Proof. See Section 6. ◻

Proof. See Section 6. ◻

Proof. By Lemma 1, it is enough to show that \(h\) has a purely discrete spectrum. For \(h \in S_{\mathbb{R}^n}\), see [35]. For \(h \in S^D_\Omega \cup S_{\mathbb{T}^n}\), Weyl’s essential spectrum theorem applied to the corresponding Laplacian operators gives the purely discreteness of the spectrum. ◻

Proof. An eigenfunction of \(h\) is a solution of an elliptic PDE, whose smoothness is a well-studied problem. The hypoellipticity of Schrödinger operators on \({\mathbb{R}^n}\) [36], the elliptic regularity on a compact, smooth-boundary domain with the Dirichlet boundary [33] and the elliptic regularity on \({\mathbb{T}^n}\) [37] give the smoothness of the eigenfunction of \(h\) in each case. ◻

Proof. See Appendix 9. ◻

Proof. See Section 10. ◻

Proof. See Section 11. ◻

Proof. We have \(h = -\Delta^D_\Omega + tV_1+(1-t)V_2\). By the domain monotonicity (Lemma 9), we have \(\lambda_1(h)\le \lambda_1(h^D_{\Omega'})\), where \(h^D_{\Omega'}:=-\Delta^D_{\Omega'} +V_1\big|_{\Omega'}\) is a Dirichlet Schrödinger operator on \(\Omega'\) with the potential given as the restriction of \(V_1\).

By Lemma 8, we have \[\begin{align} \lambda_1(h^D_{\Omega'}) \le \lambda_1(-\Delta^D_{\Omega'} +\max_{x\in \Omega'} V_1(x)) =\lambda_1(-\Delta^D_{\Omega'}) +\max_{x\in \Omega'} V_1(x), \end{align}\] since \(V_1(y)\le \max_{x\in \Omega'}V(x)\) at all \(y\in\Omega'\). Therefore, we have \[\begin{align} \lambda_1(h)\le \lambda_1(h^D_{\Omega'}) \le \lambda_1(-\Delta^D_{\Omega'}) +\max_{x\in\Omega'} V_1(x). \end{align}\] ◻

Proof of Theorem 1. We define functions \(V_{init},V_{final}, W_t:B^\infty_{L/2} \rightarrow\mathbb{R}_{\ge 0}\) for \(t \in [0,T]\) to be \[\begin{align} V_{init} &:= b \sum_{i\in[n]} \left( 1 - \text{Cut}_{\frac{1}{4\sqrt n},\frac{1}{4\sqrt n}}\!\left(\left| x_i \right| \right) \right) , \\ V_{final} &:= \text{Sat}_{c,1}\!\circ\! V(x), \\ W_t(x) &: = \left(1- \frac{t}{T}\right) V_{init}(x) + \frac{t}{T}V_{final} (x). \end{align}\]

Also define time-dependent Hamiltonians on \(B:=B^2_{r+1}, \mathbb{T}^n :=\mathbb{R}^n/L\mathbb{Z}^n\), and \(\mathcal{N}^n\) \[\begin{align} H_{B}(t) &: = -\Delta^D_B + W_{B,t} , \\ H_{\mathbb{T}^n}(t) &: = -\Delta_{\mathbb{T}^n} + W_{\mathbb{T}^n,t}, \\ H_{Q}(t) &:= \; \; K_Q + \sum_{y \in \mathcal{N}} |y\rangle W_t\big({yL}/{N}\big) \langle y |, \end{align}\] where \(W_{B,t}:= W_t|_B\) is the restriction of \(W_t\) on \(B\), and \(W_{\mathbb{T}^n,t}(x) := W_t(x')\) for \(x '\in[-L/2,L/2]^n\), such that \(x =x'\mod L\mathbb{Z}^n\). Note that \(H_Q(t)\) is defined consistently as in Algorithm 1.

We use the notation \(g:={3\pi^2}/{(2(r+1))^2}\) for the rest of the proof.

6.2.0.2 Step 1:

show \(\text{gap}(H_{B}(t))\ge g\) and \(2(\lambda_1(H_B(t))+1)\le E_0\) for all \(t\in[T]\). At an arbitrary time \(t \in [0,T]\), we know from Theorem 4 that \[\begin{align} \text{gap}(H_B(t)) \ge \; \frac{3\pi^2}{(2(r+1))^2}=g, \end{align}\] since the diameter of \(B\) is \(2(r+1).\)

To upper bound \(\lambda_1(H_B(t))\), we use the fact that \(B\) includes the box \(B':=B^\infty_{1/2\sqrt{n}}\), where the potential is at most \(a\). We define \[\begin{align} H_{B'}(t):= -\Delta^D_{B'} + W_{t}|_{B'} \end{align}\]

By the domain monotonicity (Lemma 9), we have \(\lambda_1(H_B(t))\le \lambda_1(H_{B'}(t))\). We have \(W_{t}|_{B'}(x) \le a\) for \(x \in B'\), by the condition \(B^2_1 \subset V^{-1}_{a}\). Therefore, \[\begin{align} \lambda_1(H_B (t)) &\le \lambda_1(H_{B'}(t)) \\ &\le \lambda_1(-\Delta^D_{B'}) + a \\ &= n(n+3) \pi^2 + a\\ &= E_0/10. \end{align}\]

6.2.0.3 Step 2:

show \(\text{gap}(H_{\mathbb{T}^n}(t))\ge 0.99g\), and \(2(\lambda_1(H_{\mathbb{T}^n}(t))+1)\le E_0\) for all \(t\in[T]\). We first show that \(H_B(t)\) and \(H_{{\mathbb{T}^n}}(t)\) are \((E_0,\sigma)\)-equivalent, and then, transfer the large gap from \(B\) to \(\mathbb{T}^n\) via Lemma 14.

For all \(x\) with \(\|x\|_2\ge r\), we have \(V_{init}(x) >b\) (due to the condition \(V^{-1}_b\subset B^2_r\)) and \[\begin{align} V_{final}(x) = \text{Sat}(V(x))\ge \min(V(x), c) > b, \end{align}\] where we denote \(\text{Sat}: = \text{Sat}_{c,1}\). Therefore, we have \(W_t(x)\ge \min(V_{init}(x),V_{final}(x))> b\). By Lemma 5, it follows that \(H_{B}(t)\) and \(H_{\mathbb{T}^n}(t)\) are \((E_0,\sigma)\)-equivalent.

In addition, by Lemma 13, we have that \(\lambda_1(H_{\mathbb{T}^n}(t)) \le E_0/10 + O(\epsilon_0)\). Therefore, we have \(2(\lambda_1(H_{\mathbb{T}^n}(t) +1)\le E_0\), and Lemma 14 is applicable.

Since \(\sigma = \Theta(\epsilon_0/r^4 E_0^{1.5}) = O(g^2/E_0^{1.5})\), by Lemma 14, we have \[\begin{align} \text{gap}(H_{\mathbb{T}^n}(t)) \ge \text{gap}(H_{B}(t)) - 0.01g \ge 0.99g. \label{eq:main95torus95gap} \end{align}\tag{6}\]

6.2.0.4 Step 3:

show that \(H_Q(t)\ge0.98g\) for all \(t\in[0,T]\). Again, we show that \(H_{{\mathbb{T}^n}}(t)\) and \(H_{Q}(t)\) are \((E_0,\sigma)\)-equivalent and then apply Lemma 14 to lower bound the gap.

To argue equivalence between \(H_{\mathbb{T}^n }(t)\) and \(H_{Q}(t)\) via Lemma 6, we need an upper bound on \(\log_\partial(W_{\mathbb{T}^n,t},\mathbb{T}^n, m)\). We first show that the function \(W_{\mathbb{T}^n,t}\) is smooth at all \(t\in[0,T]\), which amounts to showing smoothness of \(V_{final}\) and \(V_{init}\). For \(x\) with \(\|x\|_\infty < L/2\), \(\text{Sat}\circ V\) is smooth because it is a composition of two smooth functions. For \(x\) with \(\|x\|_\infty = L/2\), we have \(V(x) \ge c +1\) by the condition \(V^{-1}_{c+1}\subset B^\infty_{L/2}\); hence \(\text{Sat}(V(x))\) is constant, and all of its derivatives vanish. Also, \(V_{init}\) is smooth, because each \(\text{Cut}(|x_i|)\) is smooth on the torus, where we denote \(\text{Cut}:=\text{Cut}_{1/4\sqrt n,1/4\sqrt n}\). Therefore, \(W_{\mathbb{T}^n,t}\) is smooth on the torus at all \(t\).

By Condition 2, we have \(V_{final} (\mathbb{T}^n)\subset [ 0, c +1]\), and we have \(V_{init}(\mathbb{T}^n) \subset [0,n b]\) by definition. Therefore, \(W_{\mathbb{T}^n,t}(\mathbb{T}^n) \subset [0,O(c + nb )].\)

We bound the smoothness factor of \(W_{\mathbb{T}^n,t }\), using Lemma 20, \[\begin{align} \log_\partial( W_{\mathbb{T}^n,t},\mathbb{T}^n, m) &= \log_\partial( (1 -t/T) V_{init} + (t/T)V_{final}, \mathbb{T}^n,m) \\ & \le \log 2 + \log_{\partial}(V_{init}, \mathbb{T}^n, m) + \log_{\partial}(V_{final} , \mathbb{T}^n, m). \end{align}\] By Lemmas 20 and 22, we get \[\begin{align} \log_{\partial}(V_{init} , \mathbb{T}^n, m) & \;\le \; \log n + \log b +\log_\partial (\text{Cut}(|x_i|) , \mathbb{T}^n, m) \\ & \;= \; \log n + \log b +\log_\partial (\text{Cut}(|x_i|) , \mathbb{R} , m) \\ & \;= \; \log O (bn^{1.5} m^4). \end{align}\] Also, by Lemma 21, \[\begin{align} \log_{\partial}(V_{final} , \mathbb{T}^n, m) & \;\le \; 2 \log m + \log_\partial (\text{Sat}, [0, c+1], m ) + \log_\partial (V, [-L/2,L/2]^n, m ) \\ & \;\le \; \log (\Theta(m^2 \cdot m^4\cdot p(m))). \end{align}\] Combining the two smoothness factors gives \[\begin{align} \log_\partial( W_{\mathbb{T}^n,t},\mathbb{T}^n, m) \;\le \;\log(\Theta(bn^{1.5} m^{10} p(m)). \end{align}\] Therefore, by Lemma 6, encoding each spatial dimension with \[\begin{align} N \ge \Theta \left( \frac{L^2 \cdot b^2 n^{11.5} (p(2n))^2\cdot (V_{\max } + E_0^6/\sigma^7)^{1.5} }{\epsilon} \right) \end{align}\] dimensional quantum space gives that \(H_{\mathbb{T}^n}(t)\) and \(H_Q(t)\) are \((E_0,\sigma)\)-equivalent.

We have established that \(2(\lambda_1(H_{\mathbb{T}^n}(t))+1)\le E_0\) in Step 2. Therefore, by Lemma 14, it follows that \[\begin{align} \text{gap}(H_{Q}(t))\ge 0.98g. \end{align}\]

6.2.0.5 Step 4:

show \(|\lambda_0(H_{\mathbb{R}^n}(T))-\lambda_0(H_{Q}(T))|\le 0.02\epsilon_0\). By Lemma 5, the condition \(V^{-1}_b \subset B^2_r\) gives that \(H_{\mathbb{R}^n}(T)\) and \(H_{\mathbb{T}^n}(T)\) are \((E_0,\sigma)\)-equivalent. Lemma 12 implies that \[\begin{align} |\lambda_0(H_{\mathbb{R}^n}(T) ) -\lambda_0(H_{\mathbb{T}^n}(T) )| \le \sigma \le 0.01\epsilon_0. \end{align}\]

By Step 3, \(H_{\mathbb{T}^n }(T)\) and \(H_Q(T)\) are \(\sigma\)-equivalent under our choice of parameters. Lemma 12 implies that \[\begin{align} |\lambda_0(H_{\mathbb{T}^n}(T) ) -\lambda_0(H_{ Q}(T) )| \le 2\sigma \le 0.01\epsilon_0. \end{align}\] Therefore, by the triangle inequality, we have \[\begin{align} |\lambda_0(H_{\mathbb{R}^n} (T)) - \lambda_0(H_Q(T))| \le 0.02 \epsilon_0 \end{align}\]

6.2.0.6 Step 5:

finish the proof.

