# What is the required differentiability of the solutions to do QFT?

Given a real scalar field satisfying:

$$P\psi=(\square_{g}+m^{2})\psi=0$$

on a globally hyperbolic spacetime ($M,g_{ab}$). One can construct a $C^{*}$-algebra $A(M,g)$ ("the minimal algebra") which encodes the canonical commutation relations together with the dynamics.

The way to construct the algebra is to take the abstract objects $\psi(f)$(interpreted as smeared quantum fields $\int_{M}f\psi\nu_{g}$) where $f\in C^{\infty}_{o}(M)$ and generate the free $C^{*}$-algebra with identity $I$ imposing the following:

• $\psi(f)^{*}=\psi(f)$
• $\psi(\lambda f+\mu g)=\lambda \psi(f)+\mu \psi(g)$
• $\psi(\square_{g}+m^{2})f)=0$ for all $f\in C^{\infty}_{o}(M)$
• $[\psi(f),\psi(g)]=iE(f,g)I$

where $E(f,g)=\int_{M}fEg\mu_{g}$ and we have defined the map $E=G_{+}+G_{-}$ where $G_{+},G_{-}$ are the advanced and retarded fundamental solutions.

In other words, the algebra $A(M,g)$ consist of the polynomials in the abstract $\psi(f)$. Moreover, the elemnts of $A(M,g)$ corresponds to sum of suitably smeared products of fields at different points,

$$\psi(f_{1})...\psi(f_{n})=\int_{M}...\int_{M}\psi(x_{1})...\psi(x_{n})f_{1}(x_{1})...f_{n}(x_{n})\nu_{g}...\nu_{g}$$

Then the GNS representation of this algebra constitutes a quantization of the scalar field.

My question is:

The definition of the map $E$, requires to have advance and retarded fundamental solutions. Although this is possible always in the globally hyperbolic case. Is it necessary to have that degree of regularity of the solutions in general?

With this I mean, formally green operators solve the equations in a distributional sense $$PG=\delta$$ and prove the problem is well-posed and allow to see the Green's function as propagator.

Nevertheless, one can prove using energy inequalities that the Cauchy problem of the scalar field is well-posedness in other functional spaces (for example $L^{2},H^{k}$) But there is no need to build a Green operator. Can one still do QFT in this cases?

Would the difference in regularity affect the predictions of the theory or any degree of regularity gives the same physical information?

I think one must at least require $H^{1}$ solutions in order to have a well defined notion of energy momentum. Is this correct?

In principle it is not necessary to restrict the classical space to the set of solutions of the classical dynamics.

The usual "basic" requirement for the classical space $X$ is the following:

$(X,b)$ is a couple with $X$ a real vector space and $b:X\times X\to \mathbb{R}$ a non-degenerate skew-symmetric bilinear form (symplectic form).

This is sufficient to ensure the uniqueness, up to $*$-isomorphisms, of the $C^*$ algebra generated by the set of bosonic Weyl operators ("exponential of the fields") $\{W(x),x\in X\}$, where $\forall x,y\in X$:

• $W^*(x)=W(-x)$;
• $W(x)W(y)=e^{\frac{i}{2}b(x,y)}W(x+y)$.

In some papers (Slawny 1971) an even more minimal requirement is considered, namely that the classical space is an abelian group with a bicharacter (if I remember correctly).

So if you choose $L^2$ or $H^k$ as your classical space (possibly constrained to just the solutions of free KG in a suitable sense), and endow it with a symplectic form that is compatible then you are ok. The symplectic form is everything you actually need, and so it is not necessary to define the "full scalar product" $E$.

In addition, the exponential form of the CCR is the correct formulation for bosons, since in general there are states (non-regular) wrt which the GNS construction does not yield a differentiable Weyl operator, i.e. no generator (field operator $\psi$). These non-regular states are related to infrared problems.

• Thanks for your answer. Would the different function spacesgive a noticeable difference in the quantum theory or are all the theories going to be equivalent? – yess Jun 25 '15 at 18:31
• @yess The CCR algebra is unique, up to $*$-isomorphisms, only once the space $(X,b)$ is fixed. To different spaces correspond different algebras (e.g. if $X$ is $\mathbb{R}^d$ with the usual symplectic form you get QM; if it is $L^2(\mathbb{R}^d)$ you get the algebra of time-zero fields in a bosonic scalar field theory...). Also, as you probably know, fixed an infinite dimensional space $(X,b)$, there uncountably many irreducible representations of that CCR algebra that are unitarily inequivalent. – yuggib Jun 25 '15 at 18:42