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In QFT literature one wants to look at $n-$point correlation functions of "operators" inserted at $x$ say, $\cal{O}(x)$ and if $\phi_i$ are the fields then the quantity one has in mind is written as, $<\cal{O}(x)\phi_1(x_1)\phi_2(x_2)..\phi_n(x_n)>$ and this is defined as a path-integral. Typically there are going to be short-distance singularities if any of the two $x_i$ start coinciding. (and that was the topic of my last question)

Though I have done numerous calculations of calculating such correlation functions in various QFTs, it remains unclear to me as to at the fundamental level where does one draw the line between an "operator" and a "field". After quantization aren't all fields actually operators or more precisely Hilbert space operator valued fields on the space-time? This is the confusion that I would like to clarify here.

To start off let me cite some of the definitions that I have seen in this regard.

  • For the quantities labelled as $\phi$, as used above, one seems to use two terms - "local functionals" and "Wightman fields", namely,

    • A "local functional" at $x$ is a function of the fields and finitely many derivatives of the fields evaluated at $x$, like $\phi(x)$, $\phi^2(x)$ but NOT $\phi(x) + \phi(2x)$

    • "Wightman fields" $\phi(x)$ are distributions on the Minkowski space ($V$) with values in the space of operators on the subspace $\cal{D} \subset \cal{H}$ (the Hilbert space of multiparticle states). This means that for any Schwartz function $f$ on $V$, $\phi(f)$ is an honest operator $\phi(f)$ on $\cal{D}$.

It is not clear to me whether "local functionals" and "Wightman fields" are the same things of if there is a natural way to pass between the two things above but I feel that in literature these terms are used interchangeably.

  • For the quantities $\cal{O}$ I seem to see two statements,

    • that $\cal{O}(x)$ does not act as an operator (or an operator valued distribution) on any reasonable subspace of the Hilbert space.

    • that since one wants to deal with products of operators at different space-time points, its better to think in terms of "smeared" quantities like $\cal{O}(f) = \int f(x)\cal{O}(x) d^nx $ where $f$ is a compactly supported smooth function on $V$. Then the statement is that, "..the product $\cal{O}(f)\cal{O}(f')$ exists in the sense of correlation functions if and only if the "operator" $\cal{O}(x)$ (an operator valued distribution) is actually an honest operator i.e matrix elements of $\cal{O}(f)$ are matrix elements of some operator on $\cal{D}$.."

Like in the first pair of points, here too there seems to be some interchangeability between the notion of $\cal{O}(x)$ and $\cal{O}(f)$ and they seem quite analogous to the corresponding $\phi(x)$ and $\phi(f)$! Where is the difference?

What exactly is an "operator valued distribution" that $\cal{O}(x)$ usually perhaps is not (first statement of the above pair) but it seem to have to be if products like $\cal{O}(f)\cal{O}(f')$ have to be defined (last statement of the above pair)?

It would be very helpful if someone can disentangle the above (and may be also my linked previous question!) and explain the difference between the notions of a "local functional", "Wightman field" and "operator".

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"After quantization aren't all fields actually operators or more precisely Hilbert space operator valued fields on the space-time?" formally it is not always true. Consider Chern-Simnons on 3-fold M^3. Hilbert-space picture works only when M^3=N^2\times T. I.e. when you start with Lagrangian formalism you do NOT need to have time axes selected - but in a Hilbert space this is a MUST. So my attitude to this - that even if you can develop Hilbert space formalism - you are not obliged to do this, so you do NOT need to think of fields as operators. –  Alexander Chervov Jan 30 '12 at 13:18
    
In your definition of "Wightman fields" I do not see how "x" is entering. Am I missing something ? –  Alexander Chervov Jan 30 '12 at 13:20
    
About O(x) and O(f) - I never seen notation O(f) - but it seems if you take f=delta( at point x) then O(x) = O(f) , so probably just O(f) more general than O(x). In general O(f) may NOT be local operator –  Alexander Chervov Jan 30 '12 at 13:25
    