In the first step of the algorithm, each register measures the ground state with probability \(1- O(1/n)\). Therefore \(|\Psi_{init}\rangle\) is correctly prepared with probability \(\Omega(1)\).

We showed in Step 3 that the spectral gap of \(H_Q(t)\) is at least \(0.98\cdot3\pi^2/(2(r+1))^2\). By Adiabatic Theorem 3, after evolving \(|\Psi_{init}\rangle\) under the time-dependent Hamiltonian \(H_Q(t)\) for time \[T = O\!\left( \frac{\|H_Q(0) - H_Q(T)\|^{2}}{0.01 \cdot (\min_{t \in [0,T]} \text{gap}(H_Q(t)))^3} \right) \le O\! \left( {r^6 (N^2 + c + nb)^2 } \right),\] the final state \(|\Psi_{final}\rangle\) is \(0.01\)-close to the ground state of \(H_Q(T)\). The evolution time \(T\) is polynomial in the complexity parameters and so is the norm \(\|H_Q(t)\|\). Therefore, any off-the-shelf Hamiltonian simulation algorithm, such as [30], simulates \(H_Q\) in polynomial time.

The Hamiltonian simulation results in a state that is 0.02-close to the ground state of \(H_Q(T)\), which we measure in the final step of the Algorithm with probability \(\ge 96\%\). The energy we measure deviates from \(\lambda_0(h)\) by at most \(0.02\epsilon_0\) (Step 4 of this proof) plus the error from the phase estimation, which can be made negligible. ◻

Proof of Theorem 2. We specify the parameters \(a,b,c,E, \sigma,r, L\) in Theorem 1, and show that they are polynomials in \(n,m, R\). First, we set \[\begin{align} a = 0, \quad r = 2R,\quad L= 3R, \quad E = \Theta(n^2) , \qquad \sigma = \Theta(\epsilon_0/r^4E_0^{1.5}). \end{align}\]

Condition 2 of Theorem 1 enforces that \[\begin{align} b = \Theta(E^7/\sigma^6) = \Theta(E^{16} R^{24}/ \epsilon_0^6) = \Theta(n^{32}R^{24}/\epsilon_0^6) \end{align}\] and at the same time, \(V^{-1}_b \subset B^2_r\). We satisfy these conditions by choosing a sufficiently small \(\mu\) in [eq:DL-potential].

Note that, for any \(x\in{\mathbb{R}^n}\) such that \(\|x\|=R\), there exists \(j\in[m]\) such that \(a_j\cdot x - b_j \ge 0\). Therefore, for \(x\) such that \(\|x\| = 2R\), we have \[\begin{align} V(x) \ge \frac{E}{\mu^6}\text{Bar}_\epsilon(R) = \Theta\left(\frac{n^2 R}{\mu^6}\right), \end{align}\] and for \[\begin{align} \label{eq:DL95sigma95first} \mu = O \left(\frac{\epsilon_0}{n^{5}R^{23/6} }\right), \end{align}\tag{7}\] we have \(V^{-1}_b \subset B^2_r\).

On the other hand for \(\sigma,\) it should be chosen that \(|\lambda_0(-\Delta^E_\Omega) - \lambda_0(h)| \le O(\epsilon_0)\). By 48 , we define the expanded region \(\Omega\) with \(\epsilon = O(\epsilon_0/n^2)\) to obtain \[\begin{align} |\lambda_0(h^D) - \lambda_0(-\Delta^D_\Omega)| = \epsilon_0/3. \end{align}\] By Lemma 12, we need \((E,O(\epsilon_0))\)-equivalence between \(h\) and \(h^D\) to obtain \[\begin{align} |\lambda_0(h) - \lambda_0(h^D)| =\epsilon_0/3, \end{align}\] where \(E=\Theta(\lambda_0(h)) \le \Theta(n^2)\). By Lemma 24, setting \[\begin{align} \label{eq:DL95sigma95second} \mu = O\left(\frac{\epsilon_0}{E (mn^2)^{1/3}}\right) = O\left(\frac{\epsilon_0}{( m n^8)^{1/3}}\right) \end{align}\tag{8}\] gives \((E_0,O(\epsilon_0))\)-equivalence between \(h\) and \(h^D\), finally yielding \[\begin{align} |\lambda_0(-\Delta^D_\Omega) - \lambda_0(h) | = \epsilon_0/3. \end{align}\] Our algorithm estimates \(\lambda_0(-\Delta^D_\Omega)\) up to \(\epsilon_0/3\) when it estimates \(\lambda_0(h)\) up to \(\epsilon_0/3\).

To satisfy both 7 and 8 , we set \[\begin{align} \mu = O\left(\frac{\epsilon_0}{n^5 m^{1/3} R^{23/6}}\right) . \end{align}\]

We are only left with \(c\), whose upper bound we find by \[\begin{align} c \le \max_{x:\|x\|_\infty \le L} V(x) \le \max_{x:\|x\|_2 \le \sqrt n L} V(x) \le \sum_{j\in[m]} \frac{E}{\mu^6} \sqrt n L = O\left( \frac{ELm\sqrt n}{\mu^6 }\right) \end{align}\] Therefore, each parameter in Theorem 1 is polynomial in \(n,m,R, \epsilon_0\).

Also, since \(V\) is a sum of \(\text{Bar}_\epsilon\), which has a logarithmic smoothness factor by Lemma 22, the algorithm requires only polynomial time and qubits in the parameters. ◻

Proof. Let \(p:=-i\nabla\), so \(-\Delta=p^2\). The conjugations are well-known to be \[T_a^\dagger p\,T_a=p,\quad T_a^\dagger V(x)\,T_a=V(x-a),\qquad M_b^\dagger p\,M_b=p+b,\quad M_b^\dagger V(x)\,M_b=V(x).\] Hence \[W_{a,b}^\dagger\,p^2\,W_{a,b}=(p+b)^2.\] Since \(W_{a,b}\) is unitary, \(\|W_{a,b}\psi\|=\|\psi\|\), and therefore \[h[W_{a,b} \psi]=\frac{\langle\psi|((p+b)^2 \psi\rangle + \langle T_a \psi | V |T_a \psi\rangle}{\langle\psi|\psi\rangle}.\] Set \[\bar p:=\left( \frac{\langle\psi | p_j \psi\rangle}{\langle\psi| \psi\rangle}\right)_{j\in[n]} \in\mathbb{R}^n,\qquad K:=-\frac{\langle\psi|\Delta\psi\rangle}{\langle\psi| \psi\rangle}<\infty.\] Expanding gives \[h[W(a,b)\psi]=\underbrace{\big(\|b\|^2+2\,b\cdot\bar p+K\big)}_{=:q(b)} \;+\;\underbrace{\int_{\mathbb{R}^n} V(x-a)\,|\psi(x)|^2\;\text{d} x}_{=:G(a)}.\] The function \(q(b)\) has Hessian \(2I_n\succeq 0\), hence is (strictly) convex in \(b\). For each fixed \(x\), the map \(a\mapsto V(x-a)\) is convex; integrating convex functions preserves convexity, so \(G(a)\) is convex in \(a\). Therefore \(h[W(a,b)\psi]=q(b)+G(a)\) is jointly convex in \((a,b)\). ◻

Proof of Lemma 4. Since \(g\le E\), it suffices to show Lemma 12 and Lemma 13. ◻

Proof. Let \(\widetilde{\mathcal{D}}_a, \widetilde{\mathcal{D}}_b\) be \((E,\epsilon)\)-truncated domains of \(h_a,h_b\) that are isomorphic to each other via a unitary \(U:\widetilde{\mathcal{D}}_a\rightarrow\widetilde{\mathcal{D}}_b\) such that \(|h_a[\widetilde{\psi}]-h_b[U\widetilde{\psi}]|\le \epsilon\) for all \(\psi \in\widetilde{D}_a\). Suppose \(\lambda_0(h_a) = h_a[\psi_0]\) for a normalized \(\psi_0 \in \mathcal{D}(h_a)\). Because \(h_a[\psi_0] < E\), there is an \(\epsilon\)-truncation of \({\psi_0}\), denoted \(\widetilde{\psi}_0\in\widetilde{\mathcal{D}}_a\). By the definition of \((E,\epsilon)\)-equivalence, we have \[\lambda_0(h_b) \le h_b[U\widetilde{\psi}_0]\le h_a[\widetilde{\psi}_0] + \epsilon\le h_a[\psi_0]+2\epsilon = \lambda_0(h_a) +2\epsilon,\] and therefore \[\begin{align} \lambda_0(h_b)- \lambda_0(h_a) \le 2\epsilon. \label{eq:ground95energy95thm} \end{align}\tag{9}\]

Similarly, let \(\mu_0\in\mathcal{D}_b\) be a vector such that \(h_b[\mu_0] =\lambda_0(h_b).\) The fact that \(\lambda_0(h_b) \le \lambda_0(h_a)+2\epsilon \le E\), for \(\epsilon \in[0,1/2]\), allows an \(\epsilon\)-truncation of \(\mu_0\), which we denote \(\widetilde{\mu}_0\in\widetilde{\mathcal{D}}_b\). By the definition of \((E,\epsilon)\)-equivalence, we have \[\begin{align} \lambda_0(h_a)\le h_a[U^{-1}\widetilde{\mu}_0]\le h_b[\widetilde{\mu}_0] + \epsilon\le h_b[\mu_0]+2\epsilon = \lambda_0(h_b) +2\epsilon, \end{align}\] and therefore \[\begin{align} \lambda_0(h_a)- \lambda_0(h_b) \le 2\epsilon. \end{align}\] Together with 9 , we prove the lemma. ◻

Proof. Let \(\widetilde{\mathcal{D}}_a, \widetilde{\mathcal{D}}_b\) be \((E,\epsilon)\)-truncated domains of \(h_a,h_b\) that are isomorphic to each other via a unitary \(U:\widetilde{\mathcal{D}}_a\rightarrow\widetilde{\mathcal{D}}_b\)

Let \(\mu_0\in \mathcal{D}(h_b)\) be a normalized vector such that \(h_b[\mu_0] = \lambda_0(h_b) .\) From Lemma 12, we know that \(h_b[\mu_0]\le \lambda_0(h_a) + 2 \sigma \le E.\) Let \(\widetilde{\mu}_0\in \widetilde{\mathcal{D}}_b\) be a \(\sigma\)-truncation of \(\mu_0\), and let \(\widetilde{\alpha}_0:=U^{-1}\widetilde{\mu}_0 \in \widetilde{\mathcal{D}}_a\). Then \(h_a[\widetilde{\alpha}_0]\) is a good approximation of \(\lambda_0(h_a)\): \[\begin{align} h_a[\widetilde{\alpha}_0] \le h_b[\widetilde{\mu}_0] +\sigma \le h_b[{\mu}_0] +2\sigma \le \lambda_0(h_a) + 4\sigma. \label{eq:alpha95energy} \end{align}\tag{12}\]

