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$:\!\!\hat\phi(x)^2\!\!:$, for example, constructed from the real Klein-Gordon quantum field.

For a Wightman field, the Wightman function $\left<0\right|\hat\phi(x)\hat\phi(y)\left|0\right>$ is a distribution, which is certainly the case for the real Klein-Gordon quantum field --- call it $C(x-y)$ in this case. In contrast, the expected value $\left<0\right|:\!\!\hat\phi(x)^2\!\!:\,:\!\!\hat\phi(y)^2\!\!:\left|0\right>$ is a product of distributions, in fact for the real Klein-Gordon quantum field it's $2C(x-y)^2$, which seems to make this normal-ordered product of quantum fields not a Wightman field. $C(x-y)^2$ is well-enough behaved off the light-cone, but on the light-cone it has a $[\delta((x-y)^2)]^2$ component.

If $:\!\!\hat\phi(x)^2\!\!:$ is not a Wightman field, then is it nonetheless in the Borchers' equivalence class of the free field? If so, why so? A (mathematically clear) citation would be nice!

Finally, if $:\!\!\hat\phi(x)^2\!\!:$ is not in the Borchers' equivalence class of the free field, because it isn't a distribution, is it nonetheless empirically equivalent to the free field at the level of S-matrix observables, as is proved to be the case for Borchers' equivalence classes (Haag, Local Quantum Physics, $\S$ II.5.5), even though it is manifestly not empirically equivalent to the free field at the level of Wightman function observables?

My reading of the Wightman fields literature of the late 1950s and 1960s is far from complete, which may be why I haven't so far found clear answers for these questions.

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If I am not wrong the product is well defined, because the 2 point funtion is boundary value of an analytic function in some tube (see Streater/Wightman) with certain bounds. More genral there is 1-1 correspondence between translation invariant tempered distributions which satisfy the spectrum condition and analytic functions in this mentioned tube and an easy way to define the product of the distribution is the product of the analytic functions which lies in the same class and therefore gives a well defined distribution on the boundary.

Really easy is the massless case in let's say 4D.

$W(x-y) \sim \frac1{(x-y+i\epsilon(x_0-y_0))^2}$

$W^2(x-y) \sim \frac1{(x-y+i\epsilon(x_0-y_0))^4}$ where the two-point function is just the boundary value of the analytic function $1/z^2$.

In 1 dimension, that means a chiral field on a light ray, it is even more drastic. There the Wick (=normal ordered) square of the free fermion is a free boson.

edit

The answer to the question "Is a normal-ordered product of free fields at a point a Wightman field?" is: YES!

For convinience I give you a reference, which is "General Principles of Quantum Field Theory" by Bogolubov, Logunov, Oksak, Todorov (1990) p. 344

Ex. 8.16 Prove that the Wick Monomials satisfy the Wightman Axioms

edit Tim: I can not comment on your yet. I did not say you can multiply general distribution which are boundary values by multiplying the function, just for a certain class but maybe I was sloppy because I thought this was standard. The $\delta$ is not in this class, because it does not fullfill a "spectrum contion", i.e. the fourier transform is supported in some convex cone with non-empty dual cone. The fourier transform of $\delta(x)$ is constant so does not fall into this class. But distributions which have this property on their fourier transformation you can multiply: see the standard textbook Reed Simon Volume 2 - page 92 Ex 4.

I hope my references clear the confusion.

