A Nonrelativistic Quantum Field Theory with Point Interactions in Three Dimensions

Abstract

We construct a Hamiltonian for a quantum-mechanical model of nonrelativistic particles in three dimensions interacting via the creation and annihilation of a second type of nonrelativistic particles, which are bosons. The interaction between the two types of particles is a point interaction concentrated on the points in configuration space where the positions of two different particles coincide. We define the operator, and its domain of self-adjointness, in terms of co-dimension-three boundary conditions on the set of collision configurations relating sectors with different numbers of particles.

Technical Lemmas

In this appendix, we spell out the details concerning the bounds on T and S. These bounds are obtained using variants of the Schur test, similar to those derived in [13], for sums of integral operators that give control on the growth in n as the number of summands increases. Applying the basic Schur test to every summand would yield a bound that grows like the number of summands. In the following lemma, we use the symmetry of the functions in ${\mathscr {H}}^{(n)}$ to obtain an improvement that is reflected in the order of the sum and the supremum in the constants $\Lambda$, $\Lambda '$ below. In the cases relevant to us, this will lead to bounds that are independent of the number of summands. We also remark that the same lemma holds for antisymmetric wavefunctions, since only the symmetry of $\left|\psi \right|^2$ is used.

Lemma 16

Let $\ell < n$ and d be positive integers, $M\in {\mathbb {N}}$, and

$$\begin{aligned} {\mathscr {J}}:=\left\{ J:\{1,\dots ,\ell \}\rightarrow \{1,\dots ,n\}\Big \vert J \text { one-to-one}\right\} . \end{aligned}$$

For every $J\in {\mathscr {J}}$ let $\kappa _J\in L^1_\mathrm {loc}({\mathbb {R}}^{dM}\times {\mathbb {R}}^{dn}\times {\mathbb {R}}^{d\ell })$ be a real, nonnegative function and let $F:{\mathbb {R}}^{d\ell }\times {\mathbb {R}}^{d\ell }\rightarrow {\mathbb {R}}^{dM}$ be measurable.

Define the operator $I:{\mathscr {D}}({\mathbb {R}}^{d(M+n)})\rightarrow {\mathscr {D}}'({\mathbb {R}}^{d(M+n)})$ by

$$\begin{aligned} (I\psi )(P,Q)=\sum _{J\in {\mathscr {J}}} \int _{{\mathbb {R}}^{d\ell }} \kappa _J(P,Q,R) \psi (P+F(Q_J,R),{\hat{Q}}_J, R) \mathrm {d}R, \end{aligned}$$

where $Q_J=(q_{J(1)},\dots ,q_{J(\ell )})=(q_{j_1},\dots , q_{j_\ell })$ and ${\hat{Q}}_J$ denotes the vector in ${\mathbb {R}}^{d(n-\ell )}$ formed by $q_1, \dots , q_n\in {\mathbb {R}}^d$ without the entries of $Q_J$. Denote by $\kappa _J^t(Q,R)$ the kernel obtained from $\kappa _J(Q,R)$ by exchanging $r_i$ with $q_{j_i}$ for $i=1, \dots ,\ell$.

If there exists a positive function $g\in L^\infty _\mathrm {loc}({\mathbb {R}}^d)$ for which the quantities

$$\begin{aligned} \Lambda := \sup _{Q\in {\mathbb {R}}^{d n}}\sum _{J\in {\mathscr {J}}} \int _{{\mathbb {R}}^{d \ell }} \prod _{i=1}^\ell \frac{g(q_{j_i})}{g(r_i)}\sup _{P\in {\mathbb {R}}^{dM}} \kappa _J (P,Q,R) \mathrm {d}R \end{aligned}$$

and

$$\begin{aligned} \Lambda ':= \sup _{Q\in {\mathbb {R}}^{d n}}\sum _{J\in {\mathscr {J}}} \int _{{\mathbb {R}}^{d\ell }}\prod _{i=1}^\ell \frac{g(q_{j_i})}{g(r_i)} \sup _{P\in {\mathbb {R}}^{dM}}\kappa _J^t(P,Q,R) \mathrm {d}R \end{aligned}$$

are finite, then I extends to a bounded operator from ${\mathscr {H}}^{(n)}=L^2({\mathbb {R}}^{dM})\otimes L^2({\mathbb {R}}^{d})^{\otimes _{\mathrm {sym}} n}$ to $L^2({\mathbb {R}}^{d(M+n)})$ with norm at most $\sqrt{\Lambda \Lambda '}$.

Proof

Since $\kappa _J$ is nonnegative, we have for any $\varphi ,\psi \in {\mathscr {D}}({\mathbb {R}}^{d(M+n)})$ with $\varphi ,\psi \in {\mathscr {H}}^{(n)}$ and any $\delta >0$

$$\begin{aligned} 0\le&\sum _{J\in {\mathscr {J}}} \int \left| \delta \varphi (P,Q)\prod _{i=1}^\ell \sqrt{\frac{g(q_{j_i})}{ g(r_i)}} - \frac{1}{\delta }\psi (P+F(Q_J,R),{\hat{Q}}_J,R) \prod _{i=1}^\ell \sqrt{\frac{ g(r_i)}{g(q_{j_i})}} \right| ^2 \\&\times \kappa _J(P,Q,R)\mathrm {d}P\mathrm {d}Q \mathrm {d}R. \end{aligned}$$

