Perturbative Algebraic Quantum Field Theory on Quantum Spacetime: Adiabatic and Ultraviolet Convergence

Abstract

The quantum structure of Spacetime at the Planck scale suggests the use, in defining interactions between fields, of the Quantum Wick product. The resulting theory is ultraviolet finite, but subject to an adiabatic cutoff in time which seems difficult to remove. We solve this problem here by another strategy: the fields at a point in the interaction Lagrangian are replaced by the fields at a quantum point, described by an optimally localized state on QST; the resulting Lagrangian density agrees with the previous one after spacetime integration, but gives rise to a different interaction hamiltonian. But now the methods of perturbative Algebraic Quantum Field Theory can be applied, and produce an ultraviolet finite perturbation expansion of the interacting observables. If the obtained theory is tested in an equilibrium state at finite temperature the adiabatic cutoff in time becomes immaterial, namely it has no effect on the correlation function at any order in perturbation theory. Moreover, the interacting vacuum state can be obtained in the vanishing temperature limit. It is nevertheless important to stress that the use of states which are optimally localized for a given observer brakes Lorentz invariance at the very beginning.

Appendices

Basic Properties of Modified Propagators

We discuss in this appendix some elementary properties enjoyed by the modified propagators used in this paper. We recall that

$$\begin{aligned} {\hat{\Delta }}_{\lambda }(p)&= {\hat{\Delta }}(p) e^{-{\lambda ^2}\langle p\rangle ^2} = -\frac{i}{(2\pi )^3} \delta (p^2+m^2) \varepsilon (p_0) e^{-{\lambda ^2}\langle p\rangle ^2}, \end{aligned}$$

(A.1)

$$\begin{aligned} {\hat{\Delta }}_{+,\lambda }(p)&= {\hat{\Delta }}_+(p) e^{-{\lambda ^2}\langle p\rangle ^2} = \frac{1}{(2\pi )^3} \delta (p^2+m^2) \theta (p_0) e^{-{\lambda ^2}\langle p\rangle ^2}, \end{aligned}$$

(A.2)

which entails

$$\begin{aligned} \Delta _{\lambda }(x)&= \frac{1}{(2\pi )^3} \int _{{{\mathbb {R}}}^3} \frac{d{\mathbf {p}}}{\omega ({\mathbf {p}})} e^{-\lambda ^2(2|{\mathbf {p}}|^2+m^2)}\sin (\omega ({\mathbf {p}})t-{\mathbf {p}}\cdot {\mathbf {x}}), \end{aligned}$$

(A.3)

$$\begin{aligned} \Delta _{+,\lambda }(x)&= \frac{1}{(2\pi )^3} \int _{{{\mathbb {R}}}^3} \frac{d{\mathbf {p}}}{2\omega ({\mathbf {p}})} e^{-\lambda ^2(2|{\mathbf {p}}|^2+m^2)}e^{-i(\omega ({\mathbf {p}})t-{\mathbf {p}}\cdot {\mathbf {x}})}, \end{aligned}$$

(A.4)

where, as customary, $\omega ({\mathbf {p}})=\sqrt{|{\mathbf {p}}|^2+m^2}$.

Proposition A.1

The distributions $\Delta _\lambda$ and $\Delta _{+,\lambda }$ given in (4.5) and in (4.6) have the following properties:

• (a) $\Delta _\lambda$ and $\Delta _{+,\lambda }$ are smooth functions which are also in $L^\infty$.

• (b) $\Delta _\lambda$ and $\Delta _{+,\lambda }$ are solutions of the equation of motion (2.3).

Proof

From the definition we have that

$$\begin{aligned} \Delta _{\lambda }(x) = \langle \Delta (G_\lambda ),G_{\lambda ,x}\rangle = \Delta (G_\lambda * G_\lambda ) (x)= \Delta (G_{\sqrt{2}\lambda })(x), \end{aligned}$$

where $G_{\lambda ,x}(y)=G_\lambda (y-x)$ and * denotes convolution. Since $\Delta$ is a map from Schwartz functions to smooth functions, we have that $\Delta (G_{\sqrt{2}\lambda })$ is a smooth function. From Eq. (A.3) we obtain

$$\begin{aligned} \Vert \Delta _\lambda \Vert _\infty \le \frac{1}{(2\pi )^3} \int _{{{\mathbb {R}}}^3} d{\mathbf {p}} \frac{1}{\omega ({\mathbf {p}})} e^{-\lambda ^2 (2|{\mathbf {p}}|^2+m^2)} \end{aligned}$$

and the right hand side of the previous inequality is finite, so that $\Delta _\lambda$ is bounded. This proves a) for $\Delta _\lambda$. To prove b) we notice that

$$\begin{aligned} (\Box -m^2)\Delta _\lambda (x) = \langle \Delta ((\Box -m^2)G_\lambda ),G_{\lambda ,x}\rangle . \end{aligned}$$

The thesis follows from the fact that $\Delta$ is a weak solution of the equation of motion. The same proofs can be applied with minor modifications to study $\Delta _{+,\lambda }$. $\quad \square$

Proposition A.2

There holds, for the Fourier transform of the modified Feynman propagator,

$$\begin{aligned} {{\hat{\Delta }}}_{F,\lambda }(p) = -\frac{i}{(2\pi )^4} \frac{1}{ p^2+m^2 -i\epsilon } e^{-\lambda ^2 (2|{\mathbf {p}}|^2+m^2)}. \end{aligned}$$

