aa r X i v : . [ phy s i c s . g e n - ph ] O c t An exactly solvable toy model
Wang X G , , and Zhang J M , Fujian Provincial Key Laboratory of Quantum Manipulation and New EnergyMaterials, College of Physics and Energy, Fujian Normal University, Fuzhou 350007,China Fujian Provincial Collaborative Innovation Center for OptoelectronicSemiconductors and Efficient Devices, Xiamen 361005, China Institute of Physics, Chinese Academy of Sciences, Beijing 100080, China
Abstract.
In an attempt to regularize a previously known exactly solvable model[Yang and Zhang, Eur. J. Phys. , 035401 (2019)], we find yet another exactlysolvable toy model. The interesting point is that while the Hamiltonian of the modelis parameterized by a function f ( x ) defined on [0 , ∞ ), its spectrum depends only onthe end values of f , i.e., f (0) and f ( ∞ ). This model can serve as a good exercise inquantum mechanics at the undergraduate level.PACS numbers: 03.65.-w, 03.65.Ge n exactly solvable toy model
1. Introduction
Recently, in studying the quench dynamics of a Bloch state in a one-dimensional tight-binding ring [1, 2, 3], an exactly solvable model was devised and its dynamics wassolved. Later, the eigenstates and eigenenergies of the model were solved in closed andneat forms [4].The model is very simple. It consists of infinitely many levels {| n i , n ∈ Z } , withthe Hamiltonian being H = ∞ X n = −∞ n ∆ | n ih n | + g ∞ X n ,n = −∞ | n ih n | . (1)The first term is diagonal. It simply means that the levels are equally spaced with thespacing being ∆. The second term is off-diagonal. Its peculiarity is that it couples twoarbitrary levels, regardless of their distance in energy, with a constant strength g . Thismakes it a rank-1 operator ‡ , a fact which was believed to be responsible for the exactsolvability of the model.It should be instructive to review how this ideal, contrived model (1) arises in amore realistic problem. In [1, 2, 3], we studied such a dynamical problem. The settingis a tight-binding lattice [5] with the Hamiltonian H tb = − L − X m =0 ( | m ih m + 1 | + | m + 1 ih m | ) . (2)Here {| m i} are a set of orthonormal, localized single-particle orbits. The Hamiltoniansimply means that the particle can hop between two adjacent sites. We take the periodicboundary condition so that | i = | L i . Equivalently, one can imagine that the sites arearranged into a ring. The eigenstates of H tb are simply plane waves (Bloch states) onthe ring, | k i = L − X m =0 | m ih m | k i = 1 √ L L − X m =0 e i πkm/L | m i , (3)where k is an integer. Apparently, | k i = | k + L i . The corresponding eigenenergy is ε ( k ) = − πk/L ).Now consider such a problem. Suppose initially the particle is in a generic Blochstate, i.e., | φ ( t = 0) i = | k i i , where k i is an arbitrary integer. It is stationary with respectto H tb . However, at t = 0, we quench it by adding a potential to an arbitrary site of thering. Without of loss of generality (since all the sites are on an equal footing), let thissite be m = 0. The Hamiltonian is now H tb + H pot , with H pot = V | ih | , (4) ‡ In linear algebra, a p × q rank-1 matrix is the product of a p -component column vector anda q -component row vector. Here, the off-diagonal term can obviously be written as g | u ih u | , with | u i = P n ∈ Z | n i . n exactly solvable toy model V is the potential strength. The initial state is no longer an eigenstate of thepost-quench Hamiltonian and it starts evolving. Many interesting phenomena ensue.For example, it is found that the survival probability of the initial state, namely |h φ ( t = 0) | φ ( t ) i| , display cusps periodically in time.To account for these phenomena, we notice two salient features of the problem.First, it is observed numerically that at any time, only those Bloch states with energyclose to that of the initial Bloch state are significantly populated. That is, onlythose Bloch states | k i with k ≃ k i or k ≃ − k i are actively participating in thedynamics. These states are almost equally spaced in energy, with the gap being( ∂ε/∂k ) | k = k i = (4 π/L ) sin(2 πk i /L ). Second, as can be easily verified, the quenchpotential H pot , which induces the nontrivial dynamics, couples two arbitrary Bloch stateswith a constant strength. Specifically, h k | H pot | k i = L − X m =0 h k | m ih m | H pot | m ih m | k i = 1 L L − X m =0 h m | H pot | m i e i π ( k − k ) m/L = VL , (5) regardless of the values of k , .It is these two features that motivated the construction of the model (1), whichsuccessfully accounted for all the dynamical phenomena observed in the original tight-binding model.
