# Wightman quantum field theory

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Curator: Raymond Frederick Streater

Wightman quantum field theory (also known as Axiomatic quantum field theory) refers to a branch of mathematical physics that studies relativistic quantum field theories satisfying Wightman's axioms.

## Why Rigour Is Needed in Relativistic Quantum Field Theory

In a relativistic classical scalar field theory, one introduces a function $$\phi(\mathbf{x},t)\ ,$$ evaluated at each time $$t\in \mathbb{R}$$ to be a function of the point of space, $$\mathbf{x} \in \mathbb{R}^3\ .$$ The dynamics of such a field, that is, how the function depends on time, must be consistent with special relativity. For a free field of mass $$m\geq 0$$ the dynamics is given by the free wave equation $\frac{1}{c^2}\frac{\partial^2}{\partial t^2}\phi(\mathbf{x},t)-\Delta\phi(\mathbf{x},t)+ m^2\phi(\mathbf{x},t)=0.$ Here, $$c$$ is the speed of light, and $$\Delta=\sum_{j=1}^{3}\frac{\partial}{\partial x^j}\frac{\partial}{\partial x^j}\ .$$ We may choose units, $$x^0=ct,x^j\ ,$$ $$j=1,2,3$$ of time and space. Then the Minkowski metric becomes $g\equiv(g_{\mu\nu})_{\mu,\nu=0,\cdots,3}=\left(\begin{array}{cccc} 1&0&0&0\\ 0&-1&0& 0\\ 0& 0&-1&0\\ 0&0&0&-1\end{array}\right)\;.$ The time, position $$\{t,\mathbf{x}\}$$ are contravariant coordinates, $$x\equiv(x^\mu)=(x^0,x^j)\ ,$$ $$j=1,2,3\ ;$$ the covariant coordinates $$x_{\mu}\ ,$$ $$\mu=0,1,2,3$$ are defined to be $$x_\mu=g_{\mu\nu}x^\nu\ ,$$ where we use the summation convention of summing over repeated coordinates, $$\nu=0,1,2,3\ .$$ We also adopt the convention of writing the derivative $$\partial/\partial x^\mu$$ as $$\partial_\mu$$ and $$\partial/\partial x_\mu$$ as $$\partial^\mu\ ,$$ $$\mu=0,1,2,3\ .$$ Then the equation of motion takes the manifestly Lorentz-covariant form $\partial^\mu\partial_\mu\phi+m^2\phi=0.$

It happens that this equation can be obtained from a Lagrangian, $$L_0\ ,$$ equal to $L_0=\frac{1}{2}\int \left\{\partial^\mu\phi\partial_\mu\phi-m^2\phi^2\right\}d^4x.$

A popular way to introduce a theory of an interacting field is to add to the free Lagrangian $$L_0$$ a Lorentz invariant term $$L_1\ ,$$ which is usually a polynomial in $$\phi$$ of degree higher than two. For example, $L_1=\frac{\lambda}{4}\int\phi^4(x)d^4x.$ Then the dynamics of the interacting field obeys the non-linear equation $\tag{1} \partial^\mu\partial_\mu\phi+m^2\phi-\lambda\phi^3=0.$

When $$\phi(x)$$ is assumed to be a real-valued field, then this equation has smooth solutions for any smooth bounded initial conditions at time $$t=0\ .$$ The field at time zero, $$\phi(0,\mathbf{x})$$ and its canonical conjugate, the time-derivative $$\dot{\phi}(0,\mathbf{x})\ ,$$ determine $$\phi(t,\mathbf{x})$$ for all time $$t\ .$$ At any time, say $$t=0\ ,$$ there is a Poisson bracket between $$\phi$$ and $$\dot{\phi}$$ which states that $\left\{\phi(0,\mathbf{x}),\dot{\phi}(0,\mathbf{y}) \right\}=\delta(\mathbf{x-y})$ This relation also holds for the Lorentz-transformed field $$\phi_\Lambda(x):=\phi(\Lambda^{-1}x)\ ;$$ here, $$\Lambda$$ is an element of the Lorentz group, that is, a four-by-four real matrix obeying the Lorentz condition $\tag{2} \Lambda^Tg\Lambda=g.$

The Dirac quantisation of $$\phi$$ means that we should try to find an operator field $$\phi(t,\mathbf{x})\ ,$$ whose commutator with $$\dot{\phi}(t,\mathbf{y})$$ is a Dirac delta-function. This means that $$\phi(t,\mathbf{x})$$ cannot be a function of $$\mathbf{x}\ ;$$ it must itself be a distribution in the sense of Schwartz, for example. However, if so then the expression $$\phi(t,\mathbf{x})^3\ ,$$ which appears in the equation (1), is infinite, or at least, undefined: Schwartz does not give us the definition of non-linear functions of distributions. A much more indirect definition of the dynamics is therefore needed for quantum field theory. To attempt to make $$\phi$$ satisfy a non-linear equation leads to the occurrence of divergences in the calculation. This is the reason why infinities are found in Feynman-Dyson theory. All such problems were solved for the free field; by 1956 Wightman had defined space-time derivatives of free fields, and also constructed polynomials of free fields and their derivatives. He found that to make sense of these procedures, it was enough to smear the field with an infinitely smooth function of Schwartz class $$\mathcal{S}(\mathbb{R}^4)$$ in space-time.

