A Gaussian Process Related to the Mass Spectrum of the Near-Critical Ising Model

Abstract

Let $\Phi ^h(x)$ with $x=(t,y)$ denote the near-critical scaling limit of the planar Ising magnetization field. We take the limit of $\Phi ^h$ as the spatial coordinate y scales to infinity with t fixed and prove that it is a stationary Gaussian process X(t) whose covariance function K(t) is the Laplace transform of a mass spectral measure $\rho$ of the relativistic quantum field theory associated to the Euclidean field $\Phi ^h.$X and K should provide a useful tool for studying the mass spectrum; e.g., the small distance/time behavior of the covariance functions of $\Phi ^h$ and X(t) shows that $\rho$ is finite but has infinite first moment.

Appendices

Appendix A: $\Phi ^h$ Paired with a 1-d Delta Function

In [5], it was shown that $\Phi ^h(f)$ is a well-defined random variable for any f in the Sobolev space $\mathcal {H}^{-3}(\mathbb {R}^2).$ This was later generalized in [13] to any f in the Besov space $\mathcal {B}_{p,q}^{-\frac{1}{8}-\epsilon ,\text {loc}}(\mathbb {R}^2)$ where $\epsilon >0$ and $p,q\in [1,\infty ].$ But the test function $1_{[-L,L]}(y)\delta _s(t)$ is in neither of those two spaces. The next lemma justifies the pairing of $\Phi ^h$ with such a test function.

Lemma A

For any $\epsilon >0,$ let $g_{\epsilon }$ be the probability density function of $N(0,\epsilon )$ (i.e., Gaussian with mean 0 and variance $\epsilon ).$ For any $f\in L^1(\mathbb {R})\cap L^2(\mathbb {R}),$ we have that

$$\begin{aligned} \{\Phi ^h(f(y)g_{\epsilon }(t)); \epsilon >0\} \text { is a Cauchy sequence in }L^2. \end{aligned}$$

(53)

Remark A

Lemma A holds for more general probability density functions than $g_{\epsilon };$ we choose a Gaussian distribution to simplify the proof.

Proof

For $f\in L^1(\mathbb {R})\cap L^2(\mathbb {R}),$ let

$$\begin{aligned} h(u):=\int _{\mathbb {R}} f(y)f(u+y) dy. \end{aligned}$$

(54)

Our assumption on f implies $|h(u)|\le \Vert f\Vert _2$ for each $u\in \mathbb {R}.$ Therefore,

$$\begin{aligned} \int _{\mathbb {R}}|h(u)|H(0,u)du<\infty , \end{aligned}$$

(55)

where the last inequality follows from the exponential decay of H(0, u) for large |u| and $H(0,|u|)=O(|u|^{-1/4})$ as $|u|\downarrow 0$ (see Theorem 2).

In the next calculation we use the fact that the random variables $\Phi ^h(f(y)g_{\epsilon }(t))$ and $\Phi ^h(f(y)g_{\tilde{\epsilon }}(t))$ have the same expectation, which can be proved with an argument analogous to that used in the proof of Theorem 2 to show that $\mathbb {E}^h\left( \Phi ^h\left( f_{\epsilon }(t,y)\right) \right)$ is independent of $\epsilon$:

$$\begin{aligned}&\mathbb {E}\left[ \Phi ^h(f(y)g_{\epsilon }(t))-\Phi ^h(f(y)g_{\tilde{\epsilon }}(t))\right] ^2\nonumber \\&\quad =\mathbb {E}\left[ \Phi ^h\left( f(y)g_{\epsilon }(t)-f(y)g_{\tilde{\epsilon }}(t)\right) \right] ^2\nonumber \\&\quad = \int _{\mathbb {R}^2}\int _{\mathbb {R}^2}\left[ f(y_1)g_{\epsilon }(t_1)-f(y_1)g_{\tilde{\epsilon }}(t_1)\right] \left[ f(y_2)g_{\epsilon }(t_2)-f(y_2)g_{\tilde{\epsilon }}(t_2)\right] \nonumber \\&\qquad \times H(t_2-t_1, y_2-y_1)dt_2 dy_2 dt_1 dy_1. \end{aligned}$$

(56)

