Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\phi$ be a scalar field and then I see the following expression (1) for the square of the normal ordered version of $\phi^2(x)$.

$$T(:\phi^2(x)::\phi^2(0):) ~=~ 2<0|T(\phi(x)\phi(0))|0>^2 $$ $$ ~+~ 4<0|T(\phi(x)\phi(0))|0>:\phi(x)\phi(0): ~+~ :\phi^2(x)\phi^2(0): \tag{1}$$

It would be great if someone can help derive the above expression - may be from scratch - and without outsourcing to Wick's theorem - and may be help connect as to why the above is related (equal?) to the Wick's theorem?

  • Isn't the above also known as OPE (Operator Product Expansion)? If yes, then is there at all any difference between OPE and Wick's theorem? Is there a systematic way to derive such OPEs?

  • Can one help extend this to Fermions?

share|cite|improve this question

migrated from Apr 21 '12 at 13:29

Apologies but this question seems to be asking "please teach me Wick's theorem" - and it moreover says "do it without teaching me Wick's theorem". Have you tried to study Wick's theorem? At least ? The standard pedagogical treatment answers all your questions. The Wick's theorem is the systematic way to construct such identities that you're looking for. Fermions differ by some signs only. I find it questionable whether copying/rephrasing sections from standard textbook material is a good way to use this server and people's time. – Luboš Motl Apr 21 '12 at 11:57
@LubošMotl stack exchange's philosophy is to have answers to as many physics questions as possible in one place, regardless of where else there might be a solution. I find it questionable whether complaining about questions I find boring is a good way to use this server and people's time. – Anarchist Birds Worship Fungus Jun 23 at 13:03
Dear Anarchist, as you can see through the number "6" next to my comment, most users agree with me, however. It is extremely important on this server to have quality control both on answers as well as on questions. If the server were completely overwhelmed by very unintelligent questions or questions that are self-contradictory or easy to be answered by a straightforward Google search, it would mean that the server becomes a useless redundant cesspool and top users' brains are wasted. Maybe you wouldn't care but the people who care about this server and have a reputation here do care. – Luboš Motl Jun 23 at 13:13
up vote 10 down vote accepted

As Lubos Motl mentions in a comment, for all practical purposes, OP's sought-for eq. (1) is proved via Wick's Theorem.

It is interesting to try to generalize Wick's Theorem and to try to minimize the number of assumptions that goes into it. Here we will outline one possible approach.

I) Assume that a family $(\hat{A}_i)_{i\in I}$ of operators $\hat{A}_i\in{\cal A}$ lives in a (super) operator algebra ${\cal A}$

  1. with (super) commutator $[\cdot,\cdot]$, and

  2. with center $Z({\cal A})$.


  1. the index $i\in I$ runs over an index set $I$ (it could be continuous), and

  2. the index $i$ contains information, such as, e.g., position $x$, time instant $t$, annihilation/creation label, type of field, etc., of the operator $\hat{A}_i$.

II) Assume that $$ \forall i,j\in I~: \qquad [\hat{A}_i,\hat{A}_j] ~\in~Z({\cal A}). $$

III) Assume that there are given two ordering prescriptions, say $T$ and $::$. Here $T$ and $::$ could in principle denote any two ordering prescriptions, e.g. time order, normal order, radial order, etc. This means that the index set $I$ is endowed with two strict total orders, say, $<$ and $\prec$, respectively, such that

  1. The $T$ symbol is (graded) multilinear wrt. supernumbers.

  2. $ T(\hat{A}_{\pi(i_1)}\ldots\hat{A}_{\pi(i_n)})~=~\pm T(\hat{A}_{i_1}\ldots\hat{A}_{i_n} )$ is (graded) symmetric, where $\pi\in S_n$ is a permutation of $n$ elements.

  3. $ T(\hat{A}_{i_1}\ldots\hat{A}_{i_n} )~=~\hat{A}_{i_1}\ldots\hat{A}_{i_n}$ if $i_1 > \ldots > i_n$. [A similar condition should hold for the second ordering $(::,\prec)$.]

