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Finite-dimensional Feynman Diagrams


Feature Column Archive

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 calculationto combinatorics.

Wick's theorem

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

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 onQuantum 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 termwill be $ (1/n!)(1/2^n)(\sum A^{-1}_{i,j}b^ib^j)^n$. This term is a homogeneous polynomialin the bi of degree 2n

Differentiating ktimes a homogeneous polynomialof degree 2nand evaluating at zero will give zero unless k = 2n. So the job is to analyzethe result of 2ndifferentiations 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$.

  • n = 1.

    $\displaystyle \partial_2 \partial_1(\frac{1}{2}\sum A^{-1}_{i,j}b^ib^j) = $$\displaystyle \partial_2 ( \sum_j A^{-1}_{1,j}b^j ) = $A-11,2,

    using the symmetry of the matrix A-1.The same calculation showsthat

    $ \partial_1 \partial_1(\frac{1}{2}\sum A^{-1}_{i,j}b^ib^j) = A^{-1}_{1,1}~.$

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

  • n = 2

    $\displaystyle \partial_4\partial_3\partial_2\partial_1(1/2!)(1/2^2)(\sum A^{-1}_{i,j}b^ib^j)^2 = $

    $\displaystyle \partial_4\partial_3\partial_2(1/2)(\sum A^{-1}_{i,j}b^ib^j)( \sum A^{-1}_{1,j}b^j) =$

    $\displaystyle \partial_4\partial_3[(\sum A^{-1}_{2,j}b^j)( \sum A^{-1}_{1,j}b^j) +(1/2)(\sum A^{-1}_{i,j}b^ib^j)A^{-1}_{1,2}] = $

    $\displaystyle \partial_4[A^{-1}_{2,3}( \sum A^{-1}_{1,j}b^j)+(\sum A^{-1}_{2,j}b^j)A^{-1}_{1,3}+(\sum A^{-1}_{3,j}b^j)A^{-1}_{1,2}] = $

    $\displaystyle A^{-1}_{2,3}A^{-1}_{1,4} + A^{-1}_{2,4}A^{-1}_{1,3} + A^{-1}_{3,4}A^{-1}_{1,2}~.$

    Similarly:

    $\displaystyle \partial_4\partial_3\partial_1\partial_1~~{\rm gives~~}2A^{-1}_{1,4}A^{-1}_{1,3} + A^{-1}_{3,4}A^{-1}_{1,1}~.$

    $\displaystyle \partial_4\partial_1\partial_1\partial_1~~{\rm gives~~} 3A^{-1}_{1,4}A^{-1}_{1,1}~.$

    $\displaystyle \partial_4\partial_4\partial_1\partial_1~~{\rm gives~~} 2 A^{-1}_{1,4}A^{-1}_{1,4} + A^{-1}_{4,4}A^{-1}_{1,1}~.$

    $\displaystyle \partial_1\partial_1\partial_1\partial_1~~{\rm gives~~} 3 A^{-1}_{1,1}A^{-1}_{1,1}~.$