Definition The space $$\mathcal{S}:=\mathcal{S}(\mathbb{R}^4)$$ consists of infinitely differentiable real functions of four variables, which go to zero at infinite infinitely faster than any power of Euclidean distance. Schwartz furnished $$\mathcal{S}$$ with a topology; see, for example, the books by Schwartz (Schwartz 1949,1951). A tempered distribution is then a continuous linear map, $$\mathcal{S}\rightarrow\mathbb{C}\ .$$

Wightman then showed that, for the free field and also the Wick ordered powers of the free field, the smeared field $$\phi(f)=\int\phi(x)f(x)d^4x$$ is a well-defined operator on a domain in Fock space. All such models constructed so far give a unit scattering matrix. How can we do better?

## The Wightman Axioms for a Scalar Field

Wightman wanted the quantised field to follow von Neumann's idea of quantum theory. Thus, the ideas of Dirac, involving bras and kets, is not followed. A Hilbert space over the complex numbers is a vector space with a scalar product, which is complete in the norm topology. So for his first axiom, he postulated this:

AXIOM I. The states of a quantum theory are normalised vectors in a separable Hilbert space, $$\mathcal{H}\ ,$$ two such that differ by a complex phase giving rise to the same state.

We denote the scalar product between two vectors, $$\Phi$$ and $$\Psi\ ,$$ by $$\langle\Phi,\Psi\rangle\ .$$ If we had a real Hilbert space, this would be a bilinear form, but in a complex Hilbert space it is complex. It obeys the condition $$\langle\Phi,\Psi\rangle=\langle\Psi,\Phi\rangle^*$$ for all $$\Phi$$ and $$\Psi\ ;$$ here, $$z^*=\bar{z}$$ denotes the complex conjugate. We adopt the convention that it is linear in the second vector, $$\Psi\ ,$$ and antilinear in the first, $$\Phi\ ;$$ such a form is called sesquilinear; it is a scalar product if it satisfies $$\langle\Phi,\Phi\rangle>0$$ unless $$\Phi=0\ .$$

The great success of Wigner in analysing the representations of the Poincaré group should be used. However, it was known since 1956 that invariance under space reflections, called parity, does not hold for elementary particles; Since 1961 it was known that time reversal is not an exact symmetry either. So Wightman postulated

AXIOM II. The space $$\mathcal{H}$$ carries a continuous unitary representation, $$(a,\Lambda)\mapsto U(a,\Lambda)$$ of the restricted orthochronous Poincaré group $$\mathcal{P}_+^\uparrow\ .$$ In $$\mathcal{H}\ ,$$ there exists a vector, unique up to a phase, (called the vacuum state, and denoted by $$\Psi_0$$) that is invariant under all $$U(a,\Lambda)$$ and for all other vectors $$\Psi\in\mathcal{H}$$ the energy is positive: the spectral resolution of $$P^0\ ,$$ the self-adjoint generator of the Abelian one-parameter group of time-translations, is non-negative.

Here, $$\mathcal{P}_+^\uparrow$$ is the Lie group of linear transformations of space-time, $$x\mapsto \Lambda x+a\ ,$$ such that equation (2) holds, as well as $$\det\Lambda=1$$ (the restricted condition) and $$\Lambda^{00}>0$$ (the orthochronous condition). The last two conditions rule out the parity operator and time reversal as necessarily invariances of the dynamics, though they might be. The condition of positivity of the energy, combined with the unitary operators $$U(0,\Lambda)$$ implementing the Lorentz transformations, leads to the spectral condition: the simultaneous spectrum of $$P^\mu$$ lies in the closed forward light cone $$V^+\ :$$

Definition $V^+:=\{p\in\mathbb{R}^4:p^\mu p_\mu\geq 0,\;\;p^0\geq 0\}.$

To find suitable conditions on the quantised field-operators, Wightman noted that for the quantised free field $$\phi\ ,$$ one finds that $$\phi(f):=\int f(x)\phi(x)d^4x$$ is unbounded; for an unbounded operator, one must specify its domain: this is the set, say $$D\ ,$$ of vectors on which it is defined. The free vacuum (which is the Fock vacuum) is in the domain of the free field, and $$\phi(f)$$ can also be multiplied by another such operator, say $$\phi(g)\ ,$$ or any product of such, to get an operator with the Fock vacuum in its domain. Moreover, the set $$D$$ of vectors of the form of sums of monomials $$\phi_0(f_1)\ldots\phi_0(f_n)\ ,$$ $$n=0,1,2\ldots\ ,$$ span Fock space. We see that $$D$$ is dense in the Hilbert space of states. Thus, Wightman requires a scalar field $$\phi$$ to obey the

AXIOM IIIa: The vacuum of the theory, $$\Psi_0\ ,$$ lies in the domain of any $$\phi(f)$$ and for any $$f_1,\ldots f_n\in\mathcal{S}\ ,$$ we have $$\phi(f_1)\phi(f_2) \ldots\phi(f_n)\Psi_0\in D\ .$$ One takes $$D$$ to be the complex span of such vectors, as we vary $$n$$ and the test-functions. The operator $$\phi(f)$$ is symmetric on $$D$$ for each $$f\in\mathcal{S}\ .$$ Moreover, the field transforms under $$U(a,\Lambda)$$ according to $\tag{3} U(a,\Lambda)\phi(f)U^{-1}(a,\Lambda)=\phi(f_{a,\Lambda}).$

Here, the test-function transforms as the inverse element of the group: $f_{a,\Lambda}(x):=f(\Lambda^{-1}(x-a)).$

Wightman did not postulate the essential self-adjointness of $$\phi(f)$$ on the domain $$D\ ,$$ replacing this by the weaker axiom that $$\phi(f)$$ is symmetric: for each pair of vectors $$\Phi,\Psi\in D\ ,$$ we have $$\langle\Psi,\phi(f)\Phi\rangle=\langle\phi(f)\Psi,\Phi\rangle\ .$$

AXIOM IIIb: For any pair of vectors $$\Phi$$ and $$\Psi$$ in $$D\ ,$$ the map $$\mathcal{S}\rightarrow\mathbf{C}$$ given by $$f\mapsto\langle\Phi,\phi(f)\Psi\rangle$$ is continuous.