Proof

Computing the inverse Fourier transform along $p_0$ of Eq. (A.2) one gets

$$\begin{aligned} {{\Delta }}_{+,\lambda }(t,{\mathbf {p}}) = \frac{1}{(2\pi )^{7/2}} \frac{e^{-i\omega (\mathbf {p})t} e^{-\lambda ^2(2\omega (\mathbf {p})^2-m^2)}}{2\omega (\mathbf {p})}. \end{aligned}$$

The corresponding partial Fourier transform of the modified Feynman propagator is

$$\begin{aligned} {{\Delta }}_{F,\lambda }(t,{\mathbf {p}}) = \theta (t) {{\Delta }}_{+,\lambda }(t,{\mathbf {p}}) + \theta (-t){{\Delta }}_{+,\lambda }(-t,-{\mathbf {p}}). \end{aligned}$$

Inserting an ultraviolet regulator, its Fourier transform along t gives

$$\begin{aligned} \hat{{\Delta }}_{F,\lambda }(p_0,{\mathbf {p}})&= \frac{1}{\sqrt{2\pi }}\int _0^\infty dt\; e^{-\epsilon t} e^{ip_0 t}\hat{{\Delta }}_{+,\lambda }(t,{\mathbf {p}}) \\&\quad + \frac{1}{\sqrt{2\pi }}\int _0^\infty dt\; e^{-\epsilon t} e^{-ip_0 t}\hat{{\Delta }}_{+,\lambda }(t,-{\mathbf {p}}) \\&= \frac{1}{(2\pi )^4}\frac{e^{-\lambda (2\omega ({\mathbf {p}})^2-m^2)}}{2\omega ({\mathbf {p}}) } \int _0^\infty dt\; \left( e^{-\epsilon t +ip_0 t -i\omega ({\mathbf {p}}) t } + e^{-\epsilon t -ip_0 t -i\omega ({\mathbf {p}}) t } \right) \\&= \frac{1}{(2\pi )^4} \frac{e^{-\lambda ^2 (2\omega ({\mathbf {p}})^2-m^2)}}{2\omega ({\mathbf {p}}) } \frac{2(\epsilon +i \omega ({\mathbf {p}}))}{(\epsilon +i \omega ({\mathbf {p}}))^2 +p_0^2} \\&= \frac{i}{(2\pi )^4} e^{-\lambda ^2 (2\omega ({\mathbf {p}})^2-m^2)} \frac{1}{\epsilon ^2 +2i\epsilon \omega ({\mathbf {p}}) -\omega ({\mathbf {p}})^2 +p_0^2} \\&= \frac{-i}{(2\pi )^4} \frac{1}{ p^2+m^2 -i\epsilon } e^{-\lambda ^2 (2|{\mathbf {p}}|^2+m^2)}, \end{aligned}$$

as requested. $\quad \square$

Bounds for the Modified Propagators

In this appendix, we discuss some bounds valid for the propagators used in this paper. The first proposition is about the decay in space while the second is about the decay in time.

Proposition B.1

For each fixed $a >0$ the propagators $\Delta _{F,\lambda }$ and $\Delta _{+,\lambda }$ satisfy the estimates

$$\begin{aligned} |D(t,{\mathbf {x}})|\le C e^{-m |{\mathbf {x}}|}, \qquad D\in \{\Delta _{F,\lambda },\Delta _{+,\lambda }\}, \qquad t\in (-a,a) \end{aligned}$$

where $C > 0$ is a constant (depending on a and $\lambda$). Furthermore, the function $t\mapsto \Delta _{+,\lambda }(t,{\mathbf {x}})$ is an entire analytic function. For $Im \,t<0$, it holds that

$$\begin{aligned} |\Delta _{+,\lambda }(t,{\mathbf {x}})|\le C e^{-m r}, \qquad Re \,t\in (-a,a), \qquad r = \frac{1}{2}\left( {|\text {Im}\,t|+|{\mathbf {x}}|}\right) \end{aligned}$$

where $C>0$ is the same constant as above. Similarly, the function $t\mapsto \Delta _{\beta ,\lambda }(t,{\mathbf {x}})$ constructed with the thermal two-point function $\Delta _{\beta ,\lambda }$, is an entire analytic function and for $\text {Im}\,t<0$ it holds that

$$\begin{aligned} |\Delta _{\beta ,\lambda }(t,{\mathbf {x}})|\le C e^{-\frac{m}{2}|{\mathbf {x}}|}\left( \frac{e^{-\frac{m}{2}|Im \,t|}}{1-e^{-\beta m}}+\frac{e^{-\frac{m}{2}(\beta -|Im \,t|)}}{1-e^{-\beta m}}\right) , \qquad Re \,t\in (-a,a), \end{aligned}$$

(B.1)

where $C>0$ is the same constant as above.

