Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

This is a follow-up question to What are Wightman fields/functions

Ok, so based on my reading, the field operators of a theory are understood to be operator-valued distributions, that is, to be integrated over a smooth function: e.g. $\hat\phi(f)≡\int dx\, f(x) \hat\phi(x)$. But if the smooth function has support over such a large region in space-time that if I try to compute time-ordered products, I run into some ambiguity. How do I define the time-ordering symbol for Wightman functions?

share|improve this question
Is there anything wrong with this definition : - $T(\hat\phi_1(f_1)..\hat\phi_n(f_n))≡\int dx_1..dx_n\, f_1(x_1)..f_n(x_n) T(\hat\phi_1(x_1)..\hat\phi_n(x_n))$ ? –  user10001 Nov 18 '12 at 0:56
@dushya: That's a circular definition. The question really is how to define time-ordering when you're thinking of the field as a distribution, in which case $\hat{\phi}(x_1)$ isn't defined. –  user1504 Nov 18 '12 at 2:35
@user1504 you are right, but is this one fine ? :- $$T(\hat\phi_1(f_1)..\hat\phi_n(f_n))≡ Lim_{\{\epsilon_i\}\rightarrow0}\int dx_1..dx_n\, f_1(x_1)...f_n(x_n) T(\hat\phi_1(g_{x_1,\epsilon_1})...\hat\phi_n(g_{x_n,\epsilon_n}))$$ where for each $i$, $g_{x_i,\epsilon_i}(y)$ converges to $\delta(y-x_i)$ when $\epsilon_i$ is taken to zero. –  user10001 Nov 18 '12 at 3:48
@dushya: That one will work, although it takes a bit of work to show that it does. You're asserting that the product of Wightman fields has a kernel. –  user1504 Nov 18 '12 at 13:13
add comment

2 Answers

up vote 6 down vote accepted

@ArnoldNeumaier and @dushya have both pointed out correct solutions, but I want to elaborate a bit.

The easy approach is the one dushya suggested. (You can also do what Arnold Neumaier suggests: First define the time-ordered product of operator-valued distributions, and then take expectation values.)

Begin by recalling how Wightman functions are defined. The correlation function of $n$ smeared fields is a multilinear functional $(f_1,...,f_n) \mapsto \langle vac | \hat{\phi}(f_1) ... \hat{\phi}(f_n)|vac\rangle$. You can use the Schwarz nuclear theorem to prove this VEV has a kernel $W$ function such that

$\langle vac |\hat{\phi}(f_1) ... \hat{\phi}(f_n)|vac\rangle = \int f(x_1)...f(x_n) W(x_1,...,x_n) dx_1...dx_n$

This $W$ is a Wightman function. Morally, $W$ is $\langle vac | \hat{\phi}(x_1)...\hat{\phi}(x_n)|vac\rangle$.

Now, if you want to define a time-ordered correlation function, all you have to do is permute the arguments in the Wightman function.

There's a somewhat more conceptual approach available, via analytic continuation from Minkowski signature to Euclidean and then back.

You can prove that the Wightman functions $W(x_1,...x_n)$ are boundary values of an analytic function defined on (a domain in) the product of $n$-copies of the complexification of Minkowski space. If you restrict this analytic function to the Euclidean subspace, the function you get -- known as a Schwinger function -- is permutation-invariant. (It has to be, because it's the output of a Euclidean functional integral constructed with only commuting variables.)

So where did the ordering of operators go? It's in the analytic continuation. The Schwinger functions are analytic, but not entire. The analytic continuation of a Schwinger function has one branch for each of the possible orderings of the arguments. So you can get the time-ordered correlation functions by continuing the Wightman functions from Minkowski space to Euclidean space and then looping back along a different branch.

The basic picture above -- Wightman functions and their analyicity -- is explained beautifully in Streater & Wightman's book PCT, Spin, Statistics, and All That. It's also in the 2nd of Kazhdan's IAS lectures. (Although the heavy lifting needed to show that the time-ordered Wightman functions are well-defined seems to be in Epstein & Eckmann's Time-ordered products and Schwinger functions.) It's also used regularly in particle physics texts, when computing propagators in momentum space. (Have a look at the contour integrals in Peskin & Schroeder's discussion of the Klein-Gordon propagator.)

share|improve this answer
This is not correct. Since $\phi(f)$ and $\phi(g)$ do not commute unless $f$ and $g$ have causally independent support, $W$ is not symmetric in the coordinates, while the time-ordered expectation is. One gets Wightman from Schwinger by Osterwalder-Schrader theory, but not the time-ordered version. The naive version o the latter that you describe has ambiguities whenever two arguments coincide, which can be resolved only by appropriate renormalization (distribution splitting). –  Arnold Neumaier Nov 18 '12 at 16:19
@ArnoldNeumaier: I didn't say the Minkowski signature W was symmetric. I said that the Euclidean signature Schwinger function was symmetric. And one can certainly recover the time-ordered Wightman functions from Schwinger. See page 9 of lecture 2 of Kazhdan. –  user1504 Nov 18 '12 at 16:25
Well, you said that $W$ was morally the vacuum expectation value of $T$, which would make it symmetric. - Kazhdan's definition of the time-ordered Wightman functions requires an additional twist compared to what you described. I wonder if his limit is always well-defined as a distribution on $(R^d)^n$. Do you have a reference for the proof? –  Arnold Neumaier Nov 18 '12 at 16:34
@ArnoldNeumaier: That was a typo (copy-paste from earlier draft)! Thanks for pointing it out! –  user1504 Nov 18 '12 at 16:35
+1 after correction. –  Arnold Neumaier Nov 18 '12 at 16:39
show 3 more comments

The time-ordering operator is well-defined rigorously through the Epstein-Glaser approach of distribution splitting. You can read about this in Scharf's book on QED, or http://arxiv.org/abs/arXiv:0906.1952. See also the summary in http://de.wikipedia.org/wiki/FQFT

share|improve this answer
add comment

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.