# Test functions in wightman axioms

1. Sep 25, 2015

### naima

According to wikipedia AQFT needs test functions so that the fields are distributions smeared on these functions. I'd want to know what are these test functions.
I read in Haag's book that they are fast decreasing functions defined on space time. They belong to the set S of Schwartz functions. As they are defined on space time have they to decrease along t as they decrease along a space direction when itgoes to infinity?
Have you links with an answer to this question?
Thanks

2. Sep 26, 2015

### OptikNerv81

A function is called a Schwartz function if it goes to zero as faster than any inverse power of [PLAIN]http://mathworld.wolfram.com/images/equations/SchwartzFunction/Inline3.gif, [Broken] as do all its derivatives. That is, a function is a Schwartz function if there exist real constants http://mathworld.wolfram.com/images/equations/SchwartzFunction/Inline4.gif such that

where the multi index notation has been used for and [PLAIN]http://mathworld.wolfram.com/images/equations/SchwartzFunction/Inline6.gif. [Broken]

Last edited by a moderator: May 7, 2017
3. Sep 26, 2015

### OptikNerv81

The set of all Schwartz functions is called a Schwartz space and is denoted by [PLAIN]http://mathworld.wolfram.com/images/equations/SchwartzFunction/Inline7.gif. [Broken] It can also be proven that the Fourier transform gives a one-to-one and onto correspondence between http://mathworld.wolfram.com/images/equations/SchwartzFunction/Inline8.gif and [PLAIN]http://mathworld.wolfram.com/images/equations/SchwartzFunction/Inline9.gif, [Broken] where the pointwise product is taken into the convolution product and vice versa. The Fourier transform has a fixed point in [PLAIN]http://mathworld.wolfram.com/images/equations/SchwartzFunction/Inline10.gif, [Broken] which is the function [PLAIN]http://mathworld.wolfram.com/images/equations/SchwartzFunction/Inline11.gif, [Broken] the Gaussian function. Its image under the Fourier transform is the function http://mathworld.wolfram.com/images/equations/SchwartzFunction/Inline12.gif (times some factors of [PLAIN]http://mathworld.wolfram.com/images/equations/SchwartzFunction/Inline13.gif). [Broken]

Instead of [PLAIN]http://mathworld.wolfram.com/images/equations/SchwartzFunction/Inline14.gif, [Broken] one can also consider [PLAIN]http://mathworld.wolfram.com/images/equations/SchwartzFunction/Inline15.gif. [Broken] It consists of functions http://mathworld.wolfram.com/images/equations/SchwartzFunction/Inline16.gif that go to zero, as [PLAIN]http://mathworld.wolfram.com/images/equations/SchwartzFunction/Inline17.gif, [Broken] faster than any inverse power of http://mathworld.wolfram.com/images/equations/SchwartzFunction/Inline18.gif ([PLAIN]http://mathworld.wolfram.com/images/equations/SchwartzFunction/Inline19.gif). [Broken] It is well known that the Fourier transform carries http://mathworld.wolfram.com/images/equations/SchwartzFunction/Inline20.gif onto [PLAIN]http://mathworld.wolfram.com/images/equations/SchwartzFunction/Inline21.gif, [Broken] where http://mathworld.wolfram.com/images/equations/SchwartzFunction/Inline22.gif is the [PLAIN]http://mathworld.wolfram.com/images/equations/SchwartzFunction/Inline23.gif-torus, [Broken] defined as the direct product of http://mathworld.wolfram.com/images/equations/SchwartzFunction/Inline24.gif copies of the circle [PLAIN]http://mathworld.wolfram.com/images/equations/SchwartzFunction/Inline25.gif. [Broken]

Last edited by a moderator: May 7, 2017
4. Sep 27, 2015

### strangerep

I suspect the responses so far are not really answering your question, and I've been hoping that someone else would chime in.

I will just say that this issue bugs me too. If the test functions go to 0 at asymptotic times, i.e., $|t|\to\infty$, then it's hard to see how physically nontrivial asymptotic fields can be in the theory. I had assumed that part of the answer lies in using only causal Poincare reps, such that certain quantum numbers are conserved. I.e., what comes in at the infinite past must account correctly with what goes out in the infinite future. Hence the fields can be nontrivial at asymptotic times.

