Undamped Bloch Oscillations in the $U\rightarrow \infty$ one-dimensional Hubbard model

TL;DR

This paper exactly solves the one-dimensional Hubbard model at infinite interaction strength in an electric field, revealing persistent Bloch oscillations linked to Hamiltonian periodicity, and compares it with a non-integrable continuous model showing similar behavior.

Contribution

It demonstrates the existence of undamped Bloch oscillations in the $U ightarrow \infty$ Hubbard model and relates this to Hamiltonian periodicity rather than integrability, with comparative analysis.

Findings

Undamped Bloch oscillations are extensively present in the model.

Charge current exhibits dissipationless behavior similar to Bloch oscillations in a related non-integrable model.

The persistence of oscillations is linked to Hamiltonian periodicity, not integrability.

Abstract

The $U\rightarrow +\infty$ one-dimensional Hubbard model in an electric field has be exactly solved, with an emphasis on the charge current. It is found that undamped Bloch oscillations extensively exist in the system. Such conclusion has also been discussed for more general cases and we find that it is closely related to the temporal periodicity of the model Hamiltonian in electric field, rather than to the integrability of the model. As a comparison, we have also studied a model of electrons with $\delta$-function interactions in continuous space, which is closely related to the Hubbard model, but is non-integrable; and we find that the charge current strangely shows a dissipationless behaior which is comparable with the undamped Bloch oscillations.

Equations

$$H=-t_0\sum _{j,\sigma }[e^{i\frac{e}{\hslash c}A(t)}{C}_{j\sigma }^{†}{C}_{j+1\sigma }+\text{H.c.}]+U\sum _j{n}_{j\uparrow }{n}_{j\downarrow },$$

$$H=-t_0\sum _{j,\sigma }[e^{i\frac{e}{\hslash c}A(t)}{\hat{C}}_{j\sigma }^{†}{\hat{C}}_{j+1\sigma }+\text{H.c.}],$$

$$\hat{j}=-c\frac{\partial H}{\partial A}.$$

$$|\mathbf{j}⟩⨂|\mathbf{\sigma }⟩=|j_1,j_2,\cdots j_N⟩⨂|{\sigma }_1{\sigma }_2\cdots {\sigma }_N⟩,$$

$$T=\sum _{\mathbf{j},\mathbf{\sigma }}|\mathbf{j}⟩⨂|\mathbf{\sigma }⟩⟨\mathbf{j},\mathbf{\sigma }|,$$

$$\begin{gathered} \sum _{\sigma }{\hat{C}}_{j\sigma }^{†}{\hat{C}}_{j+1\sigma }\to T\sum _{\sigma }{\hat{C}}_{j\sigma }^{†}{\hat{C}}_{j+1\sigma }{T}^{†} \\ =\text{sideset}{}^{\prime }{\sum }_{{\mathbf{j}}_i}\widetilde{|{\mathbf{j}}_i⟩}⟨{\mathbf{j}}_i|⨂\sum _{\mathbf{\sigma }}|\mathbf{\sigma }⟩⟨\mathbf{\sigma }|\equiv f_j^{†}{f}_{j+1}I, \end{gathered}$$

$$\begin{gathered} \sum _{\sigma }{\hat{C}}_{L\sigma }^{†}{\hat{C}}_{1\sigma }\to T\sum _{\sigma }{\hat{C}}_{L\sigma }^{†}{\hat{C}}_{1\sigma }{T}^{†} \\ =f_L^{†}{f}_1\sum _{\mathbf{\sigma }}\widetilde{|\mathbf{\sigma }⟩}⟨\mathbf{\sigma }|\equiv f_L^{†}{f}_{1}P, \end{gathered}$$

$$\sum _{\sigma }{\hat{C}}_{1\sigma }^{†}{\hat{C}}_{L\sigma }\to T\sum _{\sigma }{\hat{C}}_{1\sigma }^{†}{\hat{C}}_{L\sigma }{T}^{†}=f_1^{†}{f}_L{P}^{-1},$$

$$\begin{gathered} \tilde{H}=-t_0\sum _{j=1}^{L-1}\left[e^{i\theta (t)}{f}_j^{†}{f}_{j+1}I+\text{H.c.}\right] \\ -t_0{e}^{i\theta (t)}{f}_L^{†}{f}_{1}P-t_0{e}^{-i\theta (t)}{f}_1^{†}{f}_L{P}^{-1}, \end{gathered}$$