After expanding the square, the quadratic term in $\varphi$ can be estimated by

$$\begin{aligned} \delta ^2\int \bigg (\sum _{J\in {\mathscr {J}}}\int \kappa _J(P,Q,R)\prod _{i=1}^\ell \frac{g(q_{j_i})}{g(r_i)}\mathrm {d}R \bigg ) \vert \varphi (P,Q) \vert ^2 \mathrm {d}P\mathrm {d}Q \le \delta ^2 \Lambda \Vert \varphi \Vert ^2_{L^2}. \end{aligned}$$

For the term with $\left|\psi \right|^2$, changing variables to ${\hat{Q}}'_J={\hat{Q}}_J$, $Q'_J=R$, $R'=Q_J$, and using the permutation symmetry of $\psi$, gives

$$\begin{aligned}&\sum _{J\in {\mathscr {J}}} \int \kappa _J(P,Q,R) \vert \psi (P+F(Q_J,R),{\hat{Q}}_J,R) \vert ^2 \prod _{i=1}^\ell \frac{g(r_i)}{ g(q_{j_i})}\mathrm {d}P\mathrm {d}Q \mathrm {d}R\\&\quad = \sum _{J\in {\mathscr {J}}} \int \kappa _J^t(P,Q',R') \vert \psi (P+F(R',Q_J'), Q') \vert ^2 \prod _{i=1}^\ell \frac{ g(q'_{j_i})}{ g(r_i')}\mathrm {d}P\mathrm {d}Q' \mathrm {d}R'. \end{aligned}$$

Using first the Hölder inequality in P and then changing variables to $P'=P+F(R',Q_J')$, we can bound this by

$$\begin{aligned}&\sum _{J\in {\mathscr {J}}} \int \left( \sup _{P\in {\mathbb {R}}^{dM}} \kappa _J^t(P,Q',R')\right) \vert \psi (P', Q') \vert ^2 \prod _{i=1}^\ell \frac{ g(q'_{j_i})}{ g(r_i')}\mathrm {d}P'\mathrm {d}Q' \mathrm {d}R'\\&\quad \le \Lambda ' \left\| \psi \right\| _{{\mathscr {H}}^{(n)}}^2. \end{aligned}$$

Together, these estimates imply that

$$\begin{aligned}&2 {Re}(\langle \varphi , I \psi \rangle ) \\&\quad = \sum _{J\in {\mathscr {J}}} \int \kappa _J(P,Q,R) 2 {Re}({\overline{\varphi }}(P,Q)\psi (P+F(Q_J,R),{\hat{Q}}_J,R)) \mathrm {d}P\mathrm {d}Q \mathrm {d}R \\&\quad \le \delta ^2 \Lambda \Vert \varphi \Vert ^2_{L^2} + \frac{\Lambda '}{\delta ^2}\Vert \psi \Vert ^2_{{\mathscr {H}}^{(n)}}. \end{aligned}$$

The same holds true for the negative of the real part of $\langle \varphi , I\psi \rangle$, by replacing $\psi$ by $-\psi$, and the imaginary part, replacing $\psi$ by $\mathrm {i}\psi$. This yields

$$\begin{aligned} \sup _{\begin{array}{c} \Vert \varphi \Vert _{L^2}=1 \\ \Vert \psi \Vert _{{\mathscr {H}}^{(n)}}=1 \end{array}} |\langle \varphi , I \psi \rangle | \le \frac{1}{ 2} \sqrt{2(\delta ^2 \Lambda ^2 + \Lambda '^2/\delta ^2)} , \end{aligned}$$

so I is bounded from ${\mathscr {H}}^{(n)}$ to $L^2({\mathbb {R}}^{d(M+n)})$. Choosing $\delta =\sqrt{\Lambda '/\Lambda }$ gives the bound on the norm $\left\| I \right\| \le \sqrt{\Lambda \Lambda '}$. $\square$

Since the operator T of Lemma 8 is not bounded on ${\mathscr {H}}^{(n)}$, we need to slightly adapt the technique of Lemma 16 for this case.

Lemma 17

There exists a constant C such that for all $n\in {\mathbb {N}}$ and $\psi \in {\mathscr {H}}^{(n)} \cap H^{1}({\mathbb {R}}^{3(M+n)})$

$$\begin{aligned} \left\| T_\mathrm {od}\psi \right\| _{{\mathscr {H}}^{(n)}} \le C \left\| \psi \right\| _{H^1({\mathbb {R}}^{3(M+n)})}. \end{aligned}$$

Proof

Recall the definition of $T_\mathrm {od}$ from Eq. (20)

$$\begin{aligned} \widehat{T_\mathrm {od} \psi }(P,K) =&-\frac{1}{(2\pi )^{3}}\bigg ( \sum _{\mu ,\nu =1}^M \sum _{i=1}^n \int \frac{{\hat{\psi }}(P-e_\mu \xi + e_\nu k_i,{\hat{K}}_i,\xi )}{L(P-e_\mu \xi , K,\xi )} \mathrm {d}\xi \end{aligned}$$

(39)

$$\begin{aligned}&+ \sum _{\mu \ne \nu =1}^M \int \frac{{\hat{\psi }}(P-e_\mu \xi + e_\nu \xi ,K)}{L(P-e_\mu \xi , K,\xi )}\mathrm {d}\xi \bigg ). \end{aligned}$$

(40)