By the change of variables $u=y_2-y_1, v=t_2-t_1, w=y_1, x=t_1,$ this equals

$$\begin{aligned} \int _{\mathbb {R}^2}h(u)H(v,u)\left[ G_{\epsilon ,\epsilon }(v)+G_{\tilde{\epsilon },\tilde{\epsilon }}(v)-G_{\epsilon ,\tilde{\epsilon }}(v)-G_{\tilde{\epsilon },\epsilon }(v)\right] dv du, \end{aligned}$$

(57)

where h(u) is as in (54) and $G_{\epsilon ,\tilde{\epsilon }}:=g_{\epsilon }*g_{\tilde{\epsilon }}$ is the convolution (we have used the fact that $g_{\epsilon }$ is even). Since $g_{\epsilon }$ is the density of $N(0,\epsilon ),$ we have

$$\begin{aligned} G_{\epsilon ,\tilde{\epsilon }}=g_{\epsilon +\tilde{\epsilon }}. \end{aligned}$$

(58)

Another change of variables and using the explicit formula for $g_{\epsilon }$ give that (57) equals

$$\begin{aligned} \int _{\mathbb {R}^2} \frac{h(u)}{\sqrt{2\pi }}e^{-v^2/2}\left[ H(\sqrt{2\epsilon }v,u)+H(\sqrt{2\tilde{\epsilon }}v,u)-2H(\sqrt{\epsilon +\tilde{\epsilon }}v,u)\right] dv du. \end{aligned}$$

(59)

Note that by the monotonicity of H, 

$$\begin{aligned} \left| [H(\sqrt{2\epsilon }v,u)+H(\sqrt{2\tilde{\epsilon }}v,u)-2H(\sqrt{\epsilon +\tilde{\epsilon }}v,u)\right| \le 4H(0,u). \end{aligned}$$

(60)

For any $\eta >0,$ by (55), one may choose $\xi >0$ and $N>\xi$ such that

$$\begin{aligned}&4\int _{ \{u:|u|<\xi \text { or } |u|>N\} }\int _{\mathbb {R}}\frac{|h(u)|}{\sqrt{2\pi }}e^{-v^2/2}H(0,u) dv du \nonumber \\&\quad + 4\int _{\mathbb {R}}\int _{\{v: |v|>N\}}\frac{|h(u)|}{\sqrt{2\pi }}e^{-v^2/2}H(0,u) dv du<\eta /2. \end{aligned}$$

(61)

Since

$$\begin{aligned}&\left| H(\sqrt{2\epsilon }v,u)+H(\sqrt{2\tilde{\epsilon }}v,u)-2H(\sqrt{\epsilon +\tilde{\epsilon }}v,u)\right| \nonumber \\&\quad = \left| H(0,\sqrt{u^2+2\epsilon v^2})+H(0,\sqrt{u^2+2\tilde{\epsilon } v^2})-2H(0,\sqrt{u^2+(\epsilon +\tilde{\epsilon }) v^2})\right| \end{aligned}$$

is uniformly continuous on $\{(u,v):\xi \le |u|\le N, |v|\le N\},$ we have for all small enough $\epsilon$ and $\tilde{\epsilon },$

$$\begin{aligned}&\int _{\{u:\xi \le |u|\le N\}}\int _{\{v:|v|\le N\}}\frac{|h(u)|}{\sqrt{2\pi }}e^{-v^2/2}\Big | H(\sqrt{2\epsilon }v,u)+H(\sqrt{2\tilde{\epsilon }}v,u)\nonumber \\&\quad -\, 2H(\sqrt{\epsilon +\tilde{\epsilon }}v,u)\Big | dvdu < \eta /2. \end{aligned}$$

(62)

This combined with (59)–(61) completes the proof of the lemma. $\square$

Appendix B: Large Distance/Time Behavior of H and K

In this Appendix, we first derive the large t behavior of $\hat{H}(t)$ [recall the definition of $\hat{H}$ in (37)], based on the conjecture that the mass spectrum has an upper gap $(m_1,m_1+\epsilon )$ for some $\epsilon >0.$ Then, based on this behavior, we prove the (first order) large time behavior for K.

