Math 262A: Lecture 12

In Lecture 11, we were getting to the exciting part of this meandering topics course, where two seemingly disparate pieces of mathematics (Laplace’s Principle, the Exponential Formula) collide and produce something new (Feynman diagrams). Let’s review the two ingredients.

Take an interval $[a,b] \subseteq \mathbb{R}$ which may be finite or infinite, and let $S \colon [a,b] \to \mathbb{R}$ be a smooth function which attains a global minimum at a unique interior point $c \in (a,b),$ and which is such that the integral

$\int_a^b e^{-S(x)} \mathrm{d}x$

is finite.

Theorem (Laplace Principle): The quantity

$I(\hbar) = \sqrt{\frac{S''(c)}{2\pi\hbar}} e^{\frac{1}{\hbar}S(c)} \int_a^b e^{-\frac{1}{\hbar}S(x)} \mathrm{d}x$

extends to a smooth function of $\hbar \in [0,\infty),$ with $I(0)=1.$

The utility of this result is the following. Let

$\sum\limits_{j=0}^\infty a_j\hbar^j$

be the Maclaurin series of the smooth function $I(\hbar)$ . Taylor’s theorem with remainder tells us that for any nonnegative integer $k$ , we have

$\left| I(\hbar) - \sum\limits_{j=0}^ka_j\hbar^j \right| \leq M_k \hbar^{k+1},$

where $M_k \geq 0$ depends on $k$ but not on $\hbar$ . In other words, we have the $\hbar \to 0$ asymptotic expansion

$I(\hbar) = \sum\limits_{j=0}^k a_j \hbar^j + o(\hbar^k),$

where the error term is uniform in $\hbar$ . Often times this is used to get the $N \to \infty$ asymptotic expansion of an integral of the form

$\int_a^b e^{-NS(x)} \mathrm{d}x,$

in which case setting $\hbar = 1/N$ the Laplace principle gives us the $N \to \infty$ asymptotics

$\sqrt{\frac{NS''(c)}{2\pi}} e^{NS(c)} \int_a^b e^{-NS(x)} \mathrm{d}x = 1+ \frac{a_1}{N} + \dots + \frac{a_k}{N^k} + o\left(\frac{1}{N^k} \right).$

The prototypical example is Stirling’s approximation to $N!$ via the Euler integral.

Now we recall the Exponential Formula, which tells us how to compute the exponential of a formal power series

$S(x) = \sum\limits_{d=1}^\infty s_d \frac{x^d}{d!},$

or equivalently how to compute the Maclaurin series of $e^S$ when $S$ is a smooth function with derivatives $s_d=S^{(d)}(0)$ such that $S(0)=0.$ Writing

$e^{S(x)} = 1 + \sum\limits_{d=1}^\infty h_d \frac{x^d}{d!},$

the Exponential Formula says that

$h_d = \sum\limits_{\pi \in \mathrm{Par}(d)} \prod\limits_{B \in \pi} s_{|B|},$

the sum being over all partitions $\pi$ of $\{1,\dots,d\},$ and the product being over the blocks $B$ of each partition $\pi.$ An alternative form of the Exponential Formula, which is more useful for our current purpose, arises when we instead write

$S(x) = \sum\limits_{d=1}^\infty p_d \frac{x^d}{d},$

so $s_d=(d-1)!p_d.$ In terms of the scaled derivatives $p_d,$ the formula for $h_d$ becomes

$h_d = \sum\limits_{\pi \in \mathrm{Par}(d)} \prod\limits_{B \in \pi} (d-1)!p_{|B|}.$

Since a finite set of cardinality $c$ can be cyclically ordered in $(c-1)!$ ways, we can rewrite the above as

