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I am considering the possible Wick contractions for the following expression: \begin{align*} \left< x ( t' )^5 x ( t'' )^5 \right> = \left< x( t' ) x( t' ) x( t' ) x( t' ) x( t' ) x( t'') x( t'') x( t'') x( t'') x( t'') \right> \end{align*} I believe there are a possible of $(10-11)!! = 9!! = 945$ possible contractions to be consider. I think there are three possible classes of diagrams to consider (see the attached figure):

(a) Connect each $x(t')$ to each $x(t'')$ which gives a total of $5!=120$ contractions

(b) Connect two $x ( t' ) $ to each other which implies the remaining three $ x ( t') $ are connected to three $ x ( t'' ) $ and the remaining two $ x( t'' ) $ are connected to each other, with multiplicity $\!^5C_2 \times 3 ! \times \!^5C_2 = 10 \times 6 \times 10 = 600$

(c) Connect four $x ( t' ) $ to each other which implies the remaining $ x ( t') $ connects to one $ x ( t'' ) $ and the remaining four $ x( t'' )$ are connected to each other, with multplicity $\!^5C_4 \times 1 ! \times \!^5C_4 = 5 \times 1 \times 5 = 25$.

However, this can't be correct since this gives a total of $120 + 600 + 25 = 745$ contractions which is not equal to the expected number of $945$ contractions. Any suggestion where I went wrong?

enter image description here

octonion
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jim
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1 Answers1

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You went wrong on the last part. There are indeed $5$ ways to pick four of the $x(t’)$, but then there are $3$ ways to contract those four in pairs. Your last term should be $15 \times 15 = 225$.

knzhou
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