[TOC]
参考
- 《Topological Insulators and Topological Superconductors》 - B. Andrei Bernevig
- Chapter-5 Magnetic Field on the Square Lattice
Chapter-5 Magnetic Field on the Square Lattice
Historically, the first system exhibiting topological behavior was the quantum Hall effect. When placed in a magnetic field large enough that Landau-level quantization becomes important, electrons exhibit a quantized Hall effect, in which the Hall conductance is an integer in units of the quantum of conductance, \(e^{2} / h\). This effect can be analyzed through different means; in the continuum limit we obtain Landau levels, with their massive degeneracy of states-and the Hall conductance measures the number of Landau levels that are occupied. However, a better understanding of the problem is obtained by understanding the problem from a lattice perspective, which is the subject of the current chapter. We present a detailed exposition of the problem of a 2-D lattice pierced by a uniform magnetic field and learn about magnetic translation generators, the magnetic translation group, the Hofstadter problem, the Diophantine equation, explicit gauge fixing, and Hall conductance on the square lattice.
5.1 Hamiltonian and Lattice Translations
We first look at spinless electrons and want to analyze the simplest nontrivial, tight-binding Hamiltonian: \[ H=T_{x}+T_{y}+\text { h.c. } \] where
- \(T_{x}, T_{y}\) are the (covariant) translation operators by one lattice constant in the \(x\) - and \(y\) directions,
Peierel's Phase
when a magnetic field is applied, take the covariant form \[ T_{x}=\sum_{m, n} c_{m+1, n}^{\dagger} c_{m, n} e^{i \theta_{m, n}^{x}, n}, \quad T_{y}=\sum_{m, n} c_{m, n+1}^{\dagger} c_{m, n} e^{i \theta_{m, n}^{y}} \] Since the Hamiltonian contains only nearest-neighbor terms, the phase factors can be consistently chosen as the integral of the external vector potential over the bond linking the nearest neighbors (as we learned in the Peierel's substitution of equation) \[ \theta_{m n}^{x}=\frac{e}{\hbar} \int_{m}^{m+1} A \cdot d x , \quad \theta_{m n}^{y}=\frac{e}{\hbar} \int_{n}^{n+1} A \cdot d y . \] We introduce the lattice derivatives: \[ \Delta_{x} f_{m n}=f_{m+1, n}-f_{m, n}, \quad \Delta_{y} f_{m n}=f_{m, n+1}-f_{m, n} . \] The lattice curl of the phase factors is related to the the flux per plaquette \(\phi_{m n}\) : \[ \begin{aligned} \operatorname{rot}_{m n} \theta &=\Delta_{x} \theta_{m n}^{y}-\Delta_{y} \theta_{m n}^{x}\\ &=\theta_{m+1, n}^{y}-\theta_{m, n}^{y}-\theta_{m, n+1}^{x}+\theta_{m, n}^{x} \\ &=\frac{e}{\hbar} \int_{\text {unit cell }} A \cdot d l \\ &=2 \pi \frac{e}{h} \int B d S\\ &=2 \pi \phi_{m n} \end{aligned} \] where
- \(\phi_{m n}\) is the number of flux quanta in units of \(h / e\).
U(1) symmetry
The lattice Hamiltonian has a \(U(1)\) gauge symmetry: \[ c_{i} \rightarrow U_{i} c_{i}, \quad e^{i \theta_{i j}} \rightarrow U_{i} e^{i \theta_{i j}} U_{j}^{-1}, \quad\left|U_{i}\right|=1 \quad \forall j \in(m, n) \] Under the preceding transformation, it is clear that the Hamiltonian remains invariant. The transformation on the hopping phases is just a gauge transformation on \(A\). The spectrum of the current problem is the famous Hofstadter butterfly.
In the weak field limit \(\phi \rightarrow 0\), it gives the Landau-level structure, which we will soon see. Before that, we want to analyze its symmetries.