Thus, Wightman says that a quantised field is an operator-valued tempered distribution. Finally, Wightman could write down the condition, which he called causality. Given $$f\in\mathcal{S}\ ,$$ let the support of f be the set $\operatorname{supp} f:=\{x\in\mathbb{R}^4:f(x)\neq 0\}.$ Then he required

AXIOM IV: Suppose that $$f,g\in\mathcal{S}$$ are such that $$\operatorname{supp} f$$ is space-like to $$\operatorname{supp} g\ ;$$ then $$\phi(f)\phi(g)=\phi(g)\phi(f)$$

Here, we say that a subset $$A$$ of $$\mathbb{R}^4$$ is space-like to $$B\ ,$$ another subset, if every point $$x$$ in $$A$$ is space-like to every point $$y$$ in $$B$$ $x\in A, y\in B$ implies that $$(x - y).(x - y):=(x - y)^\mu(x - y)_\mu$$ is negative.

This holds for the free field, and also for the local polynomials, the Wick-ordered powers, constructed by Wightman. The idea is that if we measure two quantised fields in space-like separated regions, no light signal can be created by one measurement, and interfere with the experiment in the other region, before the other experiment is completed. So the two measurements should be compatible, and the corresponding operators should commute as operators on $$\mathcal{H}\ ;$$ this is the condition that they should be simultaneously measurable. This axiom is hard to satisfy, and all known examples of such fields are, so far, derived from free fields and lead to a trivial scattering-matrix, if one proceeds in the usual way by looking at the vacuum representation.

## The Wightman Distributions

Let $$\Psi_0$$ be the vacuum of a Wightman field, $$\phi\ .$$ Then, given test-functions $$f_1,\ldots,f_n\ ,$$ we arrive at the multi-functional $\langle\Psi_0,\phi(f_1)\phi(f_2)\ldots,\phi(f_n)\Psi_0\rangle;$ this maps the $$n$$ test-functions to the complex numbers; more, the mapping is continuous, by Wightman's assumption that the field is a distribution. This holds for each $$f_i\ ,$$ given that the other test-functions are fixed. Schwartz proved, in his nuclear theorem, that then there is a unique distribution of $$4n$$ variables, valid for all test-functions $$f(x_1,x_2,\ldots,x_n)\ ,$$ which gives the same result when $$f$$ is a product: $f(x_1,x_2,\ldots,x_n)=\left(f_1\otimes f_2 \otimes \cdots\otimes f_n\right)(x_1,x_2,\ldots,x_n)\equiv f_1(x_1)f_2(x_2)\ldots f_n(x_n).$ This distribution is called the $$n$$-point Wightman distribution, $$\mathcal{W}_n\ .$$ Moreover, if the field is assumed to be a tempered distribution, then $$\mathcal{W}_n$$ is tempered.

One point of the present article is to show that the Wightman distributions, $$\mathcal{W}_n,\;n=0,1,2,\ldots$$ determine the field $$\phi$$ up to unitary equivalence; two fields with the same Wightman distributions are thus physically the same. To one of Wightman's axioms, I, II, III or IV, the $$\mathcal{W}_n,\;n=0,1,\ldots$$ must have a corresponding property. These are easy to find. Thus, AXIOM I must imply that the scalar product of any vector with itself is non-negative. Apply this to the finite sum $\Psi=\alpha_0\Psi_0+\phi(f_1)\Psi_0+\phi(f_{21})\phi(f_{22})\Psi_0+\ldots+\phi(f_{n1})\ldots\phi(f_{nn})\Psi_0.$ We get a sum of norms of the products, plus the scalar products of the products with different values of $$n\ ,$$ such as $\langle\phi(f_1)\Psi_0,\phi(f_{21})f_{22}\Psi_0\rangle=\langle\Psi_0,\phi(f_1)\phi(f_{21})\phi(f_{22})\Psi_0\rangle,$ which can be written as $$\mathcal{W}_3(f_1\otimes f_{21}\otimes f_{22})\ .$$ This gives us the non-linear conditions, for each integer $$n\geq 0$$ and all real test functions $$f_1, f_{21},f_{22}, \ldots\ :$$

Theorem 1 $\tag{4} \sum_{j=0,k=0}^n\mathcal{W}_{j+k}(f_{j1}\otimes\cdots \otimes f_{jj}\otimes f_{k1}\otimes\cdots\otimes f_{kk})\geq 0.$