Let \(\psi_0\in \mathcal{D}(h_a)\) be a normalized vector such that \(h_a[\psi_0] = \lambda_0(h_a)\). We show that \(\|\widetilde{ \alpha}_0 - \psi_0\|\) is small. Since \(h_a\) has a purely discrete spectrum, its eigenvectors form a complete set of basis. Therefore, we can write \[\widetilde{\alpha}_0 = \sqrt{1- t^2} \psi_0 + t {\psi}_0^\perp\] for some \(t\ge 0\) and a normalized vector \({\psi}_0^\perp \perp \psi_0\) in \(\mathcal{D}(h_a)\). We are assuming that \(\lambda_0(h_a)\) has a nonzero gap \(g\), so we have \(h_a[\psi_0^\perp] \ge \lambda_1(h_a) = \lambda_0(h_a) + g.\) Since \(\langle \psi_0|h_a|\psi_0^\perp\rangle = 0\), we have \[\begin{align} h_a[\widetilde{\alpha}_0] = (1- t^2) \lambda_0(h_a) + t^2 h_a[{\psi}_0^\perp] \ge (1- t^2) \lambda_0(h_a) + t^2(\lambda_0(h_a) + g) = \lambda_0(h_a) + t^2 g. \label{eq:alpha95energy95lowerbound} \end{align}\tag{13}\] Combining 12 with 13 , we get \[t^2 \le 4\sigma/g.\] Because \(1 - \sqrt{1-t^2} = t^2/({1+\sqrt{1-t^2}})\le t^2\), we have \[\begin{align} \|\widetilde{\alpha}_0 - \psi_0\| = \sqrt{(1 - \sqrt{1-t^2})^2 +t^2} \le \sqrt{ t^4 + t^2} \le \sqrt{2}t \le 2\sqrt {2 \sigma/g}. \label{eq:gap95alpha95distance} \end{align}\tag{14}\]

Let \(\psi_1\in \mathcal{D}(h_a)\) be a normalized eigenvector such that \(h_a[\psi_1] = \lambda_1(h_a)\). Since \(\lambda_1(h_a) < E\), there exists an \(\sigma\)-truncation of \(\psi_1\), which we denote \(\widetilde{\psi}_1 \in \widetilde{\mathcal{D}}_a\). We know that \(\psi_1\) is orthogonal to \(\psi_0\). We now show that \(\widetilde{\beta}_1:= U\widetilde{\psi}_1\) is approximately orthogonal to \(\mu_0\), which in turn will approximately lower bound \(\lambda_1(h_b)\) by \(h_1[\widetilde{\beta}_1]\): \[\begin{align} |\langle\widetilde{\beta}_1|\mu_0\rangle| &\;=\; |\langle\widetilde{\beta}_1|((|\mu_0\rangle -|\widetilde{\mu}_0\rangle) +| \widetilde{\mu}_0\rangle)| \nonumber \\ &\;\le\; |\langle\widetilde{\beta}_1| \widetilde{\mu}_0\rangle | + \|\langle\widetilde{\beta}_1| \| \cdot \| |\mu_0\rangle -|\widetilde{\mu}_0\rangle \| \nonumber \\ &\;\le\; |\langle\widetilde{\beta}_1|\widetilde{ \mu}_0\rangle| + \sigma \tag{15}\\ &\;=\; |\langle U^{-1} \widetilde{\beta}_1|U^{-1}\widetilde{ \mu}_0 \rangle| + \sigma \nonumber \\ &\;=\; |\langle \widetilde{\psi}_1|\widetilde{ \alpha}_0 \rangle| + \sigma \nonumber \\ &\;\le\; |\langle {\psi}_1|\widetilde{ \alpha}_0 \rangle| + 2\sigma \tag{16}\\ &\;\le\; |\langle {\psi}_1|\psi_0 \rangle| + 2\sigma + 2\sqrt{2\sigma/g} \tag{17} \\ &\; = \; 2\sigma + 2\sqrt{2\sigma/g} \tag{18} \end{align}\] Here, 15 and 16 follow from Cauchy-Schwarz and the norm condition of \(\sigma\)-truncation, and 17 follows from 14 . Since \(\epsilon, E^{-1}, g/E<1\), we have \[\begin{align} \label{eq:gaps95trunc95sigma95choice} \sigma = O(\epsilon g/E^{1.5}) =O(e \cdot E^{-0.5} \cdot g/E) \le O(\min(\epsilon,\sqrt{\epsilon/E}, \epsilon g/E)). \end{align}\tag{19}\] It follows that 18 is less than 1.

Therefore, we can write \(\widetilde{\beta}_1 = s\mu_0 + \sqrt{1- |s|^2} {\mu}_0^\perp\) for some \(s\in \mathbb{C}\) such that \(|s|\le 2\sigma + 2\sqrt{2\sigma/g}\) and a normalized vector \({\mu}_0^\perp\in \mathcal{D}(h_b)\) such that \(\mu_0^\perp \perp \mu_0\). Then we have \[\begin{align} (1-|s|^2)h_b[\mu_0^\perp] \; \le \; |s|^2 \lambda_0(h_b) + (1 - |s|^2)h_b[\mu_0^\perp] &\;=\; h_b[\widetilde{\beta}_1]\\ &\;=\; h_b[U\widetilde{\psi}_1] \\ &\;\le\; h_a[\widetilde{\psi}_1] + \sigma \\ &\;\le\; h_a[\psi_1] + 2\sigma \\ &\;=\; \lambda_1(h_a) + 2\sigma, \end{align}\] and therefore \[\begin{align} \lambda_1(h_b) \le h_b[\mu_0^\perp] &\;\le\; \frac{\lambda_1(h_a) +2\sigma}{1-|s|^2} \nonumber \\ &\;\le\; \frac{\lambda_1(h_a) + 2\sigma}{1- (2\sigma + 2\sqrt{2\sigma/g})^2} \tag{20} \\ & \;\le \;2{\lambda_1(h_a)+1} \nonumber\\ &\;\le\; E , \tag{21} \end{align}\] by 19 the condition 10 . From Inequalities 19 and 20 , we have \[\begin{align} \lambda_1(h_b) - \lambda_1(h_a) &\;\le \; \frac{(2\sigma +2 \sqrt{2\sigma/g})^2\lambda_1(h_a) + 2\sigma}{1-(2\sigma +2 \sqrt{2\epsilon/g})^2} \nonumber\\ &\;\le \;(2\sigma +2 \sqrt{2\sigma/g})^2E + \frac{ 2\sigma}{1-(2\sigma +2 \sqrt{2\sigma/g})^2} \nonumber\\ &\;\le \;\epsilon . \label{eq:thm95gap95lambda1hb-lambda1ha} \end{align}\tag{22}\]

Now we upper bound \(\lambda_1(h_a) - \lambda_1(h_b)\) in a similar manner. Let \(\mu_1\in \mathcal{D}(h_b)\) be a normalized eigenvector of \(h_b\) with the second lowest eigenvalue \(\lambda_1(h_b).\) From 21 , we know that there exists an \(\sigma\)-truncation of \(\mu_1\), denoted \(\widetilde{\mu}_1 \in \widetilde{\mathcal{D}}_a\). Let \[\begin{align} \widetilde{\alpha}_1 \;: =\; U^{-1}\widetilde{\mu}_1 \;= \; t'\psi_0 + \sqrt{1-t'^2}\psi_0'^\perp, \end{align}\] for some \(t'\) such that \(|t'|\in [0,1]\) and \(\psi_0'^\perp \in \mathcal{D}(h_a)\) such that \({\psi_0'} ^\perp \perp \psi_0\). By 14 and the norm condition of \(\epsilon\)-truncation, \[\begin{align} |t'| &\;=\; |\langle\psi_0| \widetilde{\alpha}_1\rangle| \nonumber\\ &\;\le\; |\langle\widetilde{\alpha}_0| \widetilde{\alpha}_1\rangle| + 2\sqrt{2\sigma/g} \nonumber\\ &\;=\; |\langle\widetilde{\mu}_0| \widetilde{\mu}_1\rangle| + 2\sqrt{2\sigma/g} \nonumber\\ &\;\le\; |\langle{\mu}_0| {\mu}_1\rangle| +2\sigma + 2\sqrt{2\sigma/g} \nonumber\\ &\;=\; 2\sigma + 2\sqrt{2\sigma/g}. \nonumber \end{align}\] By similar arguments as before, we get \[\begin{align} (1 - |t'|^2)\lambda_1(h_a) \le |t'|^2\lambda_0(h_a) + (1 - |t'|^2)\lambda_1(h_a) = h_a[\widetilde{\alpha}_1] \le h_b[\widetilde{\mu}_1] + \sigma \le \lambda_1(h_b) + 2\sigma \end{align}\] and, \[\begin{align} \frac{\lambda_1(h_b) + 2\sigma}{ 1- (2\sigma + 2\sqrt{2\sigma/g})^2} \ge \lambda_1(h_a). \end{align}\] By similar calculation that led to 22 , we get \[\begin{align} \lambda_1(h_a) - \lambda_1(h_b) \le \epsilon, \nonumber \end{align}\] and combining with 22 gives \[\begin{align} |\lambda_1(h_a) - \lambda_1(h_b) | &\;\le \;\epsilon. \nonumber \end{align}\] ◻

Proof. We have \(\sigma = O(g)\), since \(\sigma \le \Theta(g\cdot g/E\cdot 1/E^{0.5})\) and \(g\le \lambda_1(h_a) < E\). By Lemma 12, \(\sigma\le \Theta(g)\) implies that \[\begin{align} |\lambda_0(h_a) - \lambda_0(h_b)| \le 0.01g/2. \end{align}\] By Lemma 13, \(\sigma\le \Theta(g\cdot g/E^{1.5})\) implies that \[\begin{align} |\lambda_1(h_a) - \lambda_1(h_b)| \le 0.01g/2. \end{align}\] Therefore, the triangle inequality gives \[\begin{align} |\text{gap}(h_a) - \text{gap}(h_b) | \le |\lambda_0(h_a) - \lambda_0(h_b)| + |\lambda_1(h_a) - \lambda_1(h_b)| \le 0.01g. \end{align}\] ◻

Proof of Lemma 5. Let us denote \(L^2(B\subset \Omega_i)\) to be the space of \(L^2\) functions on \(\Omega_i\in\{\mathbb{R}^n, B, \mathbb{T}^n\}\) that is supported on \(B\subset \Omega_i\). By Lemma 16, \(L^2(B\subset \Omega_i)\) is an \((E,\epsilon)\)-truncated domain for \(h_i\) for all \(i\in\{1,2,3\}\). The two \((E,\epsilon)\)-truncated spaces \(L^2(B\subset \Omega_i)\) and \(L^2(B\subset \Omega_j)\) are identified by the unitary \(U: L^2(B\subset \Omega_i)\rightarrow L^2(B\subset \Omega_j)\), such that \[\begin{align} U\psi(x) =\begin{cases} \psi(x) \qquad &\|x\|_2\le r+1 \\ 0 \qquad & \|x\|_2> r+1. \end{cases} \end{align}\]

We have \(h_i[\psi] = h_j[U\psi]\) because \(V_i(x) = V_j(x)\) when \(x \in B.\) Therefore, \(h_i\) and \(h_j\) are \(\epsilon\)-equivalent. ◻

Proof. The proof is by Markov’s inequality. In each case of \(\Omega \in \{\mathbb{R}^n, B, \mathbb{T}^n\}\), we have \[\begin{align} \langle \psi| (-\Delta)|\psi\rangle \;= \; -\int_\Omega \overline{\psi}\nabla \cdot \nabla \psi \;\text{d} x \;= \;\int_\Omega |\nabla \psi|^2 \;\text{d} x \;\ge \; 0, \end{align}\] by integration by parts. Since \(\psi\) is normalized, we have \[\begin{align} E \;= \;h[\psi] \; = \int_\Omega \overline{\psi}(x) (-\Delta + V(x) ) \psi(x) \;\text{d} x \;\ge \int |\psi(x) |^2 V(x) \;\text{d} x \;\ge \frac{E}{\sigma^2}\int_{\|x\| \ge r} |\psi(x )|^2 \;\text{d} x \end{align}\] Therefore, \[\begin{align} \sigma^2 \; \ge \; \int _{\|x\| \ge r} |\psi(x)|^2 \;\text{d} x. \end{align}\] Now we can bound \[\begin{align} \|\psi - f_r\psi\|^2 & \;= \;\int \;\;\;\;\;(1 - f_r(x))^2 |\psi(x)|^2 \;\text{d} x \\ & \;= \; \int_{\| x\| \ge r} (1 - f_r(x))^2 |\psi(x)|^2 \;\text{d} x \\ & \;\le \;\int_{\|x\| \ge r} |\psi(x)|^2 \;\text{d} x \\ & \;\le \;\sigma^2. \end{align}\] ◻