$$\tilde{j}=-cT\frac{\partial H}{\partial A}{T}^{†}=-c\frac{\partial \tilde{H}}{\partial A}.$$

$$s=\{|{\mathbf{\sigma }}_{s_1}⟩,|{\mathbf{\sigma }}_{s_2}⟩,\cdots ,|{\mathbf{\sigma }}_{s_{m_s}}⟩\},$$

$$\begin{gathered} |{\chi }_{k_s}⟩=1/\sqrt{m_s}[\exp (i{k}_s)|{\mathbf{\sigma }}_{s_1}⟩+\exp (i2k_s)|{\mathbf{\sigma }}_{s_2}⟩ \\ +\cdots +\exp (i{m}_s{k}_s)|{\mathbf{\sigma }}_{s_{m_s}}⟩], \end{gathered}$$

$$\widetilde{H}=\sum_{s}\sum_{k_{s}}\bigg{[}-t_{0}e^{i\theta(t)}\bigg{(}\sum_{j=1}^{L-1}f^{{\dagger}}_{j}f_{j+1}+f^{{\dagger}}_{L}f_{1}e^{ik_{s}}\bigg{)}+\text{H.c.}\bigg{]}\\ \bigotimes{}(|\chi_{k_{s}}\rangle\langle\chi_{k_{s}}|),$$

$$-t_0{e}^{i\theta (t)}{a}_{j+1}=a{a}_j,\text{ j=N},$$

$$-t_0{e}^{i\theta (t)}\exp (i{k}_s)a_1=a{a}_N.$$

$$a_N=\frac{a}{-t_0{e}^{i\theta (t)}}{a}_{N-1}=\cdots =(\frac{a}{-t_0{e}^{i\theta (t)}}{)}^{N-1}{a}_1,$$

$$q=k_s/L+2n\pi /L,n=0,1,2,\cdots ,L-1.$$

$$f_q^{†}=\frac{1}{\sqrt{L}}\sum _{j=1}^L\exp (iq{r}_j)f_j^{†}.$$

$${ϵ}_q(t)=-2t_0\cos \left[q+\theta (t)\right].$$

$$h_s=\sum _q{ϵ}_q(t)f_q^{†}{f}_q.$$

$$|\mathbf{q}⟩\equiv f_{q_1}^{†}{f}_{q_2}^{†}\cdots f_{q_N}^{†}|0⟩,$$

$$|{\Psi }_{\mathbf{q}s}⟩=|\mathbf{q}⟩⨂|{\chi }_{k_s}⟩,$$

$$E_{\mathbf{q}}(t)={ϵ}_{q_1}(t)+{ϵ}_{q_2}(t)+\cdots +{ϵ}_{q_N}(t).$$

$$i\hslash \frac{\partial }{\partial t}|\Psi (t)⟩=\tilde{H}|\Psi (t)⟩=E_{\mathbf{q}}(t)|\Psi (t)⟩,$$

$$|\Psi(t)\rangle=\exp\Big{[}\frac{-i}{\hbar}\int_{0}^{t}E_{\mathbf{q}}(t)dt\Big{]}|\Psi_{\mathbf{q}s}\rangle.$$

$$\begin{gathered} j(t)=⟨\Psi (t)|\tilde{j}|\Psi (t)⟩=-c⟨\Psi (t)|\frac{\partial \tilde{H}}{\partial A}|\Psi (t)⟩ \\ =-c\frac{\partial }{\partial A}⟨\Psi (t)|\tilde{H}|\Psi (t)⟩=-c\frac{\partial E_{\mathbf{q}}(t)}{\partial A}, \end{gathered}$$

$$j(t)=B_1\cos \left(\frac{e}{\hslash }Et\right)-B_2\sin \left(\frac{e}{\hslash }Et\right),$$

$$B_1=\frac{2t_{0}e}{\hslash }\sum _{m=1}^N\sin q_m;B_2=\frac{2t_{0}e}{\hslash }\sum _{m=1}^N\cos q_m.$$

$$[{\zeta }^{†},H]=0.$$

$$H(t+\tau )=H(t),$$

Full text

I Introduction

Though one-dimensional (1D) Hubbard model can be exactly solved with Bethe ansatz Bt1 ; Bt2 , the discussion of its response to electric field does not benefit much from this. In the simplest treatment, the electric field can be introduced to the model via a time-dependent Peierls phase factor PS ; PS2 , even though, an exact addressing of the response behavior of the system generally is still impossible.