Proof

We discuss the decay properties of $\Delta _{+,\lambda }$. The definition (4.7) implies that the same result will then hold for $\Delta _{F,\lambda }$. In view of the rotation invariance of $\Delta _{+,\lambda }(t,{\mathbf {x}})$ we just need to analyze the decay in the direction ${\mathbf {x}} = (x^1,0,0)$ for large $x^1$. To this end, we recall that

$$\begin{aligned} \Delta _{+,\lambda }(t,{\mathbf {x}}) = \frac{e^{-\lambda ^2 m^2}}{(2\pi )^3} \int _{{\mathbb {R}}^3} \frac{ e^{-2\lambda ^2 |{\mathbf {p}}|^2}e^{-it \sqrt{|{\mathbf {p}}|^2+m^2}} e^{i{\mathbf {p}}{\mathbf {x}}}}{2\sqrt{|{\mathbf {p}}|^2+m^2}} d{\mathbf {p}}. \end{aligned}$$

(B.2)

We observe that the integrand, seen as a function of $p_1$ is analytic in the strip $\text {Im}(p_1)\in (-m,m)$: actually the cut of $\sqrt{|{\mathbf {p}}|^2+m^2}$ as well as the poles of $1/\sqrt{|{\mathbf {p}}|^2+m^2}$ are contained in the region $|\text {Im}\,p_1|\ge \sqrt{m^2+|p_2|^2+|p_3|^2}$. Furthermore, the integrand vanishes for large $p_1$ in that strip. Hence we consider, for $\epsilon \in (0,m)$, with $q:=m-\epsilon$,

$$\begin{aligned} e^{x^1 (m-\epsilon )}\Delta _{+,\lambda }(t,{\mathbf {x}})&= \frac{e^{-\lambda ^2 m^2}}{(2\pi )^3} \int _{{\mathbb {R}}^3} \frac{ e^{-2\lambda ^2 |{\mathbf {p}}|^2}e^{-it \sqrt{|{\mathbf {p}}|^2+m^2}} e^{i({\mathbf {p}}{\mathbf {x}}-i(m-\epsilon ) x^1)}}{2\sqrt{|{\mathbf {p}}|^2+m^2}} d{\mathbf {p}}\\&= \frac{e^{-\lambda ^2 m^2}}{(2\pi )^3} \int _{{\mathbb {R}}^3} \frac{ e^{-2\lambda ^2 (|{\mathbf {p}}|^2+2iqp_1-q^2)}e^{-it \sqrt{|{\mathbf {p}}|^2+2iqp_1-q^2+m^2}} e^{i{\mathbf {p}}{\mathbf {x}}}}{2\sqrt{|{\mathbf {p}}|^2+2iqp_1-q^2+m^2}} d{\mathbf {p}} \end{aligned}$$

where in the last equality we have used the residue theorem to move the integration in $p_1$ from the real line to the line $\mathrm {Im}\, p_1 = q$. We obtain then the following bound

$$\begin{aligned} \left| e^{x^1 (m-\epsilon )}\Delta _{+,\lambda }(t,{\mathbf {x}})\right|&\le \frac{e^{\lambda ^2 (2q^2-m^2)}}{(2\pi )^3} \int _{{\mathbb {R}}^3} \frac{ e^{-2\lambda ^2 |{\mathbf {p}}|^2}e^{|t|\left[ (|{\mathbf {p}}|^2+m^2-q^2)^2+4q^2p_1^2\right] ^{\frac{1}{4}} }}{\sqrt{p_2^2+p_3^2}} d{\mathbf {p}} \nonumber \\&\le \frac{e^{\lambda ^2 m^2}}{(2\pi )^3}\int _{{\mathbb {R}}^3} \frac{ e^{-2\lambda ^2 |{\mathbf {p}}|^2}e^{a\left[ (|{\mathbf {p}}|^2+m^2)^2+4m^2p_1^2\right] ^{\frac{1}{4}} }}{\sqrt{p_2^2+p_3^2}} d{\mathbf {p}} =: C, \end{aligned}$$

(B.3)

where the constant C does not depend on $\epsilon$ nor on t for $t\in (-a,a)$. By the same argument we can also estimate $e^{-x^1(m-\epsilon )}\Delta _+(t,{\mathbf {x}})$ thus concluding the first part of the proof.

The analyticity of the function $t\mapsto \Delta _{+,\lambda }(t,{\mathbf {x}})$ is manifest in (B.2). To prove the second estimate stated in the proposition we observe that for $q=\frac{m}{2}$,

$$\begin{aligned} 0\le \sqrt{|{\mathbf {p}}|^2-q^2+m^2}-\frac{m}{2} \le \text {Re}\sqrt{|{\mathbf {p}}|^2+2iqp_1-q^2+m^2} -\frac{m}{2} \end{aligned}$$

hence, for positive large $x^1$, positive large $\beta$ and real t, adapting the estimate (B.3) we get

$$\begin{aligned} \left| e^{(\beta +x^1) \frac{m}{2}}\Delta _{+,\lambda }(t-i\beta ,{\mathbf {x}})\right| \le \frac{e^{\lambda ^2 m^2}}{(2\pi )^3}\int _{{\mathbb {R}}^3} \frac{ e^{-2\lambda ^2 |{\mathbf {p}}|^2}e^{a\left[ (|{\mathbf {p}}|^2+m^2)^2+4m^2p_1^2\right] ^{\frac{1}{4}} }}{\sqrt{p_2^2+p_3^2}} d{\mathbf {p}} =: C \end{aligned}$$

with the same constant as before.