2. A more regular model
From the above, specifically (5), we see that the constant coupling comes from a veryspecial potential, which is supported on a single site. For a generic, more extendedpotential, the coupling cannot be a constant. Actually, from (5), we see that thecoupling strength between two Bloch states is just the Fourier transform of the quenchpotential. Now by the Riemann-Lebesgue lemma [6], if the quench potential is moreextended, its Fourier transform would decay to zero in the high frequency limit. We arethus motivated to “regularize” the off-diagonal coupling term in (1) and consider thefollowing Hamiltonian H f = ∞ X n = −∞ n ∆ | n ih n | + ∞ X n ,n = −∞ f ( n − n ) | n ih n | , (6)where f ( x ) is a real, even function defined on the real line satisfying the conditionlim x →∞ f ( x ) = 0.The coupling term is now no longer a rank-1 perturbation, and the approach in[4] fails. But it is still regular. Actually, it is in the form of a Toeplitz matrix § and § A Toeplitz matrix, or a diagonal-constant matrix, is a structured matrix in which each descendingdiagonal from left to right is constant. See B¨ottcher A and Grudsky S M 2012 Toeplitz Matrices,Asymptotic Linear Algebra, and Functional Analysis (Birkh¨auser, Basel). n exactly solvable toy model f . This observationsuggests the Fourier transform k , and we solve the eigenvalues as ε m = m ∆ + f (0) , m ∈ Z . (7)Here it is interesting that while the Hamiltonian (6) is defined with the function f as aparameter as a whole, its spectrum depends only on the end value f (0).The Fourier transform approach works also for the original Hamiltonian (1). Wethus find that for the function f we can drop the condition that it converges to zero as x → ±∞ , and require just that the limit exists. In this case, the spectrum is ε m = m ∆ + f (0) − f ( ∞ ) + ∆ π arctan (cid:18) πf ( ∞ )∆ (cid:19) , m ∈ Z . (8)It depends only on the end values f (0) and f ( ∞ ). For the original Hamiltonian (1), f (0) = f ( ∞ ) = g , and we recover the results in [4].
3. Analytical solution of the eigenvalue problem
First of all, as in [4], we notice that the off-diagonal part is invariant under the translation | n i → | n + 1 i , which implies that the energy spectrum should be equally spaced withthe equal gap being ∆. Formally, let us define the translation or raising operator T = P ∞ n = −∞ | n + 1 ih n | . We claim that if | ψ i is an eigenstate of H f with eigenvalue ε , then T | ψ i is an eigenstate with eigenvalue ε + ∆. ¶ This is because we have thecommutation relation [ H f , T ] = ∆ T and hence H f T | ψ i = ( T H f + ∆ T ) | ψ i = ( T ε + ∆ T ) | ψ i = ( ε + ∆) T | ψ i . (9)To determine the spectrum, we have to consider the Schr¨odinger equation H f | ψ i = ε | ψ i . Assuming that | ψ i = P n | n ih n | ψ i = P n a n | n i , we have εa n = n ∆ a n + ∞ X m = −∞ f ( n − m ) a m . (10)The second term on the right hand side is in the form of (discrete) convolution. Thissuggests the Fourier transform, A k = ∞ X n = −∞ e ikn a n , F k = ∞ X n = −∞ e ikn f ( n ) . (11)For the left hand side of (10), we get εA k . For the right hand side, we get ∞ X n = −∞ e ikn ∆ na n + ∞ X n = −∞ ∞ X m = −∞ e ikn f ( n − m ) a m k As a well-known theorem, the Fourier transform of the convolution of two functions is the pointwiseproduct of their Fourier transforms. That is, two functions convolved in the real space are factorizedin the momentum space. This is of course a great simplification. ¶ At this point, one might jump to the conclusion that the spectrum of the model must be in the formof (7), as we have