Proof. By Theorem 1 and Lemma 3, there exists an orthonormal basis \(\{ \; \psi_i \;| \;h\psi_i = \lambda_i\psi _i, \;i \in \mathbb{Z}_{\ge 0} \;\}\) for the Hilbert space such that \(0\le \lambda_0\le \lambda_1 \le \lambda_2 \le \cdots\), and each \(\psi_i\) is smooth. Therefore, for a normalized vector \(\psi \in \mathcal{D}(h)\) with \(h[\psi]\le E\), we can write \[\begin{align} \psi = \sum_{i\in\mathbb{Z}_{\ge 0}} c_i\psi_i \end{align}\] for \(c_i\in\mathbb{C}\) such that \(\sum_i |c_i|^2 = 1\). We apply the Markov’s inequality on \(\{|c_i|^2\}_i\) as follows. For \(\sigma>0\), we have \[\begin{align} E\ge h[\psi] = \sum_{i \in \mathbb{Z}_{\ge 0}} |c_i|^2\lambda_i \ge \frac{E}{\sigma^2} \sum_{i: \lambda_i >E/\sigma^2} |c_i|^2 , \end{align}\] which gives us \[\begin{align} \sigma^2 \ge \sum_{i: \lambda_i >E/\sigma^2} |c_i|^2 = \|\psi - \psi^\downarrow\|^2, \; \;\;\;\;\;\text{where} \;\;\; \;\; \; \psi^\downarrow:=\sum_{i: \lambda_i \le E/\sigma^2 } c_i\psi_i. \label{eq:position95trunc95markov95ortho} \end{align}\tag{24}\] Since the spectrum of \(h\) is purely discrete, the number of \(i\) such that \(\lambda_i\le E/\sigma^2\) is finite. Otherwise, there would be an accumulation point of \(\lambda_i\)’s in the compact set \([0,E/\sigma^2]\), which is a contradiction to the fact that purely discrete spectrum accumulates only at \(\infty\). Therefore, \(\psi^\downarrow\) is smooth.

To prove the theorem, it is sufficient to prove the following claim: the vector \(\widetilde{\psi}\in \widetilde{\mathcal{D}}\) is an \(\sigma\)-truncation of \(\psi\), where \[\begin{align} \widetilde{\psi} \;& := \;\frac{f\psi^\downarrow }{\|f \psi^\downarrow \|} , \\ f(x)\;&:= \text{Cut}_{r,1}(\|x\|_2). \end{align}\] The function \(\widetilde{\psi}\) is supported on \(B\), and \(\|f\psi^\downarrow\| \le \|\psi^\downarrow\| \le \|\psi\|\). Therefore, we have \(\psi^\downarrow \in \widetilde{\mathcal{D}}\). We show that \(\widetilde{\psi}\) satisfies the norm and energy conditions of \(\sigma\)-trunction in the rest of the proof.

We first verify the norm condition of \(\sigma\)-truncation. By the triangle inequality, \[\begin{align} \| \psi - \widetilde{\psi} \| &\;\le\;\|\psi -f\psi^\downarrow\| + \|f\psi^\downarrow -f\psi^\downarrow /\|f\psi^\downarrow \|\|\;\nonumber \\ &\;=\;\|\psi -f\psi^\downarrow \| + |\|f\psi^\downarrow\| - 1| \\ &\;=\;\|\psi -f\psi^\downarrow\| + |\|f\psi^\downarrow\| - \|\psi\|| \nonumber \\ &\;\le\;2\|\psi - f\psi^\downarrow\| \\ \label{eq:thm95position95trunc95psitilde95nrom} & \; \le \; 2(\|\psi - \psi^\downarrow\| + \| \psi^\downarrow - f\psi^\downarrow\| \; ) \end{align}\tag{25}\] We have \(\|\psi - \psi^\downarrow\| \le \sigma\) due to 24 . Also, we have \(h[\psi^\downarrow] \le h[\psi]<E\), because \(h[\psi]\) is a convex combination of \(h[\psi^\downarrow]\) and \(h[\psi - \psi^\downarrow]\), and we know \(h[\psi^\downarrow]\le E/\sigma^2 \le h[\psi - \psi^\downarrow]\) by definition. By applying Lemma 15 on \(\psi^\downarrow/\|\psi^\downarrow\|\), we get \(\|\psi^\downarrow - f\psi^\downarrow\|/\|\psi^\downarrow\| \le \sigma^3.\) Therefore, we verify that \(\sigma =O(\epsilon/E)\) for a small enough constant factor satisfies the norm condition of \((E,\epsilon)\)-truncation as \[\begin{align} \|\psi - \widetilde{\psi} \| \le 2(\sigma + \sigma^3\|\psi^\downarrow\|) \le 2(\sigma + \sigma^3) = O(\epsilon/E)\le \epsilon. \label{eq:position95based95norm95condi} \end{align}\tag{26}\]

Now we prove the energy condition. Since \(\nabla f^2 =2f \nabla f = 0\) on the boundary of \(B\), we can apply the divergence theorem to \(|\psi^\downarrow|^2\nabla f^2\) on \(B\) and get \[\begin{align} 0 =\oint_{\partial B} (|\psi^\downarrow|^2\nabla f^2 )\cdot \frac{ x }{\|x\|} \; \text{d}^{n-1}x &= \int_{B} \nabla \cdot( |\psi^\downarrow|^2\nabla f^2) \; \text{d}^{n}x \\ &= \int_{B} 2\text{Re}(\overline{\psi^\downarrow} \nabla \psi^\downarrow) \cdot \nabla f^2 + |\psi^\downarrow|^2\Delta f^2 \; \text{d}^{n}x. \end{align}\] Since \(f\) is 0 outside the ball, \[\begin{align} -\int_{\Omega} \;\;\text{Re}(\overline{\psi^\downarrow}\nabla \psi^\downarrow) \cdot \nabla f^2 \;\text{d} x & \;=\; \nonumber \\ -\int_{B} \text{Re}( \overline{\psi^\downarrow}\nabla \psi^\downarrow) \cdot \nabla f^2 \;\text{d} x & \;= \;\frac{1}{2} \int_{B} |\psi^\downarrow|^2\Delta f^2 \; \text{d} x \nonumber \\ & \;= \;\frac{1}{2} \int_{\Omega} \;\; |\psi^\downarrow|^2\Delta f^2 \; \text{d} x \;= \;\langle \psi^\downarrow |\Delta f^2| \psi^\downarrow\rangle/2 \label{eq:divergence}. \end{align}\tag{27}\] Since \(h[\cdot]\) always gives a real number, we have \[\begin{align} &\;\;\;\;\; \;\|f\psi^\downarrow\| \;h[f\psi^\downarrow] \\ \nonumber & \;= \; \text{Re}(\|f\psi^\downarrow\| \;h[f\psi^\downarrow] ) \\ & \;= \; \text{Re}\int -f \overline{\psi^\downarrow} \Delta (f\psi^\downarrow) + f\overline{\psi^\downarrow} V f\psi^\downarrow \;\text{d}x \nonumber &\\ &\;= \; \int -f |\psi^\downarrow|^2 \Delta f - 2\text{Re}(f\overline{\psi^\downarrow} \nabla f\cdot \nabla \psi^\downarrow) - \text{Re}( f^2 \overline{\psi^\downarrow} \Delta {\psi^\downarrow}) + f^2\overline{\psi^\downarrow} V\psi^\downarrow \;\text{d}x \nonumber &\\ &\;= \; \int -f |\psi^\downarrow|^2 \Delta f - \text{Re}(\overline{\psi^\downarrow}\nabla \psi^\downarrow)\cdot \nabla f^2 - \text{Re}( f^2 \overline{\psi^\downarrow} \Delta \psi^\downarrow) + f^2\overline{\psi^\downarrow} V\psi^\downarrow \;\text{d}x \nonumber &\\ &\;= \; \int |\psi^\downarrow|^2 ( - f\Delta f + \Delta f^2/2) + \text{Re}(f^2 \overline{\psi^\downarrow}(- \Delta +V) \psi^\downarrow) \;\text{d}x \nonumber &\\ &\;= \; \langle \psi^\downarrow | \|\nabla f\|^2 |\psi^\downarrow\rangle + \text{Re}\langle f^2\psi^\downarrow|h\psi^\downarrow\rangle ,\label{eq:thm95position95trunc95hcmu} \end{align}\tag{28}\] where the second to the last equation follows from 27 , and the last equation is by applying the identity \(\Delta f^2 = 2|\nabla f|^2 + 2f \Delta f\).

We will show that the first term of 28 is small, and the second term is approximately \(\langle\psi^\downarrow|h\psi^\downarrow\rangle.\) To upper bound the first term, note that \(\nabla f\) is nonzero only outside \(B^2_r\) where we show \(\psi^\downarrow\) has a low weight. We define a multiplication operator \(p_r:\mathcal{H} \rightarrow \mathcal{H}\) such that for \(\phi \in \mathcal{H},\) \[\begin{align} (p_r\phi)(x) = \begin{cases} \phi(x) \; & \; \text{if } \;\|x\|_2 \le r, \\ 0 \; & \; \text{if } \;\|x\|_2> r. \end{cases} \end{align}\] We apply Lemma 15 on \(\psi^\downarrow/\| \psi^\downarrow\|\) and \(p_r\), and obtain \[\begin{align} \|(1-p_r)\psi^\downarrow\| = \left\|(1-p_r) \frac{\psi^\downarrow}{ \|\psi^\downarrow\|} \right\| \|\psi^\downarrow\| \le \sigma^3 \|\psi^\downarrow\| \le \sigma^3, \end{align}\] which is equivalent to \[\begin{align} \int_{x: \|x\|_2\ge r} |\psi^\downarrow(x) |^2 \;\text{d} x \le \sigma^6. \end{align}\] Therefore, we have \[\begin{align} \langle \psi^\downarrow | \|\nabla f\|^2 |\psi^\downarrow\rangle = \int_{x : \|x\|_2\ge r} \| \nabla f(x)\|^2 |\psi^\downarrow(x)|^2 \;\text{d} r \;\le \sigma^6 \max_{x: \|x\|_2\ge r}\left|\nabla f(x) \right|^2 . \label{eq:thm95position95trunc95nablaf95upperbound} \end{align}\tag{29}\] We compute the gradient in the polar coordinate with the radius \(t = \|x\|_2\) and get \[\begin{align} \max_{\|x\|\ge r} \|\nabla f(x)\|^2 &= \max_{t\ge r} \left(\frac{\text{d} \text{Cut}_{r,1}(t)}{\text{d} t}\right)^2 = \left(\frac{ \text{Bump}(t - r) }{\int_r^{r+1} \text{Bump}( {y - r}) \;\text{d} y } \right)^2 \le \left( \frac{\max_{t \in [0,1]} \text{Bump}(t) }{\int_0^1 \text{Bump}(t) \text{d} t } \right)^2 \\ &= {C}, \end{align}\] where \(C\) is a universal constant. Applying this upper bound to 29 gives \[\begin{align} \langle\psi^\downarrow|\|\nabla f\|^2|\psi^\downarrow\rangle \le {\sigma^6 C}{}. \label{eq:thm95position95trunc95first95term} \end{align}\tag{30}\]

Now we show that the second term of 28 is close to \(\langle \psi^\downarrow | h\psi^\downarrow\rangle\). By the Cauchy-Schwarz inequality, \[\begin{align} |\langle f^2\psi^\downarrow|h\psi^\downarrow\rangle -\langle \psi^\downarrow | h\psi^\downarrow\rangle | \; = \; |(\langle f^2\psi^\downarrow| -\langle \psi^\downarrow |)|h\psi^\downarrow\rangle | \; \le \; \|f^2 \psi^\downarrow - \psi^\downarrow\|\|h\psi^\downarrow\| \label{eq:thm95position95trunc95secondterm} \end{align}\tag{31}\] By applying Lemma 15 on \(\psi^\downarrow/\|\psi^\downarrow\|\) with respect to \(1 -f^2\), we bound the first norm \[\begin{align} \|f^2\psi^\downarrow - \psi^\downarrow\| \le \sigma^3 \| \psi^\downarrow\| \le \sigma^3. \label{eq:thm95position95trunc95secondterm95firstterm} \end{align}\tag{32}\] The second norm is bounded as: \[\begin{align} &\|h\psi^\downarrow\| \;=\; \sqrt{\sum_{i: \lambda_ i \le E/\sigma^2}\lambda_i^2 |c_i|^2} \le \frac{E}{\sigma^2}. \label{eq:thm95position95trunc95secondterm95secondterm} \end{align}\tag{33}\]