The analyticity of the function $t\mapsto \Delta _{\beta ,\lambda }(t,{\mathbf {x}})$ holds because of the presence of the $e^{-2\lambda ^2 |{\mathbf {p}}|^2}$ in its Fourier expansion. The last estimate can be obtained in a similar way starting from (5.6)

$$\begin{aligned} \Delta _{\beta ,\lambda }(t,{\mathbf {x}})= & {} \frac{e^{-\lambda ^2 m^2}}{(2\pi )^3} \int _{{\mathbb {R}}^3} \frac{ e^{-2\lambda ^2 |{\mathbf {p}}|^2} e^{i{\mathbf {p}}{\mathbf {x}}}}{2\sqrt{|{\mathbf {p}}|^2+m^2}} \left( e^{-it \sqrt{|{\mathbf {p}}|^2+m^2}}b_+(\sqrt{|{\mathbf {p}}|^2+m^2}) \right. \\&\left. +\,e^{it \sqrt{|{\mathbf {p}}|^2+m^2}}b_-(\sqrt{|{\mathbf {p}}|^2+m^2}) \right) d{\mathbf {p}} \end{aligned}$$

and bounding the Bose factors $b_\pm$ as follows

$$\begin{aligned} b_+(\sqrt{|{\mathbf {p}}|^2+m^2})= & {} \frac{1}{1-e^{-\beta \sqrt{|{\mathbf {p}}|^2+m^2}}} \le \frac{1}{1-e^{-\beta m}},\\ b_-(\sqrt{|{\mathbf {p}}|^2+m^2})= & {} \frac{1}{e^{\beta \sqrt{|{\mathbf {p}}|^2+m^2}}-1} \le \frac{e^{-\beta {m}}}{1-e^{-\beta m}}\le \frac{e^{-\beta \frac{m}{2}}}{1-e^{-\beta m}}. \end{aligned}$$

$\quad \square$

Proposition B.2

Consider a compact set $\Lambda \subset \Sigma _0$. The propagators $\Delta _{+,\lambda }$, $\Delta _{\beta ,\lambda }$ satisfy the estimates

$$\begin{aligned} |D(t-iu,{\mathbf {x}})|\le \frac{C}{1+|t|^{\frac{3}{2}}}, \qquad D\in \{\Delta _{+,\lambda },\Delta _{\beta ,\lambda }\}, \end{aligned}$$

valid for $t \in {{\mathbb {R}}}$, $u\in [0,\beta ]$ and ${\mathbf {x}}\in \Lambda$, where $C > 0$ is a constant which depends on $\lambda$ and $\Lambda$ but not on $\beta$ for $\beta > \frac{1}{m}$. Furthermore,

$$\begin{aligned} |\Delta _{\beta ,\lambda }(t,{\mathbf {x}})-\Delta _{+,\lambda }(t,{\mathbf {x}})|\le \frac{C}{1+|t|^{\frac{3}{2}}}e^{-\beta m}, \quad (t,{\mathbf {x}}) \in {{\mathbb {R}}}\times \Lambda , \end{aligned}$$

where C is the same constant as before.

Proof

Consider

$$\begin{aligned} \Delta _{+,\lambda }(t-iu, {\mathbf {x}}) := \frac{e^{-\lambda ^2 m^2}}{(2\pi )^3} \int _{{\mathbb {R}}^3} e^{-2\lambda ^2 |{\mathbf {p}}|^2} e^{i{\mathbf {p}}\cdot {\mathbf {x}}} \frac{e^{-i t \sqrt{|{\mathbf {p}}|^2+m^2}}}{2\sqrt{|{\mathbf {p}}|^2+m^2}} e^{- u \sqrt{|{\mathbf {p}}|^2+m^2}} d{\mathbf {p}}. \end{aligned}$$

Assuming for the moment ${\mathbf {x}}=0$, $u=0$ and $t>0$ we get

$$\begin{aligned} \Delta _{+,\lambda }(t,0)&= \frac{e^{-\lambda ^2 m^2}}{(2\pi )^3} \int _{{\mathbb {R}}^3} e^{-2\lambda ^2 |{\mathbf {p}}|^2} \frac{e^{i t \sqrt{|{\mathbf {p}}|^2+m^2}}}{2\sqrt{|{\mathbf {p}}|^2+m^2}} d{\mathbf {p}} \\&= \frac{e^{-\lambda ^2 m^2}}{4\pi ^2} \frac{e^{imt}}{|t|^{\frac{3}{2}}}\int _{0}^\infty e^{-2\lambda ^2 \frac{w}{t}(\frac{w}{t}+2m) } {e^{iw}} \sqrt{w}\sqrt{2m+\frac{w}{t}} d w \end{aligned}$$

where we have integrated over the angular degrees of freedom of the momentum ${\mathbf {p}}$, and we have operated a change of coordinates $w=t(\sqrt{|{\mathbf {p}}|^2+m^2}-m)$. We have now to estimate an integral of the form

$$\begin{aligned} I=\int _{0}^\infty {e^{-iw}} \sqrt{w} g\left( \frac{w}{t}\right) d w \end{aligned}$$

where g is a rapidly decreasing smooth function. We have that

$$\begin{aligned} I=\lim _{\epsilon \rightarrow 0^+} g(0) \int _{0}^\infty {e^{-(i+\epsilon )w}} \sqrt{w} d w + \lim _{\epsilon \rightarrow 0^+}\int _{0}^\infty {e^{-(i+\epsilon )w}} \sqrt{w} \left( g\left( \frac{w}{t}\right) - g(0) \right) d w \end{aligned}$$

the first integrals which contributes to I and the corresponding limit $\epsilon \rightarrow 0$ can be computed, we get

$$\begin{aligned} I= g(0) \frac{\sqrt{\pi }}{2 i^{\frac{5}{2}}} + \lim _{\epsilon \rightarrow 0^+}\frac{1}{(i+\epsilon )^2} \int _{0}^\infty \left( e^{-(i+\epsilon )w}-1\right) \frac{d^2}{d w^2 } \sqrt{w} \left( g\left( \frac{w}{t}\right) - g(0) \right) d w \end{aligned}$$

where in the second contribution we written ${e^{-(i+\epsilon )w}} = \frac{1}{(i+\epsilon )^2}\frac{d^2}{d w^2} \left( e^{-(i+\epsilon )w}-1\right)$ and we have integrated by parts two times. The second contribution can be rewritten as follows with a $0<\delta <\frac{1}{2}$.