T r ( H f ) = P m ∈ Z [ m ∆ + f (0)] = P m ∈ Z ε m . But this identity is derived for a finitematrix. A mindless generalization to the infinite case is dangerous. Actually, it is even a nontrivialproblem to make sense of the trace of an infinite matrix. n exactly solvable toy model − i ∆ ∂∂k ∞ X n = −∞ e ikn a n + ∞ X n − m = −∞ e ik ( n − m ) f ( n − m ) ∞ X m = −∞ e ikm a m = − i ∆ ∂∂k A k + F k A k . (12)We thus get the equation of A k , εA k = − i ∆ ∂∂k A k + F k A k . (13)Note that this is a Dirac-type equation, in the sense that the Hamiltonian on the righthand side is linear in the momentum.Let us first consider the special case of f ( ∞ ) = 0. In this case, as long as f decayssufficiently fast, F is a regular function of k , and we can solve A as A k = A exp (cid:20) i ∆ Z k dq ( ε − F q ) (cid:21) . (14)The value of ε is determined by the boundary condition of A = A π , which is equivalentto 2 πm ∆ = 2 πε − Z π F k dk (15)for some integer m . But the integral here is simply 2 πf (0). We thus obtain theeigenvalues as in (7). They are indeed equidistant. But what is interesting is thatthere is only dependence on the end value f (0).In the more general case of f ( ∞ ) = 0, we can decompose the function into twoparts as f = [ f − f ( ∞ )] + f ( ∞ ). The Fourier transform of f is then in the form of F k = F (1) k + 2 πf ( ∞ ) X m ∈ Z δ ( k − πm ) , (16)where the first term comes from f − f ( ∞ ), i.e., F (1) ( k ) = P n ∈ Z e ikn [ f ( n ) − f ( ∞ )],while the delta functions from the constant part. We still have (13). But now A isdiscontinuous at k = 0. Integrating from 0 − to 0 + , we get0 = − i ∆( A + − A − ) + 2 πf ( ∞ ) Z + − δ ( k ) A ( k ) dk = − i ∆( A + − A − ) + πf ( ∞ )( A + + A − ) , (17)i.e., A + = ∆ − iπf ( ∞ )∆ + iπf ( ∞ ) A − = exp (cid:26) − i arctan πf ( ∞ )∆ (cid:27) A − . (18)On the other hand, integrating from 0 + to 2 π − , we get A − = A π − = A + exp (cid:20) i ∆ Z π dk (cid:16) ε − F (1) k (cid:17)(cid:21) = A + exp (cid:20) i ∆ 2 π ( ε − f (0) + f ( ∞ )) (cid:21) . (19) n exactly solvable toy model δ function, as a distribution, is supposed to act ona smooth function, yet A is discontinuous at k = 0 as we see in (18) a posterior . Toarrive at the second line, we defined the product between the delta function and theHeaviside function, δ ( k ) θ ( k ), to be δ ( k ). This seems reasonable, as we can “derive”it as δθ = θ ′ θ = ( θ ) ′ = θ ′ = δ . But a second thought would lead also to δθ = θ ′ θ = θ ′ θ n = n +1 ( θ n +1 ) ′ = n +1 θ ′ = n +1 δ for any positive integer n . This ambiguityshows that the product δθ is not well defined as a distribution and the treatment in(17) is questionable. A different approach is to consider the delta function in (16) as thelimit of some ordinary function. Instead of (18), this would lead to the linking condition A + = exp " i ∆ Z + − dq ( ε − F q ) A − = exp (cid:20) − πif ( ∞ )∆ (cid:21) A − , (20)and eventually result in the spectrum of (7) rather than that of (8).We note that in the literature dealing with the relativistic Dirac equation [7], bothapproaches exist. In [8, 9], the first approach was taken, which was criticized in [10],where the second approach was favored. But actually both approaches are equally good.Mathematically, the most appropriate way to deal with the delta potential is by usingthe theory of self-adjoint extension [11, 12, 13]. The effect of the delta potential is toimpose a twisted boundary condition A + = e iφ A π − = e iφ A − . The question is justwhat values of φ will make the momentum operator − i∂/∂k a self-adjoint operator + .The answer is, as can be easily checked, that any real number φ will do [11, 12, 13].Therefore, the linking conditions (18) and (20) are just two among the infinitely manypossibilities. In a particular problem, whether (18) or (20) or any other value of φ isrealized cannot be determined a priori , but can only be inferred from some physicalobservables (like the spectrum or