Lie–Schwinger Block-Diagonalization and Gapped Quantum Chains

Abstract

We study quantum chains whose Hamiltonians are perturbations by bounded interactions of short range of a Hamiltonian that does not couple the degrees of freedom located at different sites of the chain and has a strictly positive energy gap above its ground-state energy. We prove that, for small values of a coupling constant, the spectral gap of the perturbed Hamiltonian above its ground-state energy is bounded from below by a positive constant uniformly in the length of the chain. In our proof we use a novel method based on local Lie-Schwinger conjugations of the Hamiltonians associated with connected subsets of the chain.

Appendix

Lemma A.1

For any $1\le n \le N$

$$\begin{aligned} \sum _{i=1}^{n} P^{\perp }_{\Omega _i} \ge \mathbb {1}-\bigotimes _{i=1}^{n} P_{\Omega _i}=:\,\Big (\bigotimes _{i=1}^{n} P_{\Omega _i}\Big )^{\perp } \end{aligned}$$

(A.1)

where $P^{\perp }_{\Omega _i}=\mathbb {1}-P_{\Omega _i}$.

Proof

We call $P_{vac}:=\bigotimes _{i=1}^{n} P_{\Omega _i}$ acting on $\mathcal {H}^{(n)}:=\bigotimes _{i=1}^{n} \mathcal {H}_i$. We define

$$\begin{aligned} A_n:=\sum _{j=1}^{n}P_{\Omega _j}^{\perp }+P_{vac}\,. \end{aligned}$$

(A.2)

Notice that all operators $P_{\Omega _j}^{\perp }$ and $P_{vac}$ commute with each other and are orthogonal projectors. Therefore we deduce that

$$\begin{aligned} \text {spec}(A_n)\subseteq \{0,1,2,\dots , n+1\}\,. \end{aligned}$$

(A.3)

We will show that

$$\begin{aligned} \text {Range}\,A_n=\mathcal {H}^{(n)}\,. \end{aligned}$$

(A.4)

If (A.4) holds then $0\notin \text {spec}(A_n)$. By (A.3) it then follows that

$$\begin{aligned} A_n\ge \mathbb {1}\,. \end{aligned}$$

(A.5)

Thus, we are left with proving (A.4).

• (i) Assume that $\psi$ is perpendicular to the range of $A_{n}$, and let $P_{\Omega _j}^{\perp }\psi =:\phi _j$. Then, since $\psi \perp \text {Range}\, A_n$, we have that

  $$\begin{aligned} 0=\langle \psi , A_n \psi \rangle =\sum _{i=1\,;\,i\ne j}^{n}\langle \psi \,,\, P_{\Omega _i}^{\perp }\psi \rangle +\langle \psi \,\, P_{vac}\psi \rangle +\langle \psi \,,\, P_{\Omega _j}^{\perp }\psi \rangle \ge \langle \psi \,,\, P_{\Omega _j}^{\perp }\psi \rangle \nonumber \\ \end{aligned}$$

  (A.6)

  but

  $$\begin{aligned} \langle \psi \,,\, P_{\Omega _j}^{\perp }\psi \rangle =\langle P_{\Omega _j}^{\perp }\psi \,,\, P_{\Omega _j}^{\perp }\psi \rangle =\langle \phi _j\,,\,\phi _j\rangle \end{aligned}$$

  (A.7)

  where we have used that $P_{\Omega _j}^{\perp }$ is an orthogonal projector. We conclude that $\phi _j=0$ for all j.

• (ii) Let $\psi \perp \text {Range}\,A_n$. Then, by (i),

  $$\begin{aligned} \psi =\Big (\bigotimes _{j=1}^{n}\,(P_{\Omega _j}^{\perp }+P_{\Omega _j})\Big )\,\psi = (\bigotimes _{j=1}^{n}P_{\Omega _j})\,\psi =P_{vac}\psi \end{aligned}$$

  (A.8)

  and

  $$\begin{aligned} 0=\langle \psi \,,\, A_n\,\psi \rangle =\langle \psi \,,\, P_{vac}\,\psi \rangle =\langle \psi \,,\,\psi \rangle \quad \Rightarrow \quad \psi =0\,. \end{aligned}$$

  (A.9)

Thus, $\text {Range}\, A_n=\mathcal {H}^{(n)}$, and (A.4) is proven . $\quad \square$

From Lemma A.1 we derive the following bound.

