4.2 Wick’s theorem
Evaluating the right-hand side of the Gell-Mann and Low reduction formula (4.16) turns out to be rather cumbersome. One has to compute a series of time-ordered products
\(\seteqnumber{0}{4.}{16}\)\begin{flalign} \expect {0}{\TO \Big (\Phi _0(x_1)\cdots \Phi _0(x_n) \, e^{\ii \,S_{\mathrm {int}}[\Phi _0]}\Big )}{0} = \sum _{m\geq 0}\frac {\ii ^m }{m!} \expect {0}{\TO \Big (\Phi _0(x_1)\cdots \Phi _0(x_n) \,\big (S_{\mathrm {int}}[\Phi _0]\big )^m\Big )}{0}\quad , \end{flalign} where we recall that the interaction term \(S_{\mathrm {int}}[\Phi _0]\) is a polynomial in the free quantum fields \(\Phi _0\) and hence it adds to the number of fields under the time-ordered product. This means that, even for computing the interacting time-ordered \(n\)-point functions \(\expect {\Omega }{\TO \big (\Phi (x_1)\cdots \Phi (x_n)\big )}{\Omega }\) for a very low \(n\), we have to know the free time-ordered \(n\)-point functions for a much larger \(n\), which is determined by the form of the interaction and by the order \(m\) of the perturbative expansion we wish to reach. For example, if we take \(S_{\mathrm {int}}[\Phi _0] = -\frac {\lambda }{4!}\int _{\bbR ^{d}} \big (\Phi _0(x)\big )^4\,\dd x\) to be a quartic interaction and want to determine the interacting time-ordered \(2\)-point function to order \(\lambda ^2\), we have to compute
\(\seteqnumber{0}{4.}{17}\)\begin{flalign} \nn \expect {0}{\TO \Big (&\Phi _0(x_1)\,\Phi _0(x_2) \, e^{\ii \,S_{\mathrm {int}}[\Phi _0]}\Big )}{0} = \expect {0}{\TO \Big (\Phi _0(x_1)\,\Phi _0(x_2)\Big )}{0}\\ \nn &+ \frac {-\ii \,\lambda }{4!} \int _{\bbR ^d} \expect {0}{\TO \Big (\Phi _0(x_1)\,\Phi _0(x_2)\,\big (\Phi _0(y)\big )^4\Big )}{0}\,\dd y\\ &+ \frac {1}{2!} \,\bigg (\frac {-\ii \,\lambda }{4!}\bigg )^2\, \int _{\bbR ^{2d}} \expect {0}{\TO \Big (\Phi _0(x_1)\,\Phi _0(x_2)\,\big (\Phi _0(y_1)\big )^4\,\big (\Phi _0(y_2)\big )^4\Big )}{0}~\dd y_1\,\dd y_2 + \mathcal {O}(\lambda ^3)\quad . \end{flalign} This involves the \(2\), \(6\) and \(10\)-point functions, and when we would like to know the order \(\lambda ^3\) contribution then we also need the \(14\)-point function. (I think the pattern becomes clear, isn’t it?) So what we need are techniques to determine the time-ordered \(n\)-point functions \(\expect {0}{\TO \big (\Phi _0(x_1)\cdots \Phi _0(x_n)\big )}{0}\) of the free theory for arbitrary \(n\).