Proposition A

Suppose that $\tilde{\rho }\left( (m_1,m_1+\epsilon )\right) =0$ for some $\epsilon >0$ where $\tilde{\rho }$ is defined in (29). Then there exists a constant $C_3\in (0,\infty )$ such that

$$\begin{aligned} \lim _{t\rightarrow \infty }\frac{\hat{H}(t)}{t^{-1/2}e^{-m_1t}}=C_3:=\tilde{\rho }\left( \{m_1\}\right) \sqrt{\pi /(2m_1)}. \end{aligned}$$

(63)

Proof

Using the Källén–Lehmann spectral representation [see (29) and (30)], we have for $t>0$ that

$$\begin{aligned} \hat{H}(t)= \int _{0}^{\infty } \int _{-\infty }^{\infty }\frac{e^{-\sqrt{m^2+p^2}t}}{2\sqrt{m^2+p^2}}dpd\tilde{\rho }(m). \end{aligned}$$

(64)

Recall that the support of $\tilde{\rho }$ is in $[m_1,\infty ).$ Under the assumption that $\tilde{\rho }\left( (m_1,m_1+\epsilon )\right) =0$ for some $\epsilon >0,$ we have

$$\begin{aligned} \hat{H}(t)= \tilde{\rho }\left( \{m_1\}\right) \int _{-\infty }^{\infty }\frac{e^{-\sqrt{m_1^2+p^2}t}}{2\sqrt{m_1^2+p^2}}dp+\int _{m_1+\epsilon }^{\infty } \int _{-\infty }^{\infty }\frac{e^{-\sqrt{m^2+p^2}t}}{2\sqrt{m^2+p^2}}dpd\tilde{\rho }(m). \end{aligned}$$

(65)

By the change of variables $y=\sqrt{1+p^2/m^2},$ we get

$$\begin{aligned} \int _{-\infty }^{\infty }\frac{e^{-\sqrt{m^2+p^2}t}}{2\sqrt{m^2+p^2}}dp&=\int _{1}^{\infty }\frac{e^{-mty}}{\sqrt{y^2-1}}dy\end{aligned}$$

(66)

$$\begin{aligned}&=\int _{1}^{\infty }\frac{e^{-mty}}{\sqrt{2(y-1)}}dy+\int _{1}^{\infty }\left[ \frac{1}{\sqrt{y^2-1}}-\frac{1}{\sqrt{2(y-1)}}\right] e^{-mty}dy. \end{aligned}$$

(67)

Using the inequality

$$\begin{aligned} \sqrt{y+1}-\sqrt{2}\le \frac{y-1}{2\sqrt{2}}\quad \text {for all } y\ge 1, \end{aligned}$$

(68)

we have

$$\begin{aligned} \left| \int _{1}^{\infty }\left[ \frac{1}{\sqrt{y^2-1}}-\frac{1}{\sqrt{2(y-1)}}\right] e^{-mty}dy\right|&\le \frac{1}{8}\int _1^{\infty }\sqrt{y-1}e^{-mty}dy\nonumber \\&=\frac{1}{8}t^{-3/2}e^{-mt}\int _0^{\infty }y^{1/2}e^{-my}dy. \end{aligned}$$

(69)

On the other hand, the change of variables $u=\sqrt{t(y-1)}$ gives

$$\begin{aligned} \int _1^{\infty }\frac{e^{-mty}}{\sqrt{2(y-1)}}dy&=(2t)^{-1/2}e^{-mt}\int _0^{\infty }2e^{-mu^2}du\nonumber \\&=\sqrt{\pi /(2m)}t^{-1/2}e^{-mt}. \end{aligned}$$

(70)

Combining (66)–(70), we get that

$$\begin{aligned} \lim _{t\rightarrow \infty }\frac{\int _{-\infty }^{\infty }\frac{e^{-\sqrt{m_1^2+p^2}t}}{2\sqrt{m_1^2+p^2}}dp}{t^{-1/2}e^{-m_1t}}=\sqrt{\pi /(2m_1)}. \end{aligned}$$

(71)

By (66) and (70), we have

$$\begin{aligned} \int _{m_1+\epsilon }^{\infty } \int _{-\infty }^{\infty }\frac{e^{-\sqrt{m^2+p^2}t}}{2\sqrt{m^2+p^2}}dpd\tilde{\rho }(m)&\le \int _{m_1+\epsilon }^{\infty } \int _{1}^{\infty }\frac{e^{-mty}}{\sqrt{2(y-1)}}dyd\tilde{\rho }(m)\nonumber \\&= \int _{m_1+\epsilon }^{\infty }\sqrt{\pi /(2m)}t^{-1/2}e^{-mt}d\tilde{\rho }(m). \end{aligned}$$

(72)