First, we look at the symmetries of this Hamiltonian. We immediately see that the covariant translation operators do not commute; for example, when we act on a single-particle state at site \((m, n),\left|\psi_{m n}\right\rangle=c_{m, n}^{\dagger}|0\rangle\), we find \[ \begin{array}{l} T_{x} T_{y}\left|\psi_{i j}\right\rangle=T_{x} c_{i, j+1}^{\dagger} e^{i \theta_{i j}^{y}}|0\rangle=e^{i \theta_{i, j+1}^{x}}+i \theta_{i j}^{y} c_{i+1, j+1}^{\dagger}|0\rangle \\ T_{y} T_{x}\left|\psi_{i j}\right\rangle=e^{i \theta_{i j}^{x}+i \theta_{i+1, j}^{y}} c_{i+1, j+1}^{\dagger}|0\rangle, \\ T_{y} T_{x}\left|\psi_{m n}\right\rangle=e^{i 2 \pi \phi_{n n}} T_{\chi} T_{y}\left|\psi_{m n}\right\rangle \end{array} \] Even in an external, constant magnetic field, we see that the Hamiltonian is not translationally invariant with the original translation operators because the two do not commute with each other (and hence do not commute with the Hamiltonian, which is a sum of the two). Even though the magnetic field is translationally invariant, the gauge potential is not. A gauge transformation is required to make the Hamiltonian translationally invariant, but we will see that we cannot maintain the translational symmetry of the original lattice.
magnetic translation operators
We can find operators that commute with the Hamiltonian, \(\hat{T}_{x}, \hat{T}_{y}-\) they are called the magnetic translation operators and are defined by \[ \hat{T}_{x}=\sum_{m, n} c_{m+1, n}^{\dagger} c_{m, n} e^{i x_{n, n}^{x},}, \quad \hat{T}_{y}=\sum_{m, n} c_{m, n+1}^{\dagger} c_{m n} e^{i \chi_{m, n}^{y}} \] We could have guessed the form of these operators on physical grounds: they have to be onebody operators because we are solving a one-body Hamiltonian. We also knew that they have to have phases that are different from the original ones, because those do not commute. We find the phases \(\chi\) by requiring that the operators commute with the Hamiltonian, which is a sum of the translation operators in the \(x\) - and \(y\) - directions.
For \(\hat{T}_{x}\), \[ \left[T_{x}, \hat{T}_{x}\right]=\sum_{n m} c_{m+2, n}^{\dagger} c_{m, n} e^{i\left(x_{n n}^{x}+\theta_{m+1, n}^{x}\right.}\left[e^{i\left(x_{m+1, n}^{x}+\theta_{m, n}^{x}-\chi_{n n}^{x}-\theta_{m+1, n}^{x}\right)}-1\right] \] To obtain zero, we hence require \[ \Delta_{\chi} \chi_{m n}^{\chi}=\Delta_{x} \theta_{m n}^{x} \] We also want another commutator to vanish: \[ \left[\hat{T}_{x}, T_{y}\right]=0 \] which, once worked out, imposes the constraint \[ \Delta_{y} \chi_{m n}^{\chi}=\Delta_{x} \theta_{m n}^{y}=\Delta_{y} \theta_{m n}^{\chi}+2 \pi \phi_{m n} . \] We