A suitable technique for this task is provided by Wick’s theorem, which allows us to rewrite the time-ordered product \(\TO \big (\Phi _0(x_1)\cdots \Phi _0(x_n)\big )\) as a sum of normal ordered products (see Definition 3.1) and contractions given by the Feynman propagator. Before we state and prove Wick’s theorem in its full form, let us explore this problem for low \(n\) in order to get some intuition. For \(n=1\), we have the obvious identity
\(\seteqnumber{0}{4.}{18}\)\begin{flalign} \label {eqn:Wick1} \TO \big (\Phi _0(x_1)\big ) = \noor {\Phi _0(x_1)} =\Phi _0(x_1)\quad , \end{flalign} which tells us that both the time-ordered product and the normal ordering of \(\Phi _0(x_1)\) agree with the field operator \(\Phi _0(x_1)\) itself. Consider now the case \(n=2\). Recalling the expression (4.4b) of the free field operator in terms of annihilation and creation operators, let us write \(\Phi _0(x) = a(x) + a^\dagger (x)\) to denote its decomposition into the annihilating and the creating part. Explicitly,
\(\seteqnumber{0}{4.}{19}\)\begin{flalign} \label {eqn:anncredecomposition} a(x) := \int _{\bbR ^{d-1}} \frac {1}{\sqrt {2\omega _{\mathbf {k}}}}~ a(\mathbf {k}) \,e^{\ii \,k\,x}\,\frac {\dd \mathbf {k}}{(2\pi )^{d-1}}\quad ,\quad a^\dagger (x):= \int _{\bbR ^{d-1}} \frac {1}{\sqrt {2\omega _{\mathbf {k}}}}~ a^\dagger (\mathbf {k})\, e^{-\ii \,k\,x}\,\frac {\dd \mathbf {k}}{(2\pi )^{d-1}}\quad . \end{flalign} For \(x_1^0\geq x_2^0\), we can then write
\(\seteqnumber{1}{4.21}{0}\)\begin{flalign} \nn \TO \big (\Phi _0(x_1)\,\Phi _0(x_2)\big ) &= \Phi _0(x_1)\,\Phi _0(x_2)\\ \nn &= a(x_1)\,a(x_2) + a(x_1)\,a^\dagger (x_2) + a^\dagger (x_1)\,a(x_2) + a^{\dagger }(x_1)\,a^\dagger (x_2)\\ \nn &= a(x_1)\,a(x_2) + \big [ a(x_1), a^\dagger (x_2)\big ] + a^\dagger (x_2)\,a(x_1) + a^\dagger (x_1)\,a(x_2) + a^{\dagger }(x_1)\,a^\dagger (x_2)\\ &=\noor {\Phi _0(x_1)\,\Phi _0(x_2)} + \big [ a(x_1), a^\dagger (x_2)\big ] = \noor {\Phi _0(x_1)\,\Phi _0(x_2)} + W_2(x_1,x_2) \quad , \end{flalign} where \(W_2\) is the \(2\)-point Wightman function (3.49b), while for \(x_2^0 \geq x_1^0\) we find
\(\seteqnumber{1}{4.21}{1}\)\begin{flalign} \TO \big (\Phi _0(x_1)\,\Phi _0(x_2)\big ) = \noor {\Phi _0(x_2)\,\Phi _0(x_1)} + W_2(x_2,x_1) \quad . \end{flalign} Since the normal ordering \(\noor {\Phi _0(x_1)\,\Phi _0(x_2)} = \noor {\Phi _0(x_2)\,\Phi _0(x_1)}\) is symmetric under the exchange of \(x_1\) and \(x_2\), we find the following case-independent identity
\(\seteqnumber{0}{4.}{21}\)\begin{flalign} \TO \big (\Phi _0(x_1)\,\Phi _0(x_2)\big ) = \noor {\Phi _0(x_1)\,\Phi _0(x_2)} + \Delta _F(x_1-x_2)\quad , \end{flalign} where \(\Delta _F\) is the Feynman propagator (3.56b). Introducing the shorthand notation
\(\seteqnumber{0}{4.}{22}\)\begin{flalign} \Phi _i \,:=\,\Phi _0(x_i)\quad ,\quad (\Delta _F)_{ij}\,:=\, \Delta _{F}(x_i-x_j)\quad , \end{flalign} we can write this more compactly as
\(\seteqnumber{0}{4.}{23}\)\begin{flalign} \label {eqn:Wick2} \TO \big (\Phi _1\,\Phi _2\big ) = \noor {\Phi _1\,\Phi _2} + (\Delta _F)_{12}\quad . \end{flalign} Playing the same game for \(n=3\), you will find that