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  • $\begingroup$ I think you can't take the product of limits like that. $W(x-y)$ is the limit as $\epsilon\rightarrow 0$, which in other terms is a delta function. In your expression for $W(x-y)^2$, there's a question how we should take the distinct limits of the two different $\epsilon$'s. There are mathematical constructions that make products of generalized functions possible, up to a point, for example an approach due to Colombeau, but that puts us outside Wightman's axiomatization. The limitations on what is and is not OK when we take products of generalized functions are non-trivial. $\endgroup$
    – Peter Morgan
    Jan 27, 2011 at 0:36
  • $\begingroup$ Normally this is studied in microlocal analysis, but for this kind of functions one don't need this big machinery. I'll search the ref $\endgroup$
    – Marcel
    Jan 27, 2011 at 10:07
  • $\begingroup$ Voilà: Reed Simon Volume 2 - page 92 Ex 4. $\endgroup$
    – Marcel
    Jan 27, 2011 at 10:10
  • $\begingroup$ And no the limit for $\epsilon \rightarrow 0$ is not a delta function. The Commutator function is like a delta function on the light cone (Huygens principle). $\endgroup$
    – Marcel
    Jan 27, 2011 at 10:17
  • $\begingroup$ Thanks, Marcel, I'll look at your references as soon as I can. $\endgroup$
    – Peter Morgan
    Jan 27, 2011 at 15:31
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the real free bosonic quantum field with mass m > 0 is a Wightman field, because one can define it as an operator valued distribution that satisfies the Klein-Gordon equation in the sense that $ \phi(\Box f + m^2 f) = 0 $ for all test functions f. The product $\phi(x) \phi(x)$ is then simply not defined for well known reasons (you don't have to look at the two point functions, already the field itself is not well defined). I assume that you know all this and your question was if physicists know a trick around this problem: As far as I know there is no such trick within the Wightman framework.

In order to make sense of this "field" I think we would need to leave the Wightman axioms and allow more severe singularities of two point functions on the diagonal, which are usually described via operator product expansions.

Addendum: When we have a representation of a distribution as a boundary value of an analytic function, it is not true that we can consistently define e.g. the square of this distribution as the boundary value of the square of the analytic function (if this would work, we would have found a solution to a problem where our math friends have already proven that there is no solution, namely to define $\delta^2(x)$). The reason for this is that we are actually talking about an equivalence class of analytic functions instead of one unique analytic function (equivalence means equal up to an analytic function that has an analytic extension to the boundary). Multiplication is not well defined on the equivalence classes. For more details see the buzzword "hyperfunction".

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  • $\begingroup$ Although I somewhat agree with your sentiment, it's not clear whether "Leaving the Wightman axioms" is on the table, insofar as we would then have to provide an alternative, motivated axiomatization (at a comparable level of abstraction -- not, in other words, the Haag-Kastler axiomatization). If we "allow more severe singularities", that would put us into a different order of mathematics, but it seems that people who have worked within the Wightman axioms do take normal-ordered products of quantum fields not to be too singular to work with within the Wightman axioms. $\endgroup$
    – Peter Morgan
    Jan 27, 2011 at 0:26
  • $\begingroup$ Your Addendum seems to me very sensible. I can't fit my response, with much LaTeX, into 600 characters, however. (1) Ordinarily, we smear $\hat\phi(x)$ to give $\hat\phi_f=\int\hat\phi(x)f(x)d^4x$. Suppose we similarly construct $\hat\Phi_f=\int:\!\!\hat\phi(x)^2\!\!:f(x)d^4x$. If two test functions $f(x),g(x)$ have space-like or time-like separated supports, assuming(!) we take $\int\delta(x^2)^2\cdot ZERO d^4x=0$, we can work out $\int f(x)C(x-y)^2g(y) d^4x$ fairly happily. If $f(x),g(x)$ have light-like separated supports, not so much. We'll see if it will let me do a 2nd comment... $\endgroup$
    – Peter Morgan
    Jan 27, 2011 at 1:00
  • $\begingroup$ (2) If $f(x),g(x)$ have light-like separated supports, the Fourier transform of $C(x-y)^2$ is a convolution, which, with the test functions, we can integrate fairly happily, IF there isn't a resonance between the frequencies in the Fourier space representations of the two test functions. In these terms, the resonance is what makes $:\!\!\hat\phi(x)^2\!\!:$ not a distribution, because a distribution is not supposed to care what test functions we give it. In a Physics point of view, a resonance is sort-of-OK, but it's also what one expects to see in an interacting theory, not in a free field. $\endgroup$
    – Peter Morgan
    Jan 27, 2011 at 1:11
  • $\begingroup$ Hi there, this software is unfit to host ongoing discussions, so I'll be brief: Making sense of $\phi(x) \phi(x)$ as a Wightman field would, that's my educated guess, result in the very first construction of an interacting Wightman theory. I doubt that it can be done. Products of distributions can be defined, however, under certain circumstances. Note that in the two point function for a Wightman field the product is actually a tensor product (you feed every factor one test function separately). $\endgroup$
    – Tim van Beek
    Jan 27, 2011 at 9:02

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