Since we are not interested in the exact dependence of the norm of $T_\mathrm {od}$ on M and m, we will just estimate the operator for fixed indices $\mu , \nu$ and $m=\tfrac{1}{2}$. Set $\kappa (K,\xi )=\frac{1}{n+1+K^2+\xi ^2}$. To bound the sum over i, we argue as in Lemma 16 and obtain for $\psi \in {\mathscr {H}}^{(n)} \cap H^{1}({\mathbb {R}}^{3(M+n)})$, $\varphi \in {\mathscr {H}}^{(n)}$ and $0<\varepsilon <\frac{1}{2}$ the inequality

$$\begin{aligned}&\sum _{i=1}^n \int \frac{2{Re}\left( {\overline{\varphi }}(P,K) {\hat{\psi }}(P-e_\mu \xi + e_\nu k_i,{\hat{K}}_i,\xi )\right) }{n+1+ (P-e_\mu \xi )^2 +K^2 + \xi ^2}\mathrm {d}P \mathrm {d}K \mathrm {d}\xi \\&\quad \le \sum _{i=1}^n \int \kappa (K,\xi ) \bigg (\kappa (K,\xi )^{-1/2+\varepsilon } |{\hat{\psi }}|^2(P-e_\mu \xi + e_\nu k_i,{\hat{K}}_i,\xi ) \frac{\big (\frac{1}{n}+\xi ^2\big )^{1+\varepsilon }}{(1+k_i^2)} \\&\qquad + \kappa (K,\xi )^{1/2-\varepsilon } | \varphi |^2(P,K) \frac{\big (1+k_i^2\big )}{{\big (\frac{1}{n}+\xi ^2\big )}^{1+\varepsilon }} \bigg )\mathrm {d}P \mathrm {d}K \mathrm {d}\xi . \end{aligned}$$

The term with $|\varphi |^2$ is bounded by

$$\begin{aligned}&\left\| \varphi \right\| ^2_{{\mathscr {H}}^{(n)}} \sup _{K\in {\mathbb {R}}^{3n}} \sum _{i=1}^n (1+k_i^2)\int \kappa (K,\xi )^{3/2-\varepsilon } (1+\xi ^2)^{-1-\varepsilon } \mathrm {d}\xi \\&\quad \le \left\| \varphi \right\| ^2_{{\mathscr {H}}^{(n)}} \sup _{K\in {\mathbb {R}}^{3n}} \frac{n+K^2}{(n+1 + K^2)} \int \frac{1}{(1+\eta ^2)^{3/2-\varepsilon }\eta ^{2(1+\varepsilon )}} \mathrm {d}\eta \\&\quad \le \Lambda \left\| \varphi \right\| ^2_{{\mathscr {H}}^{(n)}} , \end{aligned}$$

where $\Lambda$ is clearly independent of n. In the ith term with $|{\hat{\psi }}|^2$, we perform the change of variables $(k_i, \xi ) \mapsto (\eta , k_i)$ which gives a bound by

$$\begin{aligned}&\sum _{i=1}^n \int |{\hat{\psi }} |^2(P,K) \kappa (K,\eta )^{1/2+\varepsilon }\frac{\left( \frac{1}{n}+k_i^2\right) ^{1+\varepsilon }}{1+\eta ^2} \mathrm {d}P \mathrm {d}K \mathrm {d}\eta \\&\quad \le \sum _{i=1}^n \left\| \sqrt{1+K^2}\psi \right\| ^2_{{\mathscr {H}}^{(n)}} \sup _{K\in {\mathbb {R}}^{3n}} \frac{\sum _{i=1}^n\left( \frac{1}{n} + k_i^2\right) ^{1+\varepsilon }}{(1 + K^2)^{1+\varepsilon }} \int \frac{1}{(1+\eta ^2)^{1/2+\varepsilon }\eta ^{2}} \mathrm {d}\eta \\&\quad \le \Lambda ' \left\| \psi \right\| ^2_{H^1}. \end{aligned}$$

Here $\Lambda '$ is independent of n because the $(1+\varepsilon )$-norm of the vector $(1/n+q_1^2,...,1/n+q_n^2)\in {\mathbb {R}}^n$ is bounded by its 1-norm. By the argument of Lemma 16, this proves that (39) defines an operator that is bounded from $H^1$ to ${\mathscr {H}}^{(n)}$ by $\sqrt{\Lambda \Lambda '}$, which is independent of n.

The remaining operator (40) we have to estimate is

$$\begin{aligned} \psi \mapsto \int \frac{{\hat{\psi }}(P-e_\mu \xi + e_\nu \xi ,K)}{n+1+(P-e_\mu \xi )^2 +K^2 + \xi ^2}\mathrm {d}\xi \end{aligned}$$

with $\mu \ne \nu$. Since the number of these terms is independent of n, the necessary bound can be obtained by the standard Schur test. Explicitly, we have, as above,

$$\begin{aligned}&\int \frac{2{Re}\left( \overline{\varphi }(P,K) {\hat{\psi }}(P-e_\mu \xi + e_\nu \xi ,K)\right) }{n+1+(P-e_\mu \xi )^2 +K^2 + \xi ^2}\mathrm {d}\xi \mathrm {d}P \mathrm {d}K\\&\quad \le \int \frac{1}{1+(P-e_\mu \xi )^2 + \xi ^2} \bigg ( |{\hat{\psi }}|^2(P-e_\mu \xi + e_\nu \xi , K) \frac{(1+(p_\nu +\xi )^2)^{1+\varepsilon }}{(1+p_\mu ^2)^{1/2+\varepsilon }}\\&\qquad +\left|\varphi \right|^2(P,K) \frac{(1+p_\mu ^2)^{1/2+\varepsilon }}{(1+(p_\nu +\xi )^2)^{1+\varepsilon }}\bigg )\mathrm {d}\xi \mathrm {d}K \mathrm {d}P. \end{aligned}$$

