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High-energy theory

How many orders are worth computing

About 25 min if you have seen this material before, 1 hr or so if it is new to you.

Background
Almost nothing in this subject can be worked out exactly. What can be done is to write the answer as a series in a number λ\lambda that measures how weakly the particles interact, and compute the terms one at a time. Four or five terms is several years of work for a small collaboration. Each term is a sum over A picture standing for one contribution to a calculation, with vertices where particles meet and lines joining them.. A diagram is one way the process being calculated can happen, drawn as a picture, and the rule is that every way gets added up. Richard Feynman introduced them in 1948, and they are how predictions in particle physics are organised today. The version in the setup is stripped down. In the real theory each diagram stands for a separate integral and they differ from one another; here every diagram at a given order contributes the same amount, so all that is left of the calculation is counting how many there are.
A vertex is a point where four line-ends meet. The ends carry labels so they can be told apart: 
      a       b
       \     /
        \   /
         \ /
          o
         / \
        /   \
       /     \
      d       c
A diagram of order nn is built from nn vertices by joining every one of the 4n4n line-ends to exactly one other end, with nothing left dangling. A single vertex has four ends, and there are three ways to join them up: 
  a  b  c  d      a  b  c  d      a  b  c  d
  |  |  |  |      |  |  |  |      |  |  |  |
  +--+  +--+      +--|--+  |      |  +--+  |
                     +-----+      +--------+

   a-b, c-d         a-c, b-d        a-d, b-c
Each of those ties the vertex to itself twice, so all three draw the same shape: two loops meeting at a point. They still count as three. The ends are distinct, and what the calculation sums over is ways of joining ends, not distinct shapes. That is what "the number of diagrams" means here. At order 2 there are two vertices and eight ends, and a line may or may not run between the two vertices: 
  no line between the vertices          two lines between them

    a   b   c   d       e   f   g   h     a   b   c   d       e   f   g   h
    |   |   |   |       |   |   |   |     |   |   |   |       |   |   |   |
    +---+   +---+       +---+   +---+     +---+   |   |       |   |   +---+
                                                  +---|-------+   |
                                                      +-----------+
Write DnD_n for the number of order-nn diagrams, with D0=1D_0 = 1. Each vertex carries a factor −λ-\lambda, and the order-nn term is the sum over diagrams divided by n!n!, because relabelling the vertices reproduces diagrams that have already been counted. The probability of a process in particle physics is calculated from a sum of exactly this shape, one term per order, with each term standing for all the ways the process can happen using that many interactions: ∑n≥0Dn (−λ)nn!.\sum_{n \ge 0} \frac{D_n\,(-\lambda)^n}{n!}. Because λ\lambda is small, each extra order looks as though it should be a smaller correction than the last, and for the first few it is.
The parts below count the diagrams at low orders, then use the count to work out how far into the series it is worth going.
(a)
Give D1D_1, D2D_2, D3D_3 and D4D_4, as four whole numbers separated by spaces.
(b)
Of the order-2 diagrams, in how many is there at least one line joining an end of the first vertex to an end of the second? The setup draws one diagram of each kind.
(c)
Take λ=0.025\lambda = 0.025. Compute the terms of the series and add them up to and including the smallest one. Give the result to four decimal places. An independent calculation of the same quantity, one that does not use this series, gives 0.94450180.9445018.
(d)
Take λ=0.001\lambda = 0.001 instead. At which order is the term smallest?

Past 1.5 hr on this? That usually means a missing prerequisite rather than a lack of effort — worth reading up before pushing further, starting with Asymptotic expansion (Wikipedia).

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