# Metadata **Title:** Topological insulators with commensurate antiferromagnetism **Authors:** Fang, Chen; Gilbert, Matthew J.; Bernevig, B. Andrei **Year:** 2013 **Tags:** #AFTI #TI #Z2 [URL](http://dx.doi.org/10.1103/physrevb.88.085406)|[Bookends](bookends://sonnysoftware.com/85420) --- $\require{braket}$ # Introduction In this work, the authors study the topological features of noninteracting antiferromagnetic (AFM) insulators, where the period of the antiferromagnetism is commensurate with the lattice period. These insulators are classified by the presence or absnece of an emergent antiunitary symmetry $\hat{\Theta}_S$, which combines time-reversal symmetry (TRS) $\hat{\Theta}$ and translation by a vector $\pmb{D}$, i.e. $\hat{\Theta}_S = \hat{\Theta} T_{\pmb{D}}$. For AFM insulators preserving $\hat{\Theta}_S$, the authors show that there is a new type of [[Kramers Theorem|Kramers' degeneracy]] protected by this antiunitary symmetry, and a new $Z_2$ index may be defined for 3D AFM insulators (but not in lower dimensions). For 3D [[Antiferromagnetic Topological Insulator|AFM TIs]] (i.e. 3D AFM insulators with a nontrivial Z2 index), there is an odd number of gapless surface modes if and only if the surface termination also preserves this antiunitary symmetry. While the theory associated with [[Time-Reversal Symmetric Topological Insulator|time-reversal invariant topological insulators]] is well understood at this point, the presence of [[Time-Reversal Operator|time-reversal symmetry]] in a material is a major constraint. Therefore, it is important to go beyond the well-established paradigm of noninteracting time-reversal-symmetric TI to look for novel topological phases. This is because the established theory for TRS insulators cannot be applied to classify any magnetic materials as their ground states break TRS. In a pioneering [work](https://doi.org/10.1103/physrevb.81.245209) by Mong, Essin and Moore, they showed that a $Z_2$ index can be defined in an antiferromagnet where the unit cell is doubled by the magnetic structure. While both TRS and translation along a primitive basis vector are broken, their combination is preserved. It is clear that ferromagnetism always breaks a [[Kramers Theorem|Kramers' pair]] into energetically separated spin-up and spin-down states. This work shows that AFM preserves the degeneracy of almost all single-particle states, with some exceptions, under a condition. Namely, if there is a lattice translation vector $\pmb{D}$ that inverts the spin at every site. This is because if such a $\pmb{D}$ exists, then we can define the antiunitary operator mentioned earlier $\hat{\Theta}_S = \hat{\Theta} T_{\pmb{D}}$, where $\hat{\Theta}$ is [[Time-Reversal Operator|time reversal symmetry]] and $T$ is the translation operator. Note that $\hat{\Theta}_S$ is a symmetry because *both* of its constituent operators invert the magnetization so their combination recovers it. AFM insulators that are invariant