? - as the case here - this I think is conceptually different than doing an approximation on the integral $\int e^{Nf(x)}dx$ and then to say that this produces a cancelling overall factor of $\frac{1}{\sqrt{N}}$.

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@user6818: The problem is that here you have no clear at all the mathematics. It is an everyday fact that the integral of an exponential like this takes its most important contribution where the argument of the exponential has an extremum. This is an asymptotic approximation to the integral and holds also for the oscillating case (stationary phase). I urge you to read some good book about asymptotic approximations like http://www.amazon.com/Asymptotic-Expansions-Dover-Books-Mathematics/dp/0486603180 . In this particular case it is just the integral that must be evaluated.

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@Jon I can't agree with you that this is some standard fact. I haven't seen this equality in any place whereever I have seen a discussion on the method of steepest descent. I have asked various people - quite brilliant students! - around me and they too don't seem to know this - though all of us have used method of steepest descent somewhere or the other.

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