$$\begin{aligned}&\int _{0}^\infty \left( e^{-(i+\epsilon )w}-1\right) \frac{d^2}{d w^2 } \sqrt{w} \left( g\left( \frac{w}{t}\right) - g(0) \right) d w \\&\quad = \int _{0}^\infty \left( e^{-(i+\epsilon )w}-1\right) \frac{1}{w^{\frac{3}{2}-\delta }} \frac{1}{t^{\delta }}f_g\left( \frac{w}{t}\right) d w \end{aligned}$$

where $f_g\left( y\right) = \frac{1}{y^\delta }\left( y^2 g''(y) + y g'(y) -\frac{1}{4} (g(y)-g(0))\right) .$ Notice that $f_g(y)$ is a bounded function, while $\left( e^{-(i+\epsilon )w}-1\right) \frac{1}{w^{\frac{3}{2}-\delta }}$ is bounded by an absolutely integrable function uniformly in $\epsilon$. We thus have that

$$\begin{aligned} |I| \le \frac{C_1}{|t|^{\frac{3}{2}}}+\frac{{C}_2}{|t|^{\frac{3}{2}+\delta }} \end{aligned}$$

(B.4)

where $C_1$ is a linear function of g(0) and where ${C}_2$ is a linear function of the maximum of |f|. If an ${\mathbf {x}} \ne 0$, and $u\in [0,\beta ]$ is considered g(y) needs to be multiplied by

$$\begin{aligned} h_{{\mathbf {x}},u}(y)=\text {sinc}(\sqrt{y(y+2m)}|{\mathbf {x}}|) e^{-u (y+m)}. \end{aligned}$$

Notice that $h_{{\mathbf {x}},u}(y)$ is bounded by a constant uniformly for $u\in [0,\beta ]$ and ${\mathbf {x}}\in \Lambda$ and that the same holds for $f_{gh_{{\mathbf {x}},u}}$. Therefore $C_1$ and $C_2$ can be chosen independent of $u\in [0,\beta ]$ and ${\mathbf {x}}\in \Lambda$. If on the other hand $\Delta _{\beta ,\lambda }$ is considered we can analyze separately the positive and negative contribution in a similar way. In particular we will have to multiply g by suitable functions related to the Bose factors

$$\begin{aligned} b^+_{{\mathbf {x}},u}(y)= & {} \text {sinc}(\sqrt{y(y+2m)}|{\mathbf {x}}|) \frac{e^{-u (y+m)}}{1-e^{-\beta (y+m)}},\qquad \\ b^-_{{\mathbf {x}},u}(y)= & {} \text {sinc}(\sqrt{y(y+2m)}|{\mathbf {x}}|) \frac{e^{u (y+m)}}{e^{\beta (y+m)}-1}. \end{aligned}$$

Again, we have that $gb^{\pm }_{{\mathbf {x}},u}(0)$ and $f_{gb^\pm _{{\mathbf {x}},u}}$ are bounded uniformly for $u\in [0,\beta ]$, $\beta \ge \frac{1}{m}$ and ${\mathbf {x}}\in \Lambda$. Hence, the constant $C_1$, ${C}_2$ can be chosen independently on $u\in [0,\beta ]$, ${\mathbf {x}}\in \Lambda$ and $\beta$ for large $\beta$. This proves the first part of the proposition. To analyze the second part, we observe that

$$\begin{aligned} \Delta _{\beta ,\lambda }(t,{\mathbf {x}})-\Delta _{+,\lambda }(t,{\mathbf {x}})= & {} \frac{e^{-\lambda ^2 m^2}}{(2\pi )^3} \int _{{\mathbb {R}}^3} d{\mathbf {p}} \frac{ e^{-2\lambda ^2 |{\mathbf {p}}|^2} e^{i{\mathbf {p}}{\mathbf {x}}}}{2\sqrt{|{\mathbf {p}}|^2+m^2}} \left( e^{-it \sqrt{|{\mathbf {p}}|^2+m^2}}+ \right. \\&\left. +e^{it \sqrt{|{\mathbf {p}}|^2+m^2}} \right) \frac{e^{-\beta \sqrt{|{\mathbf {p}}|^2+m^2}}}{1-e^{-\beta \sqrt{|{\mathbf {p}}|^2+m^2}}}. \end{aligned}$$

We can now repeat the analysis done in the first part of the proof for the positive and negative contribution separately. We need furthermore to multiply the function g discussed above with

$$\begin{aligned} h(y) = e^{-\beta m } \frac{e^{-\beta y}}{1-e^{-\beta (y+m)}} \end{aligned}$$

the corresponding $f_{gh}$ is bounded by a constant which multiplies $e^{-\beta m}$ for $\beta m > 1$. This observation allows us to conclude the proof. $\quad \square$

Return to Equilibrium on Noncommutative Spacetime

The decay for large values of t of $\Delta _{\beta ,\lambda }(t,{\mathbf {x}})$ stated in Proposition B.1 implies the following proposition, which is a clustering condition similar to the one stated in Proposition 3.3 in [17] for fields on a commutative spacetime.

