# A toy model in 0-d QFT

### Questions

1. For any positive integer $$r$$, compute $$(\frac{d}{dY})^r e^{(Y^2)}| _{Y=0}.$$ The answer should directly relates to a counting problem about Feynman diagrams.

2. Is there a tutorial for how Feynman diagrams work in this context? I look forward to an answer a lot, since the question has been reduced to the simplest form. Thank you!

EDIT: 3. Turns out the answer to the first question is trivial considering its Taylor expansion. So a better question should be what's the benefit of solving the first one by the combinatorial way.

### Context

I am a math student trying to learn QFT and Feynman Diagrams using Hori et al's Mirror Symmetry. Much to my surprise, even a toy model in 0-dimension the theory is already complicated (for me).

On a point, a function is just a number, so integration over all functions reduces to an ordinary integral. I am looking at the following particular toy model:

$$\int dX e^{-S(X)},$$

where $$S(X) = \frac{1}{2}X^2 + i\epsilon X^3$$, and am focusing on the perturbation with small $$\epsilon$$, which reduces to the computation $$(\frac{d}{dY})^r e^{(Y^2)}| _{Y=0}.$$

Basic attempts show that this is a combinatorial problem, which I have no idea how to solve. This is where the book introduces Feynman diagrams, claiming that they help computing the value at zero of the $$r$$-th derivative above.

However, the explanation is not clear to me. I don't know what the book means by "choosing pairs", "contracting", and "propagators". I have tried other lecture notes online, but all of what I have found use physics terminologies making the situation more complicated.

• Hermite polynomials – abx Oct 12 '19 at 16:50
• There are two different aspects one needs to master to be fully conversant with the language of Feynman diagrams. 1) how does one evaluate a single diagram 2) how to precisely define the sum over all diagrams. For 1), a diagram is just a graphical encoding of a contraction of tensors which generalize matrix algebra. Even $\Delta=b^2-4ac$ from high school can be learned that way see, e.g., arxiv.org/abs/math/0411110 For 2), the best mathematical formalism for that is that Joyal's theory of species, see, e.g., emis.de/journals/SLC/wpapers/s49abdess.pdf – Abdelmalek Abdesselam Oct 12 '19 at 19:47
• What you are looking at is the Wick theorem. It has many views, one of which is integration by parts (this should relate to your combinatorial interpretation) – lcv Oct 13 '19 at 13:38
• I don't understand your follow-up question no. 3. Wasn't the whole point of this exercise to learn a new method? Isn't that a benefit? – Michael Engelhardt Oct 13 '19 at 16:47
• Regarding edit 3, I think the short answer is that there really isn't a good reason to do it for this integral; the point is to introduce the concept so it can be applied to more complicated integrals of the same "shape". – lambda Oct 13 '19 at 19:51

Draw a point from which $$r$$ lines emanate (a "vertex"). The ends of the lines are associated with the derivatives. Now let the derivatives act. Two things can happen:

a.) The derivative acts on the exponential. Represent this by attaching a dot to the end of the line (now the derivative is gone, but there's a factor $$Y$$ associated with the end of the line, symbolized by the dot).

b.) The derivative acts on one of the dots, i.e., one of the $$Y$$ prefactors you've generated by previous applications of derivatives. Represent this by attaching the line you're considering to one of the dots at the end of another line, erasing the dot (that factor $$Y$$ is now gone, having acted with the derivative) - so altogether you now have a loop emanating from the vertex and going back into it.

Finally, set $$Y=0$$. This means that all diagrams which still have dots are erased.

So, your combinatorial problem is to find all different diagrams you can construct this way; in effect, in your example, pairing off lines emanating from the vertex to form loops. There are some numerical factors to take care of, here, just the factors 2 from taking derivatives of $$Y^2$$.

In the exponential, you can think of the $$Y^2$$ as two dots connected by a line - then this incorporates a.) into the notion used in b.) of "derivatives erasing dots". And you can incorporate the aforementioned factors 2 this way - you have a choice of which dot you're attaching and erasing.

Although it might be a bit facile of me to say, since I'm familiar with the physics parlance, don't be afraid of trying to understand the physics tutorials on this. As far as these diagrammatics go, you don't need to understand the physics behind "propagators", "vertices" and the like. You just need a rudimentary dictionary of what these things are in the graphs. I already mentioned "vertex". "Propagators" are just the lines. The dots are "sources". When you connect two lines, erasing the dot, you're "contracting" ...

So let's complete the exercise: Of course, for odd $$r$$, the result is zero - there are no valid diagrams. For even $$r$$, there are $$r!/(2^{r/2} (r/2)!)$$ ways of pairing off lines coming from the vertex. Multiplied by a factor $$2^{r/2}$$ from taking $$r/2$$ derivatives of the exponential, as mentioned above, yields $$r!/(r/2)!$$. As a check, these are indeed the absolute values of the constant terms in the Hermite polynomials $$H_r$$.

• Thank you! I followed the procedure you gave and understood finally. – Student Oct 15 '19 at 20:20

Here is how I understand the appearance of diagrams in computing that integral, which may not be how the book does it.

We can formally expand as a series in $$\epsilon$$ and write $$\int_{-\infty}^\infty dX\,e^{-S(X)} = \sum_{n \ge 0} \frac{(-i\epsilon)^n}{ n!} \int_{-\infty}^\infty dX\,e^{-\frac{1}{2}X^2}X^{3n}$$ so the problem reduces to computing the integral on the RHS. This is somewhat well-known to vanish for odd $$n$$ and to be equal to $$\sqrt{2\pi}$$ times the double factorial $$(3n - 1)!! = (3n - 1) (3n - 3) \cdots 1$$ when $$n$$ is even.

Combinatorially, the double factorial counts perfect matchings (partitions into blocks of size 2) of a set of size $$3n$$. Given a perfect matching of the set $$\{1, \dots, 3n\}$$, we can construct a $$3$$-regular graph as follows:

• The vertices are $$\{1, \dots, n\}$$.
• To each vertex $$j$$, attach three "half-edges" labelled $$\{3j-2, 3j-1, 3j\}$$.
• Join half-edges together according to the perfect matching to form edges.

Thus we can interpret the coefficient of $$\epsilon^n$$ as counting cubic graphs of size $$n$$. (Though the way we've labelled the half-edges means there is some overcounting to deal with.) Thinking of it this way might not seem very motivated, but for slightly more complicated integrals the benefit is easier to see. For instance, if we added an $$\epsilon^4$$ term as well, we'd get something counting graphs where vertices can have either degree 3 or 4. The Isserlis–Wick theorem is the generalization of that "well-known" fact about double factorials to the multivariate setting, and can be used to interpret multivariate integrals of this type as counting graphs with decorated edges.