similarly have \[ \left[\hat{T}_{y}, T_{x}\right]=0, \quad\left[\hat{T}_{y}, T_{y}\right]=0 \] which gives the mirror constraints to the preceding constraints: \[ \begin{array}{l} \Delta_{x} \chi_{m, n}^{y}=\Delta \theta_{m n}^{y} \quad\left(=\Delta_{x} \theta_{m, n}^{y}-2 \pi \phi_{m n}\right), \\ \Delta_{y} \chi_{m, n}^{y}=\Delta_{y} \theta_{m, n}^{y} . \end{array} \] The constraints can be solved, with the solutions \[ \chi_{m n}^{\chi}=\theta_{m n}^{x}+2 \pi n \phi_{m, n}, \quad \chi_{m n}^{y}=\theta_{m n}^{y}-2 \pi m \phi_{n n} . \] We have now found operators that commute with \(H\) but do not commute between themselves: \[ \left[\hat{T}_{x}, \hat{T}_{y}\right] \neq 0 \] To obtain the commutators of the two preceding operators,we use a useful trick: we take the action of the operators on the single particle state at site \((i, j)\). \[ \begin{array}{l} \hat{T}_{x} \hat{T}_{y} c_{i, j}^{\dagger}|0\rangle=\sum_{m n} e^{i\left(x_{n n}^{x}+x_{i j}^{y}\right)} c_{m+1, n}^{\dagger} c_{m, n} c_{i, j+1}^{\dagger}|0\rangle=e^{i\left(x_{i, j+1}^{x} x_{i j}^{x}\right)} c_{i+1, j+1}^{y}|0\rangle, \\ \hat{T}_{y} \hat{T}_{x} c_{i, j}^{\dagger}|0\rangle=\sum_{m n} e^{i\left(x_{n n}^{y}+x_{i j}^{x}\right)} c_{m, n+1}^{\dagger} c_{m, n} c_{i+1, j}^{\dagger}|0\rangle=e^{i\left(x_{i+1, j}^{y}+x_{i j}^{x}\right)} c_{i+1, j+1}^{\dagger}|0\rangle . \end{array} \] They do not commute.
Landau gauge
In general, we can find a combination of the translation operators that does commute. This combination depends on the gauge that we pick. We select the Landau gauge \[ A_{y}=B x=2 \pi \phi m \] where
- \(\phi\) is the uniform flux per plaquette.
Hence, \(\theta_{m n}^{y}=\) \(2 \pi \phi m(n+1-n)=2 \pi \phi m\), and we have \[ \hat{T}_{x} \hat{T}_{y}=e^{i 2 \pi \phi} \hat{T}_{y} \hat{T}_{x} \] By successive application of \(\hat{T}_{x}\) to the preceding commutation, we find that \(\hat{T}_{x}^{q} \hat{T}_{y}=e^{i 2 q \pi \phi} \hat{T}_{y} \hat{T}_{x}^{q}\) and, in general, no power of the magnetic translation operators commutes with the other translation operator. However, for the special case of rational flux per plaquette, \(\phi=p / q\), where \(p, q\) are relatively prime, we then find that \[ \hat{T}_{x}^{q} \hat{T}_{y}=\hat{T}_{y} \hat{T}_{x}^{q} \] We hence have two operators, \(\hat{T}_{x}^{q}, \hat{T}_{y}\), which commute between themselves and commute with the Hamiltonian; hence, they define a new set of good quantum numbers.
We hence have two operators, \(\hat{T}_{x}^{q}, \hat{T}_{y}\), which commute between themselves and commute with the Hamiltonian; hence, they define a new set of good quantum numbers.