To the level of linear response, many studies have concluded that 1D Hubbard model shows ideal conductance In1 ; In2 ; In3 ; In4 ; and such behavior is closely related to the integrability of the model, which always results in a nonzero Drude weight in cases away from half filling.

While to the level of nonlinear response, less is known about this model. Generally, it is believed that the integrability of a model still plays an important role cc ; cc1 in a nonlinear response: for integrable models, the charge current can resemble the Bloch oscillations as in the noninteracting case; however, the appearance of electric field may break integrability of the model itself, resulting in a damped current oscillation.

Actually, as for Hubbard model, Eckstein and Werner 1 , using dynamical mean-field theory, have shown numerically that, different values of the on-site Coulomb repulsion $U$ of electrons can lead to different damped-oscillating behaviors of charge current. And the oscillating feature would even disappear if $U$ exceeds some critical value. It seems that the damping of Bloch oscillations is inevitable in the presence of on-site $U$. Such picture has been further supported by Mandt 2 via a variational solution of the Boltzmann equation for the Hubbard model.

Nevertheless, since the exact solution of the Hubbard model in electric field is lack even in the 1D case, it is still an open problem that how the model responses an electric field nonlinearly. However, we find that 1D Hubbard model in the $U\rightarrow+\infty$ limit provides a good platform to address this question. We will show that in such case, using a method of unitary transformation, the model is still solvable when the electric field appearing.

We will show that the behavior of charge current indeed can resemble Bloch oscillations, but such behavior is closely related to the temporal periodicity of the Hamiltonian in electric field, rather than to the integrability of the model. And this conclusion can be extended to more general models and cases.

Especially, to see more about the effect of integrability and correlation interactions on the response of a system to electric field, we will show in the appendix a non-integrable model of electrons with $\delta$-function interactions, of which, however, the conductivity can be exactly determined.

II Model and transformation

We consider an $N$-electron system on a periodic ring with $L$ sites, and we can treat the constant electric field as the temporal derivative of a time-dependent vector potential along the ring, $E=-1/cA_{t}(t)$ or ${A}(t)=-cEt$. The model Hamiltonian then can be written as

[TABLE]

where $C^{{\dagger}}_{j\sigma}$ ($C_{j\sigma}$) is the creation (annihilation) operator of electron at site $j$ and spin $\sigma=\uparrow$ or $\downarrow$. And $n_{j\sigma}$ stands for the particle number operator as usual. We use the periodic boundary condition $C^{{\dagger}}_{j+L\sigma}=C^{{\dagger}}_{j\sigma}$ and $C_{j+L\sigma}=C_{j\sigma}$. For simplicity, we set the lattice constant $a=1$.

In the limit $U\rightarrow+\infty$, which we are interested in, the double occupation of a site by $\uparrow$ and $\downarrow$ electrons is prohibited, and the Hamiltonian is formally reduced to

[TABLE]

where $\hat{C}^{{\dagger}}_{j\sigma}=(1-n_{j\bar{\sigma}})C^{{\dagger}}_{j\sigma}$ and $\hat{C}_{j\sigma}=(1-n_{j\bar{\sigma}})C_{j\sigma}$ as usual. Then, we can concentrate our discussion in the subspace without double occupations. And the basis states involved can be written in the form $|\mathbf{j},\bm{\sigma}\rangle=C^{{\dagger}}_{j_{1}\sigma_{1}}C^{{\dagger}}_{j_{2}\sigma_{2}}\cdots C^{{\dagger}}_{j_{N}\sigma_{N}}|0\rangle\equiv|j_{1}\sigma_{1},j_{2}\sigma_{2},\cdots j_{N}\sigma_{N}\rangle$, in which $N$ electrons with spins $\sigma_{1}$, $\sigma_{2}$, $\cdots$, $\sigma_{N}$ occupy sites $j_{1}$, $j_{2}$, $\cdots$, $j_{N}$ respectively, where $1\leq j_{1}<j_{2}<\cdots<j_{N}\leq L$, to guarantee that no double occupation occurs. And the charge current operator PS2 ; co can be written as

[TABLE]

Without electric field, the wave function of 1D $U\rightarrow+\infty$ Hubbard model has a spin-charge separated form, as is well-known Bt2 . However, when electric field appearing, our problem becomes a time-dependent one, and still requires a careful discussing.