Corollary A.2

For $i+r\le N$, we define

$$\begin{aligned} P^{(+)}_{I_{r,i}}:=\Big (\bigotimes _{k=i}^{i+r}P_{\Omega _{k}}\Big )^{\perp }\,. \end{aligned}$$

(A.10)

Then, for $1\le l \le L \le N-r$,

$$\begin{aligned} \sum _{i=l}^{L}P^{(+)}_{I_{r,i}}\le (r+1) \sum _{i=l}^{L+r} P^{\perp }_{\Omega _i} \,. \end{aligned}$$

(A.11)

Proof

Lemma A.1 says that

$$\begin{aligned} \sum _{j=i}^{i+r} P^{\perp }_{\Omega _{j}}\ge \Big (\bigotimes _{k=i}^{i+r}P_{\Omega _{k}}\Big )^{\perp }\,. \end{aligned}$$

(A.12)

By summing over i, from $i=l$ up to L, the l-h-s of (A.12), for each j we get no more than $r+1$ terms of the type $P^{\perp }_{\Omega _{j}}$ and the inequality in (A.11) follows . $\quad \square$

Lemma A.3

Assume that $t>0$ is sufficiently small, $\Vert V^{(k,q-1)}_{I_{r,i}}\Vert \le \frac{8}{(r+1)^2}\,t^{\frac{r-1}{3}}$, and $\Delta _{I_{k,q}}\ge \frac{1}{2}$. Then, for arbitrary N, $k\ge 1$, and $q\ge 2$, the inequalities

$$\begin{aligned} \Vert V^{(k,q)}_{I_{k,q}}\Vert \le 2\Vert V^{(k,q-1)}_{I_{k,q}}\Vert \, \end{aligned}$$

(A.13)

$$\begin{aligned} \Vert S_{I_{k, q}}\Vert \le C\cdot t \cdot \Vert V^{(k,q-1)}_{I_{k,q}}\Vert \end{aligned}$$

(A.14)

hold true for a universal constant C. For $q=1$, $V^{(k,q-1)}_{I_{k,q}}$ is replaced by $V^{(k-1,N-k+1)}_{I_{k,1}}$ on the right side of (A.13) and (A.14).

Proof

In the following we assume $q\ge 2$; if $q=1$ an analogous proof holds. We recall that

$$\begin{aligned} V^{(k,q)}_{I_{k,q}}:= \sum _{j=1}^{\infty }t^{j-1}(V^{(k,q-1)}_{I_{k,q}})^{diag}_j \, \end{aligned}$$

(A.15)

and

$$\begin{aligned} S_{I_{k,q}}:=\sum _{j=1}^{\infty }t^j(S_{I_{k,q}})_j\ \end{aligned}$$

(A.16)

with

$$\begin{aligned} (V^{(k,q-1)}_{I_{k,q}})_1:=V^{(k,q-1)}_{I_{k,q}} \end{aligned}$$

and, for $j\ge 2$,

$$\begin{aligned}&(V^{(k,q-1)}_{I_{k,q}})_j\, \end{aligned}$$

(A.17)

$$\begin{aligned}&\quad :=\sum _{p\ge 2, r_1\ge 1 \dots , r_p\ge 1\,; \, r_1+\dots +r_p=j}\nonumber \\&\qquad \frac{1}{p!}\text{ ad }\,(S_{I_{k,q}})_{r_1}\Big (\text{ ad }\,(S_{I_{k,q}})_{r_2}\dots (\text{ ad }\,(S_{I_{k,q}})_{r_p}(G_{I_{k,q}}))\dots \Big )\end{aligned}$$

(A.18)

$$\begin{aligned}&\qquad +\,\sum _{p\ge 1, r_1\ge 1 \dots , r_p\ge 1\,; \, r_1+\dots +r_p=j-1}\nonumber \\&\qquad \frac{1}{p!}\text{ ad }\,(S_{I_{k,q}})_{r_1}\Big (\text{ ad }\,(S_{I_{k,q}})_{r_2}\dots (\text{ ad }\,(S_{I_{k,q}})_{r_p}(V^{(k,q-1)}_{I_{k,q}}))\dots \Big )\,. \end{aligned}$$