\(\seteqnumber{0}{4.}{24}\)\begin{flalign} \label {eqn:Wick3} \TO \big (\Phi _1\,\Phi _2\,\Phi _3\big ) = \noor {\Phi _1\,\Phi _2\,\Phi _3}\,+ (\Delta _F)_{12} \,\noor {\Phi _3} + (\Delta _F)_{13}\,\noor {\Phi _2} + (\Delta _F)_{23} \,\noor {\Phi _1} \quad , \end{flalign} and for \(n=4\) you will find that
\(\seteqnumber{0}{4.}{25}\)\begin{flalign} \nn \TO \big (\Phi _1\,\Phi _2\,\Phi _3\,\Phi _4\big ) &= \noor {\Phi _1\,\Phi _2\,\Phi _3\,\Phi _4}\,+ (\Delta _F)_{12} \,\noor {\Phi _3\,\Phi _4} + (\Delta _F)_{13} \,\noor {\Phi _2\,\Phi _4} + (\Delta _F)_{14} \,\noor {\Phi _2\,\Phi _3} \\ \nn &\quad + (\Delta _F)_{23} \,\noor {\Phi _1\,\Phi _4} + (\Delta _F)_{24} \,\noor {\Phi _1\,\Phi _3} + (\Delta _F)_{34} \,\noor {\Phi _1\,\Phi _2} \\ &\quad + (\Delta _F)_{12}\,(\Delta _F)_{34} + (\Delta _F)_{13}\,(\Delta _F)_{24} + (\Delta _F)_{14}\,(\Delta _F)_{23}\quad .\label {eqn:Wick4} \end{flalign} Observe that there emerges the following pattern: The time-ordered product of \(n\) field operators seems to be equal to the normal ordering of the field operators, plus all possible contractions with the Feynman propagator, plus all possible double contractions with the Feynman propagator, and so on. This is indeed the case. To give a precise statement and proof of this result, we need some additional notation. Given two positive integers \(n\) and \(q\), we denote by \(\mathcal {P}_n^{q}\) the set of all collections \(\{\{i_1,j_1\},\dots , \{i_q,j_q\}\}\) of \(q\) pairs of elements in \(I_n := \{1,\dots ,n\}\), such that any of the \(2q\) elements \(i_1,\dots ,i_q,j_1,\dots ,j_q\in I_n\) are distinct. By convention, we set \(\mathcal {P}_n^0\) to be the set with a single element given by the empty collection \(\{\}\). Note that \(\mathcal {P}_n^q = \emptyset \) whenever \(2q>n\) because in this case there are not enough elements in \(I_n=\{1,\dots ,n\}\) for the \(2q\) elements to be distinct. We denote elements of \(\mathcal {P}_n^{q}\) by the shorthand notation \(\underline {\{i,j\}} := \{\{i_1,j_1\},\dots , \{i_q,j_q\}\}\) and write \(I_n\setminus \underline {\{i,j\}}:= I_n\setminus \bigcup _{l=1}^q\{i_l,j_l\}\) for the complement in \(I_n = \{1,\dots ,n\}\) of all the pairs.
-
Theorem 4.1 (Wick’s theorem for time-ordered products). With the notations established above, the following identity holds true for all positive integers \(n\)
\(\seteqnumber{0}{4.}{26}\)\begin{equation} \label {eqn:Wickn} \TO \bigg (\prod _{k=1}^n \Phi _k\bigg ) = \sum _{q=0}^{\lfloor \frac {n}{2}\rfloor } \sum _{\underline {\{i,j\}}\in \mathcal {P}_n^{q}} \,\prod _{l=1}^q (\Delta _F)_{i_lj_l}~~\noor {\prod _{k\in I_n\setminus \underline {\{i,j\}} } \Phi _k } \quad , \end{equation}
where we recall that \(\lfloor \frac {n}{2}\rfloor \) denotes the greatest integer less than or equal to \(\frac {n}{2}\).
-
Remark 4.2. I have to admit that this formula looks a bit heavy, but I don’t see any better way to write it down in a closed form. What is usually presented in the physics literature is a less precise, but easier to read and digest, formula of the form
where the contraction symbols mean insertions of Feynman propagators, i.e.
. By direct inspection you can convince yourself that this formula is the same as (4.27).