under $\hat{\Theta}_S$ are referred to as $\hat{\Theta}_S$-symmetric insulators. In [[Time-Reversal Symmetric Topological Insulator|TRS-TIs]], there exists the so-called [[Bulk-Boundary Correspondence|bulk-boundary correspondence]] between the bulk $Z_2$ invariant and the existence of gapless boundary modes. When the $Z_2$ number is non-trivial, there exists an odd number of gapless Dirac modes at the boundary of the system. As discussed earlier, there is an extra condition for [[Antiferromagnetic Topological Insulator|AFM TI]]: the boundary must also preserve the antiunitary symmetry $\hat{\Theta}_S$. --- # Kramer's Degeneracy in an AFM Insulator: Proof by Symmetries ## The model We assume that the electrons in an AFM insulator can be modeled by the following Hamiltonian: $ \hat{H} = \hat{H}_0 + \hat{H}_M $ where $ \hat{H}_0 = \sum_{\pmb{k}, \alpha, \beta, \tau, \tau'}(\pmb{k}) c^\dagger_{\alpha \tau}(\pmb{k}) c_{\beta \tau'}(\pmb{k}) $ is the [[Time-Reversal Operator|time-reversal symmetric]] hopping part and $ \hat{H}_M = \sum_{\pmb{r}, \alpha, \tau, \tau'} [\pmb{J}(\pmb{r}) \cdot \sigma_{\tau\tau'}] c^\dagger_{\alpha \tau}(\pmb{r}) c_{\alpha \tau'}(\pmb{r}) $ is the part coupled to a Zeeman field $\pmb{J}(\pmb{r})$. Greek indices here denote orbital degrees of freedom inside the unit cell except $\tau$ which denotes the spin degree of freedom. We additionally assume that $\pmb{a}_{i = 1, \ldots, d}$ are the basis vectors of the $d$-dimensional lattice and $\pmb{b}_{i = 1, \ldots, d}$ are the corresponding reciprocal lattice basis vectors. Note that the AFM field $\pmb{J}(\pmb{r})$ breaks the lattice translation symmetry, but as long as its periods are commensurate with the lattice vectors, it simply enlarges the unit cell of the crystal lattice. This enlarged cell is known as the magnetic unit cell, and its basis vectors are $\pmb{a}^M_{i = 1, \ldots, d}$, which are a linear combination with integer coefficients of the $\pmb{a}_i$. The corresponding magnetic Brillouin zone has reciprocal lattice vectors $\pmb{a}^M_{i = 1, \ldots, d}$ such that $\pmb{a}_i^M \cdot \pmb{b}_i^M = 2 \pi \delta_{ij}$. ## Review: Kramers' Degeneracy in a TRS System The [[Time-Reversal Operator|time-reversal operator]] in the single-fermion sector of the Hilbert space is defined as: $ \hat{\Theta} = (i \sigma_y) K $ where $K$ is the conjugation operator and $\sigma_y$ is the Pauli matrix. Note that $\hat{\Theta}$ is antiunitary and $ \hat{\Theta}^2 = -1 $ These two properties lead to each single-particle state being at least doubly degenerate, i.e. [[Kramers Theorem]]. In translationally invariant systems, we know that $\hat{\Theta}$ sends an eigenstate with momentum $\pmb{k}$ into a state with momentum $-\pmb{k}$. Note that at centers and corners of the Brillouin zone we have $ -\pmb{k}_{inv} = \pmb{k}_{inv} \mod \pmb{G} $ These are known as the [[Time-Reversal Invariant Momenta|time-reversal invariant momenta (TRIM)]]. Note that