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.

Theorem 1   (Wick's theorem)

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

where the sum is taken over all pairings $(i_{p_1},i_{p_2}), \dots, (i_{p_{m-1}},i_{p_m})$ of $i_1, \dots, i_m$.

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 sum running from 1 to $d$), using the series expansion $\exp x = 1 + x + x^2/2! + x^3/3! + \cdots.$ 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 $b^i$ of degree $2n$.

Differentiating $m$ times a homogeneous polynomial of degree $2n$ and evaluating at zero will give zero unless $k=2m$. 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

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

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 $\partial_i$.

• $n=1$.
  $$\partial_2 \partial_1(\frac{1}{2}\sum A^{-1}_{i,j}b^ib^j) =$$
  $$\partial_2 ( \sum_j A^{-1}_{1,j}b^j ) =$$
  $$A^{-1}_{1,2}~,$$

  using the symmetry of the matrix $A^{-1}$. The same calculation shows that $\partial_1 \partial_1(\frac{1}{2}\sum A^{-1}_{i,j}b^ib^j) = A^{-1}_{1,1}~.$

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

• $n=2$.
  $$\partial_4\partial_3\partial_2\partial_1(1/2!)(1/2^2)(\sum A^{-1}_{i,j}b^ib^j)^2 =$$
  $$\partial_4\partial_3\partial_2 (1/2)(\sum A^{-1}_{i,j}b^ib^j)( \sum A^{-1}_{1,j}b^j) =$$
  $$\partial_4\partial_3 [(\sum A^{-1}_{2,j}b^j)( \sum A^{-1}_{1,j}b^j) +(1/2)(\sum A^{-1}_{i,j}b^ib^j)A^{-1}_{1,2}] =$$
  $$\partial_4[A^{-1}_{2,3}( \sum A^{-1}_{1,j}b^j)+(\sum A^{-1}_{2,j}b^j)A^{-1}_{1,3} +(\sum A^{-1}_{3,j}b^j)A^{-1}_{1,2}] =$$
  $$A^{-1}_{2,3}A^{-1}_{1,4} + A^{-1}_{2,4}A^{-1}_{1,3} + A^{-1}_{3,4}A^{-1}_{1,2}~.$$

  Similarly:

  $$\partial_4\partial_3\partial_1\partial_1(1/2!)(1/2^2)(\sum A^{-1}_{i,j}b^ib^j)^2 = 2A^{-1}_{1,4}A^{-1}_{1,3} + A^{-1}_{3,4}A^{-1}_{1,1}~.$$
  $$\partial_4\partial_1\partial_1\partial_1(1/2!)(1/2^2)(\sum A^{-1}_{i,j}b^ib^j)^2 = 3A^{-1}_{1,4}A^{-1}_{1,1}~.$$
  $$\partial_4\partial_4\partial_1\partial_1(1/2!)(1/2^2)(\sum A^{-1}_{i,j}b^ib^j)^2 = 2 A^{-1}_{1,4}A^{-1}_{1,4} + A^{-1}_{4,4}A^{-1}_{1,1}~.$$

  And

  $$\partial_1\partial_1\partial_1\partial_1(\sum A^{-1}_{i,j}b^ib^j)^2 = 3 A^{-1}_{1,1}A^{-1}_{1,1}~.$$