The requirement that the field $$\phi(f)\ ,$$ for $$f$$ real, is symmetric on the domain $$D\ ,$$ leads to the property that for each $$n\geq 0\ ,$$ we have \begin{align} \mathcal{W}_n(f_1\otimes\cdots\otimes f_n)&= \langle\Psi_0,\phi(f_1)\cdots\phi(f_n)\Psi_0\rangle\\ &=\langle\phi(f_n)\cdots\phi(f_1)\Psi_0,\Psi_0\rangle\\ &=\langle\Psi_0,\phi(f_n)\cdots\phi(f_1)\Psi_0\rangle^*\\ &=\mathcal{W}_n(f_n\otimes\cdots\otimes f_1)^*. \end{align} That is, we have

Theorem 2 For each $$n$$ and each $$f_1,\cdots f_n$$ in $$\mathcal{S}\ ,$$ we have $\mathcal{W}_n(f_1\otimes \cdots\otimes f_n)=\mathcal{W}_n(f_n\otimes\cdots \otimes f_1)^*.$

The transformation law of the field under the Poincaré group, by the unitary operator $$U(a,\Lambda)$$ leads to the Lorentz invariance of the distributions $$\mathcal{W}_n\ ,$$ and their invariance under space-time translations. For, $$U(a,\Lambda)$$ obeys $\langle U(a,\Lambda)\Psi,U(a,\Lambda)\Phi\rangle=\langle\Psi,\Phi\rangle$ for any two vectors in $$\mathcal{H}\ .$$ In particular, this holds for every pair of vectors of the form $$\Psi=\Psi_0$$ and $$\Phi= \phi(f_1)\phi(f_2)\ldots\phi(f_n)\Psi_0\ .$$ The assumed law of transformation of the field $$\phi(f)$$ given by AXIOM III then says that

\begin{align} U(a,\Lambda)\phi(f_1)\ldots\phi(f_n)\Psi_0&=U(a,\Lambda)\phi(f_1)U^{-1}(a\Lambda)U(a,\Lambda)\phi(f_2)U^{-1}(a,\Lambda) \ldots U(a,\Lambda)\Psi_0\\ &=\phi(f_{1,a,\Lambda})\ldots \phi(f_{n,a,\Lambda})\Psi_0, \end{align} where we have used the invariance of the vacuum, $$\Psi_0\ .$$ Take the scalar product of this equation with $$\Psi_0\ ,$$ to get the Poincaré- invariance of $$\mathcal{W}_n\ :$$ \begin{align} \mathcal{W}_n(f_1\ldots f_n)&=\langle\Psi_0,\phi(f_1)\ldots\phi(f_n)\Psi_0\rangle\\ =\langle\Phi,\Psi_0\rangle&=\langle\Phi,U^*(a,\Lambda)\Psi_0\rangle\\ &=\langle U(a,\Lambda)\Phi,\Psi_0\rangle\\ &=\langle\phi(f_{1,a,\Lambda}\ldots\phi(f_{n,a,\Lambda})\Psi_0\rangle\\ &=\mathcal{W}_n(f_{1,a,\Lambda}\ldots f_{n,a,\Lambda}). \end{align} That is, the distributions $$\mathcal{W}_n$$ are invariant under Poincaré transformations of the test-functions.

The group-action of space-time translation acts on the Wightman distributions by duality of its action on the test-functions. Indeed, we see that \begin{align} \mathcal{W}_n\left(f_1\otimes f_2\ldots\otimes f_n\right)&= \langle\Psi_0,\psi(f_1)\ldots\psi(f_n)\Psi_0\rangle\\ &=\langle U(a)\Psi_0,U(a)\phi(f_1)\phi(f_2)\ldots\phi(f_n)\Psi_0\rangle\\ &=\langle\Psi_0,U(a)\phi(f_1)U(a)^{-1}\;U(a)\phi(f_2)U(a)^{-1}\ldots U(a)\phi(f_n)U(a)^{-1}\Psi_0\rangle\\ &=\mathcal{W}_n\left(f_{1,a}\otimes f_{2,a}\ldots\otimes f_{n,a}\right). \end{align} This means that for each non-negative integer $$n\ ,$$ $$\mathcal{W}_n$$ is invariant under the translation group. It can therefore be written as a distribution depending only on the $$n-1$$ difference variables, $$x_1-x_2,x_2-x_3,\ldots, x_{n-1}-x_n\ .$$ In particular, there exists a constant, $$C$$ say, such that $\langle\Psi_0,\phi(f)\Psi_0\rangle=C\int f(x)d^4x.$ That is, the one-point function $$\langle\Psi_0,\phi(x)\Psi_0\rangle$$ is equal to the constant $$C\ .$$ We may subtract $$C$$ from the field, to get another field in which the one-point function is zero.

To summarise, we get

Theorem 3 The $$\mathcal{W}_n$$ are invariant under the action of the special orthochronous Poincaré group.

The spectral properties of the translation group, assumed in AXIOM II, tell us that the Fourier transform of each $$\mathcal{W}_n$$ has a restricted spectrum. The Fourier transform $$\hat{f}$$ of a test function $$f\in\mathcal{S}$$ is the function $\hat{f}(p)=(2\pi)^{-2}\int f(x)e^{ip_\mu x^\mu}d^4x.$ The inverse Fourier transform of a function, say $$\hat{f}(p)=g(p)\ ,$$ is denoted by $$\check{g}(x)\ .$$ It is given by $$\check{g}(x)=(2\pi)^{-2}\int g(p)e^{-ip_\mu x^\mu}=f(x)\ ,$$ Schwartz showed that the Fourier transform of $$\mathcal{S}(\mathbb{R}^d)$$ is the same space, in any dimension $$d\ ,$$ and that the transform is continuous in the topology given to $$\mathcal{S}\ .$$