Therefore,

$$\begin{aligned}&\limsup _{t\rightarrow \infty } \left[ \int _{m_1+\epsilon }^{\infty } \int _{-\infty }^{\infty }\frac{e^{-\sqrt{m^2+p^2}t}}{2\sqrt{m^2+p^2}}dpd\tilde{\rho }(m)\right] /\left[ t^{-1/2}e^{-m_1t}\right] \nonumber \\&\quad \le \limsup _{t\rightarrow \infty } \int _{m_1+\epsilon }^{\infty }\sqrt{\pi /(2m)}e^{-(m-m_1)t}d\tilde{\rho }(m). \end{aligned}$$

(73)

It is not hard to prove that the last limit is 0 by using the fact that $\int _{m_1}^{\infty }(1/m)d\tilde{\rho }(m)<\infty$ (see Corollary 1). This, (65) and (71) complete the proof of the proposition. $\square$

The power-law correction to the exponential decay in (63) is usually known as Ornstein–Zernike decay (see, e.g., [3, Sect. VII]). It may be possible to prove (63) directly without the assumption of an upper mass gap. Based on this, we have the following large time behavior of K.

Proposition B

Suppose (63) holds. Then we have

$$\begin{aligned} \lim _{t\rightarrow \infty }\frac{K(t)}{e^{-m_1t}}=C_3\sqrt{2\pi /m_1}=\pi \tilde{\rho }\left( \{m_1\}\right) /m_1. \end{aligned}$$

(74)

We need the following ancillary lemma.

Lemma B

Let $m>0.$ For any $\alpha ,\beta$ satisfying $0<\beta<1/2<\alpha <3/4,$ we have

$$\begin{aligned} \lim _{t\rightarrow \infty }\frac{\int _0^{\infty }(u^2+t^2)^{-1/4}e^{-m\sqrt{u^2+t^2}}du}{e^{-mt}}=\lim _{t\rightarrow \infty }\frac{\int _{t^{\alpha }}^{t^{\beta }}e^{-m\sqrt{u^2+t^2}}du}{t^{1/2}e^{-mt}}=\sqrt{\frac{\pi }{2m}}. \end{aligned}$$

(75)

Proof

We first note that

$$\begin{aligned} \int _0^{t^{\beta }}(u^2+t^2)^{-1/4}e^{-m\sqrt{u^2+t^2}}du&\le \int _0^{t^{\beta }}t^{-1/2}e^{-m\sqrt{u^2+t^2}}du\nonumber \\&\le t^{-1/2}\int _0^{t^{\beta }} e^{-m\left( t^{-1/2}u+\sqrt{1-t^{-1}}t\right) }du\nonumber \\&=\frac{1}{m}e^{-m\sqrt{t^2-t}}\left[ 1-e^{-mt^{\beta -1/2}}\right] \nonumber \\&\le t^{\beta -1/2}e^m e^{-mt}, \end{aligned}$$

(76)

where we have used the bound (which can be proved by the Cauchy–Schwarz inequality)

$$\begin{aligned} \sqrt{u^2+t^2}\ge t^{-1/2}u+\sqrt{1-t^{-1}}t\quad \text {for all } u,\quad t>0 \end{aligned}$$

(77)

in the second inequality, and

$$\begin{aligned} 1-e^{-y}\le y\quad \text {for all }y\ge 0,\quad \sqrt{t^2-t}\ge t-1\quad \text { for all } t\ge 1 \end{aligned}$$

(78)

in the last inequality. Using the same inequalities, one can show that

$$\begin{aligned} \int _{t^{\alpha }}^{\infty }(u^2+t^2)^{-1/4}e^{-m\sqrt{u^2+t^2}}du&\le \int _{t^{\alpha }}^{\infty }t^{-1/2}e^{-m\sqrt{u^2+t^2}}du\nonumber \\&\le t^{-1/2}\int _{t^{\alpha }}^{\infty } e^{-m\left( t^{-1/2}u+\sqrt{1-t^{-1}}t\right) }du\nonumber \\&=\frac{1}{m}e^{-m\sqrt{t^2-t}}e^{-mt^{\alpha -1/2}}\nonumber \\&\le \frac{e^m}{m}e^{-mt^{\alpha -1/2}}e^{-mt}. \end{aligned}$$

(79)

Combining (76) and (79), we get

$$\begin{aligned} \lim _{t\rightarrow \infty }\frac{\int _0^{\infty }(u^2+t^2)^{-1/4}e^{-m\sqrt{u^2+t^2}}du}{e^{-mt}}=\lim _{t\rightarrow \infty }\frac{\int _{t^{\beta }}^{t^{\alpha }}(u^2+t^2)^{-1/4}e^{-m\sqrt{u^2+t^2}}du}{e^{-mt}} \end{aligned}$$