As has be noted previously 89 , all the terms in Eq. (2), except the ones with $\hat{C}^{{\dagger}}_{L\sigma}\hat{C}_{1\sigma}$ or $\hat{C}^{{\dagger}}_{1\sigma}\hat{C}_{L\sigma}$, when acting on a state, can only change the sites of the $N$ electrons, without any alteration in the spin sequence; that is, they can only change a state $|j_{1}\sigma_{1},j_{2}\sigma_{2},\cdots j_{N}\sigma_{N}\rangle$ to something like $|j^{\prime}_{1}\sigma_{1},j^{\prime}_{2}\sigma_{2},\cdots j^{\prime}_{N}\sigma_{N}\rangle$. Actually, without electric filed, basing on this point, the $U\rightarrow+\infty$ Hubbard model can be mapped to a set of Hamiltonians of spinless fermions, and can be exactly solved 89 ; 90 . Here, we also base our discussion on this point, but use a method of unitary transformation.

Firstly, we introduce a new kind of basis states, in which the spin and charge are completely independent,

[TABLE]

where $|\mathbf{j}\rangle=|j_{1},j_{2},\cdots j_{N}\rangle$ denotes a state that $N$ spinless but charged fermions occupy sites $j_{1}$, $j_{2}$, $\cdots$, $j_{N}$, and it is required that $1\leq j_{1}<j_{2}<\cdots<j_{N}\leq L$, as before; by introducing the $j$-site creation (annihilation) operator $f^{{\dagger}}_{j}$ ($f_{j}$) of these fermions, such state can be expressed as $|\mathbf{j}\rangle=f^{{\dagger}}_{j_{1}}f^{{\dagger}}_{j_{2}}\cdots f^{{\dagger}}_{j_{N}}|0\rangle$. And $|\bm{\sigma}\rangle=|\sigma_{1}\sigma_{2}\cdots\sigma_{N}\rangle$ is the configuration of $N$ sequenced spins. Obviously, the direct product states defined by Eq. (4) have a one-to-one correspondence with $|\mathbf{j},\bm{\sigma}\rangle$. We can introduce a transformation

[TABLE]

of which the unitary characteristic $T^{{\dagger}}T=\mathbf{1}$ can be easily verified. Then, we have $T|\mathbf{j},\bm{\sigma}\rangle=|\mathbf{j}\rangle\bigotimes|\bm{\sigma}\rangle$. Under such transformation, the form of all the terms in Eq. (2), except the ones with $\hat{C}^{{\dagger}}_{L\sigma}\hat{C}_{1\sigma}$ and $\hat{C}^{{\dagger}}_{1\sigma}\hat{C}_{L\sigma}$, can be easily obtained. For $j\neq N$,

[TABLE]

where $|\mathbf{j}_{i}\rangle$ is a state with site $j$ empty and site $j+1$ occupied by a spinless fermion. And $\widetilde{|\mathbf{j}_{i}\rangle}$ is same to $|\mathbf{j}_{i}\rangle$, except with site $j+1$ empty and site $j$ occupied. “$\Sigma^{\prime}$” represents a sum over these states, covering all the cases in which a spinless fermion hops from site $j+1$ to site $j$. With the creation and annihilation operators of spinless fermions, such sum can be rewritten as $f^{{\dagger}}_{j}f_{j+1}$ AQ . And the spin part, due to the unchanged sequence of spins, is equivalent to a unit operator of spin configurations, which we have denoted by $I$.

