Integrability in the Chiral Model of Magic Angles

Magic angles in the chiral model of twisted bilayer graphene are parameters for which the chiral version of the Bistritzer–MacDonald Hamiltonian exhibits a flat band at energy zero. We compute the sums over powers of (complex) magic angles and use that to show that the set of magic angles is infinite. We also provide a new proof of the existence of the first real magic angle, showing also that the corresponding flat band has minimal multiplicity for the simplest possible choice of potentials satisfying all symmetries. These results indicate (though do not prove) a hidden integrability of the chiral model.


Introduction and statement of results
When two sheets of graphene are stacked on top of each other and twisted, it has been observed that at certain angles, coined the magic angles, the composite system becomes superconducting.In this article, we study the chiral limit of the Bistritzer where the parameter α is proportional to the inverse relative twisting angle.After a simple rescaling, the potential is a smooth and periodic function satisfying U (z + a ℓ ) = ωU (z), U (ωz) = ωU (z), and U (z) = U (z), ( where ω = e 2πi/3 and a ℓ = 4 3 πiω ℓ .The simplest example of such a potential and our canonical choice of U is (1.2) Even though the potential U (z) is only periodic with respect to Γ = 4πi(ωZ ⊕ ω 2 Z) the first property implies that the matrix potential, and thus D(α), commutes with the translation operator L a w(z) := ω a 1 +a 2 0 0 1 w(z + a), a ∈ 1 3 Γ, (1.3) where w ∈ C 2 and a = 4 3 πi(ωa 1 + ω 2 a 2 ), a j ∈ Z.We note that if Γ * is the dual (reciprocal) lattice of Γ, then 3Γ * is the dual lattice of 1 3 Γ.When moving to functions with values in C 4 = C 2 × C 2 (on which H(α) acts) we extend the action of L a to an action on each C 2 component.We then consider the Floquet spectrum of defined by (H k (α) − E j (α, k))w j (α, k) = 0, where eigenvalues of positive energy are labelled with j ≥ 1 in ascending order, as a self-adjoint operator on with the domain given by H ∩ H 1 (C/Γ) such that This Hamiltonian is an effective one-particle model which exhibits perfectly flat bands at magic angles.This appearance of perfectly flat bands in the chiral limit was explained by Tarnopolsky, Kruchkov and Vishwanath [TKV19] with the help of Jacobi theta functions 1 .An equivalent spectral theoretic characterization of magic angles was then provided in [Be*22]: if we define the following compact Birman-Schwinger operator where H is defined in (1.5).In other words, the spectrum of T k 0 is independent of k 0 ∈ C \ (3Γ * − {0, i}) and characterizes the values of α ∈ C at which the Hamiltonian exhibits a flat band at zero energy.Since the parameter α is inherently connected with the twisting angle, we shall refer to α's at which (1.7) occurs as magic and denote their set by A ⊂ C.
The analysis of magic angles is therefore reduced to a spectral theory problem involving a single compact non self-adjoint operator.Since even non-trivial non self-adjoint compact operators do not necessarily have non-zero eigenvalues, the existence of a parameter α at which the Hamiltonian exhibits a flat band at zero energy is non-trivial.In [Be*22] the existence of such a complex parameter α ∈ C \ {0} was first concluded by showing that tr H (T 4 k ) = 8π/ √ 3 which implied existence of a non-zero eigenvalue 2 .This result was improved by a computer-assisted proof [WaLu21] in which Watson and Luskin used the complex-analytic characterization of magic angles from [TKV19] to prove existence of the first real magic angle and obtained explicit bounds on its position.
In this article, we exhibit a general form of traces of powers of T k .This suggests a hidden integrability of the Hamiltonian H(α) for potentials satisfying (1.1), as all traces 1 As was pointed out to us by Alex Sobolev a similar argument appeared in the work of Dubrovin and Novikov [DuNo80] who studied magnetic Hamiltonians on tori.
2 In [Be*22] we considered the trace on L 2 (C/Γ; C 2 ) which gave this answer multiplied by 9. exhibits special arithmetic properties.With our current techniques, we do not have explicit control on the full set of traces which would imply a complete understanding of all magic angles.These are already visible in the regular but evasive structure of the set of of magic α, A ⊂ C -see Figure 2.
In addition, we are able to express the rational numbers q ℓ ∈ Q in terms of a finite sum involving residues of rational functions which is fully presented in Theorem 4. A generalization of Theorem 1 which extends this result to more general potentials U (1.1) is presented in Theorem 5.As we show in §6, it is already possible to conclude directly from Theorem 1 that Theorem 2. Let U = U 0 with U 0 as in (1.2).There exist infinitely many magic α's, that is, This theorem will follow from the more general Theorem 6 and the observation that by the aforementioned explicit computation tr H (T 4 k ) = 8π/ √ 3 for U = U 0 there is at lest one complex magic angle.We then focus on real magic angles.Since the operator T 2 k is Hilbert-Schmidt, we can use the regularized determinant to study real magic α.Compared with the initial approach proposed in [TKV19], this approach has two advantages.Unlike the series expansion in [TKV19,WaLu21], the regularized determinant is an entire function with explicit error bounds in terms of the Hilbert-Schmidt norm.In addition, the Taylor coefficients of the determinant are polynomials of traces as in Theorem 1.This leads to Theorem 3. The chiral Hamiltonian with U = U 0 and U 0 as in (1.2), exhibits a flat band of multiplicity 2 at a real magic α * ∈ (0.583, 0.589), which is minimal, in the sense that the Hamiltonian does not possess a flat band for any α satisfying |α| where the counting |•| respects multiplicities.In particular, the flat bands of multiplicity 2 are uniformly gapped from all other bands.
Remark.Compared with results in [WaLu21] which require floating-point arithmetic, our proof of existence relies only on exact symbolic computations, the exact evaluation of residues to compute traces of powers of T k and the summation of finitely many matrix entries to estimate the Hilbert-Schmidt norm.