But perhaps someone else can give a better answer?

5. Sep 27, 2015

### naima

Thank you.
It is never said that the test functions have to be interpreted as wave functions. a test function f is associated to an operator. When it acts on the vacuum of the Hilbert space, it gives a vector V_0. As we use Heisenberg picture this vector does not evolve. if we translate the test function in time -> f(x, t-a) we get another test function associated to another operator (and a new V_a).
The aim of the wightman axioms is to get standard QFT. Can we say that these V_t may be interpreted as wave functions and that they do not equate the test functions?

6. Sep 27, 2015

### naima

When i translate the test function $f -> f_a = f(x, t - a)$ the field $\phi$ associates an operatoe $\phi (f_a)$ which acts on H. As the field is covariant under the U(a) action of the poincaré group , we could also associate to the translated test function the operator
$U(a)^{\dagger} \phi (f) U(a)$ (is the dagger on the left or on the right?)
but the third Wightman axiom tells us that these two operators are the same!
This operator does not tend to zero when t -> infinity
Am i right?

7. Sep 27, 2015

### strangerep

Last edited: Sep 28, 2015
8. Sep 28, 2015

### rubi

What naima said is more or less correct. The test functions don't represent any physical object or a quantum wave function. Their task is rather to probe the value of a field in the region of their support. A (quantum) field is supposed to be a function that assigns (an operator) a real number to each point of spacetime. However, this turns out to be too much to ask for, especially in quantum theory. We would like to have operators $\hat\phi(x)$ for every point $x$ in spacetime and the physical information is supposed to be given by the expectation value $\left<\Psi,\hat\phi(x)\Psi\right>$ (for instance). However, we can't do that. Unfortunately, the field theoretic CCR/CAR algebras turn out to be distributional ($[\hat\phi(x),\hat\pi(y)] = i\hbar\delta(x-y)$) and one way to overcome this difficulty is to use operator-valued distributions instead of just operators. The downside is that we can't compute $\left<\Psi,\hat\phi(x)\Psi\right>$ directly anymore, so we can't ask the question: "What is the expectation value of the field at a point $x$ of spacetime?" Instead, we can take a test function $f$ that is sharply peaked at $x$ and ask: "What is the expectation value of the integral of the field with a test function that is sharply peaked at $x$?" The answer would then be $\left<\Psi,\hat\phi[f]\Psi\right>$. So the test functions are really just a replacement of the spacetime coordinates that are needed in order to make the mathematical formulation rigorous. The correct interpretation of $\left<\Psi,\hat\phi[f]\Psi\right>$ would be that if we actually had well-defined operators $\hat\phi(x)$ for spacetime points $x$, then $\left<\Psi,\hat\phi[f]\Psi\right>$ would be equal to $\int f(x) \left<\Psi,\hat\phi(x)\Psi\right>\mathrm d x$. The physical information about the value of the field $\phi$ in all spacetime points is really contained in the state $\Psi$.

For example: If there was an operator $\hat\phi(x)$, then there might be a state $\Psi_1$ such that $\left<\Psi_1,\hat\phi(x)\Psi_1\right> = \delta(x)$. The rigorous version of that is that there is a state $\Psi_1$ such that for all test functions $f$, we have $\left<\Psi_1,\hat\phi[f]\Psi_1\right> = f(0)$. Another example is that if there was an operator $\hat\phi(x)$, then there might be a state $\Psi_2$ such that $\left<\Psi_2,\hat\phi(x)\Psi_2\right> = e^{i k x}$ (i.e. the state resembles a plane wave). The rigorous way to express this is $\left<\Psi_2,\hat\phi[f]\Psi_2\right> = \mathcal{F}^{-1}[f](k)$, where $\mathcal F$ is the fourier transform (and i omitted some factors of $2\pi$).

Last edited: Sep 28, 2015
9. Sep 28, 2015

### naima

Thank you;
I understand that and i agree. Could you explain what is the distribution defined on the points of space time allowed by the Schwartz nuclear theorem. It is used to define the wightman functions.