The term with $\varphi ^2$ is bounded using

$$\begin{aligned}&\int \frac{\left|\varphi \right|^2(P,K)}{1+(P-e_\mu \xi )^2 + \xi ^2} \frac{(1+p_\mu ^2)^{1/2+\varepsilon }}{(1+(p_\nu +\xi )^2)^{1+\varepsilon }} \mathrm {d}\xi \mathrm {d}P \mathrm {d}K\\&\quad \le \left\| \varphi \right\| _{{\mathscr {H}}^{(n)}}^2 \sup _{P\in {\mathbb {R}}^{3M}} (1+p_\mu ^2)^{1/2+\varepsilon } \int \frac{\mathrm {d}\xi }{\left( 1+\frac{1}{2} p_\mu ^2 +2 (\xi -\frac{1}{2} p_\mu )^2\right) (1+(p_\nu +\xi )^2)^{1+\varepsilon }} \\&\quad \le \left\| \varphi \right\| _{{\mathscr {H}}^{(n)}}^2 \sup _{P\in {\mathbb {R}}^{3M}} (1+p_\mu ^2)^{1/2+\varepsilon } \int \frac{\mathrm {d}\xi }{\left( 1+\frac{1}{2} p_\mu ^2 +2 \xi ^2\right) (1+\xi ^2)^{1+\varepsilon }}\\&\quad \le \left\| \varphi \right\| _{{\mathscr {H}}^{(n)}}^2 2\int \frac{\mathrm {d}\eta }{(1+4\eta ^2)\eta ^{2+2\varepsilon }}, \end{aligned}$$

where we have used the Hardy–Littlewood inequality. After changing variables to $Q=P-e_\mu \xi + e_\nu \xi$, the argument for the term with $|\psi |^2$ is essentially the same, and we conclude as before. $\square$

Lemma 18

Let $\varepsilon >0$, $\psi \in {\mathscr {H}}^{(n)} \cap H^\varepsilon ({\mathbb {R}}^{3(M+n)})$ and $R\psi =- L^{-1}T G \psi$. Then

$$\begin{aligned} S_\mathrm {sing}\psi := \lim _{r\rightarrow 0}\sum _{\mu =1}^M \frac{1}{4\pi }\int \limits _{S^2} \left( \sqrt{n+1}\left( R\psi \right) (X,Y,x_\mu +r\omega )+\gamma _m \log (r)\psi (X,Y)\right) \mathrm {d}\omega \end{aligned}$$

exists in ${\mathscr {H}}^{(n)}$.

Proof

Let $R_\mathrm {d}=- L^{-1}T_\mathrm {d} G$, $R_\mathrm {od}=- L^{-1}T_\mathrm {od} G$.

In this proof, we will focus on the exact form of the logarithmic divergence and discard some regular terms. Precise estimates of these will be given in the proof of Lemma 19.

When expanded, the expression for $\sqrt{n+1} R_\mathrm {d}=- \sqrt{n+1} L^{-1}T_\mathrm {d} G$ will contain a sum over $\nu \in \{1,\dots , M\}$ coming from $T_\mathrm {d}$ (see Eq. (19)), and a second sum over pairs $(\lambda ,i)\in \{1,\dots , M\}\times \{1,\dots n+1\}$ coming from the creation operator in G. We then want to evaluate this expression on the plane $\{x_\mu =y_{n+1}\}$. For a fixed set of indices, we can write the corresponding term using the Fourier transform, obtaining an expression similar to (35). This is, up to a prefactor,

$$\begin{aligned} \int \frac{\sqrt{n+2+ \frac{1}{2m +1} p_\nu ^2 + \frac{1}{2m} {\hat{P}}_\nu ^2+ K^2}}{L(P,K)^2 }\mathrm {e}^{\mathrm {i}X P + \mathrm {i}Y K} {\hat{\psi }}(P+ e_\lambda k_i, {\hat{K}}_i) \mathrm {d}P \mathrm {d}K. \end{aligned}$$

(41)

To analyse the behaviour as $x_\mu -y_{n+1}\rightarrow 0$, it is instructive to set $Q=P+e_\mu k_{n+1}$. The function $\psi$ then appears as

$$\begin{aligned} {\hat{\psi }}(Q-e_\mu k_{n+1} + e_\lambda k_i, {\hat{K}}_{i}). \end{aligned}$$

The operator defined by (41) is of a very different nature for $(\mu ,n+1)=(\lambda ,i)$ and all other cases. In the case of equality, it is essentially a Fourier multiplier by the value of the $k_{n+1}$-integral, since then ${\hat{\psi }}$ no longer depends on this variable. This is singular as $x_\mu -y_{n+1}\rightarrow 0$, because the integral is not absolutely convergent. One the other hand, if $(\mu ,n+1)\ne (\lambda ,i)$, then this is an integral operator that can be bounded on $L^2$ by the Schur test and depends continuously on $x_\mu -y_{n+1}$. We prove uniform bounds in the particle number on this operator in Lemma 19. Consider now the singular terms, with $(\mu ,n+1)=(\lambda ,i)$. For given $\mu$, there are still M of these, indexed by $\nu$. For $\nu \ne \mu$, the integral over $k_{n+1}$ can be rewritten as

$$\begin{aligned} \int \frac{\mathrm {e}^{\mathrm {i}\rho (y_{n+1}-x_\mu )} \sqrt{n+2+\frac{1}{2m+1}\sigma ^2 + \frac{2m+1}{2m} \rho ^2 + \frac{1}{2m+1}p_\nu ^2 + \frac{1}{2m}{\hat{P}}_{\mu ,\nu }^2 + {\hat{K}}_{n+1}^2}}{\left( n+1 +\frac{1}{2m+1}\sigma ^2 + \frac{2m+1}{2m} \rho ^2+ \frac{1}{2m}{\hat{P}}_\mu ^2 + {\hat{K}}^2_{n+1}\right) ^2}\mathrm {d}\rho . \end{aligned}$$