While for $\hat{C}^{{\dagger}}_{L\sigma}\hat{C}_{1\sigma}$ or $\hat{C}^{{\dagger}}_{1\sigma}\hat{C}_{L\sigma}$, since when acting on a basis, they will give a cyclic permutation of the spin configuration to the left or right respectively, we have

[TABLE]

and

[TABLE]

where $\widetilde{|\bm{\sigma}\rangle}$ represents a spin configuration obtained via a left cyclic permutation of $|\bm{\sigma}\rangle$, that is, if we take $|\bm{\sigma}\rangle=|\sigma_{1}\sigma_{2}\cdots\sigma_{N}\rangle$, then $\widetilde{|\bm{\sigma}\rangle}=|\sigma_{2}\cdots\sigma_{N}\sigma_{1}\rangle$. And the sum over all these configurations can be replaced by the left or right cyclic permutation operator, which we have denoted by $P$ and $P^{-1}$ respectively.

Then the final form of the Hamiltonian after the unitary transformation (5) is

[TABLE]

where $\theta(t)=eA(t)/(\hbar c)=-eEt/\hbar$. We will find that our discussion can be substantially reduced with such new form of Hamiltonian.

And noting that $T$ is independent of $t$ or $A(t)$, the charge current operator under this transformation becomes

[TABLE]

III Solutions

In Eq. (6), the spin-part operators $P$, $P^{-1}$ and $I$ do not cause any trouble for our discussion, since they commute with each other and the eigenstate can be obtained easily. For any spin configuration $|\bm{\sigma}_{s_{1}}\rangle=|\sigma_{1}\sigma_{2}\cdots\sigma_{N}\rangle$, we introduce $|\bm{\sigma}_{s_{2}}\rangle=P|\bm{\sigma}_{s_{1}}\rangle$, $|\bm{\sigma}_{s_{3}}\rangle=P|\bm{\sigma}_{s_{2}}\rangle=P^{2}|\bm{\sigma}_{s_{1}}\rangle$, $\cdots$, till some integer $m_{s}\leq N$, for which the resulted state repeats $|\bm{\sigma}_{s_{1}}\rangle$ for the first time, namely, $|\bm{\sigma}_{s_{m_{s}+1}}\rangle=P^{m_{s}}|\bm{\sigma}_{s_{1}}\rangle=|\bm{\sigma}_{s_{1}}\rangle$. Obviously, $m_{s}$ is directly related to the detailed form of $|\bm{\sigma}_{s_{1}}\rangle$. These configurations form a subset as

[TABLE]

from which $m_{s}$ eigenstates of $P$ or $P^{-1}$ can be obtained as follows,

[TABLE]

where $k_{s}=2\pi m/m_{s}$, and $m=0,1,2,\cdots,m_{s}-1$. We have $P|\chi_{k_{s}}\rangle=\exp(ik_{s})|\chi_{k_{s}}\rangle$, $P^{-1}|\chi_{k_{s}}\rangle=\exp(-ik_{s})|\chi_{k_{s}}\rangle$. Furthermore, we obtain the representation $P=\sum_{s}\sum_{k_{s}}\exp(ik_{s})|\chi_{k_{s}}\rangle\langle\chi_{k_{s}}|$, $P^{-1}=\sum_{s}\sum_{k_{s}}\exp(-ik_{s})|\chi_{k_{s}}\rangle\langle\chi_{k_{s}}|$ and $I=\sum_{s}\sum_{k_{s}}|\chi_{k_{s}}\rangle\langle\chi_{k_{s}}|$, where “$\sum_{s}$” is a sum over all the possible subsets. Then, the Hamiltonian (6) is reduced to

[TABLE]

which is already in a diagonal form now as far as the spin part is concerned. What we need to do next is just to diagonalize the spinless-fermion part inside “[ ]”, which we can denote by $h_{s}$, associating with each $|\chi_{k_{s}}\rangle$. Introducing $h_{0}=-t_{0}e^{i\theta(t)}(\sum_{j=1}^{L-1}f^{{\dagger}}_{j}f_{j+1}+f^{{\dagger}}_{L}f_{1}e^{ik_{s}})$ to write $h_{s}=h_{0}+h_{0}^{\dagger}$, and noting $[h_{0},h_{0}^{\dagger}]=0$, we find that both $h_{0}$ and $h_{0}^{\dagger}$, and hence $h_{s}$, can be written in a diagonalized form simultaneously. To do this, we assume a form of instantaneous eigenstates for $h_{0}$ as $|\varphi\rangle=\sum_{j=1}^{L}a_{j}f^{{\dagger}}_{j}|0\rangle$. Letting $h_{0}|\varphi\rangle=a|\varphi\rangle$ yields