$h_d = \sum\limits_{\sigma \in \mathrm{S}(d)} \prod\limits_{C \in \sigma} p_{|C|},$

where the sum is over all permutations $\sigma$ of $\{1,\dots,d\}$ and the product is over the cycles $C$ of each permutation $\sigma.$ Observe that this product does not depend on the internal structure of the cycles $C$ of $\pi,$ but only their sizes, so that any two permutations of the same cycle type (i.e. any two permutations which are conjugates of one another) give the same contribution to the sum. That is to say, we may rewrite the above as

$h_d = \sum\limits_{\alpha \vdash d} |C_\alpha| p_\alpha,$

where the sum is over all Young diagrams $\alpha,$ and for each diagram $C_\alpha \subseteq \mathrm{S}(d)$ is the conjugacy class of permutations of cycle type $\alpha$ and

$p_\alpha = \prod\limits_{I=1}^{\ell(\alpha)} p_{\alpha_i}.$

We now combine the Laplace Principle with the (permutation form of) the Exponential Formula to calculate the asymptotic expansion coefficients $a_j$ . To simplify the computation, we will make several assumptions on the Laplace side, none of which are essential — these can all be eliminated without too much difficulty and the end result is more or less the same. First, we assume the integration is over the whole real line. Second, we assume that the critical point $c$ where the unique minimum of $S$ occurs is $c=0$ , and that $S(c)=0.$ Finally, let us assume that the positive number $S''(0)$ is equal to one.

With these assumptions, we have

$I(\hbar)=\frac{1}{\sqrt{2\pi\hbar}}\int_\mathbb{R} e^{-\frac{1}{\hbar}S(x)}\mathrm{d}x,$

and the Maclaurin series of $S$ has the form

$\frac{x^2}{2} + \sum\limits_{d=3}^\infty p_d\frac{x^d}{d!}.$

with $p_d:=\frac{S^{(d)}(0)}{(d-1)!}.$ We factor the integrand as

$e^{-\frac{1}{\hbar}S(x)} = e^{-\frac{1}{\hbar}\frac{x^2}{2}} e^{-\frac{1}{\hbar}(S(x)-\frac{x^2}{2})},$

and by the Exponential Formula the Maclaurin series the exponential of the non-quadratic part of the action $S$ is

$1+\sum\limits_{d=1}^\infty \left(\sum\limits_{\alpha \vdash d} \frac{|C_\alpha|}{d!} (-1)^{\ell(\alpha)}p_\alpha \hbar^{-\ell(\alpha)} \right)x^d$

with $p_1=p_2=0$ . We get the Maclaurin series of $I$ by integrating this term-by-term against the Gaussian density:

$1+\sum\limits_{d=1}^\infty \left(\sum\limits_{\alpha \vdash d} \frac{|C_\alpha|}{d!} (-1)^{\ell(\alpha)}p_\alpha \hbar^{-\ell(\alpha)} \right)\frac{1}{\sqrt{2\pi\hbar}}\int_\mathbb{R}x^d\mathrm{d}x = 1+\sum\limits_{d=1}^\infty \sum\limits_{\alpha \vdash d} \frac{t_d|C_\alpha|}{d!} (-1)^{\ell(\alpha)}p_\alpha \hbar^{\frac{d}{2}-\ell(\alpha)}$

where we used the exact evaluation

$\frac{1}{\sqrt{2\pi\hbar}}\int_\mathbb{R} x^d \mathrm{d}x = t_d\hbar^{\frac{d}{2}},$

with $t_d$ the number of fixed point free involutions in the symmetric group $\mathrm{S}(d)$ .