The Fourier transform of a tempered distribution is got by testing $$\mathcal{W}_n$$ with the inverse Fourier transform of the test-function: thus, the definition of $$\mathcal{\hat{W}}_n$$ is that $\mathcal{\hat{W}}_n\left(\hat{f}_1\otimes\ldots\otimes \hat{f}_n\right):=\mathcal{W}\left(f_1\otimes\ldots\otimes f_n\right).$

By the spectral theorem, the space-time translation group can be written as $U(a)=\int_{\mathbb{R}^4}e^{-i p\cdot a}d\mu(p).$ Here, $$a=(a^0,\mathbf{a})$$ is the four-vector of translation, and $$\mu$$ is the spectral measure, which is zero unless $$p$$ lies in the closed forward cone, by AXIOM II. Note that this is a condition on the energy of the states; it does NOT mean that the operator $$\phi(f)$$ is zero whenever $$\hat{f}(p)$$ is zero if $$p$$ is in the closed forward light-cone, $$V^+\ .$$ It means, for example, that $$\phi(f)\Psi_0=C\Psi_0=0$$ in that case. Similarly, for each $$n\ ,$$ the state $$\phi(f_{1})\ldots\phi(f_n)\Psi_0$$ is zero unless $$\hat{f}_1(p_1)\ldots\hat{f}_n(p_n)$$ has a non-zero value in the cone $$p_1+\ldots+p_n\in V^+\ .$$ The same can then be said for the scalar product of such vectors with the vacuum, which leads to the requirement that the total sum of all momenta should be zero, if the function $$\mathcal{W}_n$$ is to be non-zero.

Theorem 4 In any Wightman theory, for each integer $$n>1$$ the distribution $$\mathcal{W}_n$$ has a Fourier transform which is zero unless the vectors $$p_j$$ obey $p_n\in V^+, p_n+p_{n-1}\in V^+, p_n+p_{n-1}+p_{n-2}\in V^+,\ldots p_n+p_{n-1}+\ldots p_2\in V^+.$ Moreover, the value is zero unless $$\sum_{j=1}^np_j=0\ .$$

In any Wightman theory, the vacuum must be unique. This property was not included in the original paper (Wightman 1956). One can show that it implies (Hepp et al. 1961).

Theorem 5 In any Wightman theory, the distributions $$\mathcal{W}_n$$ obey the cluster property.

The cluster property requires that for any integer $$j\ ,$$ such that $$0<j<n\ ,$$ we have for any 3-vector $$\mathbf{a}\ ,$$ $\tag{5} \lim_{\|\mathbf{a}\|\rightarrow\infty}\mathcal{W}_n\left(f_1\otimes\cdots\otimes f_j\otimes f_{j+1,\mathbf{a}}\otimes\cdots \otimes f_{n,\mathbf{a}}\right)= \mathcal{W}_j\left(f_1\otimes\cdots\otimes f_j\right)\mathcal{W}_{n-j}\left(f_{j+1}\otimes\cdots\otimes f_n\right).$

Finally, it is obvious that causality has the consequence:

Theorem 6 Let $$f_j$$ and $$f_{j+1}$$ be such that the support of $$f_j$$ is space-like to the support of $$f_{j+1}\ .$$ Then for all $$n\geq 1\ ,$$ $\mathcal{W}_n\left(f_1\otimes \cdots\otimes f_j\otimes f_{j+1}\otimes\cdots\otimes f_n\right)= \mathcal{W}_n\left(f_1\otimes\cdots\otimes f_{j+1}\otimes f_j\otimes\cdots f_n\right).$

The point of this series of results is that, given a set of distributions $$\mathcal{W}_n,\; n=0,1,2,\ldots$$ obeying theorems 1, ..., 6, then there exists a separable Hilbert space, and a Wightman field $$\phi$$ acting on it, such that the Wightman axioms hold for $$\phi\ .$$ Moreover, two such fields, having the same set of $$\mathcal{W}$$-distributions, are unitarily equivalent. This is proved in the next section. This result is comforting to some: quantum field theory is reduced to the problem of finding a set of distributions, $$\mathcal{W}_n\ ,$$ $$n=1,2,\ldots$$ with some specific properties.

## The Wightman Reconstruction Theorem

It has been remarked by I. E. Segal that the problem of finding a quantum field $$\phi$$ with a given set of $$\mathcal{W}_n$$ is rather like that of constructing a representation of a $$C^*$$-algebra from a state; this refers to the Gelfand-Naimark-Segal theorem (Lanford, personal communication). The main difference is that in Wightman theory the operators, generally, are unbounded. Borchers remarked (Borchers 1962) that the Wightman fields, $$\phi(f), f\in\mathcal{S}\ ,$$ generate a *-algebra over the complex numbers. Thus, we form monomials $$\phi(f_1)\ldots\phi(f_n)\ ,$$ since the domain $$D$$ of $$\phi(f)$$ is mapped into itself by the action of $$\phi(f)\ .$$ We include the identity, $$I\ ,$$ formally the case when $$n=0\ .$$ Below, we shall denote this product by the symbol $$\otimes\ ;$$ that is, $(f\otimes g)(x,y):=f(x)g(y).$ Since $$D$$ is linear, we may also add such operators, and multiply them by complex numbers, to get the general complex polynomial in the smeared fields. These operators have adjoints: $[\phi(f_1)\ldots\phi(f_n)]^*=\phi(f_n)\ldots\phi(f_1).$ Here, we have limited the test-functions to taking real values. These manipulations give the definitions, product by a complex number, products of operators, sums of such, and conjugation, needed to define the *-algebra of Wightman fields. Let us call this algebra $$\mathcal{A}\ ,$$ and denote by $$A$$ or $$B$$ an element of $$\mathcal{A}\ .$$ We call an element $$B$$ of the algebra positive if it is of the form $$B=A^*\otimes A$$ for some element $$A\in\mathcal{A}\ .$$