This can be analysed starting from Eq. (36), by replacing $\sigma ^2$ with the appropriate expression and the prefactor $\sqrt{(2m+2)/(2m+1)}$ by $\sqrt{(2m+1)/(2m)}$. The term for $\mu =\nu$ has the same prefactor as (36), so the total prefactor of the divergent term $\log \left|x_\mu -y_{n+1} \right|$ is

$$\begin{aligned} -\frac{1}{(2\pi )^3}\left( \frac{2m}{2m+1}\right) ^{3}\left( \frac{ 2\sqrt{m(m+1)}}{2m+1} + (M-1) \right) . \end{aligned}$$

(42)

For $R_\mathrm {od}$, we have a sum over $(\nu ,\lambda ,i)\in \{1,\dots , M\}^2\times \{1,\dots ,n+2\}$ with $(\nu ,n+2)\ne (\lambda ,i)$ coming from $T_\mathrm {od}$, Eq. (20), and a sum over $(\omega ,j)$, with $\omega \in \{1,\dots , M\}$ and $i\ne j\in \{1,\dots , n+2\}$, coming from the creation operator. All the summands can be written in a form similar to (37), proportional to

$$\begin{aligned} \int \frac{\mathrm {e}^{\mathrm {i}P X+\mathrm {i}{\hat{K}}_{n+2} {\hat{Y}}_{n+2} }{\hat{\psi }}( P-e_\nu k_{n+2}+e_\lambda k_i+e_\omega k_j, {\hat{K}}_{i,j}) }{L( P,{\hat{K}}_{n+2})L(P-e_\nu k_{n+2},K)L(P-e_\nu k_{n+2}+e_\lambda k_i,{\hat{K}}_{i})} \mathrm {d}P \mathrm {d}K , \end{aligned}$$

(43)

where $K=(k_1,\dots , k_{n+2})$ and L denotes the operator on the space with the number of particles corresponding to the dimension of its argument (which is either $n+1$ or $n+2$). Here, ${\hat{\psi }}$ occurs in the form (with $Q=P+e_\mu k_{n+1}$)

$$\begin{aligned} \psi (Q-e_\mu k_{n+1} - e_\nu k_{n+2} + e_\lambda k_i+ e_\omega k_j, {\hat{K}}_{i,j}). \end{aligned}$$

As in the case $n=0$, the corresponding operator is singular only when the first argument of $\psi$ equals Q. Since $(\nu ,n+2)\ne (\lambda ,i)$, this only happens for $(\nu ,{n+2})=(\omega ,j)$ and $(\mu ,{n+1})=(\lambda ,i)$. In the singular case, the operator is a Fourier multiplier by a function of $(Q, {\hat{K}}_{n+1,n+2})$ proportional to

$$\begin{aligned}&\int \frac{1}{n+1+\frac{1}{2m}(Q - e_\mu k_{n+1})^2 + {\hat{K}}^2_{n+2}} \frac{1}{n+1+\frac{1}{2m}(Q -e_\nu k_{n+2})^2 + {\hat{K}}^2_{n+1}} \\&\quad \times \frac{1}{n+2+\frac{1}{2m}(Q - e_\mu k_{n+1}-e_\nu k_{n+2})^2 + K^2} \mathrm {d}k_{n+1} \mathrm {d}k_{n+2}. \end{aligned}$$

For fixed $\mu$, this gives one term with $\mu =\nu$ that behaves exactly like expression (37) for $M=1$, $n=0$. The $M-1$ terms with $\mu \ne \nu$ give a prefactor of $\log (r)$ that exactly cancels the $(M-1)$-term in Eq. (42), so $\sqrt{n+1}R\psi$ has the same singularity for $x_\mu -y_{n+1}\rightarrow 0$ as in the case $M=1$, $n=0$ that was treated in the proof of Proposition 13.

This proves that the sum of the singular terms and $\gamma _m \log (r) \psi$ converges as $r\rightarrow 0$. Convergence of the remaining terms is implied by the bounds of Lemma 19, as argued in Lemma 9. $\square$

Lemma 19

Let $S_\mathrm {reg}$ be given by (34) and $S_\mathrm {sing}=S-S_\mathrm {reg}$ be the operator from Lemma 18. There exists a constant C such that for all $\varepsilon >0$, $n\in {\mathbb {N}}$ and $\psi \in {\mathscr {H}}^{(n)}\cap H^\varepsilon ({\mathbb {R}}^{3(M+n)})$

$$\begin{aligned} \left\| S\psi \right\| _{{\mathscr {H}}^{(n)}} \le C\left( \frac{1}{\varepsilon }\left\| \psi \right\| _{H^\varepsilon } + (1+\log (n+1)) \left\| \psi \right\| _{{\mathscr {H}}^{(n)}}\right) . \end{aligned}$$

(44)

Proof

For the regular part $S_\mathrm {reg}$, we have

$$\begin{aligned} \left\| S_\mathrm {reg}\psi \right\| _{{\mathscr {H}}^{(n)}} \le&\, M \left\| a(\delta _{x_1})L^{-1} \right\| _{{\mathscr {H}}^{(n+1)}\rightarrow {\mathscr {H}}^{(n)}}\\&\times \left\| \left( (T+c_0)L^{-1}(T+c_0) G_T-c_0 G \right) \psi \right\| _{{\mathscr {H}}^{(n+1)}}. \end{aligned}$$