dynamics of the system). In our specific case, ourprevious works [1, 2, 3, 4] indicate that it is the linking condition (18) that is realized.This will be verified again in the next section.In hindsight, it is not difficult to recognize the reason why the spectrum dependsonly on the end values of f . We note that (13) is a Dirac-type equation for a chargedparticle constrained in a ring which is pierced by a magnetic field. The Hamiltonian is K = − i ∆ ∂/∂k + F k , where k ∈ [0 , π ) is the angular variable and F k is the vector field.Actually, H f in (6) is just the representation of K in the basis of the plane waves on thering, i.e., if we identify level | n i with the basis function e ink / √ π .As is well-known and easily checked, (13) is invariant under the gauge transform + Although non-self-adjoint operators are interesting too. They should not bother us here. For thoseinterested, see Cuenin J C, Laptev A and Tretter C 2014 Eigenvalue estimates for non-selfadjoint Diracoperators on the real line Annales Henri Poincar´e n exactly solvable toy model A k → A k e iϕ ( k ) , F k → F k + ∆ ∂ϕ∂k , (21)where ϕ ( k ) is an arbitrary, smooth, and 2 π -periodic function of k . Under this gaugetransform, the value of f at a generic point f ( n ) = 12 π Z π F k e − ink dk (22)of course changes. However, the end values f (0) and f ( ∞ ) are invariant, as Z π (cid:18) ∂ϕ∂k (cid:19) e − ink dk (cid:12)(cid:12)(cid:12) n =0 = ϕ (2 π ) − ϕ (0) = 0 , (23) Z π (cid:18) ∂ϕ∂k (cid:19) e − ink dk (cid:12)(cid:12)(cid:12) n →∞ = 0 . (24)Here the second equation is due to the Riemann-Lebesgue Lemma. Conversely, any two f ’s with the same end values can be converted into each other with a gauge transform,which leaves the spectrum invariant.
4. Numerical simulation
It is instructive to check the analytic results above numerically. To do so, we have totruncate the infinite-dimensional Hilbert space to a finite one. Let us keep the 2 N + 1levels in the middle, i.e., {|− N i , . . . , | N i} . The truncated Hamiltonian is (∆ = 1 below) H ( N ) f = N X n = − N n ∆ | n ih n | + N X n ,n = − N f ( n − n ) | n ih n | . (25)Here for the function f , we shall take (just for the sake of simplicity) f ( n ) = g α | n | + g , (26)where g , are two real constants and | α | ≤ H ( N ) f can be readily diagonalized numerically and the eigenvaluescan be denoted and ordered as E − N < . . . < E < . . . < E N . It is no wonder thatthe eigenvalues in the middle of the spectrum, which are least affected by the finitenessof the truncated Hilbert space, are indeed almost equally spaced. The only concern iswhether the offset agrees with the analytic predictions. We shall therefore focus on theparticular eigenvalue E , which should suffer least from the finite-size effect.We note that the decaying factor α defines a characteristic length λ = − / ln | α | .Now for E , there exist two limiting cases. If N ≫ λ , H ( N ) f resembles the ideal, infinite-dimensional H f , and it is expected that E ≃ g + 1 π arctan( πg ) , (27)according to (8). On the contrary, if N ≪ λ but still N ≫
1, all the off-diagonalcouplings are close to g + g , and H ( N ) f resembles H in (1) with g = g + g . We thus n exactly solvable toy model Figure 1. (Color online) Crossover behavior of the central eigenvalue E of thetruncated Hamiltonian H ( N ) f as defined in (25) and (26). The parameters are∆ = g = g = 1. In both panels, the solid line corresponds to the exact value of E , while the dashed and dotted lines to the analytic predictions in (27) and (28),respectively. Note how the solid line approaches the two analytic values at the twoends. expect that E ≃ π arctan[ π ( g + g )] . (28)These limiting behaviors are verified in Fig. 1. In Fig. 1(a), we fix the value of α to α = 0 .