Degenerate
In the gauge chosen here, \(\hat{T}_{y}=\sum_{m, n} c_{m, n+1}^{\dagger} c_{m, n}\) is the noncovariantized translation operator, whereas \(\hat{T}_{x}=\sum_{m, n} c_{m+1, n}^{\dagger} c_{m, n} e^{i 2 \pi n \phi}\). The action of these two operators on single-particle states is that of translation by one lattice constant in the \(y\) -direction and by \(q\) lattice constants in the \(x\) -direction: \[ \begin{array}{l} \hat{T}_{y} c_{i, j}|0\rangle=c_{i, j+1}^{\dagger}|0\rangle, \quad \hat{T}_{x}^{n} c_{i, j}^{\dagger}|0\rangle=e^{i 2 \pi j \phi} c_{i+n, j}^{\dagger}|0\rangle \\ \hat{T}_{x}^{q} c_{i, j}^{\dagger}|0\rangle=e^{i 2 \pi q \phi} c_{i+q, j}^{\dagger}|0\rangle=e^{i 2 \pi p} c_{i+q, j}^{\dagger}|0\rangle=c_{i+q, j}^{\dagger}|0\rangle \end{array} \] The new translational unit cell in the \(x\) -direction is called the magnetic unit cell, and it is \(q\) times larger than the usual unit cell. Because \(\hat{T}_{x}^{q}\) and \(\hat{T}_{y}\) commute with the Hamiltonian, it means that the eigenstates have quantum numbers under these operators. Since these operators perform translations, it is then clear that we can index the state by momentum quantum numbers. The Bloch conditions then become \[ H| k \rangle=E( k )| k \rangle \quad \hat{T}_{x}^{q}| k \rangle=e^{i k_{x} q a}| k \rangle \quad \hat{T}_{y}| k \rangle=e^{i k_{y}}| k \rangle, \] where, because the unit cell is \(q\) times larger in the \(x\) -direction, the magnetic \(BZ\) is \(q\) times smaller: \(0 \leq k_{x} \leq 2 \pi / q, \quad 0 \leq k_{y} \leq 2 \pi\). From here we can prove that the Landau-level problem on a lattice has a \(q\) -fold degeneracy at different wavevectors, i.e., the bulk bands have the same energy at \(q\) different wavevectors on the \(y\) -axis. Assume \(\left|k_{x}, k_{y}\right\rangle\) is an eigenstate of the Hamiltonian. Because the Hamiltonian commutes with \(\hat{T}_{x}, \hat{T}_{y}\), we have that \(\hat{T}_{x}\left|k_{x}, k_{y}\right\rangle\) is also an eigenstate of the Hamiltonian. However, because \(\hat{T}_{x, y}\) do not commute between themselves, \(\hat{T}_{x}\left|k_{x}, k_{y}\right\rangle\) cannot be an eigenstate of the Hamiltonian at the same wavevector. Instead, \[ \hat{T}_{y} \hat{T}_{x}\left|k_{x}, k_{y}\right\rangle=e^{-i 2 \pi \phi} \hat{T}_{x} \hat{T}_{y}\left|k_{x}, k_{y}\right\rangle=e^{i\left(k_{y}-2 \pi \phi\right)} \hat{T}_{x}\left|k_{x}, k_{y}\right\rangle . \] The eigenvalue under \(\hat{T}_{y}\) of \(\hat{T}_{x}\left|k_{x}, k_{y}\right\rangle\) is \(k_{y}-2 \pi \phi\), which leads us to conclude that \(\hat{T}_{x}\left|k_{x}, k_{y}\right\rangle \sim\) \(\left|k_{x}, k_{y}-2 \pi \phi\right\rangle .\) The eigenstates \(\left|k_{x}, k_{y}-2 \pi \phi\right\rangle\) and \(\left|k_{x}, k_{y}\right\rangle\) have identical energy. Because \(\phi=\) \(p / q\) are rational and \(p, q\) are relatively prime, we have that the spectrum is \(q\) -fold degenerate, corresponding to the application of \(\hat{T}_{x} q\) times.
5.2 Diagonalization of the Hamiltonian
Diagonalization of the Hamiltonian of a 2-D Lattice in a Magnetic Field
We would like to make use of the newly learned fact that the true unit cell is \(q\) times larger than the initial lattice unit cell to diagonalize the Hamiltonian.