In order to determine the expansion coefficients $a_j,$ we now want to arrange the sum

$1+\sum\limits_{d=1}^\infty \sum\limits_{\alpha \vdash d} \frac{t_d|C_\alpha|}{d!} (-1)^{\ell(\alpha)}p_\alpha \hbar^{\frac{d}{2}-\ell(\alpha)}$

as a power series in $\hbar.$ At this point, it is not even clear that the series involves only nonnegative integer powers of $\hbar.$ First of all, the presence of $\frac{d}{2}$ in the exponent of $\hbar$ makes it look like we might have half-integer powers, but this is not the case since $t_d=0$ whenever $d$ is an odd number. Next, the actual exponent of $\hbar$ is $\frac{d}{2}-\ell(\alpha),$ which seems like it could be negative, but in fact since $p_1=p_2=0$ the quantity $p_\alpha$ vanishes if $\alpha \vdash d$ has more than $\frac{d}{3}$ rows. Combining these conditions, we see that the first non-zero exponent of $\hbar$ occurs at $d=6$ , and the only $\alpha \vdash 6$ which contributes to the sum is $\alpha=(3,3),$ the diagram with two rows each of length $3,$ so that in this case we have

$\hbar^{\frac{d}{2}-\ell(\alpha)}=\hbar^{3-2}=\hbar^1.$

Another thing which is not quite obvious is that this is the only situation which produces a power of $1$ in the exponent of $\hbar,$ or even that there are finitely many such situations. Let us see why this is the case. In order for this exponent to arise, we must have that $\alpha \vdash d$ is a Young diagram with an even number of cells such that $\ell(\alpha)=\frac{d}{2}-1.$ Since all rows of $\alpha$ have length at least three, we get the inequality

$\frac{d}{2}-1 \leq \frac{d}{3} \implies d \leq 6.$

So, the only term in the sum which is linear in $\hbar$ is

$\frac{1}{6!}t_6|C_{(3,3)}|p_3^2,$

and we conclude that

$a_1=\frac{1}{6!}t_6|C_{(3,3)}|p_3^2.$

We can explicitly evaluate the “universal” part of this formula, i.e. the part that has no dependence one $S.$ As previously discussed, the number of fixed point free involutions in $S(6)$ is

$t_d=(5-1)!! = 5 \cdot 3 \cdot 1 = 15.$

Next, the number of permutations in $S(6)$ whose disjoint cycle decomposition is of the form $(*\ *\ *)(*\ *\ *)$ is

$\frac{{6 \choose 3} \cdot 2 \cdot 2}{2} = 20 * 2=40,$

because there are ${6 \choose 3}$ ways to choose the elements in cycles and $2$ ways to cyclically arrange each, and this overcounts by a factor of $2$ since the cycles are unordered. We thus conclude that

$a_1 = \frac{40*15}{720}p_3^2 = \frac{5}{6}p_3^2.$

(Note: I think I made a mistake here, because the subleading correction in Stirling’s formula is supposed to be $\frac{1}{12},$ which is ten times smaller than $\frac{5}{6}$ ).

Clearly, to continue this calculation to find all the coefficients $a_j,$ we need a better way to organize this information. This can be done pictorially, and is the simplest example of the use of Feynman diagrams in perturbation theory. The number $t_d|C_\alpha|$ is equal to the number of pairs $(\gamma,\varphi) \in \mathrm{S}(d) \times \mathrm{S}(d)$ such that $\gamma$ is a fixed point free involution and $\varphi \in C_\alpha$ is a permutation of cycle type $\alpha$ . As we discussed in Lecture 11, such pairs of permutations are referred to as combinatorial maps, or rotation systems, because they can be converted into topological maps, which are graphs drawn on surfaces. So, graphs on surface are the Feynman diagrams of zero dimensional quantum field theory. We will pick this up in Lecture 13, where we will give the full diagrammatic interpretation of the expansion coefficients $a_j$ . After that, the plan is to tackle the same problem when we replace integration over $\mathbb{R}$ with integration over the real vector space $\mathrm{H}(N)$ of $N \times N$ Hermitian matrices; in this situation we can actually distinguish the genus of the maps which arise. After that we will generalize further to integrating over the real vector space of selfadjoint elements in a finite-dimensional von Neumann algebra.

Math 262A: Lecture 12

Published by Jonathan Novak

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