Plugging 32 , and 33 into 31 gives \[\begin{align} |\langle f^2\psi^\downarrow|h\psi^\downarrow\rangle -\langle \psi^\downarrow | h\psi^\downarrow\rangle | \; \le \; E\sigma \end{align}\] Since \(\langle \psi ^\downarrow|h\psi^\downarrow\rangle\) is real, we have \[\begin{align} |\text{Re}\langle f^2\psi^\downarrow|h\psi^\downarrow\rangle -\langle \psi^\downarrow |h\psi^\downarrow\rangle| \le |\langle f^2\psi^\downarrow|h\psi^\downarrow\rangle -\langle \psi^\downarrow | h\psi^\downarrow\rangle | , \end{align}\] and therefore \[\begin{align} \text{Re}\langle f^2\psi^\downarrow|h\psi^\downarrow\rangle \;\le \;& \langle \psi^\downarrow|h\psi^\downarrow\rangle + \sigma E \nonumber \\ \;= \;& h[\psi^\downarrow]\|\psi^\downarrow\|^2 + \sigma E \nonumber \\ \;\le \;& h[\psi] + \sigma E . \label{eq:thm95position95trunc95second95term} \end{align}\tag{34}\] By 28 , 30 , and 34 , we get \[\begin{align} \|f\psi^\downarrow\| \;h[f\psi^\downarrow] &\;\le \; h[\psi] + \sigma E + {\sigma^6 C}{} . \end{align}\] The triangle inequality implies \[\begin{align} \|f\psi^\downarrow\| \;\ge \;\|\psi^\downarrow\| -\|(1 - f) \psi^\downarrow\| \;\ge \; 1 - \sigma - \sigma^3. \end{align}\] Therefore, we have \[\begin{align} h[\widetilde{\psi}] = h[f\psi^\downarrow] \le \frac{ h[\psi] + \sigma E+ {\sigma^6 C}{} }{ 1 - \sigma - \sigma^3} \le h[\psi] + \epsilon \end{align}\] for \(\sigma = O(\epsilon/E)\). ◻

Proof of Lemma 6. By Lemma [lem:freq95trunc95torus] and Lemma 18, \[\begin{align} {\widetilde{\mathcal{D}}}_{\mathbb{T}^n,f} &:= \text{span}\left\{ \; \omega_k \;| \;k\in \mathbb{Z}^n, \;\|k\|_2\le K \right\}, \\ {\widetilde{\mathcal{D}}}_{Q} &:= \text{span}\{ \;|p_k\rangle \;| \; k\in \mathbb{Z}^n,\|k\|_2\le K \} \end{align}\] are \((E,\epsilon)\)-truncated domains of \(h_{\mathbb{T}^n}\) and \(h_Q\), respectively, provided that \[\begin{align} K = \Omega\left(\frac{L (E+V_{\max})^{3/2}) }{ \epsilon }\right). \end{align}\] By Lemma 19, the unitary \(U:\widetilde{\mathcal{D}}_{\mathbb{T}^n,f}\rightarrow {\widetilde{\mathcal{D}}}_{Q}\) defined by \(U \omega_k= |p_k\rangle\) satisfies \(|h[\psi] - h[U\psi]|\le \epsilon\) for all \(\psi\in\widetilde{\mathcal{D}}_{\mathbb{T}^n,f}\), provided that \[\begin{align} N\ge 4 K \ge 16\left(\frac{L p(2n) }{2\pi} \right)^2 \frac{1}{e^{1/n}}. \end{align}\] Therefore, \[\begin{align} N = 4K = \Omega\left( L^2 (E + V_{\max})^{1.5}(p(2n))^2/\epsilon \right) \ge \Omega\left( \max \left\{ {L (E+V_{\max})^{1.5}) }/{ \epsilon } , ({L p(2n) } )^2 /{e^{1/n}} \right\} \right) \end{align}\] gives that \(h_{\mathbb{T}^n}\) and \(h_Q\) are \((E,\epsilon)\)-equivalent. ◻

Proof. Let \(\psi \in \mathcal{D}(h)\) be a normalized vector with energy \(h[\psi] \le E\). We show that there exists an \((E,\epsilon)\)-truncation of \(\psi\) in \(\widetilde{ \mathcal{D}} _{\mathbb{T}^n, f}\). Since \(\psi\) is also in \(L^2(\mathbb{T}^n)\), \(\psi\) has the Fourier decomposition \[\begin{align} \psi = \sum_{k\in \mathbb{Z}^n} \widehat{\psi}_k\omega_k. \end{align}\] Since \(\langle \psi |V|\psi\rangle \ge 0\), we have \[\begin{align} E \ge \langle \psi | (-\Delta)\psi\rangle = \sum_{k \in \mathbb{Z}^n} \frac{4\pi^2\|k\|_2^2 \widehat{\psi}_k}{L^2} \langle\psi| \omega_k \rangle = \sum_{k \in \mathbb{Z}^n} \frac{4 \pi^2 \|k\|^2 }{L^2} | \widehat{\psi}_k |^2 . \label{eq:thm95frequency95trunc95kinetic} \end{align}\tag{35}\] Given the parameter \(K\), we define low- and high-frequency vectors \[\begin{align} \psi^\downarrow \; &: = \sum_{\|k\|\le K} \widehat{\psi}_k\omega_k \in \widetilde{\mathcal{D}}_{\mathbb{T}^n, f} , \\ \psi^\uparrow \; &: = \sum_{\|k\| > K} \widehat{\psi}_k\omega_k = \psi - \psi^\downarrow. \end{align}\] Since \(\{\omega_k\}_k\) forms a complete orthonormal basis set in \(L^2(\mathbb{T}^n)\), Parseval’s theorem gives \(\sum_{k \in \mathbb{Z}^n} |\widehat{\psi}_k |^2 = 1\). By the Markov’s inequality on \(|\widehat \psi_k|^2\), we obtain \[\begin{align} \|\psi^\uparrow\|^2 \;= \; \|\psi^\downarrow - \psi \|^2 \;= \;\sum_{\|k\| \ge K} |\widehat{\psi}_k |^2 \;\le \;\frac{EL^2}{4\pi^2 K^2} :=\sigma^2 \label{eq:thm95frequency95trunc95sigma} \end{align}\tag{36}\] where \(\sigma \ge 0.\) We claim that the vector \(\widetilde{\psi}:=\psi^\downarrow/\|\psi^\downarrow\|\) is an \(\epsilon\)-truncation of \(\psi.\) For the norm condition: \[\begin{align} \left\|\frac{\psi^\downarrow}{\|\psi^\downarrow\|} - \psi \right\| & \; \le \; \left\|\frac{\psi^\downarrow}{\|\psi^\downarrow\|} - \psi^\downarrow \right\| + \| \psi ^\downarrow - \psi \| \nonumber \\ & \;= \; \left|\| \psi\| -{\|\psi^\downarrow\|} \right| + \| \psi ^\downarrow - \psi \| \nonumber \\ & \;\le \; 2 \| \psi ^\downarrow - \psi \| \nonumber \\ &\; \le \; 2\sigma. \label{eq:thm95frequency95trunc95norm} \end{align}\tag{37}\] We choose \(\sigma =O( {\epsilon}/{(E + V_{\max} )} )\), so that \(2\sigma\le \epsilon\).

For the energy condition of \((E,\epsilon)\)-truncation, we first bound the Laplacian term. From 35 , we have \[\begin{align} \langle \psi^\downarrow |(-\Delta)|\psi^\downarrow\rangle - \langle \psi |(-\Delta)|\psi\rangle = - \sum_{k :\|k\|_2 > K} \frac{4\pi^2\|k\|^2 }{L^2}|\widehat{\psi}_k |^2 \le 0. \label{eq:thm95frequency95trunc95torus95diff95kinetic} \end{align}\tag{38}\] In order to upper bound \(\langle \widetilde{ \psi}|V|\widetilde{\psi}\rangle - \langle \psi|V|\psi\rangle\), we use the fact \(\| V|\psi\rangle \| \;\le \;V_{\max} \; \| \psi \|.\) By linearity and norm considerations, we have \[\begin{align} | \langle \psi^\downarrow|V|{\psi}^\downarrow\rangle - \langle \psi|V|\psi\rangle| & \; = \; |\langle \psi^\downarrow|V|{\psi}^\downarrow\rangle - (\langle \psi^\downarrow | + \langle \psi^\uparrow | )V (| \psi^\downarrow \rangle + | \psi^\uparrow \rangle ) | \nonumber \\ & \;= \; | - \langle \psi^\downarrow|V|{\psi}^\uparrow\rangle - \langle \psi^\uparrow|V|{\psi}^\downarrow\rangle - \langle \psi^\uparrow|V|{\psi}^\uparrow\rangle | \nonumber \\ & \;\le \; V_{\max} (\|\psi^\downarrow \| \|\psi^\uparrow \| + \|\psi^\uparrow \| \|\psi^\downarrow\| +\|\psi^\uparrow \| \|\psi^\uparrow \| ) \nonumber \\ & \;\le \; V_{\max}(2\sigma + \sigma^2). \label{eq:thm95frequency95trunc95torus95diff95potential} \end{align}\tag{39}\] Combining 38 and 39 gives \[\begin{align} \langle \psi^\downarrow |h|\psi^\downarrow\rangle - \langle \psi |h|\psi\rangle & \;= \; \langle \psi^\downarrow |(-\Delta)|\psi^\downarrow\rangle - \langle \psi |(-\Delta)|\psi\rangle + \langle \psi^\downarrow |V|\psi^\downarrow\rangle - \langle \psi |V|\psi\rangle \\ & \;\le \;V_{\max}(2\sigma + \sigma^2). \end{align}\] By the fact that \(\|\psi^\downarrow\| \ge \|\psi\| - \|\psi^\downarrow - \psi\| \ge 1 - \sigma\), we have \[\begin{align} h[\psi^\downarrow] \;= \; \frac{\langle \psi^\downarrow |h|\psi^\downarrow\rangle}{\| \psi^\downarrow\|^2} & \;\le \; \frac{h[\psi] + V_{\max}(2\sigma + \sigma^2 )}{\|\psi^\downarrow\|^2} \nonumber\\ & \;\le \; \frac{h[\psi] + V_{\max}(2\sigma + \sigma^2)}{( 1- \sigma)^2} \nonumber\\ & \;= \; h[\psi] + \frac{h[\psi](2\sigma - \sigma^2) + V_{\max}(2\sigma + \sigma^2)}{( 1- \sigma)^2} \nonumber \\ & \;\le \; h[\psi] + \frac{E(2\sigma - \sigma^2) + V_{\max}(2\sigma + \sigma^2)}{( 1- \sigma)^2} \nonumber \\ & \;\le \; h[\psi] + \frac{2\sigma + \sigma^2 }{( 1- \sigma)^2} (E + V_{\max}) \nonumber \\ & \;\le \;\epsilon \nonumber, \end{align}\] where the last line is by our choice \(\sigma =O( {\epsilon}/{(E + V_{\max} )} )\). By 36 , we have \(\sigma \le L\sqrt{E + V_{\max}} /( 2\pi K)\). Therefore, we have that \(\widetilde{ \mathcal{D}} _{\mathbb{T}^n, f}\) is an \((E,\epsilon)\)-truncated domain, when we have \[\begin{align} K =\Omega\left(\frac{L (E+V_{\max})^{3/2}) }{ \epsilon }\right). \end{align}\] ◻