Any vector $$\Psi$$ in $$D$$ defines the expectation functional on $$\mathcal{A}$$ given by the expectation value $A\mapsto \langle\Psi,A\Psi\rangle.$ This is linear and moreover, positive, in the sense that the expectation value of any positive element is non-negative. The expectation of the element $$I$$ is 1. Those familiar with the theory of GNS (Gelfand, Naimark and Segal) with recognise that the expectation functional has properties of state on the algebra. In particular, the vacuum expectation values $$\mathcal{W}_n,\; n=1,2,\ldots$$ define a state on $$\mathcal{A}\ .$$ The Borchers algebra consists of unbounded operators, while the GNS algebra is an algebra of bounded objects. To compensate, the $$\mathcal{W}_n$$ are taken to be continuous in the topology of $$\mathcal{S}\ ,$$ applied to each test-function. We can copy the proof of GNS, to obtain the existence of a Hilbert space, and a field $$\phi(f)$$ acting on a dense domain, given a set $$\mathcal{W}_n\ ,$$ $$n=0,1,2\ldots$$ obeying the positivity condition (4), theorem 1.

Theorem 7 Let $$\mathcal{W}_n,\; n=0,1,\ldots$$ obey theorem 1. Then there exists a separable Hilbert space, say $$\mathcal{H}\ ,$$ and vector $$\Psi_0\in\mathcal{H}\ ,$$ a dense domain $$D\subset\mathcal{H}$$ containing $$\Psi_0\ ,$$ and a field $$f\mapsto\phi(f)$$ with the domain of $$\phi(f)$$ being $$D\ ,$$ such that $$\mathcal{W}_n,\; n=0,1,2,\ldots$$ are the vacuum expectations of products of $$\phi(f_1)\ldots\phi(f_n)\ .$$ Moreover, if $$\mathcal{H}^\prime,\Psi_0^\prime,D^\prime\phi^\prime$$ is another space, vector, domain and field with the same values of $$\mathcal{W}_n\ ,$$ $$n=0,1,2,\ldots\ ,$$ then there is a unitary operator $$V:\mathcal{H}\rightarrow\mathcal{H}^\prime$$ such that $$\phi^\prime(f)=V\phi(f)V^{-1}$$ and $$\Psi_0^\prime=V\Psi_0\ .$$

Proof. To get the Hilbert space, we first note that the algebra $$\mathcal{A}$$ is itself a vector space. The general element of $$\mathcal{A}$$ can be expressed as $\tilde{\Psi}_n(f)=f_0I+f_1+f_{21}\otimes f_{22}+\cdots +f_{n1}\otimes\cdots \otimes f_{nn}.$ Here, $$f$$ means the collection of functions $$f_0,f_1,f_{21},f_{22},\cdots,f_{nn}\ ,$$ where $$f_0$$ is a complex constant, and $$f_{ij}$$ is in $$\mathcal{S}(\mathbb{R}^4)$$ for $$1\leq i,j\leq n\ .$$ The given functionals $$\mathcal{W}_n,\; n=0,1, \ldots$$ define a sesquilinear form on $$\mathcal{A}\ ,$$ by $\langle\tilde{\Psi}_m(f),\tilde{\Psi}_n(g)\rangle=\sum_{m,n}\sum_{i,k=0}^{m,n}\mathcal{W}_{i+k}\left(\otimes_{j=0}^i \bar{f}_{ij}\otimes_{\ell=0}^kg_{\ell k}\right).$ This nearly defines a scalar product on $$\mathcal{A}\ .$$ For example, the sesquilinear form applied to the two elements $$f_0+f_1+f_{21}\otimes f_{22}$$ and $$g_1+g_{21}\otimes g_{22}$$ is \begin{align} \bar{f}_0\mathcal{W}_1\left(g_1\right)&+&\bar{f}_0\mathcal{W}_2\left(g_{12}\otimes g_{22}\right)+\mathcal{W}_2\left(\bar{f}_1\otimes g_1\right)\\ &+&\mathcal{W}_3\left(\bar{f}_1\otimes g_1\otimes g_2\right)+\mathcal{W}_3\left(\bar{f}_{21}\otimes\bar{f}_{22}\otimes g_1\right)\\ &+& \mathcal{W}_4\left(\bar{f}_{21}\otimes\bar{f}_{22}\otimes g_{21}\otimes g_{22}\right). \end{align} By theorem 1, this is positive semi-definite: we may define a non-negative function on positive elements of $$\mathcal{A}$$ by $\|A^*\otimes A\|:=\langle A,A\rangle.$ This is the positive part of the sesquilinear form $A,B\mapsto \langle A,B\rangle.$ This is not quite a scalar product; it is NOT positive definite, only positive semi-definite: it is possible for $$\langle A,A\rangle$$ to be zero for a non-zero element $$A\ .$$ The set $$\mathcal{A}_0:=\{A:\langle A,A\rangle =0\}$$ is, in fact, a vector subspace of $$\mathcal{A}\ .$$ Indeed, if $$A\in\mathcal{A}_0$$ and $$B\in\mathcal{A}_0\ ,$$ then we have for any complex numbers $$\lambda$$ and $$\mu\ :$$ \begin{align} |\langle \lambda A+\mu B,\lambda A +\mu B\rangle|&=|\lambda|^2\|A^*\otimes A\|+|\mu|^2\|B^*\otimes B\|+\bar{\lambda}\mu\langle A,B\rangle\\ &+\lambda\bar{\mu}\langle B,A\rangle\\ &=\bar{\lambda}\mu\langle A,B\rangle+\lambda\bar{\mu}\langle B,A\rangle\\ &\leq |\lambda|\;|\mu|(\langle A,A\rangle)^{1/2}(\langle B,B\rangle)^{1/2} \qquad \mbox{by the Schwarz inequality}\\ &=0. \end{align} We may therefore define the quotient space, $$\mathcal{A}/\mathcal{A}_0\ ;$$ this is the set of equivalence classes of elements, where two elements are equivalent if their difference lies in the subspace $$\mathcal{A}_0\ .$$ We take the quotient space to be the domain $$D$$ in a Hilbert space, $$\mathcal{H}\ ,$$ which we define to be the completion of $$D$$ in the topology defined by the norm.

