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Finite-dimensional Feynman Diagrams

4. Wick's Theorem

Calculating high-order derivatives of a function like $ \exp({\scriptstyle\frac{1}{2}}{\bf b}^tA^{-1}{\bf b})$ can be very messy. A useful theorem reduces the calculation to combinatorics.

Wick's theorem

$\displaystyle \frac{\partial}{\partial b^{i_1}}\cdots \frac{\partial}{\partial ...
...1}_{\textstyle i_{p_1},i_{p_2}} \cdots
A^{-1}_{\textstyle i_{p_{m-1}},i_{p_m}},$

where the sum is taken over all pairings $ (i_{p_1},i_{p_2}), \dots, (i_{p_{m-1}},i_{p_m})$ of   i1, ..., im

Wick's theorem is proved (a careful counting argument) in texts on quantum field theory. The most detailed explanation is in S. S. Schweber, An Introduction to Relativistic Quantum Field Theory, Evanston, IL, Row, Peterson 1961.

Let us calculate a couple of examples.

To begin, it is useful to write $ \exp({\scriptstyle\frac{1}{2}}{\bf b}^tA^{-1}{\bf b})$ with $ {\bf b}^tA^{-1}{\bf b} = \sum A^{-1}_{i,j}b^ib^j$ (the sums running from 1 to d) using the series expansion   exp x = 1 + x +x2/2 +x3/3! ... . The typical term will be $ (1/n!)(1/2^n)(\sum A^{-1}_{i,j}b^ib^j)^n$. This term is a homogeneous polynomial in the bi of degree 2n

Differentiating k times a homogeneous polynomial of degree 2n and evaluating at zero will give zero unless k = 2n. So the job is to analyze the result of 2n differentiations on $ (1/n!)(1/2^n)(\sum A^{-1}_{i,j}b^ib^j)^n$.

The differentiation carried out most frequently in these calculations is

$\displaystyle \frac{\partial}{\partial b^{k}}(\frac{1}{2}\sum_{i,j=1}^d A^{-1}_{i,j}b^ib^j) = \sum_{i=1}^d A^{-1}_{i,k}b^i,$

where we use the symmetry of the matrix A-1, a direct consequence of the symmetry of A.

In what follows $ \frac{\partial}{\partial b^i}$ will be abbreviated as $ \partial_i$.

Note that (1,2) and (2,1) count as the same pairing.

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