(80)

provided the limit on the LHS exists. It is easy to see that

$$\begin{aligned} (t^{2\alpha }+t^2)^{-1/4}\int _{t^{\beta }}^{t^{\alpha }}e^{-m\sqrt{u^2+t^2}}du&\le \int _{t^{\beta }}^{t^{\alpha }}(u^2+t^2)^{-1/4}e^{-m\sqrt{u^2+t^2}}du\nonumber \\&\le (t^{2\beta }+t^2)^{-1/4}\int _{t^{\beta }}^{t^{\alpha }}e^{-m\sqrt{u^2+t^2}}du. \end{aligned}$$

(81)

This and (80) prove the first equality in the lemma provided the limit exists. By the Taylor expansion, one can show that there exist constants $C_4, C_5\in (0,\infty )$ such that for any $u\in [t^{\beta }, t^{\alpha }]$

$$\begin{aligned} 1+\frac{u^2}{2t^2}-C_4\left( \frac{u}{t}\right) ^4\le \sqrt{1+(u/t)^2}\le 1+\frac{u^2}{2t^2}-C_5\left( \frac{u}{t}\right) ^4. \end{aligned}$$

(82)

Therefore,

$$\begin{aligned} e^{mC_5 t^{4\beta -3}}\int _{t^{\beta }}^{t^{\alpha }}e^{\frac{-mu^2}{2t}}du&\le \int _{t^{\beta }}^{t^{\alpha }}\exp \left( mt\left[ 1-\sqrt{1+(u/t)^2}\right] \right) du\nonumber \\&\le e^{mC_4 t^{4\alpha -3}}\int _{t^{\beta }}^{t^{\alpha }}e^{\frac{-mu^2}{2t}}du. \end{aligned}$$

(83)

By our assumption on $\alpha$ and $\beta ,$ this implies

$$\begin{aligned} \lim _{t\rightarrow \infty }\frac{\int _{t^{\alpha }}^{t^{\beta }} e^{-m\sqrt{u^2+t^2}}du}{t^{1/2}e^{-mt}}=\lim _{t\rightarrow \infty }\frac{\int _{t^{\beta }}^{t^{\alpha }}e^{\frac{-mu^2}{2t}}du}{t^{1/2}} \end{aligned}$$

(84)

if the limit on the LHS exists. Since

$$\begin{aligned}&\int _0^{t^{\beta }}e^{\frac{-mu^2}{2t}}du\le t^{\beta }\text { and } \nonumber \\&\int _{t^{\alpha }}^{\infty }e^{\frac{-mu^2}{2t}}du\le \int _{t^{\alpha }}^{\infty }\exp \left( \frac{-m(t^{\alpha })^{1/\alpha }u^{2-1/\alpha }}{2t}\right) du<\infty , \end{aligned}$$

(85)

we have

$$\begin{aligned} \lim _{t\rightarrow \infty }\frac{\int _{t^{\beta }}^{t^{\alpha }}e^{\frac{-mu^2}{2t}}du}{t^{1/2}}=\lim _{t\rightarrow \infty }\frac{\int _{0}^{\infty }e^{\frac{-mu^2}{2t}}du}{t^{1/2}}=\sqrt{\frac{\pi }{2m}}, \end{aligned}$$

(86)

where we have used the fact that $(1/\sqrt{2\pi t/m})e^{-mu^2/(2t)}$ is the density of N(0, t/m). $\square$

Proof of Proposition B

The limit (63) implies that, for any fixed $\epsilon >0,$

$$\begin{aligned} (C_3-\epsilon )r^{-1/2}e^{-m_1r}\le \hat{H}(r)\le (C_3+\epsilon )r^{-1/2}e^{-m_1r} \text { for all large }r. \end{aligned}$$

(87)

Lemma B and the definition of K [see (8)] imply that

$$\begin{aligned} (C_3-\epsilon )\sqrt{\frac{2\pi }{m}}\le \lim _{t\rightarrow \infty }\frac{K(t)}{e^{-m_1t}}\le (C_3+\epsilon )\sqrt{\frac{2\pi }{m}}. \end{aligned}$$

(88)

This completes the proof of the proposition by sending $\epsilon \downarrow 0$. $\square$