Let us choose the Landau gauge from before \[ A_{y}=2 \pi \Phi m \] The expanded Hamiltonian then is \[ H=\sum_{m, n}-t_{a} c_{m+1, n}^{\dagger} c_{m, n}-t_{b} c_{m, n+1}^{\dagger} c_{m, n} e^{i 2 \pi \Phi m}+\text { h.c. } \] Fourier-transform \[ c_{m, n}=\frac{1}{(2 \pi)^{2}} \int_{-\pi}^{\pi} d k_{x} \int_{-\pi}^{\pi} d k_{y} e^{i k_{x} m+i k_{y} n} c_{c_{k_{x}, k_{y}}} \] where
- for now \(-\pi \leq k_{x}, k_{y} \leq \pi\),
- and require \(c_{k_{x}+2 \pi j, k_{y}+2 \pi l}=c_{k_{x}, k_{y}} .\)
In a sign that we have not used the full translational symmetry of the model, the Fourier-transformed Hamiltonian has - due to the magnetic field-coupling between different \(k\) -sectors and is not diagonal in \(k\) : \[ H=-\int_{-\pi}^{\pi} \frac{d k_{x}}{2 \pi} \int_{-\pi}^{\pi} \frac{d k_{y}}{2 \pi}\left[t_{a} \cos \left(k_{x}\right) c_{k_{x}, k_{y}}^{\dagger} c_{k_{x}, k_{y}}+t_{b} e^{-i k_{y}} c_{k_{x}+2 \pi \Phi, k_{y}}^{\dagger} c_{k_{x}, k_{y}}+\text { h.c. }\right] \] The Fourier transform of the Hamiltonian mixes \(\left(k_{x}, k_{y}\right) \rightarrow\left(k_{x} \pm 2 \pi \Phi, k_{y}\right)\). To diagonalize our Hamiltonian, we need to find a momentum space where there is no mixing.
To diagonalize our Hamiltonian, we need to find a momentum space where there is no mixing.
If \(\Phi=\) \(p / q\), where \(p\) and \(q\) are relatively prime, we see that the Hamiltonian can be broken in several sectors by making the Brillouin zone \(q\) times smaller than the initial one in the \(x\) -direction (corresponding to an enlarged magnetic unit cell made out of \(q\) initial unit cells): \[ H=\frac{1}{(2 \pi)^{2}} \int_{-\pi / q}^{\pi / q} d k_{x}^{0} \int_{-\pi}^{\pi} d k_{y} \hat{H}_{k_{x}^{0}, k_{y}}, \] where (made \(k_{x}=k_{x}^{0}+2 \pi \Phi n\)) \[ \begin{aligned} \hat{H}_{k_{x}^{0}, k_{y}}=& \sum_{n=0}^{q-1}\left\{-2 t_{a} \cos \left(k_{x}+2 \pi \Phi n\right) c_{k_{x}+2 \pi \Phi n, k_{y}}^{\dagger} c_{k_{x}+2 \pi \Phi n, k_{y}}\right.\\ &-t_{b}\left(e^{-i k_{y}} c_{k_{x}+2 \pi \Phi(n+1), k_{y}}^{\dagger} c_{k_{k}+2 \pi \Phi n, k_{y}}+e^{i k_{y}} c_{k_{x}+2 \pi \Phi(n-1), k_{y}}^{\dagger} c_{k_{k}+2 \pi \Phi n, k_{y}}\right) \end{aligned} \] Note that this partition of the \(k\) -space can happen only if \(p\) and \(q\) are relatively prime. Only in this case is the covering \(k_{x}^{0}+2 \pi \Phi n\) able to reproduce the whole initial \(-\pi \leq k_{x} \leq \pi\).
For example, for the case \(p=2, q=3\), we find \[ k_{x}^{0} \in[-\pi / 3, \pi / 3], k_{x}^{0}+2 \pi \Phi \in[\pi, 5 \pi / 3]=[-\pi,-\pi / 3] \] and \[ k_{x}^{0}+4 \pi \Phi \in[7 \pi / 3,9 \pi / 3]=[\pi / 3, \pi] \] thereby showing that \(k_{x}^{0}+2 \pi \Phi n\) covers the entire range \(-\pi\) to \(\pi\).
However, if \(p, q\) are not relatively prime, this is not possible: \(p=2, q=2\), \[ k_{x}^{0} \in[-\pi / 2, \pi / 2], k_{x}^{0}+2 \pi=[-\pi / 2, \pi / 2] \] so you can never sample the sectors \([-\pi, \pi / 2],[\pi / 2, \pi]\)