[TABLE]

and

[TABLE]

From Eq. (11),

[TABLE]

combining which with Eq. (12) yields $a=a_{q}\equiv-t_{0}e^{i\theta(t)}\exp(iq)$, where

[TABLE]

For each $q$, repeatedly using (11) to determine the coefficient $a_{j}$, we finally obtain an eigenstate which can be normalized as $|\varphi_{q}\rangle=\frac{1}{\sqrt{L}}\sum_{j=1}^{L}\exp(iqr_{j})|j\rangle=f^{{\dagger}}_{q}|0\rangle$, where

[TABLE]

It can be verified that $h_{0}^{\dagger}|\varphi_{q}\rangle=a_{q}^{\ast}|\varphi_{q}\rangle$, and then $h|\varphi_{q}\rangle=(a_{q}+a_{q}^{\ast})|\varphi_{q}\rangle=\varepsilon_{q}(t)|\varphi_{q}\rangle$, with

[TABLE]

Further, we can write $h_{s}$ as $h_{s}=\sum_{q}\epsilon_{q}(t)|\varphi_{q}\rangle\langle\varphi_{q}|$ or in an operator form

[TABLE]

And using Eq. (14), it can be verified that $f^{{\dagger}}_{q}$ or $f_{q}$ is still a Fermi operator. The eigenstate of $h$ for $N$ spinless fermions at each time $t$, can be written as

[TABLE]

where we have used an $N$-dimensional vector $\mathbf{q}=(q_{1},q_{2},\cdots,q_{N})$ to simplify the formulation. Due to the anticommuting property of $f^{{\dagger}}_{q}$’s, $q_{1}\neq q_{2}\neq\cdots\neq q_{N}$. Then the $N$-electron eigenstate of $\widetilde{H}$ at time $t$ can be written as

[TABLE]

with a instantaneous eigenvalue

[TABLE]

The most important thing is that, though $\widetilde{H}$ and the instantaneous eigenvalue $E_{\mathbf{q}}(t)$ both depend on time, the eigenstate $|\Psi_{\mathbf{q}s}\rangle$ is independent of time. This permits us an exact study of the time evolving behavior of the system. Since any initial state of the system can be represented by the instantaneous eigenstates at the initial time, we only focus on the case in which the initial state is an instantaneous eigenstate of the system. Namely, we assume that at $t=0$, the state of the system, $|\Psi(0)\rangle=|\Psi_{\mathbf{q}s}\rangle$. Since $|\Psi_{\mathbf{q}s}\rangle$ is still an instantaneous eigenstate of $\widetilde{H}$ at any time latter, the evolving state $|\Psi(t)\rangle$ differs from $|\Psi(0)\rangle$ only by a time-dependent factor and can be determined via

[TABLE]

following from which,

[TABLE]

With these time evolution states, we can make a detailed discussion about the time dependent properties of the system.

Obviously, we can also apply our solving strategy to other similar models in the 1D case, such as the strong-coupling $t$-$V$ model tv and SU($N$) model of impenetrable fermions SU , to study the time evolution of states in electric field.

IV Results and discussions

The charge current, which we are most interested in, is

[TABLE]

where we have used the fact that $\frac{\partial}{\partial A}\langle\Psi(t)|\Psi(t)\rangle=0$.

Using Eq. (21), we can in principle calculate the charge current at any time $t$ for any given initial instantaneous eigenstate of the system. Noting Eqs. (15) and (19), we have

[TABLE]

where

[TABLE]

For the half-filling case, $N=L$, one can find that both $B_{1}$ and $B_{2}$ are 0, due to the complete cancellation in the sum, and hence $j(t)=0$.

However, away from half-filling, nonzero $B_{1}$ and $B_{2}$ will lead to a nonzero $j(t)$, which oscillates with a phase frequency $\omega=eE/\hbar$ and amplitude $\sqrt{B_{1}^{2}+B_{2}^{2}}$, the so-called Bloch oscillation. The Bloch oscillation now, similar to that in the noninteracting case, is an undamped one, in spit of the infinity on-site interaction $U$.

