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In Peskin & Schroeder, page 97, the following expression is given as part of the demonstration of how the $n$-point correlation function is calculated using connected diagrams:

$$\sum_{\text{connected}}\sum_{\text {all }\left\{n_{i}\right\}}\left(\begin{array}{c}\text { value of } \\ \text { connected piece }\end{array}\right) \times\left(\prod_{i} \frac{1}{n_{i} !}\left(V_{i}\right)^{n_{i}}\right)\tag{p.97}$$

where the $\sum_\text{connected}$ I have abbreviated from the original "all possible connected pieces". The text following this expression reads,

where "all $\{n_i\}$" means "all ordered sets $\{n_1, n_2, n_3, ...\}$ of non-negative integers."

I don't understand this. This expression is meant to give the value of the sum of all diagrams. A typical diagram, corresponding to a specific choice of Wick contraction, is given by (4.50) in the text. For a given connected piece, such as the left-most piece of (4.50), there are finitely many possibilities for the accompanying disconnected pieces that would complete the Wick contraction, and hence the diagram (here using the language that a "diagram" is made of several "pieces"). If I return now to thinking about the first expression that I gave above, for any given connected piece in the $\sum_\text{connected}$, there are only a handful of select possibilities for $\{n_i\}$ that we would need to sum over. This is even emphasized in the text on the previous page in the line,

In any given diagram, only finitely many of the $n_i$ will be nonzero.

So why are we then summing over all ordered sets of non-negative integers? In my view, this should be a sum again over only the possible $\{n_i\}$ that correspond to correct Wick contractions appropriate for a given connected piece.

Where am I going wrong with this logic?

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  • $\begingroup$ All combinations of $\{n_i\}$ are possible if you go up to sufficient orders in the perturbation expansion, and since the sum in your post includes (formally) all orders all combinations of $\{n_i\}$ will appear. $\endgroup$
    – AfterShave
    Feb 20, 2022 at 17:13

1 Answer 1

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  1. P&S are considering all (possibly disconnected) diagrams that contribute to the 2-point correlation function $G(x,y)$ with the condition that the 2 external points $x$ and $y$ are assumed to always be connected, i.e. belong to the same connected component. Here $(V_i)_{i\in I}$ are all possible connected vacuum bubbles. The equation mentioned by OP then expresses that $$ G(x,y)~=~G_c(x,y)e^{\sum_{i\in I} V_i}. \tag{p.97}$$

  2. Although each diagram has a finite number $$1+\sum_{i\in I} n_i$$ of connected components, there is no upper bound when we consider all diagrams.

  3. For simpler arguments for the linked cluster theorem, see e.g. this, this & this Phys.SE post.

References:

  1. M.E. Peskin & D.V. Schroeder, An Intro to QFT, 1995; p. 97.
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