On the domain $$D$$ we define the operator $$\phi(f)$$ as follows. First, we find a representative of the vector $$\Psi\in D$$ in the algebra $$\mathcal{A}\ ;$$ thus we may write the representative of $$\Psi$$ as $$A\ .$$ Then the representative of $$\phi(f)\Psi$$ is taken to be $$f\otimes A\ .$$ The only problem is that we must show that the vector that this defines is independent of the representative $$A$$ we chose; two elements $$A_1$$ and $$A_2$$ represent the same vector $$\Psi$$ say, if and only if $$A_1-A_2\in\mathcal{A}_0\ .$$ For $$\phi(f)$$ to be well defined, we would then need that $$f\otimes A_1-f\otimes A_2\in\mathcal{A}_0\ .$$ But this holds: by the symmetry of $$\phi(f)\ ,$$ AXIOM IIIa, we see that \begin{align} \|f\otimes(A_1-A_2)\|^2&=\langle f\otimes(A_1-A_2),f\otimes(A_1-A_2)\rangle\\ &=\langle f\otimes f\times(A_1-A_2),A_1-A_2\rangle\\ &\leq\|f\otimes f\otimes(A_1-A_2)\|\;\|A_1-A_2\|\\ &=0. \end{align} We have thus constructed the Hilbert space, the domain $$D\ ,$$ and the operator $$\phi(f)\ ,$$ which maps $$D$$ to $$D\ .$$ If now $$\mathcal{H}^\prime\ ,$$ $$D^\prime$$ and $$\phi^\prime$$ is another Hilbert space, dense domain and field, with the same W-functions, then we define the unitary operator $$U:D\rightarrow D^\prime$$ by the identity map on space $$\mathcal{A}\ .$$ This will map the set $$\mathcal{A}_0\ ,$$ defined using the set $$\mathcal{W}_n,\; n=0,1\ldots\ ,$$ to itself. Since the Wightman distributions $$\mathcal{W}$$ are the same for both fields, the scalar products of the powers of the field $$\phi$$ are the same as those of powers of the field $$\phi^\prime\ .$$ This then defines a unitary map from $$D$$ onto $$D^\prime\ .$$ Since this is dense in $$\mathcal{H}$$ it can be extended to a unitary map on the whole of $$\mathcal{H}\ ,$$ as claimed.

The Hilbert space thus constructed is separable; we can get a countable dense set by using a countable dense set of test-functions in $$\mathcal{S}\ .$$ This is possible since $$\mathcal{S}$$ is separable. Then the power $$n$$ of the field, $$f_1\otimes\cdots \otimes f_n\ ,$$ is a countable set, and is we get a dense set, as we take sums with rational coefficients, since by construction, $$D$$ is dense in $$\mathcal{H}\ .$$ Thus the Hilbert space is separable.

This proves the theorem. $$\Box$$

Theorem 8 Assume in addition that the Wightman distributions are invariant under the Poincaré group, $$P_+^\uparrow\ .$$ Then there exists a unitary representation $$U$$ of $$P_+^\uparrow\ ,$$ such that $$U(a,\Lambda)\Psi_0=\Psi_0\ ,$$ $$U(a,\Lambda)D=D$$ and $U(a,\Lambda)\phi(f)U^{-1}(a,\Lambda)=\phi(f_{a,\Lambda})$ for all translations $$a\in\mathbb{R}^4$$ and all Lorentz transformations $$\Lambda\ .$$

Proof. Define the map $$\tilde{U}$$ on $$\mathcal{A}\ ,$$ generated by $$f\mapsto f_{a,\Lambda}\ ,$$ being the identity on the element $$I\ .$$ This leaves all the Wightman distributions invariant, and so maps $$\mathcal{A}_0$$ to itself. It can be given, then, by a linear operator $$U(a,\Lambda)$$ on $$D=\mathcal{A}/\mathcal{A}_0\ .$$ It leaves all the scalar products of vectors in $$D$$ invariant, and so is unitary on $$D\ .$$ Since $$D$$ is dense in $$\mathcal{H}\ ,$$ $$U(a,\Lambda)$$ can be uniquely extended to a unitary operator on $$\mathcal{H}\ .$$ Since the group acts on $$\mathcal{S}\ ,$$ that is, $$(a,\Lambda)(b,M)f=(a+\Lambda b,\Lambda M)f\ ,$$ we get that $U(a,\Lambda)U(b,M)=U(a+\Lambda b,\Lambda M)$ holds for all group elements $$(a,\Lambda)$$ and $$(b,M)\ .$$ It obeys $$U(a,\Lambda)\Psi_0=1$$ since $$\tilde{U}$$ leaves $$I$$ invariant. This proves the result.