10. Sep 28, 2015

### rubi

If you have operators $\phi[f]$, you can form the vacuum expectation values $\left<\Omega,\phi[f_1]\cdots\phi[f_n]\Omega\right>$, which are guaranteed to exist by the Wightman axioms. The nuclear theorem then tells you that there is a distribution $W:\mathcal S(\mathbb R^{4n})\rightarrow\mathbb R$, such that $\left<\Omega,\phi[f_1]\cdots\phi[f_n]\Omega\right>=W[f_1\otimes\cdots\otimes f_n]$, where $(f_1\otimes\cdots\otimes f_n)(x_1,\cdots,x_n)=f_1(x_1)\cdots f_n(x_n)$ is just the Schwartz function that is defined as a product of individual Schwartz functions. So $W$ agrees with the vacuum expectation values on all products of Schwartz functions. This $W$ is called the Wightman $n$-point function. Symbolically, we often just write $\left<\Omega,\phi[f_1]\cdots\phi[f_n]\Omega\right>=\int f(x_1)\cdots f(x_n) W(x_1,\cdots,x_n)\mathrm d x_1\cdots \mathrm d x_n$ and say that $W(x_1,\cdots,x_n)$ is the Wightman function.

11. Sep 30, 2015

### naima

Thank you,
It is natural to take a Fock space as Hilbert space. Is there something in the axioms which says that test functions are associated to annihilation or creation operators? and what about their adjoints?

12. Sep 30, 2015

### samalkhaiat

No, you have to impose the asymptotic condition on the field. The asymptotic free field can be decomposed into positive and negetive frequency parts which contain creation and annihilation operators:
$$\varphi_{(as)}^{(+)}(f) \sim \int \frac{d^{3}p}{2 \omega_{p}} \hat{f}^{*}(p) a_{(as)}(p) ,$$
$$\varphi^{(-)}_{(as)}(f) \sim \int \frac{d^{3}p}{2 \omega_{p}} \hat{f}(p) a^{\dagger}_{(as)}(p) .$$

13. Sep 30, 2015

### samalkhaiat

That depends very much on what goes on the right-hand-side. Under Poincare’ transformation $T=(\Lambda , a)$ of the coordinates $$x \to \bar{x} = T x = \Lambda x + a ,$$ the finite-component field $\varphi_{r}(x)$ transforms (like classical tensor fields do) by finite-dimensional matrix representation of the Lorentz group $D(\Lambda)$:
$$\bar{\varphi}_{r}(\bar{x}) = D_{r}{}^{s}(\Lambda) \ \varphi_{s}(x) .$$ This can be rewritten as
$$\bar{\varphi}_{r}(\bar{x}) = D_{r}{}^{s}(\Lambda) \ \varphi_{s}(T^{-1}\bar{x}) , \ \ \ \ \ \ (1)$$ where $$T^{-1}=(\Lambda^{-1}, -\Lambda a) ,$$ is the inverse element of the Poincare’ transformation. Renaming the coordinates label in (1) as $x$, we find
$$\bar{\varphi}_{r}(x) = D_{r}{}^{s}(\Lambda) \ \varphi_{s}(T^{-1}x) . \ \ \ \ \ \ (2)$$
However, in QFT $\varphi_{r} (x)$ “is” an “operator”. So, it must transform, like operators do, by (infinite-dimensional) unitary representation $U(T)$ of the Poincare’ group
$$\bar{\varphi}_{r}(x) = U^{-1}(T) \ \varphi_{r}(x) \ U(T) . \ \ \ \ \ \ (3)$$ From (2) and (3), we obtain
$$U^{-1}(T) \ \varphi_{r}(x) \ U(T) = D_{r}{}^{s}(\Lambda) \ \varphi_{s}(T^{-1}x) . \ \ \ \ \ (4)$$ Since the set Poincare’ transformations $\{T\}$ form a group, equation (4) must also be satisfied by the inverse element $T^{-1} \equiv (\Lambda^{-1}, - \Lambda^{-1}a)$ :
$$U^{-1}(T^{-1}) \ \varphi_{r}(x) \ U(T^{-1}) = D_{r}{}^{s}(\Lambda^{-1}) \ \varphi_{s}(Tx) .$$
And, since the $U(T) \equiv U(\Lambda , a)$ forms a representation, i.e., $U^{-1}(T) = U(T^{-1})$, we find
$$U(T) \ \varphi_{r}(x) \ U^{-1}(T) = D_{r}{}^{s}(\Lambda^{-1}) \ \varphi_{s}(Tx) . \ \ \ \ \ (5)$$ So, you are free to use either (4) or (5).
Now, let us recall that the fields $\varphi_{r}(x)$ are not operators but operator-valued (tempered) distributions (on space-time) which become (in general unbounded) operators by smearing with “good” test functions $f \in \mathcal{S}(\mathbb{R}^{4})$ (the space of fast decreasing, functions in $\mathbb{R}^{4}$):
$$\varphi_{r}(f) = \int d^{4} x \ \varphi_{r}(x) \ f(x) . \ \ \ \ \ (6)$$