Proposition C.1

Let A, B in ${\mathcal {A}}$, consider the interacting time evolution $\alpha _{t}^V$ where $V=V_{\chi ,h}$ for some h of compact spatial support and the KMS state introduced in (5.5). The following clustering condition holds

$$\begin{aligned} \lim _{t\rightarrow \infty }\left( \omega _{\beta ,\lambda }(A\star _\lambda \alpha _t^V(B)) - \omega _{\beta ,\lambda }(A)\omega _{\beta ,\lambda }(\alpha _t^V(B))\right) = 0. \end{aligned}$$

(C.1)

Proof

Consider

$$\begin{aligned} D(t)=\left( \omega _{\beta ,\lambda }(A\star _\lambda \alpha _t^V(B)) - \omega _{\beta ,\lambda }(A)\omega _{\beta ,\lambda }(\alpha _t^V(B))\right) , \end{aligned}$$

and notice that differentiating recursively the cocycle condition the interacting time evoultion can be written as

$$\begin{aligned} \alpha _t^V(B) = \sum _{n=0}^{+\infty }i^n\int _{tS_n}dT\,[K_{t_n},[K_{t_{n-1}},\ldots ,[K_{t_1},B_t]] \end{aligned}$$

(C.2)

where the integration domain is

$$\begin{aligned} tS_n := \{(t_1,\ldots ,t_n) \in {{\mathbb {R}}}^n\,:\, 0<t_n<\cdots<t_1<t\}. \end{aligned}$$

(C.3)

Hence,

$$\begin{aligned} D(t)= & {} \sum _{n\ge 0} D_n(t) \nonumber \\:= & {} \sum _{n\ge 0} i^n\int _{t S_n} dT \,M \left( (e^{\Gamma }-1 )\left( {\tilde{A}} \otimes {[}{\tilde{K}}_{t_n},[{\tilde{K}}_{t_{n-1}} \ldots [{\tilde{K}}_{t_1},{\tilde{B}}_t]_{\beta }\ldots ]_{\beta }]_{\beta }\right) \right) \nonumber \\ \end{aligned}$$

(C.4)

where $M(A\otimes B) := \omega _\lambda m(A\otimes B)=A(0)B(0)$ (recall that $m(A\otimes B)(\phi )=A(\phi )B(\phi )$),

$$\begin{aligned} \Gamma&:= \Gamma _{\Delta _{\beta ,\lambda }} = \int _{M^2}dxdy\, \Delta _{\beta ,\lambda }(x-y) \frac{\delta }{\delta \phi (x) }\otimes \frac{\delta }{\delta \phi (y)},\\ {\tilde{A}}&:= e^{\frac{1}{2} \int dxdy (\Delta _{\beta ,\lambda }(x-y) - \Delta _{+,\lambda }(x-y)) \frac{\delta ^2}{\delta {\phi (x)}\delta {\phi (y)}} }A, \end{aligned}$$

and where

$$\begin{aligned} {[}A,B]_\beta = m(e^{\Gamma } (A \otimes B - B\otimes A)) = m((e^{\Gamma }-1) (A \otimes B - B\otimes A)). \end{aligned}$$

(C.5)

Hence, the n-th element in $D_n(t)$ in the sum in (C.4) can be expanded as a sum over ${\mathcal {G}}_{n+2}$, the set of connected oriented graphs joining $n+2$ vertices. The vertices of each graphs in ${\mathcal {G}}_{n+2}$ are in correspondence with $\{A,K_{t_n},\ldots ,K_{t_1},B\}$ and the edges with $\Gamma _{ij}$ which is $\Gamma$ applied to the i-th and j-th element of the tensor product ${{\tilde{A}}} \otimes \tilde{K}_{t_n}\otimes \cdots \otimes {{\tilde{K}}}_{t_1}\otimes {{\tilde{B}}}_t$. The only admissible graphs in this graphical expansion are connected because both in (C.4) and in (C.5) $e^{\Gamma }-1$ appears. We have that

$$\begin{aligned} D_n(t)&= \sum _{G\in {{\mathcal {G}}_{n+2}}} c_G D_{n,G}(t) \\&= \sum _{G\in {{\mathcal {G}}_{n+2}}} c_G\int _{t S_n} dT \,M \bigg (\prod _{l \in E(G)} \Gamma _{s(l),r(l)}\bigg )\left( {\tilde{A}} \otimes {\tilde{K}}_{t_n} \otimes {\tilde{K}}_{t_{n-1}} \ldots \otimes {\tilde{K}}_{t_1}\otimes {\tilde{B}}_t\right) \end{aligned}$$

where $c_G$ is a numerical factor which can be either $i^n$ or 0, E(G) denotes the set of edges of G and each edge $l\in E(G)$ is a collection of two vertices $l = (s(l),r(l))$ where $s(l),r(l) \in \{0,\ldots , n+1\}$. Expanding now ${{\tilde{K}}}_{t_j}$ as a sum of terms of the form (5.12) and A, B as a sum of terms of the form (4.9), we see that $D_{n,G}(t)$ can be written as a sum of terms analogous to (5.13). Since A, K, B are of compact support, we have that $A,K,B\in {{\mathcal {A}}}(\Sigma _{0,a})$ for a sufficiently large a. Now thanks to Proposition B.2, and to the fact that $\Delta _{+,\lambda }$, $\Delta _{F,\lambda }$ and $\Delta _{\beta ,\lambda }$ are all bounded, we have that, similarly to the proof of Proposition 5.5,