The surviving Bloch oscillations under strongly correlated interactions is interesting. As we have mentioned, there have already existed some discussions about the issue that electrons in strong correlated systems could show dissipationless transport behaviors in the linear response level In1 ; In2 ; In3 ; In4 . Our result is simple but obviously beyond the linear level, and can be viewed as a lively example when discussing the conductivity of Hubbard model.

An important question is that whether such result is a special one only in the 1D $U\rightarrow+\infty$ case. It is interesting to find more examples.

IV.1 Example for Bloch oscillations in finite $U$ cases

We want to show that undamped Bloch oscillations can still exist in finite $U$ cases. An exact discussion of this question in general sense seems impossible. However, we find that there do exist undamped Bloch oscillations for a finite $U$. A simple example is given via an analysis of the current response of the special eigenstates constructed by Yang and Zhang eta for the Hubbard model. The operator used, $\zeta^{{\dagger}}=\Sigma_{m=1}^{L}C^{{\dagger}}_{m\uparrow}C_{m\downarrow}$, which is found to be an eigen-operator of the Hamiltonian (1), for any magnitude of $U$, satisfies

[TABLE]

The operator $\zeta^{{\dagger}}$, when acting on a state, only transforms one of the spin $\downarrow$ electrons in the system to a spin $\uparrow$ one. It should be noted that if all the $N$ electrons in the system are spin $\downarrow$ ones, the on-site $U$ will play no role. And then, the time-evolution states are just the same as that in the noninteracting case, and would show the same Bloch oscillations. We can introduce a state of this kind and denote it by $|\Psi_{0}(t)\rangle$. Then, due to Eq. (23), the state $|\Psi_{m}(t)\rangle=(\zeta^{{\dagger}})^{m}|\Psi_{0}(t)\rangle$ ($m\leq N$) is also a time-evolution state of Hamiltonian (1), but belongs to a system with $m$ spin $\uparrow$ electrons and $N-m$ spin $\downarrow$ electrons. Since acting of $\zeta^{{\dagger}}$’s on $|\Psi_{0}(t)\rangle$ only changes the spin of electrons from $\downarrow$ to $\uparrow$, the characteristic of Bloch oscillations keeps. That is, time-evolution states of this kind do show undamped Bloch oscillations as that in the noninteracting case, completely disregarding the on-site $U$ in the Hubbard model. Obviously, such result is also applicable to the two- or three-dimensional (2D or 3D) case.

It should be noted that the number of eigenstates generated with the $\zeta^{{\dagger}}$ operator is directly related to the particle number $N$ and the total site number $L$ in the system, and is in a scale of $C_{L}^{N}$, which generally is very large. Hence, the existence of undamped Bloch oscillations is extensive for Hubbard model and one should be cautious of their roles when discussing the conductance of the system.

IV.2 More general cases

At first glance, the undamped Bloch oscillations we obtained are directly related to the integrability of the model. It seems that such result is hard to be extended to more general cases, since it has been proposed that breaking of integrability would lead to damping of the charge current cc . However, we want to show that the role the temporal periodicity of the model Hamiltonian playing in electric field is more important.

In an constant electric field $E$, when Peierls substitution PS is adopted, the Hamiltonian and charge current operator for Hubbard model or other similar ones, such as $t-V$ and $t-V-W$ models cc , are always periodic in time,

[TABLE]

where the period $\tau$ is directly related to the electric field, e.g., for 1D Hubbard model shown in Eq. (1), $\tau=2\pi\hbar/(eE)$ . And $A(t)=-cEt$ as usual.

The evolution of states determined by such periodic Hamiltonian has been extensively studied (see, e.g., Refs. WK1 ; WK2 ; WK3 ). Due to such periodicity, the evolution operator $U(t,t_{i})$ satisfies $U(t+\tau,t_{i}+\tau)=U(t,t_{i})$, and we have

[TABLE]

where $t^{-}\equiv t-n\tau\in[0,\tau]$.

Further, we can introduce the eigenstates and eigenvalues of $U(\tau,0)$. Since $U(\tau,0)$ is a unitary operator, $U(\tau,0)^{\dagger}U(\tau,0)=\mathbf{1}$, the modulus of its eigenvalues must be 1. For each eigenstate $|m\rangle$, the corresponding eigenvalue must be in the form of $\exp(i\phi_{m})$. Namely, we have

[TABLE]

Obviously, these eigenstates as a whole form a complete set of states for the system.