Preliminaries
From now on, we consider a potential U ∈ C ∞ (C/Γ; C) satisfying the first two symmetries of (1.1).The last symmetry U (z) = U (z) will only be needed in Corollary 5 to ensure that all traces are real.We recall that an orthonormal basis of L 2 (C/Γ; C) is given by setting We can express the potential U in this basis.A straightforward calculation gives c n e kn (2.1) Our aim is to obtain trace formulae for the powers of the compact operator T k defined by in (1.6).Since odd powers of T k have only off-diagonal components, it is clear that the traces of odd powers vanish.Thus, it is sufficient to compute the traces of powers of (2. 2) The invariance of the trace under cyclic permutations shows that it is sufficient to compute traces of powers of We shall study traces of powers of A k on smaller L 2 spaces, which we define below, for (p 1 , p 2 ) ∈ Z 2 3 , by whose C 2 -valued analogues are defined, using (1.3), as We remark that the operator (2D z − k) −1 acts diagonally on the Fourier basis and thus preserves the L 2 (p 1 ,p 2 ) spaces.On the other hand, multiplication by U (±z) does not preserve the space but one has by the translational symmetry defined in (1.1) In total, we have . This shows that we can restrict the operator A k to the subspaces L 2 (p 1 ,p 2 ) .From now on, we will denote by A k the restriction of A k to L 2 (1,1) .We then define the unitary multiplication operator The k-independence of the spectrum of T k implies then where k ∈ D(0, r)\{0}, and the last equality is meant in the sense of sets: multiplicities of elements in the top row are twice the multiplicities of elements in the bottom row.We also note that (1,1) is an analytic family of operators with compact resolvents and the spectrum is independent of k ∈ C \ 3Γ * , it follows that Spec(A k ) = Spec(A 0 ) [Ka80, Theorem 1.10].From (2.6) we obtain, as sets, with multiplicities on the left, twice the multiplicities on the right.Since k = 0 is included in the set of possible k for A k .Indeed, the set of possible values of k is (2.8) We end this preliminary section by stating and proving the main three properties we will use for our calculation.
Lemma 2.2.Consider a potential U ∈ C ∞ (C/Γ; C) satisfying the first two symmetries of (1.1) with a finite number of non zero Fourier mode in its decomposition (2.1).

Define the operator
(1,1) .For ℓ ≥ 2, one has: (2.9) e m ⟩ L 2 is a finite sum of rational fractions on the complex plane C with degree equal to −2ℓ and with (a finite number of) poles contained in (3Γ * − i) ∪ (3Γ * + i).
• For any γ ∈ Γ * and for any k / ∈ (3Γ * − i) ∪ (3Γ * − 2i), we have Proof.The first point is a consequence of the independence of the spectrum of T k in k (see [BHZ22, §2.3]) as well as the relation (2.7).
For the last two points, we prove by induction that k → A ℓ k e 3γ+i , where γ ∈ Γ * , is of the form where F ⊂ (3Γ * + i) is a finite set and R ν (k) is a sum of rational fraction of degree −2ℓ with poles located on (3Γ * − i) ∪ (3Γ * + i).Moreover, we will prove that the one has the relation The result is clear for ℓ = 0. Suppose the result true for ℓ, let's prove it holds for ℓ + 1.The main observation is that multiplication by U (±z) acts as a shift on the Fourier basis.The multiplication by U (−z) sends e ν to a linear combination of e ℓ for ℓ ∈ (3Γ * + 2i).Then applying (D(0 Multiplying by U (z) gives back a linear combination of e ν with ν ∈ (3Γ * + i).Finally, applying (D(0) − k) −1 multiplies the coefficient of e ν by (ν − k) −1 .This means that, using the induction hypothesis (2.10), where L ⊂ 3Γ * +i is a finite subset that depends only on U and a • are constants.Thus, it is clear from this formula that the induction carries on to ℓ + 1.This concludes the proof of the Lemma.□