-
Proof. We have seen above that (4.27) holds true for \(n=1,2,3,4\), see (4.19), (4.24), (4.25) and (4.26). We prove the general result by induction. Suppose that (4.27) holds true for \(n\). Consider \(n+1\) fields \(\Phi _i=\Phi _0(x_i)\) at the positions \(x_1,\dots ,x_{n+1}\in \bbR ^d\) that we may assume without loss of generality to be time-ordered. We can then write
\(\seteqnumber{0}{4.}{28}\)\begin{flalign} \label {eqn:Wickproof1} \TO \bigg (\prod _{k=1}^{n+1} \Phi _k\bigg ) = \TO \bigg (\prod _{k=1}^{n} \Phi _k\bigg ) ~\Phi _{n+1} =\bigg (\sum _{q=0}^{\lfloor \frac {n}{2}\rfloor } \sum _{\underline {\{i,j\}}\in \mathcal {P}_n^{q}} \,\prod _{l=1}^q (\Delta _F)_{i_lj_l}~~\noor {\prod _{k\in I_n\setminus \underline {\{i,j\}} } \Phi _k }\bigg )\,\Phi _{n+1}\quad . \end{flalign} Decomposing \(\Phi _{n+1} = a_{n+1} + a^\dagger _{n+1}\) into the annihilating and creating parts (4.20), we compute
\(\seteqnumber{0}{4.}{29}\)\begin{flalign} \nn \noor {\prod _{k\in I_n\setminus \underline {\{i,j\}} } \Phi _k } ~~\Phi _{n+1} &= \noor {\prod _{k\in I_n\setminus \underline {\{i,j\}} } \Phi _k} \, a_{n+1} ~+~ a^\dagger _{n+1}\, \noor {\prod _{k\in I_n\setminus \underline {\{i,j\}} } \Phi _k } ~+~ \bigg [\noor {\prod _{k\in I_n\setminus \underline {\{i,j\}} } \Phi _k },a^\dagger _{n+1}\bigg ]\\ &=\noor {\prod _{k\in I_{n+1}\setminus \underline {\{i,j\}} } \Phi _k} ~+~ \sum _{r \in I_n\setminus \underline {\{i,j\}}} (\Delta _{F})_{r (n+1)} ~ \noor {\prod _{\substack {k\in I_n\setminus \underline {\{i,j\}} \\ k\neq r}} \Phi _k }\quad . \end{flalign} To obtain the first term of the second line, we have used that the first two terms of the first line correspond to including \(\Phi _{n+1}\) into the normal ordered product. (Note the \(I_{n+1}\) instead of \(I_n\).) To obtain the second term of the second line, we have used the Leibniz rule to evaluate the commutator. Inserting this result back into (4.29), we get
\(\seteqnumber{0}{4.}{30}\)\begin{flalign} \nn \TO \bigg (\prod _{k=1}^{n+1} \Phi _k\bigg ) &= \sum _{q=0}^{\lfloor \frac {n}{2}\rfloor } \sum _{\underline {\{i,j\}}\in \mathcal {P}_n^{q}} \,\prod _{l=1}^q (\Delta _F)_{i_lj_l}~~\noor {\prod _{k\in I_{n+1}\setminus \underline {\{i,j\}} } \Phi _k }\\ \nn &\qquad ~\quad + \sum _{q=0}^{\lfloor \frac {n}{2}\rfloor } \sum _{\underline {\{i,j\}}\in \mathcal {P}_n^{q}} \sum _{r \in I_n\setminus \underline {\{i,j\}}} \,\bigg (\prod _{l=1}^q (\Delta _F)_{i_lj_l}\bigg )\,(\Delta _F)_{r (n+1)}~ \noor {\prod _{\substack {k\in I_n\setminus \underline {\{i,j\}} \\ k\neq r}} \Phi _k }\\ &=\sum _{q=0}^{\lfloor \frac {n+1}{2}\rfloor } \sum _{\underline {\{i,j\}}\in \mathcal {P}_{n+1}^{q}} \, \prod _{l=1}^q (\Delta _F)_{i_lj_l}~~\noor {\prod _{k\in I_{n+1}\setminus \underline {\{i,j\}} } \Phi _k }\quad . \end{flalign} To understand the last equality, note that an element \(\underline {\{i,j\}} = \{\{i_1,j_1\},\dots ,\{i_q,j_q\}\}\in \mathcal {P}_{n+1}^{q}\) either does or does not include \(n+1\). If it does not include \(n+1\), then the corresponding term is part of the first line, and if it does include \(n+1\) then it is part of the second line. So all terms match up in the last equality and we have completed the proof. □
Taking the vacuum expectation value of the time-ordered products in (4.27) simplifies considerably the expression on the right-hand side because \(\expect {0}{\noor {\Phi _1\cdots \Phi _k}}{0} =0\) for any number \(k\geq 1\) of field operators. This is because, by Definition 3.1, the operator \(\noor {\Phi _1\cdots \Phi _k}\) is such that all annihilation operators are to the right and all creation operators are to the left, hence each summand vanishes using either \(a(\mathbf {k})\ket {0}=0\) or its adjoint \(\bra {0}a^{\dagger }(\mathbf {k}) =0\). Let us state this observation as a corollary.