(A.19)

and, for $j\ge 1$,

$$\begin{aligned} (S_{I_{k,q}})_j:=ad^{-1}\,G_{I_{k,q}}\,((V^{(k,q-1)}_{I_{k,q}})^{od}_j)= \frac{1}{G_{I_{k,q}}-E_{I_{k,q}}}P^{(+)}_{I_{k,q}}\,(V^{(k,q-1)}_{I_{k,q}})_j\,P^{(-)}_{I_{k,q}}-h.c.\,.\nonumber \\ \end{aligned}$$

(A.20)

From the lines above we derive

$$\begin{aligned} \text {ad}\,(S_{I_{k,q}})_{r_p}(G_{I_{k,q}})= & {} \text {ad}\,(S_{I_{k,q}})_{r_p}(G_{I_{k,q}}-E_{I_{k,q}})\nonumber \\= & {} \,[\frac{1}{G_{I_{k,q}}-E_{I_{k,q}}}P^{(+)}_{I_{k,q}}\,(V^{(k, q-1)}_{I_{k,q}})_{r_p}\,P^{(-)}_{I_{k,q}}\,,\,G_{I_{k,q}}-E_{I_{k,q}}]+h.c. \nonumber \\\end{aligned}$$

(A.21)

$$\begin{aligned}= & {} -P^{(+)}_{I_{k,q}}\,(V^{(k,q-1)}_{I_{k,q}})_{r_p}\,P^{(-)}_{I_{k,q}}-P^{(-)}_{I_{k,q}}\,(V^{(k, q-1)}_{I_{k,q}})_{r_p}\,P^{(+)}_{I_{k,q}}\,. \end{aligned}$$

(A.22)

We recall definition (A.20) and we observe that

$$\begin{aligned} \Vert (S_{I_{k,q}})_j\Vert \le 2\frac{\Vert (V^{(k,q-1)}_{I_{k,q}})_j\Vert }{\Delta _{I_{k,q}}}\le 4\Vert (V^{(k,q-1)}_{I_{k,q}})_j\Vert \,\,, \end{aligned}$$

(A.23)

where we use the inductive hypothesis $\Delta _{I_{k;q}}\ge \frac{1}{2}$. (Recall that $\Delta _{I_{k;q}}$ is the gap of $G_{I_{k,q}}$ above its groundstate energy $E_{I_{k,q}}$.) Then formula (A.17) yields

$$\begin{aligned}&\Vert (V^{(k,q-1)}_{I_{k,q}})_j\Vert \nonumber \\&\quad \le \sum _{p=2}^{j}\,\frac{8^p}{p!}\sum _{ r_1\ge 1 \dots , r_p\ge 1\, ; \, r_1+\dots +r_p=j}\,\Vert \,(V^{(k,q-1)}_{I_{k,q}})_{r_1}\Vert \Vert \,(V^{(k,q-1)}_{I_{k,q}})_{r_2}\Vert \dots \Vert \,(V^{(k,q-1)}_{I_{k,q}})_{r_p}\Vert \nonumber \\&\qquad +\,2\Vert V^{(k,q-1)}_{I_{k,q}}\Vert \sum _{p=1}^{j-1}\,\frac{8^p}{p!}\,\sum _{ r_1\ge 1 \dots , r_p\ge 1\, ; \, r_1+\dots +r_p=j-1}\,\Vert \,(V^{(k,q-1)}_{I_{k,q}})_{r_1}\Vert \nonumber \\&\qquad \Vert \,(V^{(k,q-1)}_{I_{k,q}})_{r_2}\Vert \dots \Vert \,(V^{(k,q-1)}_{I_{k,q}})_{r_p}\Vert \,. \end{aligned}$$

(A.24)

From now on, we closely follow the proof of Theorem 3.2 in [DFFR]; that is, assuming $\Vert V^{(k,q-1)}_{I_{k,q}}\Vert \ne 0$, we recursively define numbers $B_j$, $j\ge 1$, by the equations

$$\begin{aligned} B_1:= & {} \Vert V^{(k,q-1)}_{I_{k,q}}\Vert =\Vert (V^{(k,q-1)}_{I_{k,q}})_1\Vert \,, \end{aligned}$$