Trace computations
We prove the following result.
(1,1) be a meromorphic family of Hilbert-Schmidt operators defined for k / ∈ (3Γ * − i) ∪ (3Γ * + i).We suppose that A k satisfies the three properties stated in Lemma 2.2.Then one has, for any ℓ ≥ 2, where the all infinite sums are in fact finite.
Proof.We want to give a semi-explicit formula for τ ℓ in terms of the residue of the rational fraction (second point in Lemma 2.2) We first start by writing, using that A ℓ k is trace-class for ℓ ≥ 2, We start with the relation which follows directly from (2.9), Here, we wrote We now use the third property stated in Lemma 2.2 to write The second property in Lemma 2.2 implies that Since we assume that ℓ ≥ 2, this justifies the exchange of integration and summation such that We make the change of variable s = t + 3n 1 and sum in n 1 to get This is equal to the limit of the integral over a parallelogram Γ n,R with sides Here, we used (3.2) to prove that the integral over the small parallel sides tends to 0. In particular, because there is only a finite number of poles, we see that, for |n| large enough, one has Using formula (3.3) as well as a partial summation, this allows us to rewrite the full trace as a telescopic sum The residue theorem shows that for n ∈ Z and R large enough, (3.5) Applying the residue theorem and using (3.4) gives (3.1).□ The consequences of this formula are summarized in the following theorem: Theorem 5. Consider a potential U ∈ C ∞ (C/Γ; C) satisfying the first two symmetries of (1.1) with finitely many non-zero Fourier modes c n ∈ Q(ω/ √ 3) appearing in the decomposition (2.1).Then for any ℓ ≥ 2, one has τ ℓ ∈ πQ(ω/ √ 3).If U also has the third symmetry of (1.1) then the traces are real and thus τ ℓ ∈ πQ/ √ 3.In particular, for all potentials satisfying all three symmetries in (1.1), including U = U 0 defined in (1.2), one has where A(U ) is the set of magic angles counting multiplicity for a potential U.
Proof.Under the hypothesis of the corollary, the function k → ⟨A ℓ k e i , e i ⟩ L 2 is a rational fraction with coefficients in Q(ω/ √ 3).This ring is actually a field as ω/ √ 3 is algebraic on Q.Now, taking partial fraction expansion of k → ⟨A ℓ k e i , e i ⟩ L 2 in Q(ω/ √ 3)(X) (the space of rational fractions with coefficients living in Q(ω/ √ 3)(X)) gives coefficients in 3).In particular, the residues of k → ⟨A ℓ k e i , e i ⟩ L 2 live in the field Q(ω/ √ 3)(X).But the uniqueness of the partial fraction expansion now gives that these are also the residues of k → ⟨A ℓ k e i , e i ⟩ L 2 in C(X).Using the trace formula stated in Theorem 4, this yields ∀ℓ ≥ 2, τ ℓ ∈ πQ(ω/ √ 3).
If we add the last symmetry of (1.1), the trace is real so that

□
This rationality condition suffices to prove that there is an infinite number of magic angles as long as there exists at least one magic angle.Theorem 6.Under the assumptions and with the same notation as in Theorem 5 one has the implication In particular, the set of magic angles for our canonical potential U 0 defined in (1.2) is infinite.Let N ≥ 0, for a tuple a = (a n ) {n;∥n∥∞≤N } , define U a to be the potential defined by (2.1).Then the above implication holds for a generic (in the sense of Baire) set of coefficients a = (a n Proof.We start by observing that since π is transcendental on Q, it is also transcendental in Q(ω/ √ 3).Now, assume by contradiction, that there exist only finitely many eigenvalues λ i ∈ C for i = 1, .., N of A 2 k .Then we define the n-th symmetric polynomial Newton identities show that this polynomial can be expressed as where e n = 0 for n > N. Theorem 1 shows that n i=1 The power m 1 • • • m n from sequences allowed in (3.6) is maximized by the unique choice m = (n, 0, . . ., 0).The Newton identities for n > N then imply that the transcendental number π is a root of a polynomial with coefficients in Q ω/ √ 3 .But then all these coefficients vanish, this is equivalent to the fact that the spectrum is empty (because of the determinant function, see (4.4)).For our particular choice of potential U 0 , the fact that tr A 2 k = 0 contradicts [Be*22, Theorem 3] so the set of magic angles is non empty, and thus infinite.Now, let a = (a n ) {n;∥n∥∞≤N } ∈ C (2N +1) 2 and assume that A(U a ) ̸ = ∅.Then, we can find an open neighbourhood of a, Ω a ∋ a, such that for coefficients b is open and dense in Ω a .Hence, the set coefficients for which 0 < |A b | < ∞ is given by m∈N q∈(Q+iQ) 2N +1 Ω q \ V m,q .It is then meagre and does not contain (Q(ω/ √ 3)) (2N +1) 2 .□