  4. In the special case where some of the $ i_1 , \ldots , i_n$ are equal${}^{\dagger}$ (wrt. the order <), then one should symmetrize in appropriate (graded) sense over the corresponding subsets. For instance, $$ T(\hat{A}_{i_1}\ldots\hat{A}_{i_n} )~=~\hat{A}_{i_1}\ldots\hat{A}_{i_{k-1}}\frac{\hat{A}_{i_k}\hat{A}_{i_{k+1}}\pm\hat{A}_{i_{k+1}}\hat{A}_{i_k}}{2}\hat{A}_{i_{k+2}}\ldots\hat{A}_{i_n}$$ if $i_1 > \ldots > i_k=i_{k+1}> \ldots > i_n$. [A similar condition should hold for the second ordering $(::,\prec)$.]

IV) It then follows from assumptions I-III that the (generalized) contractions $$ \hat{C}_{ij}~=~T(\hat{A}_i\hat{A}_j)~-~:\hat{A}_i\hat{A}_j:~\in~Z({\cal A}) $$ belong to the center $Z({\cal A})$. [By the way, physicists will often casually refer to the operators in the center $Z({\cal A})$ as $c$-numbers.] The contractions are graded symmetric $$ \hat{C}_{ij}~=~\pm \hat{C}_{ji}. $$

V) It is now a straightforward exercise to establish the corresponding Wick's Theorem $$ T(f(\hat{A})) ~=~ \exp\left(\frac{1}{2}\sum_{i,j\in I}\hat{C}_{ij}\frac{\partial}{\partial\hat{A}_j}\frac{\partial}{\partial\hat{A}_i} \right):f(\hat{A}):, $$ meaning a rule for how to re-express one ordering prescription $T(f(\hat{A}))$ [where $f$ is a sufficiently nice function of the $(\hat{A}_i)_{i\in I}$ family] in terms of the other ordering prescription $::$ and (multiple) contractions $\hat{C}_{ij}$. And vice-versa with the roles of the two orderings $T$ and $::$ interchanged. Such Wick's Theorems can now be applied successively to establish nested${}^{\ddagger}$ Wick's Theorems. These Wick's Theorems may be extended to a larger class of operators than just the $(\hat{A}_i)_{i\in I}$ family through (graded) multilinearity.

VI) Let us now assume that the operators $\hat{A}_i$ are Bosonic for simplicity. A particular consequence of a nested Wick's Theorem is the following version

$$T(:\hat{A}^2_i::\hat{A}^2_j:) ~=~ 2\hat{C}_{ij}^2 + 4 \hat{C}_{ij}:\hat{A}_i\hat{A}_j: + :\hat{A}^2_i\hat{A}^2_j:$$

of OP's sought-for eq. (1). Finally, let us mention that Wick's Theorem, radial order, OPE, etc., are also discussed in this and this Phys.SE posts.



${}^{\dagger}$ Being equal wrt. an order is in general an equivalence relation, and it is often a weaker condition than being equal as elements of $I$.

${}^{\ddagger}$ A nested Wick's Theorem (between radial order and normal order) is briefly stated in eq. (2.2.10) on p. 39 in J. Polchinski, String Theory, Vol. 1. Beware that radial order is often only implicitly written in CFT texts. I'll update the answer if I find a better reference. By the way, a side-effect/peculiarity of nested ordering symbols are discussed in this Phys.SE post.

share|cite|improve this answer
Having gone through may be 2 or 3 quite advanced courses in QFT, this still looks to me to be something quite unfamiliar and hence quite surprised at Lubos's comment. What you call as a "straightforward excercise" doesn't seem to be so at all - can you kindly fill in a few more steps! – user6818 Apr 22 '12 at 19:54
Your definition of a "contraction" also looks new - I would think that a "contraction' of two fields means the vacuum expectation value of the time-ordered product. And by Wick's theorem, I would understand the statement that the difference between the time-ordered product and the normal odered product is a sum over all possible contractions. So how is what you are doing the same as Wick's theorem? ( Lubos seems to be claiming..) – user6818 Apr 22 '12 at 19:55
I plan to update the answer soon. – Qmechanic Apr 22 '12 at 20:22
Thanks a lot! I am eagerly looking forward to a detailed update from you. – user6818 Apr 23 '12 at 15:43
I like this answer just the way it is, as it makes it very clear that Wick's theorem is not a theorem about QFT, but is rather a completely general theorem about algebras obeying a few simple axioms. The standard textbook treatment with fields and VEV's obfuscates what is really just basic algebra. – Jonathan Apr 25 '12 at 0:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.