$$\begin{aligned} |D_{n,G}(t)|&\le E \int _{t S_n} d T \prod _{l \in E(G)}\frac{1}{b+|t_{r(l)}-t_{s(l)}-2a|^{3/2}} \\&\le E' \int _{t S_n} d T \prod _{l \in E(G)}\frac{1}{(|t_{r(l)}-t_{s(l)}|+1)^{3/2}} \end{aligned}$$

where $E, E' > 0$ are suitable constants (depending on G, A, B, a but not on $\beta$), $b=(2a+1)^{3/2}$ and $t_0=t$ and $t_{n+1}=0$. In order to estimate the last integral, let $E_0(G)$ be the set of edges of G connected to the vertex with index 0 (corresponding to B), and let $l_0 \in E_0(G)$ be the one among them which is connect to the vertex with minimal index, indicated by $i_0 \in \{1,\ldots ,n+1\}$. Then, by the form of the integration domain (C.3), $(|t_{r(l_0)}-t_{s(l_0)}|+1)^{-3/2} \le (t_0-t_1+1)^{-3/2}$ and, for all $l \in E_0(G)$, $(|t_{r(l)}-t_{s(l)}|+1)^{-3/2} \le (|t_{r(l')}-t_{s(l')}|+1)^{-3/2}$ where $l'$ is an edge which connects $i_0$ with the vertex different from 0 originally attached to l. Therefore

$$\begin{aligned} \prod _{l \in E(G)}\frac{1}{(|t_{r(l)}-t_{s(l)}|+1)^{3/2}} \le \frac{1}{(t_0-t_1+1)^{3/2}} \prod _{l' \in E(G')}\frac{1}{(|t_{r(l')}-t_{s(l')}|+1)^{3/2}}, \end{aligned}$$

where $G'$ is the connected graph obtained by removing the 0 vertex and $l_0$ from G and by replacing all the edges $l \in E_0(G){\setminus }\{l_0\}$ with the corresponding $l'$ defined above. Iterating this procedure, one gets in the end

$$\begin{aligned} \prod _{l \in E(G)}\frac{1}{(|t_{r(l)}-t_{s(l)}|+1)^{3/2}} \le \prod _{i=0}^n \frac{1}{(t_i-t_{i+1}+1)^{3/2}}. \end{aligned}$$

The integral over $tS_n$ of the last product can now be estimated iteratively thanks to

$$\begin{aligned} \int _{0}^{t_{i-1}} \frac{dt_i}{(t_{i-1}-t_i+1)^{\frac{3}{2}}(t_i+1)^{\frac{3}{2}}}= & {} \frac{4t_{i-1}}{\sqrt{1+t_{i-1}}(2+t_{i-1})^2}\\ {}\le & {} \frac{4}{(t_{i-1}+1)^{\frac{3}{2}}}, \quad i=1,\ldots ,n, \end{aligned}$$

which finally implies

$$\begin{aligned} |D_{n,G}(t)| \le \frac{4^n E'}{(t+1)^{\frac{3}{2}}}, \end{aligned}$$

and therefore $|D_{n,G}(t)|$ vanishes in the limit of large t. $\quad \square$

The following proposition is a stability result for the interacting equilibrium states. Similar results are obtained for the ordinary spacetime in Theorem 2 in [8] or Theorem 3.3 in [17].

Proposition C.2

Consider $V_{\chi ,h}$ for some h of compact support. The KMS state $\omega _{\beta ,\lambda }$ of the free theory shows the following return to equilibrium property

$$\begin{aligned} \lim _{t\rightarrow \infty } \omega _{\beta ,\lambda }(\alpha _t^V(A)) = \omega _{\beta ,\lambda }^{U(i\frac{\beta }{2})}(A), \qquad A\in {\mathcal {A}}, \end{aligned}$$

where $\omega _{\beta ,\lambda }^{U(i\frac{\beta }{2})}$ is a KMS state for the interacting theory defined in (5.7) and in (5.8).

Proof

We begin by proving that $L(t)=\omega _{\beta ,\lambda }(\alpha _t^V(B))$ is a bounded function of t, order by order in perturbation theory. To this end, recalling (C.2) and operating as in (C.4), we have that

$$\begin{aligned} L(t)= \sum _{n\ge 0} L_n(t) := \sum _{n\ge 0} i^n\int _{t S_n} dT \,M \left( [{\tilde{K}}_{t_n},[{\tilde{K}}_{t_{n-1}} \ldots [{\tilde{K}}_{t_1},{\tilde{B}}_t]_{\beta }\ldots ]_{\beta }]_{\beta }\right) \end{aligned}$$

(C.6)

where as before also $L_n$ can be expanded as a sum over connected graphs

$$\begin{aligned} L_n(t)&= \sum _{G\in {{\mathcal {G}}_{n+2}}} c_G' L_{n,G}(t) \\&= \sum _{G\in {{\mathcal {G}}_{n+1}}} c_G'\int _{t S_n} dT \,M \bigg (\prod _{l \in E(G)} \Gamma _{s(l),r(l)}\bigg )\left( {\tilde{K}}_{t_n} \otimes {\tilde{K}}_{t_{n-1}} \ldots \otimes {\tilde{K}}_{t_1}\otimes {\tilde{B}}_t\right) \end{aligned}$$

for some suitable constant $c_G'$ which can also be 0 for some graphs. Therefore, using iteratively the estimate

$$\begin{aligned} \int _{0}^{t_{i-1}}\frac{dt_i}{(t_{i-1}-t_{i}+1)^{3/2}} = 2\left[ 1-\frac{1}{\sqrt{t_{i-1}+1}}\right] \le 2 \end{aligned}$$

we obtain, similarly to the proof of Proposition C.1,

$$\begin{aligned} |L_{n,G}(t)| \le C' \int _{t S_n} d T \prod _{i=1}^{n}\frac{1}{(t_{i-1}-t_{i}+1)^{3/2}} \le 2^{n} C'. \end{aligned}$$

This proves that every $L_{n}$ is bounded in t.