Before we proceed further, let me make the following remarks about (6) which are relevant for issues raised in this thread:

1) You can think of (6) as the mathematical statement of the fact that only space-time averages of the field “operators” $\varphi_{r}(x)$ are “observables”.

2) To reflect the possibility of making measurement in a finite space-time region $\mathcal{O} \subset \mathbb{R}^{4}$, the space of test functions $f(x)$, for which $\varphi_{r}(f)$ are assumed defined, is taken to be $\mathcal{D}(\mathcal{O})$, i.e., the space of all infinitely differentiable functions of compact support in space-time. Without going into detailed mathematical gibberish, we simply take the support of $f(x)$ to be the (closed + bounded = compact) set on which $f$ does not vanish. In this case, we speak of a smeared local operator, or local observable for short.

3) In order for (6) to make sense, (i) the operators $\varphi_{r}(f)$, for all $f \in \mathcal{S}(\mathbb{R}^{4})$ must have a common dense domain of definition $\mathcal{G}$ in the Hilbert space $\mathcal{H}$, i.e., a dense subspace $\mathcal{G}$ of $\mathcal{H}$ which is (ii) stable under the actions of both $\varphi_{r}(f)$ and $U(\Lambda , a)$ $$\varphi_{r}(f) \mathcal{G} \subset \mathcal{G} , \ \ \ U(T) \mathcal{G} \subset \mathcal{G} ,$$ where the term “dense” means “dense in any admissible topology $\tau$ of $\mathcal{H}$”, i.e., $$\bar{\mathcal{G}}^{\tau} = \mathcal{G}^{\perp \perp} = \mathcal{H} ,$$ where $\bar{\mathcal{G}}^{\tau}$ is the closure of $\mathcal{G}$ with respect to the topology $\tau$, and $\mathcal{G}^{\perp}$ is defined by $$\mathcal{G}^{\perp} = \{ |\Psi \rangle \in \mathcal{H}; \ \langle \Psi | \Phi \rangle = 0 \ \ \forall |\Phi \rangle \in \mathcal{G} \} .$$

4) The polynomial algebras generated by operators of the form $$\int d^{4}x_{1} \cdots d^{4}x_{n} \varphi_{r_{1}}(x_{1}) \cdots \varphi_{r_{n}}(x_{n}) f(x_{1}, \cdots , x_{n})$$ with $f \in \mathcal{S}(\mathbb{R}^{4n})$ and with $f \in \mathcal{D}(\mathcal{O}^{n})$ ($n=0,1, \cdots$) are called the field algebra $\mathcal{A}$ and the algebra of local observables $\mathcal{A}(\mathcal{O})$ respectively.

5) The linear functional which sends $f$ to $\langle \Psi | \varphi_{r}(f) | \Phi \rangle$ must be continuous with respect to the topology of $\mathcal{S}(\mathbb{R}^{4})$ for any $|\Psi \rangle , \ |\Phi \rangle \in \mathcal{H}$.

6) In a relativistic QFT with positive-definite inner product, the following three conditions are known to be equivalent with one another:
a) irreducibility of the field algebra $\mathcal{A}$,
b) cyclicity and uniqueness of the vacuum, and
c) cluster property.