(A.25)

$$\begin{aligned} B_j:= & {} \frac{1}{a}\sum _{k=1}^{j-1}B_{j-k}B_k\,,\quad j\ge 2\,, \end{aligned}$$

(A.26)

with  $a>0$  satisfying the relation

$$\begin{aligned} \frac{e^{8a}-8a-1}{a}+e^{8a}-1=1\,. \end{aligned}$$

(A.27)

Using (A.25), (A.26), (A.24), and an induction, it is not difficult to prove that (see Theorem 3.2 in [DFFR]) for $j\ge 2$

$$\begin{aligned} \Vert (V^{(k,q-1)}_{I_{k,q}})_j\Vert \le B_j\,\Big (\frac{e^{8a}-8a-1}{a}\Big )+2\Vert V^{(k,q-1)}_{I_{k;q}}\Vert \,B_{j-1}\Big (\frac{e^{8a}-1}{a}\Big )\,. \end{aligned}$$

(A.28)

From (A.25) and (A.26) it also follows that

$$\begin{aligned} B_j\ge \frac{2B_{j-1}\Vert \,V^{(k,q-1)}_{I_{k,q}}\,\Vert }{a}\,\quad \Rightarrow \quad B_{j-1}\le a\frac{B_j}{2\Vert \,V^{(k,q-1)}_{I_{k,q}}\,\Vert }\,, \end{aligned}$$

(A.29)

which, when combined with (A.28) and (A.27), yield

$$\begin{aligned} B_j\ge \Vert \,(V^{(k,q-1)}_{I_{k,q}})_j\Vert \,. \end{aligned}$$

(A.30)

The numbers $B_j$ are the Taylor’s coefficients of the function

$$\begin{aligned} f(x):=\frac{a}{2}\cdot \left( \,1-\sqrt{1- (\frac{4}{a}\cdot \Vert V^{(k,q-1)}_{I_{k,q}}\Vert ) \,x }\,\right) \,, \end{aligned}$$

(A.31)

(see [DFFR]). Therefore the radius of analyticity, $t_0$, of

$$\begin{aligned} \sum _{j=1}^{\infty }t^{j-1}\Vert (V^{(k,q-1)}_{I_{k,q}})^{diag}_j \Vert =\frac{d}{dt}\,\Big (\sum _{j=1}^{\infty }\frac{t^{j}}{j}\Vert (V^{(k,q-1)}_{I_{k,q}})^{diag}_j \Vert \Big ) \end{aligned}$$

(A.32)

is bounded below by the radius of analyticity of $\sum _{j=1}^{\infty }x^jB_j$, i.e.,

$$\begin{aligned} t_0\ge \frac{a}{4\Vert V^{(k,q-1)}_{I_{k,q}}\Vert }\ge \frac{a}{8} \end{aligned}$$

(A.33)

where we have used the assumption that $\Vert V^{(k,q-1)}_{I_{r;i}}\Vert \le \frac{8}{(r+1)^2}\,t^{\frac{r-1}{3}}$ and $(0<)t<1$. Thanks to the inequality in (A.23) the same bound holds true for the radius of convergence of the series $S_{I_{k;q}}:=\sum _{j=1}^{\infty }t^j(S_{I_{k;q}})_j\,$ . For $0<t<1$ and in the interval $(0,\frac{a}{16})$, by using (A.25) and (A.30) we can estimate

$$\begin{aligned} \sum _{j=1}^{\infty }t^{j-1}\Vert (V^{(k,q-1)}_{I_{k,q}})^{diag}_j \Vert\le & {} \frac{1}{t}\sum _{j=1}^{\infty }t^jB_j \end{aligned}$$

(A.34)

$$\begin{aligned}= & {} \frac{1}{t}\cdot \frac{a}{2}\cdot \left( \,1-\sqrt{1- (\frac{4}{a}\cdot \Vert V^{(k,q-1)}_{I_{k,q}}\Vert ) \,t }\,\right) \end{aligned}$$

(A.35)

$$\begin{aligned}\le & {} (1+C_a \cdot t )\,\Vert V^{(k,q-1)}_{I_{k,q}}\Vert \end{aligned}$$

(A.36)

for some a-dependent constant $C_a>0$. Hence the inequality in (A.13) holds true, provided that $t>0$ is sufficiently small but independent of N, k, and q. In a similar way we derive (A.14). $\quad \square$