By Lemma 8, and Proposition 12, Lemma 3, we have

$$\begin{aligned} \left\| \left( (T+c_0)L^{-1}(T+c_0) G_T-c_0 G \right) \psi \right\| _{{\mathscr {H}}^{(n+1)}} \le C (n+1)^{-1/4} \left\| \psi \right\| _{{\mathscr {H}}^{(n)}}. \end{aligned}$$

With Corollary 4, this gives the bound

$$\begin{aligned} \left\| S_\mathrm {reg}\psi \right\| _{{\mathscr {H}}^{(n)}} \le C(n+1)^{-1/4} \left\| \psi \right\| _{{\mathscr {H}}^{(n)}} . \end{aligned}$$

For the singular part $S_\mathrm {sing}$, we give quantitative improvements on the proofs of Proposition 13 and Lemma 18. Since we are not interested in the exact dependence on m here, we set the mass of the x-particles to $m=\tfrac{1}{2}$ during this proof.

We first estimate the errors made by the simplifications in the calculation of the singularity for $M=1$, $n=0$ in Proposition 13. These generalise to the corresponding calculations for arbitrary M, n in Lemma 18 in a straightforward way. The replacement made from (35) to (36) produces an error given by

$$\begin{aligned} c \int \frac{\mathrm {e}^{\mathrm {i}\sigma s +\mathrm {i}\rho r}\left( \sqrt{1+\frac{3}{2} \rho ^2 + \frac{3}{8} \sigma ^2 - \frac{1}{2} \rho \sigma }-\sqrt{\frac{3}{2}} \left|\rho \right| \right) }{\left( 1+\frac{1}{2} \sigma ^2 + 2 \rho ^2\right) ^2}{\hat{\psi }}(\sigma )\mathrm {d}\sigma \mathrm {d}\rho . \end{aligned}$$

(45)

The integrand is bounded by

$$\begin{aligned} \frac{1+ \tfrac{3}{8}\sigma ^2 + \frac{1}{2} \sigma \rho }{\left( 1+ \tfrac{3}{8}\sigma ^2 + \frac{1}{2} \sigma \rho + \tfrac{3}{2}\rho ^2\right) ^{1/2}\left( 1+\frac{1}{2}\sigma ^2 +2\rho ^2\right) ^2} \le C \frac{1+|\sigma |^\varepsilon }{\left( 1+\frac{1}{2}\sigma ^2 +2\rho ^2\right) ^{3/2+\varepsilon /2}}, \end{aligned}$$

for $0<\varepsilon <1/2$. This is integrable in $\rho$, so (45) can be evaluated at $r=0$, leading to an estimate by

$$\begin{aligned} \Vert (45) \vert _{r=0}\Vert _{{\mathscr {H}}^{(0)}} \le \frac{C}{\varepsilon }\Vert \psi \Vert _{H^\varepsilon }. \end{aligned}$$

The simplified singular part of $R_\mathrm {d}$, (36) contributes a Fourier multiplier by $\log (1+\frac{1}{\sqrt{2}} \left|\sigma \right|)$ which can be estimated on $H^\varepsilon$ in the same way. This completes the case of $R_\mathrm {d}$ for $M=1$, $n=0$. The reasoning for the divergent part and arbitrary n is essentially the same, except that there is a term growing like $\log (n+1)$ due to the n-dependence of L.

In the calculation for $R_\mathrm {od}$ with $M=1$, $n=0$, we made a simplification in replacing $(\sigma -\xi )^2$ by $\sigma ^2 + \xi ^2$ in the denominator of (37), i.e. replacing

$$\begin{aligned} \tau (\sigma , \rho , \xi )=\frac{1}{\left( 1+\frac{1}{2}\sigma ^2+2\rho ^2\right) (1 + (\sigma -\xi )^2 +\xi ^2)\left( 2+\xi ^2 + \frac{1}{2}(\sigma -\xi )^2+2(\rho +\frac{1}{2}\xi )^2\right) } \end{aligned}$$

by

$$\begin{aligned} \tau _0(\sigma , \rho , \xi )=\frac{1}{\left( 1+\frac{1}{2}\sigma ^2+2\rho ^2\right) (1 + \sigma ^2 +2 \xi ^2)\left( 1+\frac{3}{2} \xi ^2 +\frac{1}{2} \sigma ^2 +2(\rho +\frac{1}{2}\xi )^2\right) }. \end{aligned}$$

We have, with $0<\varepsilon <\frac{1}{2}$,

$$\begin{aligned} \left|\tau -\tau _0 \right| \le C \frac{|\sigma |^\varepsilon }{(1+2\rho ^2)^{1/2+\varepsilon /2}}\frac{1}{(1+\xi ^2)^{3/2}} \frac{1}{2+\xi ^2+\rho ^2}. \end{aligned}$$

(46)

And this implies that

$$\begin{aligned} \sup _{\sigma \in {\mathbb {R}}^3} (1+\sigma ^2)^{-\varepsilon /2} \int |\tau -\tau _0|(\sigma , \rho , \xi ) \mathrm {d}\rho \mathrm {d}\xi \le \frac{C}{\varepsilon }, \end{aligned}$$

which gives the desired estimate for the error as an operator from $H^\varepsilon$ to ${\mathscr {H}}^{(0)}$. The contribution of the simplified $R_\mathrm {od}$, with $\tau _0$ instead of $\tau$, can be bounded by $\log (1+\left|\sigma \right|)$, as for $R_\mathrm {d}$.