We now follow the strategy of the proof of Theorem 3.3 in  [17]. The KMS conditions implies that

$$\begin{aligned} \omega _{\beta ,\lambda }(\alpha _t^V(A)) = \omega _{\beta ,\lambda }(U(t)\star _\lambda \alpha _t(A)\star _\lambda U(t)^*) = \omega _{\beta ,\lambda }(\alpha _t(A)\star _\lambda U(t)^*\star _\lambda \alpha _{i\beta }U(t)). \end{aligned}$$

(C.7)

Now notice that the cocycle condition, the KMS condition and the time translation invariance of the state imply then that

$$\begin{aligned}&\omega _{\beta ,\lambda }(\alpha _t(A)\star _\lambda U(t)^*\star _\lambda \alpha _{s}U(t))\\&\quad = \omega _{\beta ,\lambda }(\alpha _t(A)\star _\lambda U(t)^*\star _\lambda U(s)^*\star _\lambda U(t)\star _\lambda \alpha _{t}U(s) ) \\&\quad = \omega _{\beta ,\lambda }(\alpha _{t-i\beta }U(s) \star _\lambda \alpha _t(A)\star _\lambda U(t)^*\star _\lambda U(s)^*\star _\lambda U(t)) \\&\quad = \omega _{\beta ,\lambda }(\alpha _{-i\beta }U(s) \star _\lambda A\star _\lambda \alpha _{-t}^V(U(s)^*)). \end{aligned}$$

According to the first part of the proof, we can now choose a sequence $\{t_k\}_{k\in {{\mathbb {N}}}} \subset {{\mathbb {R}}}$ converging to $+\infty$ and such that $\lim _{k \rightarrow +\infty } \omega _{\beta ,\lambda }(\alpha _{t_k}^V(A))$ exists and is finite. Passing to a subsequence, we can also assume that $N := \lim _{k \rightarrow +\infty } \omega _{\beta ,\lambda }( \alpha _{-t_k}^V(U(s)^*))$ is finite. The clustering condition (C.1) established in Proposition C.1 and the KMS condition imply that

$$\begin{aligned} \begin{aligned} \lim _{k\rightarrow \infty } \omega _{\beta ,\lambda }(\alpha _{t_k}(A)\star _\lambda U(t_k)^*\star _\lambda \alpha _{s}U(t_k))&= \omega _{\beta ,\lambda }(\alpha _{-i\beta }U(s) \star _\lambda A) \lim _{k\rightarrow \infty }\omega _{\beta ,\lambda }( \alpha _{-t_k}^V(U(s)^*))\\&= \omega _{\beta ,\lambda }( A\star _\lambda U(s) ) N. \end{aligned} \end{aligned}$$

(C.8)

The previous equality holds also for $\text {Im}s\in [0,\beta ]$: actually, we may extend the results of Proposition C.1 to U(is) following the same proof and using the bounds of Proposition B.2 which also hold when some propagator are extended in imaginary time. The limit $s\rightarrow i\beta$ together with (C.7) gives that

$$\begin{aligned} \lim _{k\rightarrow \infty } \omega _{\beta ,\lambda }(\alpha _{t_k}^V(A)) = \omega _{\beta ,\lambda }( A\star _{\lambda } U(i\beta )) {\tilde{N}} = \frac{\omega _{\beta ,\lambda }( A\star _{\lambda } U(i\beta ))}{\omega _{\beta ,\lambda }(U(i\beta ))} \end{aligned}$$

(C.9)

where in the last equality we used (C.8) with $A=1$ and $s=i\beta$, again the KMS condition and the fact that $\omega _{\beta ,\lambda }$ is normalized and hence $\omega _{\beta ,\lambda }(\alpha _t^V(1)) =1$. This shows that the limit on the left hand side of (C.9) is actually independent of the sequence $\{t_k\}$, and therefore

$$\begin{aligned} \lim _{t\rightarrow \infty } \omega _{\beta ,\lambda }(\alpha _{t}^V(A)) = \frac{\omega _{\beta ,\lambda }( A\star _{\lambda } U(i\beta ))}{\omega _{\beta ,\lambda }(U(i\beta ))}. \end{aligned}$$

The right hand side of the previous equality can be expanded as in (5.8) with the help of the KMS condition. Actually the KMS condition implies that

$$\begin{aligned} \omega _{\beta ,\lambda }( A\star _{\lambda } U(2s)) = \omega _{\beta ,\lambda }(\alpha _{-i\beta +s}U(s)\star _\lambda A\star _{\lambda } U(s)) \end{aligned}$$

which can be extended to a bounded continuous function for $\text {Im}(s)\in [0,\beta /2]$ analytic in its interior and for $s=\beta /2$ gives the desired result. $\quad \square$