7) All hell breaks loose in the indefinite-metric QFT.

8) Finally, “the good, the bad and the ugly” of the Axiomatic QFT can be found in the easy read textbook by Jan Lopuszanski (1923-2008): “An Introduction to Symmetry and Supersymmetry in Quantum Field Theory” , World Scientific, 1991.
Okay, let us go back to Eq(4) (or Eq(5)) and smear the field with a good test function $f$
$$U^{-1}(T) \ \varphi_{r}(f) \ U(T) = D_{r}{}^{s} (\Lambda) \int d^{4}x \ \varphi_{s}(T^{-1}x) \ f(x) .$$ In the left-hand-side, if we change the integration variables $T^{-1}x \to x$, we get
$$U^{-1}(T) \ \varphi_{r}(f) \ U(T) = D_{r}{}^{s} (\Lambda) \int d^{4}x \ \varphi_{s}(x) \ f(Tx) = D_{r}{}^{s} (\Lambda) \ \varphi_{s}(f_{T}) ,$$ where $f_{T}(x) \equiv f(\Lambda x + a)$. As we said before, the above equation is equivalent to
$$U(T) \varphi_{r}(f) \ U^{-1}(T) = D_{r}{}^{s} (\Lambda^{-1}) \ \varphi_{s}(f_{T^{-1}}) ,$$ with $$f_{T^{-1}}(x) \equiv f\left(\Lambda^{-1}(x - a)\right) .$$

Last edited: Sep 30, 2015
14. Oct 1, 2015

### naima

Why is there no [TEX] \varphi [/TEX] in the right size of the definition?
It appears (Unruh effect) that such a decomposition is observer dependent.How is this implemented in AQFT? The problem could be that if the observers of Unruh see differen vacua ant that in AQFT the vacuum in unique.

Last edited: Oct 1, 2015
15. Oct 1, 2015

### samalkhaiat

Why should there be a $\varphi$ on the right-hand-side? Have you taken a QFT1-course? Normally, such a course starts by solving for you the Klein-Gordon equation:
$$\phi(x) = \phi^{(+)}(x) + \phi^{(-)}(x) ,$$ where $$\phi^{(+)}(x) = (\frac{1}{2\pi})^{3/2} \int \frac{d^{3}p}{2\omega_{p}} \ e^{- i p.x} \ a(p) ,$$ and $$\phi^{(-)}(x) = (\frac{1}{2\pi})^{3/2} \int \frac{d^{3}p}{2\omega_{p}} \ e^{ i p.x} \ a^{\dagger}(p) .$$ Now, if you integrate with some test function $f(x)$, you get
$$\phi(f) = \phi^{(+)}(f) + \phi^{(-)}(f) ,$$ where
$$\phi^{(+)}(f) = (\frac{1}{2\pi})^{3/2} \int \frac{d^{3}p}{2\omega_{p}} \left( \int d^{4}x \ f(x) \ e^{- i p.x} \right) \ a(p) ,$$
$$\phi^{(-)}(f) = (\frac{1}{2\pi})^{3/2} \int \frac{d^{3}p}{2\omega_{p}} \left( \int d^{4}x \ f(x) \ e^{i p.x} \right) \ a^{\dagger}(p) .$$ Now in the last two equations, use the Fourier transforms $$\hat{f}^{*} (p) = \int d^{4}x \ f(x) \ e^{- i p.x} ,$$ $$\hat{f}(p) = \int d^{4}x \ f(x) \ e^{i p.x} ,$$ to obtain the expressions I gave you in my previous post.

The space-time symmetry-group in the question is the Poincare’ group. This means that all equivalent observers are related by Poincare’ transformations. And, in QFT1-course, you learn to prove that the above decomposition is Poincare invariant, i.e., observer independent. Why do you need to jump to accelerated observers, and ask about Unruh effect?

16. Oct 2, 2015

### naima

I appreciated your explanation for the left or right dagger!