99 (corresponding to a value of λ ≃ N increase from 10 to4500. We see indeed a crossover between the two values in (27) and (28). Similarly,in Fig. 1(b), we fix N to N = 40 and let α vary from about 0 .
01 to 1 (in this way, λ increases from about 0.2 to infinity). We see the same crossover as in Fig. 1(a) but inthe other direction.One should have noticed that here the two limits lim N →∞ and lim α → − are notinterchangeable. That is, g + 1 π arctan( πg ) = lim α → − lim N →∞ E = lim N →∞ lim α → − E = 1 π arctan[ π ( g + g )] . (29)This singularity comes essentially from that of the function α N , as we havelim α → − lim N →∞ α N = 0 = 1 = lim N →∞ lim α → − α N . (30)The situation here is somewhat similar to that of a ferromagnet below the Curie point.There, the limit of reducing the external magnetic field to zero and that of increasingthe system size to infinity do not commute [15]. If the former is taken first, the totalmagnetization would be zero; however, if the latter is taken first, the total magnetizationwould be nonzero. n exactly solvable toy model
5. Conclusions and discussions
We have tried to regularize a previously known exactly solvable model which is deemedshort of generality. In doing so, we found that it is actually an extremal member ofa class of exactly solvable models. It is the Toeplitz property instead of the rank-1property that has been generalized. Consequently, the solution method is also different.The interesting thing is that while the model is parameterized by a function f , itsspectrum depends only on the end values of this function. This characteristic is bestunderstood by noticing that the model can be realized with a charged particle (albeitwith the Hamiltonian being linear instead of quadratic in the momentum) confined ona ring which is threaded by some magnetic flux. As long as the end values are fixed,different f ’s just correspond to different choices of the gauge. But the energy spectrumof the system, as a physical observable, of course should be gauge-independent.Although deemed a toy model, we find that the current model contains aninteresting model in solid state physics as a special case. Specifically, if f = − x = ±
1, and 0 otherwise, the model (6) becomes H Stark = ∞ X n = −∞ n ∆ | n ih n | − ∞ X n = −∞ ( | n ih n + 1 | + | n + 1 ih n | ) . (31)This is just the Hamiltonian of a particle in a tilted one-dimensional tight-bindingmodel. In this system, we have the celebrated Bloch oscillation phenomenon, and theHamiltonian has been well studied [16, 17, 18]. In particular, in [17], its spectrum hasbeen solved. It is in agreement with our result (7).We note that the original model finds use in the quench dynamics of a Bloch statewith a potential non-vanishing only on a single site [1, 2, 3]. Hopefully, the currentmodel is relevant if the quench potential is a more general, extended one. But anyway,the model is of pedagogical value for students of quantum mechanics. On one hand,it adds to the short list of exactly solvable models mentioned in a typical quantummechanics course (that includes the hydrogen atom, the harmonic oscillator, etc.); onthe other hand, the solution is simple enough and illuminating. An important pointis that while the Schr¨odinger equation in the presence of a δ -potential is discussed inpossibly every quantum mechanics textbook, the Dirac counterpart is rarely mentioned.As we have seen, the latter is a little bit more tricky. Acknowledgments
The authors are grateful to K. Jin and Y. Xiang for their helpful comments. This workis supported by the National Science Foundation of China under Grant No. 11704070.
References [1] Zhang J M and Yang H T 2016 Cusps in the quench dynamics of a Bloch state EPL n exactly solvable toy model [2] Zhang J M and Yang H T 2016 Sudden jumps and plateaus in the quench dynamics of a Blochstate EPL Solid State Physics (Thomson Learning, Toronto)[6] Bochner S and Chandrasekharan K 1949
Fourier Transforms (Princeton University Press)[7] Glushkov A V 2008
Relativistic Quantum Theory: Quantum Mechanics of Atomic Systems (Astroprint, Odessa)[8] Subramanian R and Bhagwat K V 1972 The relativistic Tamm model J. Phys. C: Solid State Phys. Self-Adjoint Extensions in Quantum Mechanics (Springer, Berlin)[14] Sakurai J J 1994
Modern Quantum Mechanics (Addison-Wesley, New York)[15] Huang K
Statistical Mechanics L379[17] Hartmann T, Keck F, Korsch H J and Mossmann S 2004 Dynamics of Bloch oscillations New J.Phys.548