-
Corollary 4.3 (Wick’s theorem for time-ordered \(n\)-point functions). The free time-ordered \(2n+1\)-point function for an odd number of field operators vanishes
\(\seteqnumber{1}{4.32}{0}\)\begin{equation} \expect {0}{\TO \bigg (\prod _{k=1}^{2n+1} \Phi _k\bigg )}{0} =0\quad . \end{equation}
The free time-ordered \(2n\)-point function for an even number of field operators is given by
\(\seteqnumber{1}{4.32}{1}\)\begin{equation} \expect {0}{\TO \bigg (\prod _{k=1}^{2n} \Phi _k\bigg )}{0} = \sum _{\underline {\{i,j\}}\in \mathcal {P}^n_{2n}} \prod _{l=1}^n (\Delta _F)_{i_l j_l} \quad , \end{equation}
where we recall that \(\mathcal {P}^n_{2n}\) denotes the set of all collections \(\{\{i_1,j_1\},\dots , \{i_n,j_n\}\}\) of \(n\) pairs of elements in \(I_{2n} = \{1,\dots ,2n\}\), such that any of the \(2n\) elements \(i_1,\dots ,i_n,j_1,\dots ,j_n\in I_{2n}\) are distinct. Note that \(\mathcal {P}^n_{2n}\) is precisely the set of partitions of the set \(I_{2n} = \{1,\dots ,2n\}\) into pairs.
-
Example 4.4. The first few time-ordered \(n\)-point functions read as follows
\(\seteqnumber{0}{4.}{32}\)\begin{flalign} \nn \expect {0}{\TO \big (\Phi _1\big )}{0} &= 0\quad ,\\ \nn \expect {0}{\TO \big (\Phi _1\,\Phi _2\big )}{0} &= (\Delta _{F})_{12}\quad ,\\ \nn \expect {0}{\TO \big (\Phi _1\,\Phi _2\,\Phi _3\big )}{0} &= 0\quad ,\\ \nn \expect {0}{\TO \big (\Phi _1\,\Phi _2\,\Phi _3\,\Phi _4\big )}{0} &= (\Delta _F)_{12} \, (\Delta _F)_{34} + (\Delta _F)_{13} \, (\Delta _F)_{24} + (\Delta _F)_{14} \, (\Delta _F)_{23} \quad ,\\ \expect {0}{\TO \big (\Phi _1\,\Phi _2\,\Phi _3\,\Phi _4\,\Phi _5\big )}{0} &= 0\quad . \end{flalign} The higher time-ordered \(2n\)-point functions are considerably more involved, because there exist \(\frac {(2n)!}{2^n\,n!}\) many partitions of \(\{1,\dots ,2n\}\) into pairs. This corresponds to \(\frac {(2n)!}{2^n\,n!}\) many terms in the sum (4.32), which quickly becomes very large. For instance, for the \(6 = 2\times 3\)-point function, there are \(\frac {6!}{2^3\,3!} = 15\) many terms, and for the \(8=2\times 4\)-point function, there are \(\frac {8!}{2^4\,4!} = 105\) many terms. This high complexity suggests that it would be a good idea to use computer assistance in order to evaluate Wick’s theorem, which is indeed what QFT practitioners are doing.