Fredholm determinants and the first magic angle
In this section, we explain how to compute the first few traces from our formula and show the existence of a simple real magic angle, i.e. prove Theorem 3. From now on, our choice of potential is given by U = U 0 defined in 1.2.Here we recall some facts from [Be*22, BHZ22] needed in this paper.
We recall the following properties [Be*22, (3.24)] we see that A k reads in the new Fourier coordinates 1 3 with A k the analogous restriction to ℓ 2 (1,1) .

4.2.
Fredholm determinants.We start by defining the regularized Fredholm determinant det 2 (1 where the product respects multiplicities.We find from (2.6) that det 2 (1 ) is real-valued on the real axis.To show existence and simplicity of magic angles, in the representation, we therefore use the following Lemma which provides ab initio bounds on the Fredholm determinants and its derivatives.
) is an entire function, independent of k ∈ C, which for any n, m ∈ N 0 satisfies where , with σ j = tr A j k . (4.5) Proof.The expression (4.4) is well-defined since A k is a Hilbert-Schmidt operator and the Taylor coefficients µ j are for example stated in [Si77,(6.13)].Indeed, since Cauchy estimates for the entire function f (z) := det 2 (1 + z Âk ) show by using the growth bound (4.6) The Taylor coefficients µ j are then given by the Plemelj-Smithies formula [Si77] stated in (4.5).Since they only depend on traces σ j which are independent of k, it follows that the regularized Fredholm determinant is an entire function independent of k.Hence, it suffices to study the determinant for k = 0.
If we write A 0 = ( A 0 (n)) n∈Z 2 and let P m be the projection onto (3{−m, −m + 1, ..., m} + 1) 2 , then (4.7) The first term constitutes the Hilbert-Schmidt norm of a finite matrix which can be explicitly computed from the matrix elements using symbolic calculations, indeed To estimate the second term, we may use that the operator norm of V ± satisfies ∥V ± ∥ = 3 √ 3, therefore one has We recall that by definition (2,2) ∥ 4 .Then, a direct computation shows that in terms of . While an explicit computation shows using exact symbolic calculations |m|∞≤6 1 g(m) 2 ≤ 24 7 (4.9) Then, we may use for |m| ∞ > 6 that g(m) ≥ |m| 2 + 5 2 , such that we can estimate the remainder ), with floating point approximation, where σ 1 := lim n→∞ |i|≤n ⟨A k e i , e i ⟩ is not absolutely summable as A k is not of trace-class, computed using Theorem 4 in the version stated as Theorem 7 in the appendix.One sees that the ratio of σ p /σ p−1 ≈ 1/0.5857 2 = 2.91507, for p large, where 0.5857 is the first magic angle.
Inserting this estimate into (4.8),we find along the lines of (4.10) which shows that ∥ A 0 ∥ 2 < 11 2 .□ Using the preceding error estimate with the explicit traces in Table 1, we conclude the existence of a first real magic angle in the next Proposition.The Proposition also completes the proof of Theorem 3. Indeed, (2.8) implies together with [BHZ22, Theorem 2] the existence of a 0 gap between the two flat bands of the Hamiltonian and the remaining bands.Proof.To see that this is the first real magic angle, we first notice that the operator norm of A 0 is bounded by (1,1) →ℓ 2 (1,1) ∥ 2 = 9.
This estimate shows that α ∈ R + with 1/(α 2 ) ∈ Spec( A 0 ) satisfies α ≥ 1 3 .A finite number of traces as explicitly computed in Table 1 are then relevant to prove the existence of a magic angle.

Figure 2 .
Figure2.The set A of magic α's for which (1.7) holds, that is, the first band is flat.The positive elements of A are the reciprocals of the "physically relevant" positive angles.Potential (1.2) is responsible for the regularity of the set which seems to indicate hidden integrability.For more general potentials the distribution is more complicated -see https://math.berkeley.edu/~zworski/multi.mp4for U θ (z) = (cos 2 θ)U (z) + (sin 2 θ) 2 k=0 ω k e zω k −z ωk which satisfies the required symmetries (1.1).The animation also indicates changing multiplicities.

Table 1 .
First eight exact traces of A p k , σ p = tr(A p k