The result shows that the representation $$U(a,\Lambda)$$ of a Wightman theory is determined up to a unitary equivalence by the distributions $$\mathcal{W}_n,\; n=0,1,\cdots\ .$$

Theorem 9 Suppose that the Wightman distributions obey the cluster properties given in theorem 5 as well as the positivity (theorem 7) and invariance, theorem 8 . Then there exists a unit vector in the Hilbert space, $$\mathcal{H}$$ $$\Psi_0\ ,$$ unique up to a phase, which is invariant under $$U(a,I)\ ,$$ $$a\in\mathbb{R}^4\ .$$

Proof. The element $$\Psi_0\in D$$ coming from $$A=I$$ in invariant, since the action of $$\mathcal{P}_+^\uparrow$$ on $$\mathcal{A}$$ leaves the constants invariant. To show uniqueness, the spectral resolution $$P$$ of $$U(a,I)=\int e^{ip^\mu a_\mu}d\mu$$ contains the projection $$P_0$$ onto $$\mathbf{p}$$$$=0\ .$$ Then we see that, for $$\Psi,\Phi\in D\ ,$$ we have \begin{align} \langle\Phi, U(a,I)\Psi\rangle &=\langle\Phi,P_0\Psi\rangle + \langle\Phi,\int_{\mathbf{p}\neq 0}d\mu\, e^{ip^\mu a_\mu}\Psi\rangle \\ &\rightarrow\langle\Phi,P_0\Psi\rangle \end{align} as $$|\mathbf{a}|\rightarrow\infty\ .$$ But according to the cluster property, it should converge to $$\langle\Phi\Psi_0\rangle\langle \Psi_0,\Psi\rangle\ .$$ Since $$\Phi$$ and $$\Psi$$ can be any vectors in the dense set $$D\ ,$$ we get that $$P_0=\Psi_0\rangle\langle\Psi_0\ ,$$ as desired. $$\Box$$

The spectral properties of the $$\mathcal{W}_n,\; n=0,1\cdots$$ lead to the positivity of the energy spectrum:

Theorem 10 Let the Wightman distributions satisfy positivity and translation invariance, and vanish unless the Fourier transform $\mathcal{\hat{W}}_n\left(\hat{f}_1\otimes\cdots\otimes\hat{f}_n\right)= \mathcal{W}_n\left(f_1\otimes\cdots\otimes f_n\right)$ is zero unless \tag{6} \begin{align} &\operatorname{supp}\{\hat{f}_1\}\cap\{p_1^0\geq 0\}\neq\emptyset\\ &\operatorname{supp}\{\hat{f}_1\otimes \hat{f}_2\}\cap\{p_1^0+p_2^0\geq 0\}\neq\emptyset\\ &\ldots\\ &\operatorname{supp}\{\hat{f}_1\otimes\cdots\otimes\hat{f}_{n-1}\}\cap\{p_1^0+\cdots p_{n-1}^0\geq 0\}\neq\emptyset. \end{align}

Then the spectral function of the translation group is zero unless the energy is non-negative.

Proof. If the spectral function were non-zero at an energy $$p^0<0\ ,$$ then we could construct such a state by applying to the vacuum an element $$\phi(f_1)\cdots\phi(f_n)\ :$$ for if all such states had positive energy, so would the sum of such states, and these are dense, and we would be home. However, the norm of such a state would violate (6), and so the theory must obey the spectral condition. $$\Box$$

The condition of positive energy, combined with the Lorentz invariance of a Wightman theory, implies that the energy-momentum spectrum lies in the forward light-cone.

The final part to the Wightman reconstruction theorem uses the condition that

$$\mathcal{W}_n(x_1,\cdots,x_j,x_{j+1},\cdots,x_n) - \mathcal{W}_n(x_1,\cdots,x_{j+1},x_j,\cdots,x_n),\, \; n=2,3\cdots$$

should vanish if $$x_j-x_{j+1}$$ is spacelike, whatever the values of the other variables. This leads to the equation $$\langle\Phi,\{\phi(f)\phi(g)-\phi(g)\phi(f)\}\Psi\rangle=0$$ if the supports of $$f$$ and $$g$$ are space-like separated. Since this is true for all $$\Phi\in D\ ,$$ we get that $$\phi(f)\phi(g)\Psi=\phi(g)\phi(f)\Psi$$ for all $$\Psi\in D\ .$$ Thus, $$\phi(f)$$ obeys the causality condition.

This proves the Wightman reconstruction theorem for scalar fields.

## Applications

Wightman's axiomatic quantum field theory is the framework in which many physical properties of quantum field theories have been rigorously proven. In particular it is worth mentioning the PCT theorem (proving the invariance under combined Parity transformations, Charge conjugation and Time reversal of a Wightman QFT) as well as the Spin statistics theorem (proving a connection between the spin of a particle and the statistics it satisfies).