For general M and n, we still need to bound the evaluations of the terms in $\sqrt{n+1}R\psi$ that are regular at $x_\mu =y_{n+1}$. We will prove that the sum of these terms gives rise to a bounded operator on ${\mathscr {H}}^{(n)}$, whose norm is a bounded function of n.

For $\sqrt{n+1} R_\mathrm {d}\psi$, the regular terms are the evaluations of (41) with $(\mu ,n+1)\ne (\lambda ,i)$ at $x_\mu =y_{n+1}$. Denote by $\vartheta _{\mu ,\nu ,\lambda ,i}\psi$ the Fourier transform of this function, that is,

$$\begin{aligned} \vartheta _{\mu ,\nu ,\lambda ,i}\psi (P,{\hat{K}}_{n+1})= \frac{1}{2(2\pi )^4 2^{3/2}}\int \kappa _{\mu ,\nu }(P,K) {\hat{\psi }}(P+e_\lambda k_i-e_\mu k_{n+1},{\hat{K}}_i) \mathrm {d}k_{n+1}, \end{aligned}$$

with

$$\begin{aligned} \kappa _{\mu ,\nu }(P,K)&=\left\{ \begin{aligned}&\frac{\sqrt{n+2+ \frac{1}{2m +1} p_\nu ^2 +\frac{1}{2m} ({\hat{P}}_{\nu }-e_\mu k_{n+1})^2+ K^2}}{L(P-e_\mu k_{n+1},K)^2 } \qquad \mu \ne \nu \\&\frac{\sqrt{n+2+ \frac{1}{2m +1} (p_\mu -k_{n+1})^2+ {\hat{P}}_{\mu }^2+ K^2}}{L(P-e_\mu k_{n+1},K)^2 } \qquad \mu =\nu \end{aligned}\right. \\&\le \frac{2}{L(P-e_\mu k_{n+1},K)^{3/2}}. \end{aligned}$$

Applying, for fixed $\mu , \nu , \lambda$, Lemma 16 with kernel $\kappa _{\mu ,\nu }(P,{\hat{K}}_{n+1}, k_{n+1})$ and weight function $g(k)=1+k^2$, we obtain

$$\begin{aligned} \left\| \sum _{i=1}^{n} \vartheta _{\mu ,\nu ,\lambda ,i} \psi \right\| _{{\mathscr {H}}^{(n)}}&\le \frac{\Vert \psi \Vert _{{\mathscr {H}}^{(n)}}}{(2\pi )^4 2^{3/2}}\sum _{i=1}^{n} \int \frac{(1 +k_i^2)}{(n+1 + K^2 )^{3/2} |k_{n+1}|^2}\mathrm {d}k_{n+1}\\&\le C \Vert \psi \Vert _{{\mathscr {H}}^{(n)}}. \end{aligned}$$

The operators $\vartheta _{\mu ,\nu ,\lambda ,i}$ with $i=n+1$ are bounded by the standard Schur test (see also Lemma 17). This gives a bound on the evaluation of the regular terms in $\sqrt{n+1}R_\mathrm {d}$ that is independent of n.

For $\sqrt{n+1} R_\mathrm {od}\psi$, the regular terms are given by (43) with indices $(\nu ,n+2)\ne (\omega ,j)$ or $(\mu ,n+1)\ne (\lambda ,i)$ and $x_\mu =y_{n+1}$. Their Fourier transforms are

$$\begin{aligned}&\Theta _{\mu ,\nu ,\lambda ,\omega ,i,j} \psi (P, {\hat{K}}_{n+1,n+2})\\&\quad =-\frac{1}{(2\pi )^6} \int \frac{1}{L(P-e_\mu k_{n+1},{\hat{K}}_{n+1})L(P-e_\nu k_{n+2}-e_\mu k_{n+1}, K)}\\&\qquad \frac{{\hat{\psi }}(P-e_\mu k_{n+1} - e_\nu k_{n+2} + e_\lambda k_i + e_\omega k_j)}{L(P-e_\nu k_{n+2}-e_\mu k_{n+1}+ e_\lambda k_i, {\hat{K}}_i)}\mathrm {d}k_{n+1} \mathrm {d}k_{n+2}. \end{aligned}$$

For fixed $\mu ,\nu ,\lambda ,\omega$, there are $n(n+1)$ of these terms. For $j,i<n+1$, we apply Lemma 16 with $\ell =2$. Let $\kappa _{i,j}$ be the kernel of the operator $\Theta _{\mu ,\nu ,\lambda ,\omega ,i,j}$, then

$$\begin{aligned} \kappa _{i,j}(P,{\hat{K}}_{n+1,n+2},k_{n+1},k_{n+2})&\le \frac{1}{(n+1+{\hat{K}}_{n+2}^2)(n+2+K^2)(n+1+{\hat{K}}_i^2)}\\&\le \frac{1}{(n+1+{\hat{K}}_{n+2}^2)^{3/2}(n+1+{\hat{K}}_{i,n+1}^2)^{3/2}}. \end{aligned}$$