If you know all Wighmann functions then you can reconstruct the fields, which are densly defined (in general unbounded) operator valued distributions by the reconstruction theorem, which is basically the GNS reconstruction. So there are two sets of Wighman axioms one for the Wightmann functions and one for the fields and both set of axioms are equivalent.... –  Marcel Bischoff Jan 30 '12 at 19:42
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2 Answers 2

The similarity between what you have called "Wightman field $\phi(x)$" and "operator valued distribution $\mathcal{O}(x)$" is no accident. A Wightman field is a kind of operator valued distribution. An expression like $\phi^2(x)$, which you call a "local functional" is not defined until you specify how the product of two $\phi(x)$'s is to be renormalized (since the naive product would be divergent). But once a precise definition is given it will most likely be another example of an operator valued distribution. So in terms of terminology, there's no particular mystery. There are simply multiple names for the same concept, some names are used in more restrictive contexts, some can be used in more general contexts.

The definition of an "operator valued distribution" is also quite straightforward. The smearing you alluded to above is the defining property. A distribution is a continuous linear map from a space of test functions to a topological vector space. All the subtleties go into the choice of topology on the space of test functions and the target space. Two commonly used test function spaces in QFT are smooth functions with compact support and the Schwartz space (which consists of smooth functions with fast decay at infinity). The precise definitions for these can be found in most books on distribution theory (feel free to ask if you don't know any good ones). An operator valued distribution $\mathcal{O}$ is then a map $f \mapsto \mathcal{O}(f)$ to some space of operators acting on a Hilbert space. The choice for this space of operators also varies. Sometimes it can be taken to just be the space of bounded operators (with, say, the operator norm topology). At other times, one cannot expect the image $\mathcal{O}(f)$ to be a bounded operator. Then one could expand the space of allowed values to include unbounded operators sharing some dense domain in the Hilbert space. This is the case for Wightman fields $\phi\colon f \mapsto \phi(f)$.

As is common practice in distribution theory, the image of an operator valued distribution can be written using the notation $\int \mathcal{O}(x) f(x) dx := \mathcal{O}(f)$. Any well defined manipulation of the symbol $\mathcal{O}(x)$ must come back to this formula.

The $n$-point functions one usually computes in field theory are also easily defined using the machinery of distributions. Consider the distribution $G\colon f\otimes g \mapsto \langle \mathcal{O}_1(f) \mathcal{O}_2(g) \rangle$, where $\mathcal{O}_1$ and $\mathcal{O}_2$ are some operator valued distributions. Note that it is sufficient to define $G$ on product test functions, $(f\otimes g)(x,y) = f(x)g(y)$, from where it can be extended to all 2-argument test functions $h(x,y)$. Of course, one must also take care that the vacuum vector with respect to which this expectation value is taken must lie in the appropriate domain for the product of the possibly unbounded operators $\mathcal{O}_1(f)$ and $\mathcal{O}_2(g)$ to be defined. Now, taking advantage of the notation of the last paragraph, the same distribution can be defined using the simpler formula $G(x,y) = \langle \mathcal{O}_1(x) \mathcal{O}_2(y) \rangle$, which is how it's usually written in the physics literature.

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"What exactly is an "operator valued distribution ?""

May I be I misunderstand you question, but this seems to be very simple (although it was quite brain-damaging for me when I seen this for the first time:)

Let us consider classical mechanics - you have coordinates $q_i$.

What is field theory - it is when index $i$ become continuous so we get $q(x)$.

What is quantum mechanics - $q_i$ become operators $\hat q_i$.

What is quantum fields theory - $\hat q(x)$ become operators depending on $x$, i.e. they become operator-valued functions of $x$.

However if we recall the measure on the space of fields - which enters the definition of the field theory. You can ask on what functions this measure is supported ? In QFT the measure is supported on the generalized functions. So from this we will conclude that

$\hat q(x)$ - is not just operator valued function

but it is operator-valued distribution.

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