Proof. Let us assume that an element \(|\psi\rangle \in \mathcal{D}_{Q}\) has energy \(h[|\psi\rangle] \le E\). We write \(|\psi\rangle\) in the Fourier basis: \[\begin{align} |\psi\rangle = \sum_{k\in \mathbb{Z}^n} \widehat{\psi}_k |p_k\rangle . \end{align}\] Because \(\langle \psi |V|\psi\rangle \ge 0\), we have \[\begin{align} E \ge \langle \psi | (-\Delta_Q)|\psi\rangle = \sum_{k \in \mathbb{Z}^n} \frac{4 \pi^2 \|k\|^2 }{L^2} | \widehat{\psi}_k |^2 . \nonumber \end{align}\] After defining vectors \[\begin{align} |\psi^\downarrow\rangle \; &: = \sum_{\|k\|\le K} \widehat{\psi}_k |p_k\rangle \in \widetilde{\mathcal{D}}_{Q} , \\ |\psi^\uparrow\rangle \; &: = \sum_{\|k\| > K} \widehat{\psi}_k |p_k\rangle = \psi - \psi^\downarrow, \end{align}\] the rest of the proof is verbatim from the proof of Theorem [lem:freq95trunc95torus] and therefore omitted. ◻

Proof. For a normalized state \(\psi = \sum_{k: \|k\|_2 \le K} c_k \omega_k \in \widetilde{\mathcal{D}}_{\mathbb{T}^n,f}\) and \(|\phi\rangle :=U\psi= \sum_{k: \|k\|_2 \le K} c_k |p_k\rangle \in \widetilde{\mathcal{D}}_Q\), we prove that \[\begin{align} |h[\psi] - h_Q[\phi]| \le \epsilon. \end{align}\]

We first compute \[\begin{align} h_{{\mathbb{T}}}[\psi] \;&= \; \langle \psi|(-\Delta) | \psi \rangle + \langle \psi | V|\psi\rangle \\ \;&= \; \sum_{k:\|k\|_2\le K} |c_k|^2 \frac{4\pi^2 \|k\|_2^2}{L^2} + \sum_{\|k\|,\|k'\|\le K} \overline{c_{k'}} c_k \langle \omega_{k'}|V| \omega_k\rangle \\ \;&= \; \sum_{k:\|k\|_2\le K} |c_k|^2 \frac{4\pi^2 \|k\|_2^2}{L^2} + \sum_{\|k\|,\|k'\|\le K} \overline{c_{k'}} c_k \int_{x\in\mathbb{T}^n} \frac{V(x) e^{i 2\pi (k - k')\cdot x/L}}{L^n} \; \text{ d} x \\ \;&= \; \sum_{k:\|k\|_2\le K} |c_k|^2 \frac{4\pi^2 \|k\|_2^2}{L^2} + \sum_{\|k\|,\|k'\|\le K} \overline{c_{k'}} c_k \widehat{V}_{k' - k} , \end{align}\] and similarly, \[\begin{align} h_{Q}[\phi] \;&= \; \langle \phi|K_Q | \phi \rangle + \langle \phi | V_Q|\phi\rangle \\ \;&= \; \sum_{k:\|k\|_2\le K} |c_k|^2 \frac{4\pi^2 \|k\|_2^2}{L^2} + \sum_{\|k\|,\|k'\|\le K} \overline{c_{k'}} c_k \langle p_{k'}|V_Q| p_k\rangle \\ \;&= \; \sum_{k:\|k\|_2\le K} |c_k|^2 \frac{4\pi^2 \|k\|_2^2}{L^2} + \sum_{\|k\|,\|k'\|\le K} \overline{c_{k'}} c_k \sum_{y\in\mathbb{[}N]^n} \frac{V(yL/N) e^{i 2\pi (k - k')\cdot y/N}}{N^n} . \end{align}\] Therefore, \[\begin{align} |h[\psi] - h_Q[\phi]| & \;= \; \left |\sum_{\|k\|,\|k'\|\le K} \overline{c_{k'}} c_k \bigg( \widehat{V}_{k' - k} - \sum_{y\in\mathbb{[}N]^n} \frac{V(yL/N) e^{i 2\pi (k - k')\cdot y/N}}{N^n} \bigg) \right| \nonumber \\ & \;\le \; \sum_{\|k\|,\|k'\|\le K} |c_{k'}| |c_k| \max_{\|k\|,\|k'\| \le K} \left| \widehat{V}_{k' - k} - \sum_{y\in\mathbb{[}N]^n} \frac{V(yL/N) e^{i 2\pi (k - k')\cdot y/N}}{N^n} \right| \nonumber \\ & \;\le \; \max_{\|k\|,\|k'\| \le K} \left| \widehat{V}_{k' - k} - \sum_{y\in\mathbb{[}N]^n} \frac{V(yL/N) e^{i 2\pi (k - k')\cdot y/N}}{N^n} \right| \tag{40} \\ & \;\le \; \max_{\|k\| \le 2K} \left| \widehat{V}_{k} - \sum_{y\in\mathbb{[}N]^n} \frac{V(yL/N) e^{-i 2\pi k\cdot y/N}}{N^n} \right| \tag{41} \\ & \;= \; \max_{\|k\|_2 \le 2K} \left| \int_{\mathbb{T}^n} \frac{V (x)e^{-i2\pi k\cdot x /L} }{L^n} \;\text{d} x - \sum_{y\in\mathbb{[}N]^n} \frac{V(yL/N) e^{-i 2\pi k\cdot y/N}}{N^n} \right| \nonumber \end{align}\] where 40 is by the Cauchy-Schwarz inequality on \(|c_k|\), and 41 is by the change of variables \(k'-k\rightarrow k.\) The expression in the last line is the maximum discrepancy between the integral average and the corresponding Riemann sum average of \(V(x)e^{-i2\pi k \cdot x /L}\). The key observation for bounding this quantity is that the discrepancy vanishes whenever the function has no high-frequency Fourier components. We consider the decomposition \(V = V^\downarrow + V^\uparrow\), where \[\begin{align} V ^\downarrow(x) = \sum_{l: \|l\|_\infty \le K} \widehat {V}_l e^{i2\pi l\cdot x/L}, \\ V ^\uparrow(x) = \sum_{l: \|l\|_\infty > K} \widehat {V}_l e^{i2\pi l\cdot x/L}. \end{align}\] Then for \(k\) such that \(\|k\|_2 \le 2K\), \[\begin{align} \sum_{y\in\mathbb{[}N]^n} \frac{V^\downarrow (yL/N) e^{-i2\pi k\cdot y /N}}{N^n} \;&= \; \sum_{l: \|l\|_\infty \le K } \sum_{y\in\mathbb{[}N]^n} \frac{\widehat{V}_l e^{i2\pi l\cdot y/N} e^{-i2\pi k\cdot y /N}}{N^n} \nonumber \\ \;&= \; \sum_{l: \|l\|_\infty \le K } \widehat{V}_l \sum_{y\in\mathbb{[}N]^n} \frac{ e^{i2\pi (l-k)\cdot y/N} }{N^n} \nonumber \\ \;&= \; \sum_{\substack{l:\|l\|_\infty \le K \\ l = k \mod N} } \widehat{V}_k. \nonumber \\ \;&= \; \widehat{V}_k. \nonumber \end{align}\] The last equation holds because we enforce \(N\ge3K+1\). Since \(\|k \| _\infty \le \|k\|_2 \le 2K\) and \(\|l\|_\infty \le K\), we have \(\|k - l\|_\infty\le \|k\| _\infty + \|l\|_\infty \le 3K.\) Therefore, \(k=l \mod N\) if and only if \(k=l \in \mathbb{Z}^n\).

We conclude that \[\begin{align} |h[\psi] - h_Q[\phi]| & \;\le \; \max_{\|k\| \le 2K} \left| \widehat{V}_{k} - \sum_{y\in\mathbb{[}N]^n} \frac{V(y L/N) e^{-i 2\pi k\cdot y/N}}{N^n} \right| \\ & \;= \; \max_{\|k\| \le 2K} \left| \widehat{V}_{k} - \sum_{y\in\mathbb{[}N]^n} \frac{V^\downarrow(yL/N) e^{-i 2\pi k\cdot y/N}}{N^n} - \sum_{y\in\mathbb{[}N]^n} \frac{ V^\uparrow(yL/N) e^{-i 2\pi k\cdot y/N}}{N^n} \right| \\ & \;= \; \max_{\|k\| \le 2K} \left| \sum_{y\in\mathbb{[}N]^n} \frac{ V^\uparrow(yL/N) e^{-i 2\pi k\cdot y/N}}{N^n} \right| \\ & \;\le \; \max_{\|k\| \le 2K} \sum_{y\in\mathbb{[}N]^n} \frac{ |V^\uparrow(yL/N) | |e^{-i 2\pi k\cdot y/N}|}{N^n} \\ & \;= \; \max_{y\in[N]^n} V^\uparrow(yL/N) \; \le\; \max_{x\in[0,L]^n} V^\uparrow(x) \\ & \;= \; \max_{x\in[0,L]^n} \sum_{l:\|l\|_\infty >K} \widehat{V}_k e^{i2\pi k\cdot x/L} \\ &\; \le \;\sum_{l:\|l\|_\infty>K} |\widehat{V}_k|. \end{align}\]