Choosing again the weight function $g(k)=1+k^2$, Lemma 16 gives

$$\begin{aligned} \left\| \sum _{i=1}^{n}\sum _{i\ne j=1}^{n} \Theta _{\mu ,\nu ,\lambda ,\omega ,i,j}\psi \right\| _{{\mathscr {H}}^{(n)}} \le \frac{\left\| \psi \right\| _{{\mathscr {H}}^{(n)}}}{(2\pi )^6} \sqrt{\Lambda \Lambda '}, \end{aligned}$$

with

$$\begin{aligned} \Lambda&\le \sup _Q \sum _{i=1}^n \int \frac{(1+q_i^2)(n-1+{\hat{Q}}_i^2)}{(1+\xi ^2)(n+1+{\hat{Q}}^2+\xi ^2)^{3/2} (1+\eta ^2) (n+1+{\hat{Q}}^2_i+\eta ^2)^{3/2}}\mathrm {d}\eta \mathrm {d}\xi \\&\le \left( \int \frac{1}{\eta ^2 (1+\eta ^2)^{3/2}} \mathrm {d}\eta \right) ^2, \end{aligned}$$

and the same bound for $\Lambda '$. If $j=n+2$, $i<n+1$, we apply Lemma 16 with $\ell =1$, $\kappa _i=\kappa _{i,n+2}$ (note that $\kappa _i$ and $F=e_\lambda k_i - e_\mu k_{n+1} - e_\nu k_{n+2}$ depend on the additional variable $\xi =k_{n+2}$, but this changes nothing in the proof of Lemma 16). This gives the bound

$$\begin{aligned}&\left\| \sum _{i=1}^{n} \Theta _{\mu ,\nu ,\lambda ,\omega ,i,n+2}\psi \right\| _{{\mathscr {H}}^{(n)}}\\&\quad \le \frac{\left\| \psi \right\| _{{\mathscr {H}}^{(n)}}}{(2\pi )^6} \sup _Q \int \frac{n+Q^2}{(1+\eta ^2)(n +Q^2+\eta ^2)(n+Q^2+\eta ^2+\xi ^2)(1+\xi ^2)}\mathrm {d}\eta \mathrm {d}\xi \\&\quad \le \frac{\left\| \psi \right\| _{{\mathscr {H}}^{(n)}}}{(2\pi )^6}\left( \int \frac{1}{\eta ^2 (1+\eta ^2)^{3/2}} \mathrm {d}\eta \right) \left( \int \frac{1}{\xi ^2 (1+\xi ^2)} \mathrm {d}\xi \right) . \end{aligned}$$

The estimate for the sum with $i=n+1$, $j<n+1$ is the same. The remaining operators with $(i,j)=(n+1, n+2)$ (but restrictions on $\mu ,\nu ,\lambda ,\omega$) are again bounded by the usual Schur test. This completes the proof of the lemma. $\square$

Lemma 20

For any $\varepsilon >0$ and $n\in {\mathbb {N}}$, the operator S is symmetric on the domain $D(S)={\mathscr {H}}^{(n)} \cap H^\varepsilon ({\mathbb {R}}^{3(M+n)})$.

Proof

The operator $S_\mathrm {reg}$, defined in (34), can be written as

$$\begin{aligned} -\sum _{\mu =1}^M\sum _{\nu =1}^M a(\delta _{x_\mu }) L^{-1} \left( (T+c_0) L^{-1} (T+c_0) (L+T+c_0)^{-1} - c_0 L^{-1}\right) a^*(\delta _{x_\nu }). \end{aligned}$$

The operator $aL^{-1}:{\mathscr {H}}^{(n+1)}\rightarrow {\mathscr {H}}^{(n)}$ is bounded and $(aL^{-1})^*=L^{-1}a^*$, so the second term above is bounded and symmetric. For the first term, observe additionally that

$$\begin{aligned} L^{-1} (T+c_0) L^{-1} (T+c_0) (L+T+c_0)^{-1}=(L+T+c_0)^{-1} (T+c_0) L^{-1} (T+c_0)L^{-1}, \end{aligned}$$

by the resolvent formula. This implies that $S_\mathrm {reg}$ is bounded and symmetric.

As shown in Lemma 18, the divergent terms in $S_\mathrm {sing}$ give rise to real Fourier multipliers. These are bounded from $H^\varepsilon ({\mathbb {R}}^{3(M+n)})$ to ${\mathscr {H}}^{(n)}$ by Lemma 19 and thus symmetric on this domain. The regular terms in $S_\mathrm {sing}$ give rise to a bounded operator by the proof of Lemma 19. Since these terms are regular, the limit in the definition of $S_\mathrm {sing}$ just gives the evaluation at $y_{n+1}=x_\mu$. Denote this evaluation map by $\tau _{x_\mu }(y_{n+1})$. Then the contribution of the regular part of $R_\mathrm {d}$ to $S_\mathrm {sing}$ is

$$\begin{aligned} \sum _{\mu =1}^M\sum _{(\lambda ,i)\ne (\mu ,n+1)}\tau _{x_\mu }(y_{n+1}) L^{-1} T_\mathrm {d}L^{-1} \delta _{x_\lambda }(y_i) \psi (X,{\hat{Y}}_i). \end{aligned}$$

This defines a symmetric operator because $(\tau _x L^{-1})^* = L^{-1} \delta _x$, $T_\mathrm {d}$ is symmetric and both $T_\mathrm {d}$ and L commute with permutations of the $y_i$. Similarly, the regular terms in $R_\mathrm {od}$ are the sum of

$$\begin{aligned} \tau _{x_\mu }(y_{n+1}) L^{-1} \tau _{x_\nu }(y_{n+2}) L^{-1} \delta _{x_\lambda }(y_i) L^{-1}\delta _{x_\omega }(y_j)\psi ^{(n)}(X,{\hat{Y}}_{i,j}) \end{aligned}$$

over all indices with $i\ne j$ and $(\mu ,n+1)\ne (\lambda ,i)$ or $(\nu ,n+2)\ne (\omega ,j)$ (see also Eq. (21)). This operator is symmetric for the same reason as in the case of $R_\mathrm {d}$. $\square$