I am less convinced by what you wrote about annihilation operators. You start with the KG equation. It is related to free non interacting theories. Is phi-four theory excluded?
Reconstruction theorem show that one can construct a model in which the test functions are the one particle vectors and can act on the Fock state by annihilation plus creation.
This prove that the axioms are coherent. This does not prove that $\phi(f)$ is the sum of a creation operator and an annihilation operator.
Can we derive (at less) that$\langle \phi(f) \Omega_0 | \phi(f) \phi(f) \Omega_0 \rangle = 0$ ?

17. Oct 2, 2015

### samalkhaiat

Again, your statements and questions raise doubts about your understanding of the elementary concepts in QFT. So, may I ask you again: What course, if any, have you taken on QFT? I just want to use language appropriate to your level. Other wise anything I will say will sound like gibberish to you.
1) All asymptotic fields, $\varphi_{r}^{(as)} \equiv \varphi_{r}^{(in)}$ or $\varphi_{r}^{(out)}$, are mutually (anti)local fields. However, for non-trivial S-matrix, $\varphi_{r}^{(as)}$ cannot be (anti)local with the interacting field $\varphi_{r}$.

2) Asymptotic fields are free fields. Precisely speaking, to each discrete spectrum of $P^{2}$ which appears as a pole at $p^{2}=m_{r}^{2}$ of the momentum-space Green’s functions, there corresponds an asymptotic field satisfying the K-G equation $$(\partial^{2} + m_{r}^{2}) \varphi^{(as)}_{r} = 0 .$$

3) It is assumed (LSZ) that the asymptotic field is related to the original interacting (Heisenberg) field by the Yang-Feldman equation
$$\varphi_{r}(x) = \varphi_{r}^{(in/out)} (x) + \int d^{4} \bar{x} \ \Delta^{(R/A)}(x - \bar{x};m_{r}^{2}) \ J_{r}(\bar{x}) ,$$ where the source of the field $$(\partial^{2}+m_{r}^{2}) \varphi_{r}(x) = J_{r}(x) ,$$ contains no discrete spectrum at $p^{2}=m_{r}^{2}$.

4) Asymptotic completeness: The vacuum state $|0\rangle$ is cyclic with respect to the set of all asymptotic fields: $$\mathcal{H}^{(in)} = \mathcal{H}^{(out)} = \mathcal{H} .$$ In fact, one only needs to assume $\mathcal{H}^{(in)} = \mathcal{H}$ because the $PCT$ theorem will then implies $\mathcal{H}^{(out)} = \mathcal{H}$. Asymptotic completeness is the only known general principle which uniquely determines the representation space of the algebra $\mathcal{A}$.

Using (1), (2) and (3), the following can be proved:
5) $$[ \varphi_{r}^{(as)}(x) , \varphi_{s}^{(as)}(\bar{x}) ] \propto i \delta_{m_{r}m_{s}} \Delta (x-\bar{x};m_{r}^{2}) .$$ This means that the space of asymptotic states is a Fock space of asymptotic fields, and hence the asymptotic completeness implies the irreducibility of the asymptotic fields as well as the Heisenberg field algebra $\mathcal{A}$.

6)
$$\left( \varphi^{(as)}_{r} \right)^{+}(f) |0 \rangle = 0 .$$

18. Oct 3, 2015

### naima

You are right. I have gaps in QFT. (I will send you a private message to explain why).
But please do not change the level of your aswers. I am sure they will also help future readers. I read the scene and the subject of the drama in Łopuszański's book. (there is a typo with the accent on the cover of my book!). I will read now the following chapters and Appendix 3. Remember that the title of this thread was about the status of the test functions. Than you again.

19. Oct 5, 2015

### strangerep

Heh, yes, it is indeed good to have this stuff laid out in "the samalkhaiat way".

20. Oct 8, 2015

### naima

Could you give me a toy model (say in 1+ 1 dimension) in which a field $\phi$ maps well known test functions such as gaussians in x and t (and their polynomials) to operators (dérivatives or things like that). We could then compute probabilities!
Then having $U(\tau)$ we could see how the computed results evolve when the "time" $\tau$ grows. I write "time" because it is a translation on the real time t which appears in the test functions.

Last edited: Oct 8, 2015