We write \(x = (x_1,\dots,x_n) = (x_j, x_{ j*}) \in [0,L]^n\), \(l = (l_1,\dots,l_n) = (l_j, l_{j*}) \in \mathbb{Z}^n\) where \(x_{ j*} := (x_1,\dots, x_{j-1},x_{j+1},\dots, x_n)\), and \(l_{j*}\) is defined similarly. Then \[\begin{align} \widehat V_l &= \frac{1}{L^n} \int_{x\in\mathbb{[}0,L]^n} V(x) e^{-i2\pi l\cdot x/L} \;\text{d} x \nonumber \\ & = \frac{1}{L^n}\int_{ x_{j*} \in [0,L]^{n-1} } \;e^{-i2\pi l_{j*} \cdot \; x_{j*} /L} \int_{x_j \in [0,L]} V(x_j , x_{j*}) e^{-i2\pi l_j x_j /L} \;\text{d} x_j \;\text{d} x_{j*} \label{eq:vl95integration} \end{align}\tag{42}\] Fubini’s theorem lets us integrate the whole expression first in the variable \(x_j\) and then in the other variables sequentially. By repeatedly integrating by parts in \(x_j\) and assuming \(l_j \ne 0,\) we have, with the notation \(\partial_j^m:=\frac{\partial^m}{\partial x_j^m}\), \[\begin{align} & \int_{0}^L V(x_j , x_{j*}) e^{-i2\pi l_j x_j /L} \;\text{d} x_j \\ = & \frac{L}{-i2\pi l_j} V(x_j, x_{j*})e^{-2\pi l_j x_j/L} \Bigg|_0^L - \frac{L}{-i2\pi l_j}\int_{0}^L e^{-i2\pi l_j x_j /L} {\partial_j V(x_j,x_{j*})} \;\text{d} x_j \\ = & \frac{L}{i2\pi l_j}\int_{0}^L e^{-i2\pi l_j x_j /L} \;{\partial_j V(x_j,x_{j*})} \; \text{d} x_j \\ = & \left(\frac{L}{i2\pi l_j} \right)^2 \int_{0}^L e^{-i2\pi l_j x_j /L} \; {\partial_j^2 V (x_j,x_{j*} )} \;\text{d} x_j \\ &\vdots \\ = & \left(\frac{L}{i2\pi l_j} \right)^m \int_{0}^L e^{-i2\pi l_j x_j /L} \;\partial_j^m V(x_j,x_{j*}) \; \text{d} x_j \end{align}\] Plugging the result into 42 , we get \[\begin{align} \widehat V_l & \;= \; \frac{1}{L^n} \left(\frac{L}{i2\pi l_j} \right)^m \int_{ x \in [0,L]^n } e^{-i2\pi l\cdot x /L} \;{\partial^m_j V(x)} \; \text{d} x \end{align}\] for all \(m \in \mathbb{N}\). Since \(j\in [n]\) is arbitrary, we take \(j\) such that \(l_j = \|l\|_\infty\) and obtain \[\begin{align} | \widehat V_l | & \;\le \; \frac{1}{L^n} \left(\frac{L}{2\pi \|l\|_\infty} \right)^m \int_{ x \in [0,L]^n } | {\partial^m_j V(x)} | \; \text{d} x \\ & \;\le \; \left(\frac{L}{2\pi \|l\|_\infty} \right)^m \max_{x\in [0,L]^n} | {\partial^m_j V(x)} | \\ & \;\le \; \left(\frac{L}{2\pi \|l\|_\infty} \right)^m V_{\max} ^{(m)} \end{align}\] for all \(m \in \mathbb{N}\) where \(V^{(m)}_{\max} := \max \{|\partial_j^m V(x)| \;| \; x\in \mathbb{T}^n, j\in[n]\}\). Assuming \(m\ge n+1\), we have \[\begin{align} \sum_{l:\|l\|_\infty > K} |\widehat V_l| &\le V^{(m)}_\text{max} \sum_{\|l\|_\infty > K} \left(\frac{L}{2\pi \|l\|_{\infty}}\right)^m \\ &= V^{(m)}_\text{max} \sum_{t= K+1}^\infty \sum_{l: \|l\|_\infty = t} \left(\frac{L}{2\pi \|l\|_{\infty}}\right)^m \\ & \le V^{(m)}_\text{max} \sum_{t = K+1}^\infty 2n(2t+1)^{n-1} \left(\frac{L}{2\pi t}\right)^m \\ & \le 2n\cdot 4^{n-1} V^{(m)}_\text{max} \left(\frac{L}{2\pi }\right)^m \sum_{t =K+1}^\infty \frac{1}{t^{m-n+1 }} \\ & \le 4^{n}n V^{(m)}_\text{max} \left(\frac{L}{2\pi }\right)^m \frac{1}{(m-n) {K}^{m-n }}, \end{align}\] where the last inequality is by integration over \(t \in [K,\infty)\). We set \(m =2n\). We want the last quantity to be less than \(\epsilon,\) or equivalently \[\begin{align} n\log 4 + \log V^{(2n)}_{\max} + m \log \left( \frac{L}{2\pi} \right) - n\log K \le \log \epsilon. \end{align}\] This inequality is satisfied by choosing \(K\) so that \[\begin{align} & 2 \frac{\log V^{(2n)} _{\max}}{2n} + \log 4 + 2\log\left(\frac{L}{2\pi}\right) + \log \frac{1}{\epsilon^{1/n}} \\ \; \le \; &2 \log_\partial(V, \mathbb{T}^n, 2n) + \log 4 + 2\log\left(\frac{L}{2\pi}\right) + \log \frac{1}{\epsilon^{1/n}} \\ \; \le \; &2 \log p(2n) + \log 4 + 2\log\left(\frac{L}{2\pi}\right) + \log \frac{1}{\epsilon^{1/n}} \\ \;\le \;&\log K. \end{align}\] Therefore, we set \(K =\Theta\left(\left( p(2n) \frac{L}{2\pi}\right)^2 \frac{1}{\epsilon^{1/n}} \right)\) to get the desired bound ◻

Proof. The proof is by manipulating the maximums: \[\begin{align} \log_\partial( cf, \mathbb{T}^n, m) = &\max_{l\in[m], j\in[n],x \in\mathbb{T}^n} \frac{(\log c| \partial^l_j f(x)|)^+}{l } \\ = & \max_{l\in[m], j\in[n],x \in\mathbb{T}^n} \frac{(\log c + \log | \partial^l_j f(x)|)^+}{l } \\ \le & (\log c)^+ + \max_{l\in[m], j\in[n],x \in\mathbb{T}^n} \frac{( \log | \partial^l_j f(x)|)^+}{l } = (\log c)^+ + \log_\partial ( f, \mathbb{T}^n, m), \end{align}\] and \[\begin{align} \log_\partial( \sum_{k \in[r]} f_k, \mathbb{T}^n, m) \;=\; &\max_{l\in[m], j\in[n],x \in\mathbb{T}^n} \frac{(\log | \sum_{k\in [r]} \partial^l_j f_k(x)|)^+}{l } \\ \;\le \;& \max_{l\in[m], j\in[n],x \in\mathbb{T}^n} \frac{(\log (\sum_{k\in [r]} | \partial^l_j f_k (x)| )^+ }{l } \\ \; \le \;& \max_{l\in[m], j\in[n],x \in\mathbb{T}^n} \frac{(\log ( r\max_{k\in[r]} | \partial^l_j f_k (x)| ) )^+ }{l } \\ \; \le \;& \max_{l\in[m], j\in[n],x \in\mathbb{T}^n} \frac{(\log ( \max_{k\in[r]} | \partial^l_j f_k (x)| ) )^+ + \log r }{l } \\ \; \le \;& \log r + \max_{l\in[m], j\in[n],x \in\mathbb{T}^n} \frac{\max_{k\in[r]} (\log | \partial^l_j f_k (x)| )^+ }{l } \\ \; \le \;& \log r + \max_{k\in[r]} \max_{l\in[m], j\in[n],x \in\mathbb{T}^n} \frac{(\log | \partial^l_j f_k (x)| )^+ }{l } \\ = & \log r + \max_{k\in[r]} \log_\partial(f_k, \mathbb{T}, m) . \end{align}\] ◻

Proof. Faà di Bruno’s formula gives \(\;\forall x \in \mathbb{T}^n\) and \(\forall l \in \mathbb{N},\) \[\begin{align} \partial_j^l f(g(x)) &= \sum_{} {\frac{l!}{a_{1}!\,a_{2}!\,\cdots \,a_{l}!}}\cdot f^{(a_{1}+\cdots +a_{l})}(g(x))\cdot \prod _{t=1}^{l}\left({\frac{\partial_j^t g^{}(x)}{t!}}\right)^{a_{t}}, \end{align}\] and therefore \[\begin{align} |\partial_j^l f(g(x))| &\le \sum_{} {\frac{l!}{a_{1}!\,a_{2}!\,\cdots \,a_{l}!}} \left| f^{(a_{1}+\cdots +a_{l})}(g(x)) \right| \prod _{t=1}^{l} \left|{\frac{\partial_j^t g^{}(x)}{t!}}\right|^{a_{t}}, \end{align}\] where the sum is over \((a_1,\cdots, a_l)\in\mathbb{Z}_{\ge 0}^l\) such that \(\sum_{t=1}^l t\cdot a_t = l\). There are at most \(l^l\) such \(l\)-tuples. The first factor is at most \(l^l\). The second factor is at most \(\max_{} \{\;| f^{(l')} (y) | \; | \;l'\in [l] , y\in g(\mathbb{T}^n)\}\). The last factor is at most \(\prod_{t=1}^l | \partial_j^t g(x) |^{a_t}\). Therefore, we have \[\begin{align} |\partial^l _j f(g(x)) | &\le l^{2l} \max_{\substack{ y \in g(\mathbb{T}^n),\;l' \in[l]}} |f^{(l')} (y)| \prod_{t=1}^l | \partial_j^t g(x) |^{a_t}. \end{align}\] By taking the log of both sides, \[\begin{align} \log |\partial^l _j f(g(x)) | &\le 2l \log l + \log \max_{\substack{ y \in g(\mathbb{T}^n),\;l' \in[l]}} |f^{(l')} (y)| + \sum_{t =1}^m \log | \partial_j^t g(x) |^{a_t} \nonumber \\ & = 2l \log l + \max_{\substack{ y \in g(\mathbb{T}^n),\;l' \in[l]}} \log |f^{(l')} (y)| + \sum_{t =1}^m \log | \partial_j^t g(x) |^{a_t}. \label{eq:logpartial95composition95main95eq} \end{align}\tag{43}\] We bound the second term of 43 as follows: \[\begin{align} \;\max_{\substack{ y \in g(\mathbb{T}^n),\;l' \in[l]}} \log |f^{(l')} (y)| \nonumber \; \le & \; \max _{\;l' \in[l]} \max_{\substack{ y \in g(\mathbb{T}^n)}} (\log |f^{(l')} (y)|)^+ \nonumber \\ = & \; \max_{l' \in [l] } \;l' \; \max_{\substack{ y \in g(\mathbb{T}^n)}} \frac{(\log |f^{(l')} (y)|)^+}{l'} \nonumber \\ \le & \; \left( \max_{l'' \in [l] } \;l''\right) \;\left( \max_{\substack{ y \in g(\mathbb{T}^n) , \;l'\in[l]}} \frac{(\log |f^{(l')} (y)|)^+}{l'} \right) \nonumber \\ = & \;l \max_{\substack{ y \in g(\mathbb{T}^n),\;l' \in[l]}} \frac{(\log |f^{(l')} (y)|)^+}{l'} \\ = & \;l \log_\partial ( f, g(\mathbb{T}^n), l). \label{eq:logpartial95composition95f} \end{align}\tag{44}\] Since \(\log\) is an increasing function and \(\sum_{t=1}^l t \;a_t /l =1\), we bound the third term of 43 as follows: \[\begin{align} &\;\sum_{t =1}^l \log | \partial_j^t g(x) |^{a_t} \nonumber \\ = & \; l \sum_{t =1}^l \frac{t a_t}{l} \frac{\log | \partial_j^t g(x)| }{t} \nonumber \\ \le& \; l \max_{t\in[l]} \frac{\log | \partial_j^t g(x)| }{t} \nonumber \\ \le & \; l \max_{t\in[l],y \in \mathbb{T}^n} \frac{\log | \partial_j^t g(y)| }{t} \; \le \;l \log_\partial (g, \mathbb{T}^n, l) \label{eq:logpartial95composition95g} . \end{align}\tag{45}\] By inserting 44 and 45 into 43 and dividing by \(l\), we get \[\begin{align} \frac{\log|\partial _j^l f(g(x))|}{l} \le 2 \log l + \log_\partial (f, g(\mathbb{T}^n), l) + \log_\partial (g, \mathbb{T}^n, l). \end{align}\] Since the RHS is always nonnegative, we have \[\begin{align} \frac{(\log|\partial _j^l f(g(x))|)^+}{l} \le 2 \log l + \log_\partial (f, g(\mathbb{T}^n), l) + \log_\partial (g, \mathbb{T}^n, l). \end{align}\] The inequality still holds if we take the maximum over \(x \in \mathbb{T}^n\), \(j\in[n]\), and \(l \in [m]\) on both sides. Therefore, \[\begin{align} \log_\partial(f\circ g, \mathbb{T}^n , m) & = \\ \max_{j\in[n], l \in[m], x\in \mathbb{T}^n } \frac{(\log|\partial _j^l f(g(x))|)^+}{l} &\le 2 \log m + \max_{l \in[m]} \log_\partial (f, g(\mathbb{T}^n), l) + \max_{l \in[m]} \log_\partial (g, \mathbb{T}^n, l) \\ &= 2\log m + \log_\partial (f, g(\mathbb{T}^n),m ) + \log_\partial (g, \mathbb{T}^n, m). \end{align}\] ◻

Proof. The following is a well-known smooth function: \[f_0(x):= \begin{cases} \;\exp \left(-1/x\right) & x> 0 \\ \;0 &x\le 0. \end{cases}\] We use the fact that \(\text{Bump}(x) = f_0(x) f_0(1-x)\) and the Leibniz rule \[\begin{align} (fg)^{(m)} = \sum_{j = 0}^m \begin{pmatrix} m \\ j \end{pmatrix} f^{(j)} g^{(m-j)} \end{align}\] to get an upper bound \[\begin{align} \label{eq:bump95function95nth95diff95summation} \max_{x\in \mathbb{R}} |\text{Bump}^{(m)} (x) | \le \sum_{j = 0}^m \begin{pmatrix} m \\ j \end{pmatrix} \max_{x\in \mathbb{R}} | f_0^{(i)}(x)| \; \max_{x\in \mathbb{R}} | f_0^{(m-i)}(x) | . \end{align}\tag{46}\] We are left with upper bounding \(|f_0^{(j)}|\). Note that for \(x>0\), \(f_0'(x) =(1/x)^2 e^{-1/x} = t^2 e^{-t}\), where \(t:=1/x.\)

We prove by induction that there exists a polynomial such that \(f_0^{(j)}(x) = p_j(t) e^{-t}\), for all \(j \in \mathbb{N}\). We showed above the statement for \(j=1.\) Suppose the hypothesis is true for \(j\in \mathbb{N}\). Then for \(j+1\): \[\begin{align} f^{(j+1)}_0 (x) = \frac{dt}{dx} \frac{d}{dt} (p_{j}(t) e^{-t}) =-t^2\left(p_j'(t) - p_j(t)\right) e^{-t}:=p_{j+1}(t)e^{-t}. \end{align}\] Therefore, the induction hypothesis is true for all \(j \in \mathbb{N}\).