The time evolution of each $|m\rangle$ is given by

[TABLE]

Then we have

[TABLE]

The instantaneous energy and charge current for these states satisfy

[TABLE]

namely, both are periodic in time. Then the charge current must be a constant or keep oscillating with a period $\tau$. However, noting that

[TABLE]

we find that, to guarantee the periodicity of $E_{m}(t)$, the current $j_{m}(t)$ can not be or approach a nonzero constant.

Additionally, the average of $j_{m}(t)$ over a period gives

[TABLE]

namely, the oscillating of charge current must be an up-and-down one about zero.

Hence, for such complete set of states, the charge current, if nonzero, can only show an undamped oscillating behavior, which is very similar to Bloch oscillations in the noninteracting case.

Such conclusion is a general one, and has nothing to do with the integrability or non-integrability of the model.

As for the 1D $U\rightarrow+\infty$ Hubbard model, one can easily verify that the states shown in Eqs. (20) and (24) are the same thing.

In the appendix, we consider a 2D model of electrons with $\delta$-function interactions in electric field, which can be viewed as a version of Hubbard model in a continuous space. Such model is obviously non-integrable in the common sense. However, we find that the charge current can be exactly determined as $j(t)=j(0)+\frac{e^{2}}{m}ENt$, in an undamped form disregarding the non-integrability of the model. Such undamped behavior of charge current, exceeds our expectation once again, but can be viewed as a counterpart to the undamped Bloch oscillations of the Hubbard model, since both survive with the correlation interactions between electrons.

V Conclusion

In conclusion, when an electric field appearing, the 1D Hubbard model in the $U\rightarrow+\infty$ limit still can be exactly solved. And undamped Bloch oscillations are found to extensively exist not only in this case, but also in finite $U$ or higher-dimension cases, and hence can not be ignored on discussing the conductivity of Hubbard model.

The origin of such undamped-oscillating behavior of charge currents is mainly due to the temporal periodicity of the Hamiltonian of a model in electric field, rather than to the integrability of the model itself.

The undamped Bloch oscillations in Hubbard model, and the dissipationless charge current in the $\delta$-function-interaction model as we shown in the appendix, both can survive with the strongly correlated interactions between electrons, calling for our more caution about the effect of correlation interactions and the integrability of a model on the conductivity of the system.

Additionally, the unitary-transformation method we adopted can also be used to study the time-evolution behavior of states in an electric field for other similar 1D models such as the strong-coupling $t$-$V$ model and SU($N$) model of impenetrable fermions.

Acknowledgements.

The author acknowledges financial support from Guizhou Provincial Education Department (Grant No. QJHKY[2016]314) and Guizhou Provincial Science and Technology Department (Grant No. QKHLH[2014]7433).

Appendix A Exact conductivity for a non-integrable model

To seek more about the effect of integrability and correlation interaction of electrons on the charge current of the system, we now discuss with a $\delta$-function interaction model, which is not the Hubbard model but can be viewed as a version of Hubbard model in a continuous space. And we find that the behavior of charge currents can be exactly discussed.

For non-integrability consideration, We study the 2D case, namely, a system of electrons with $\delta$-function interactions in a torus; and one will find that the extension to other dimensions is straightforward. The Hamiltonian is formally written as

[TABLE]

where $k=(k_{x},k_{y})$ and $P_{k}=\left[\hbar k_{x}+eA(t)/c,\hbar k_{y}\right]$, i.e., the electric field is along the $x$ direction, $E=-1/cA_{t}(t)$. The involved charge current operator is

[TABLE]

Then the charge current for an evolving state $|\Psi(t)\rangle$ is

[TABLE]

and

[TABLE]

where we have used the fact that $[\widehat{j}_{x},H]=0$, and $N$ is the total number of electrons in the system. Then,

[TABLE]

where $j(0)$ is the current at $t=0$. Namely, a dissipationless charge current appears with the electric field. The non-integrability of the model and interactions between electrons do not lead to any damping. Our analysis can be easily extended to other dimensions or cases with more general particle-particle interactions.