The maximum of \(t^j e^{-t}\), for \(t>0\), is \((j/e)^j\) at the maximizer \(t = j\). Observe that this quantity increases as a function of \(j\) for \(j\ge 1.\) For an arbitrary polynomial \(p(t):=\sum_{j=0}^{\deg p} a_j t^j,\) define sum of the absolute values of its coefficients \(\| p\|_c:= \sum_{j=0}^{\deg p} |a_j|\). Then, if \(p\) does not have a constant term, \[\begin{align} \max_{t>0}| p_j(t)e^{-t}| \le \sum_{j = 1}^{\deg p_j} |a_j| (j/e)^j\le \|p\|_c (\deg p /e)^{\deg p}. \end{align}\] Therefore, \[\begin{align} \max_{t >0} \left| f_0^{(j)}(t) \right| \le \|p_j\| \left(\frac{\deg p_j}{e}\right)^{\deg p_j}. \end{align}\] From the recursive formula \(p_{j+1} (t) = t^2(p_j(t) - p'_j(t))\) with the initial condition \(p_1 = t^2\), we see that \(\deg p_j = 2j\) and \(p_j\) is without a constant term, for all \(j\in \mathbb{N}.\) From the recursive formula, we also have \[\begin{align} \|p_j\|_c &\le \|t^2 p_j\|_c + \|t^2 p'_j\|_c \\ &\le \|p_{j-1}\| + 2(j-1)\|p_{j-1}\| \\& = (2j-1)\|p_{j-1}\| \\ & \le (2j-1)(2j-3)\cdots 3\cdot 1\cdot \| p_{1}\| \end{align}\] Therefore, \(\|p_j\| <(2j)^j\), and we have \[\begin{align} \max_{x \in\mathbb{R}} |f_0^{(j)}(x)| = \max_{t>0} | p(t) e^{-t}| \le (2j^2/e)^j \le (2m^2/e)^m, \end{align}\] for \(0\le j\le m\). Finally, we can plug this inequality into 46 and use the binomial identity \(\sum_{j=0}^m \begin{pmatrix} m \\ j \end{pmatrix} = 2^m\) to get \(\max_{x\in \mathbb{R}}|\text{Bump}^{(m)}(x)| \le 2^m(4m^4/e^2)^m,\) and therefore, \[\begin{align} \log_\partial (\text{Bump}, \mathbb{R}, m ) &= \sup_{l\in[m], x \in\mathbb{R} } \frac{(\log|\text{Bump}^{(l)}(x)|)^+ }{l} \\ &\le \max_{l\in[m]}\frac{l\log 2 + l\log(4l^4/e^2) }{l} \\ &= \log O(m^4). \end{align}\]

Since \(\text{Cut}_{\alpha,\beta} '(x) = - \text{Bump}(\frac{x - \alpha}{\beta})/\int_{\alpha}^{\alpha + \beta}\text{Bump}(\frac{y-\alpha}{\beta})\text{d} y\), we have \[\begin{align} \text{Cut}_{\alpha, \beta}^{(m)} (x) = - \frac{1}{\beta^{m-1} } \text{Bump}^{(m-1)} \big(\frac{x-\alpha}{\beta} \big) \bigg/ {\int_\alpha^{\alpha + \beta} \text{Bump}\big((y-\alpha)/\beta \big) \; \text{d} y } . \end{align}\] Therefore, \[\begin{align} \log_\partial(\text{Cut}_{\alpha,\beta},\mathbb{R}, m) = \sup_{l\in[m],x\in\mathbb{R}}\frac{( \log|\text{Cut}^{(l)}_{\alpha, \beta}(x)| )^+ }{l} = \log O(1/\beta + m^4). \end{align}\] Since \(\text{Sat}_{\alpha, \beta}' = \text{Cut}_{\alpha,\beta}\) and \(\text{Bar}_\epsilon' = 1- \text{Cut}_{0,\epsilon}(x)\), we apply a similar analysis to obtain \[\begin{align} \log_\partial(\text{Sat}_{\alpha,\beta},\mathbb{R},m) = \log O(1/\beta + m^4), \\ \log_\partial(\text{Bar}_{\epsilon},\mathbb{R},m) = \log O(1/\epsilon + m^4). \end{align}\] ◻

Proof. By domain monotonicity (Lemma 9), we have \(\lambda_0(h^D) \le \lambda_0(-\Delta^D_{\Omega})\), and by potential comparison (Lemma 8), we have \(\lambda_0(-\Delta^D_{\Omega'}) \le \lambda_0(h^D)\), giving \[\begin{align} \lambda_0(-\Delta^D_{\Omega'}) \le \lambda_0(h^D) \le \lambda_0(-\Delta^D_{\Omega}). \end{align}\] Additionally, for every eigenfunction \(\psi\) of \(\Delta_{\Omega}\), we have that \(\psi'\) as an eigenfunction of \(\Delta_{\Omega'}\), where \(\psi'(x) = \psi(x/(1+\epsilon))\). Therefore, \[\begin{align} \lambda_0(-\Delta^D_{\Omega'}) = \lambda_0(-\Delta^D_\Omega)/(1+\epsilon)^2\ge \lambda_0(-\Delta^D_\Omega) -O(\epsilon\lambda_0(-\Delta^D_\Omega)), \end{align}\] which gives us a sandwich \[\begin{align} \lambda_0(-\Delta^D_{\Omega})- O(\epsilon\lambda_0(-\Delta^D_\Omega)) \le \lambda_0(h^D) \le \lambda_0(-\Delta^D_{\Omega}). \end{align}\] Hence, we have \[\begin{align} \label{eq:laplacian95nsquare} |\lambda_0(h^D) - \lambda_0(-\Delta^D_\Omega)| \le O(\epsilon\lambda_0(-\Delta^D_\Omega) ) \le O(\epsilon n^2), \end{align}\tag{48}\] where the last inequality is from the domain monotonicity (Lemma 9) of \[\begin{align} B^\infty_{1/2\sqrt n} \subset B^2_1 \subset \Omega. \end{align}\] We set \(\epsilon = O({\epsilon_0}/{n^2})\) and obtain \(|\lambda_0(h^D) - \lambda_0(-\Delta^D_\Omega)| \le O(\epsilon_0).\) ◻

Proof. We claim that \(L^2(\Omega')\) is an \((E,\epsilon)\)-truncated domain for both \(h^D\) and \(h\). The claim would prove that \(h\) and \(h^D\) are \((E,\epsilon)\)-equivalent, since the identity unitary takes \(L^2(\Omega')\) to itself, and the restrictions of \(h\) and \(h^D\) on \(L^2(\Omega')\) are identical.

For \(h^D\), we have \(\mathcal{D}(h^D)\subset L^2(\Omega')\), and therefore \(L^2(\Omega')\) is an \((E,\epsilon)\)-truncated domain for \(h^D\). Therefore, we only need to show that \(L^2(\Omega')\) is an \((E,\epsilon)\)-truncated domain for \(h\).

The derivation is almost the same as in the proof of Lemma 16, except that we use a different cutoff function \(f\). We will refer to the proof of Lemma 16, when the calculation is the same.

Suppose \(\psi\in L^2 ({\mathbb{R}^n})\) has energy \(h[\psi]\le E\). We define \[\begin{align} \psi^\downarrow:=\sum_{i: \lambda_i \le E/\mu^2 } c_i\psi_i, \end{align}\] where \(\psi_i\) is the \(i\)-th eigenvectors of \(h\) with eigenvalue \(\lambda_i\). We define sets \[\begin{align} A &:=V^{-1}([0,E/\mu^6]) , \\ B &:=V^{-1}([0,2E/\mu^6]). \end{align}\] Then we have \(A\subset B\subset \Omega'\), because of 49 .

The cutoff function \(f:= \text{Cut}_{E/\mu^6,E/\mu^6}\circ V\) satisfies that \[\begin{align} f(x) \begin{cases} = 1 & x \in A \\ \in[0,1] & x \in B\setminus A\\ = 0 & x \notin B. \end{cases} \end{align}\]

We claim that \[\begin{align} \widetilde{\psi}:=\frac{f\psi^\downarrow}{\|f\psi^\downarrow\|} \end{align}\] is an \(\mu\)-truncation of \(\psi\). The claim implies the lemma, since \(\widetilde{\psi}\in L^2 (\Omega')\).

We first show the norm condition of \(\mu\)-truncation. Since \(V(x)\ge E/\mu^6\) for all \(x\) such that \(f(x) = 0\), by Markov’s inequality (see the proof of Lemma 15), we have \[\begin{align} \|\psi^\downarrow - f\psi^\downarrow\|\le \mu^3 \|\psi^\downarrow\|. \end{align}\] By the same derivation that led to 26 , we have \[\begin{align} \|\psi - \widetilde{\psi}\| \le O(\mu) \le O(\epsilon_0). \end{align}\]

For the rest of the proof, we show the energy condition. By the same calculation that led to 28 , we have \[\begin{align} \|f\psi^\downarrow \| h[f\psi^\downarrow\| = \langle \psi^\downarrow | \|\;\nabla f\|^2 |\psi^\downarrow\rangle + \text{Re}\langle f^2 \psi^\downarrow | h\psi ^\downarrow \rangle . \end{align}\] We upper bound the two terms similarly to the proof of Lemma 16. We have \[\begin{align} \langle \psi^\downarrow | \|\;\nabla f\|^2 |\psi^\downarrow\rangle =&\int_{x \in {\mathbb{R}^n}} |\psi^\downarrow(x)|^2 \|\nabla f(x)\|^2 \;\text{d} x \\ \le & \; \max_{x \notin A} \|\nabla f(x)\|^2 \int_{x\notin A} |\psi^\downarrow(x)|^2 \;\text{d} x \\ \le & \; \max_{x \notin A} | \text{Cut}'_{E/\mu^6,E/\mu^6} ( V(x))|^2 \;\| \nabla V(x)\|^2 \mu^6 \\ \le & \; O\left(\frac{\mu^{12}}{E^2} \cdot \frac{m ^2 E^2}{\epsilon^2\mu ^{12}}\cdot \mu^6 \right) \\ = & \;O \left( \frac{m^2 n^4 \mu^{6} }{\epsilon_0^2}\right) \\ \le & \; O(\epsilon^4/ E^4). \end{align}\] Also, by the calculation that led to 34 , we have \[\begin{align} \text{Re}\langle f^2\psi^\downarrow|h\psi^\downarrow\rangle \le h[\psi] + \mu E \le h[\psi] + \epsilon_0. \end{align}\] Therefore, we have \[\begin{align} \|f\psi^\downarrow \| h[f\psi^\downarrow\| \le h[\psi] + O(\epsilon_0) + O(\epsilon_0^4 E^{4}). \end{align}\] Finally, since \[\begin{align} \|f\psi^\downarrow\| \;\ge \;\|\psi^\downarrow\| -\|(1 - f) \psi^\downarrow\| \;\ge \; 1 - \mu - \mu^3, \end{align}\] we have \[\begin{align} h[\widetilde{\psi}] = h[f\psi^\downarrow] \le \frac{ h[\psi] + O(\epsilon_0) + O(\epsilon_0^4 E^{4}) }{ 1 - \mu - \mu^3} \le h[\psi] + O(\epsilon_0) . \end{align}\] ◻