Kramers' degeneracy at these points is what makes it possible to define $Z_2$ invariants in 2D and 3D TRS insulators. To define this invariant, we define the [[Sewing Matrix|sewing matrix]] $ \mathcal{B}_{mn} = \braket{\psi_m(-\pmb{k})|\hat{\Theta}|\psi_n(\pmb{k})} $ Note that the sewing matrix is antisymmetric at all TRIM and this allows for the definition of [[Pfaffian|Pfaffians]] at the TRIM. This forms the definition of the $Z_2$ invariant given by the [[Fu-Kane Formula|Fu-Kane formula]]: $ (-1)^{\delta_0} = \prod_{\pmb{k_{inv}}} \frac{\text{pf}[\mathcal{B(\pmb{k}_{inv})}]}{\sqrt{\text{det}[\mathcal{B(\pmb{k}_{inv})}]}} $ Note that a smooth gauge is required for this definition, so that $\delta_0$ depends on the band structure in the whole BZ, not just at the TRIM. ## Kramers' degeneracy in AFM insulators Note that *any* AFM breaks [[Time-Reversal Operator|TRS]] because time-reversal reverses all spins and leaves orbital and spatial components invariant: $ \hat{\Theta} \hat{H}_M \hat{\Theta}^{-1} = - \hat{H}_M $ Thus, [[Kramers Theorem|kramers degeneracy]] in its original inception is not defined in AFM insulators. However, this does not mean that the states are non-degenerate. In fact, they may still be doubly degenerate, but they need to be connected by a different symmetry other than time-reversal, protecting them from splitting in energy. For an [[Antiferromagnetic Topological Insulator|AFM TI]] the only symmetry we can consider is translation by a magnetic superlattice vector $\pmb{L}_M$ $ \pmb{L}_M = \sum_{i=1}^3 n_i \pmb{a}^M_i, \quad n_i \in \mathbb{Z} $ $ [\hat{H}, \hat{T}_{\pmb{L}_M}] = 0 $ Thus, we can find a set of common eigenstates of these two operators. By definition of the translation operator, we have the eigenvalues of $\hat{T}_{\pmb{L}^M}$ of the form $\exp(i \pmb{k} \cdot \pmb{L}^M)$, with $k$ constrained to the first magnetic [[Brillouin Zone|BZ]]. This symmetry gives us a band structure defined in the MBZ, but does not protect any degeneracy. ==why?== In many magnetic structure, one can find a special lattice vector $\pmb{D}$ which, when translated by $\pmb{D}$, all the spins are flipped. This can be written as: $ \{\hat{T}_\pmb{D}, \hat{H}_M\} = 0 $ Note that $\pmb{D}$ is only unique up to some $\pmb{L}^M$. Now supposed that we have: $ \{\hat{T}_{\pmb{D}_1}, \hat{H}_M\} = 0 \quad \{\hat{T}_{\pmb{D}_2}, \hat{H}_M\} = 0 $ Then $ \begin{align} [\hat{T}_{\pmb{D_1}-\pmb{D}_2}, \hat{H}_M] &= [\hat{T}_{\pmb{D}_1}\hat{T}_{\pmb{D}_2}^{-1}, \hat{H}_M] \\ &= \hat{T}_{\pmb{D}_1}\hat{T}_{\pmb{D}_2}^{-1}\hat{H}_M - \hat{H}_M \hat{T}_{\pmb{D}_1}\hat{T}_{\pmb{D}_2}^{-1} \\ &= -\hat{T}_{\pmb{D}_1}\hat{H}_M \hat{T}_{\pmb{D}_2}^{-1} + \hat{T}_{\pmb{D}_1}\hat{H}_M\hat{T}_{\pmb{D}_2}^{-1} =0 \end{align} $ Thus, since $[\hat{H}_M, \pmb{L}^M] = 0$, we conclude that $ \pmb{D}_1 - \pmb{D}_2 = \pmb{L}_M $ Now, by translational invariance and time-reversal symmetry of the free part of the Hamiltonian, we get $ [\hat{H}_0, \hat{T}_{\pmb{D}}] = [\hat{H}_0, \hat{\Theta}]. $ We have: $ \begin{align} \hat{\Theta} \hat{H}_M \hat{\Theta}^{-1} &= - \hat{H}_M, \\ \hat{T}_{\pmb{D}} \hat{H}_M &= -\hat{H}_M \hat{T}_{\pmb{D}} \end{align} $ Thus $ \begin{align} \hat{\Theta} \hat{H}_M\hat{T}_{\pmb{D}} &= - \hat{H}_M \hat{\Theta} \hat{T}_{\pmb{D}}, \\ -\hat{\Theta} \hat{T}_{\pmb{D}}\hat{H}_M &= - \hat{H}_M \hat{\Theta} \hat{T}_{\pmb{D}}, \\ [\hat{H}_M, \hat{\Theta} \hat{T}_{\pmb{D}}] &= 0 \end{align} $ Thus, we get a new symmetry of this system $ \hat{\Theta}_S = \hat{\Theta} \hat{T}_{\pmb{D}} $ Note that $\hat{\Theta}_S$ is antiunitary and squares to $-\hat{T}_{2\pmb{D}}$. Now consider the following $ \begin{align} \hat{H}\ket{\pmb{k}, n} &= E_n(\pmb{k}) \ket{\pmb{k}, n} , \\ \hat{T}_{\pmb{L}^M}\ket{\pmb{k}, n} &= \exp(i \pmb{k}\cdot \pmb{L}^M) \ket{\pmb{k}, n} , \end{align} $ where $\pmb{k}$ is a wave vector in the MBZ. Thus, we have: $ \hat{H}\hat{\Theta}_S\ket{\pmb{k}, n} = E_n(\pmb{k}) \hat{\Theta}_S \ket{\pmb{k}, n}, $ where we have used the commutativity of $\theta_S$ and $\hat{H}$ shown previously. Thus, $\ket{\pmb{k}, n}$ and $\hat{\Theta}_S \ket{\pmb{k}, n}$ are eigenstates of $\hat{H}$ with the same Eigenvalue. To check whether these two states are the same or different, we compute their overlap: $ \begin{align} \braket{\pmb{k}, n|(\hat{\Theta}_S|\pmb{k},n}) &= \braket{\pmb{k}, n|(\hat{\Theta}_S^\dagger \hat{\Theta}_S \hat{\Theta}_S|\pmb{k},n}), \\ &=\braket{\pmb{k}, n|(\hat{\Theta}_S^\dagger \hat{\Theta}_S^2|\pmb{k},n}), \\ &=[(\braket{\pmb{k}, n|\hat{\Theta}_S) (\hat{\Theta}_S^2|\pmb{k},n})]^*, \\ &=(\braket{\pmb{k}, n|(\hat{\Theta}_S^{2})^\dagger) (\hat{\Theta}_S|\pmb{k},n}), \\ &=\braket{\hat{\Theta}_S^2 \pmb{k}, n|(\hat{\Theta}_S|\pmb{k},n}), \\ &=-\braket{\pmb{k}, n|\hat{T}_{-2\pmb{D}}\hat{\Theta}_S|\pmb{k},n}, \\ \end{align} $ Note that, since $[\hat{H}, \hat{\Theta}_S] = 0$ then $[\hat{H}, \hat{\Theta}_S^2] = [\hat{H}, \hat{T}_{2\pmb{D}}] = 0 = [\hat{H}, \hat{T}_{\pmb{L}^M}]$. Thus, we can conclude that $2\pmb{D}$ is a magnetic superlattice vector. Thus we get: $ \begin{align} \braket{\pmb{k}, n|(\hat{\Theta}_S|\pmb{k},n}) &= - \exp(-2 i \pmb{k} \cdot \pmb{D}) \braket{\pmb{k}, n|(\hat{\Theta}_S|\pmb{k},n}) \end{align} $ Therefore, unless $2 \pmb{k} \cdot \pmb{D} = (2n+1) \pi$, we have $\braket{\pmb{k}, n|(\hat{\Theta}_S|\pmb{k},n}) = 0$, and $\ket{\pmb{k}, n}$ and $\hat{\Theta}_S \ket{\pmb{k},n}$ are two different states. Let's take a look at how these states transform under translation by a magnetic supercell vector: $ \begin{align} \hat{T}_{\pmb{L}^M}\ket{\pmb{k}, n} &= \exp(i \pmb{k} \cdot \pmb{L}^M) \ket{\pmb{k}, n}, \\ \hat{T}_{\pmb{L}^M}\hat{\Theta}_S\ket{\pmb{k}, n} &= \hat{\Theta}_S\hat{T}_{\pmb{L}^M}\ket{\pmb{k}, n} = \hat{\Theta}_S \exp(i \pmb{k} \cdot \pmb{L}^M) \ket{\pmb{k}, n} = \exp(- i \pmb{k} \cdot \pmb{L}^M) \hat{\Theta}_S \ket{\pmb{k}, n} \end{align} $ Thus, $\hat{\Theta}_S \ket{\pmb{k},n}$ must be proportional to a state with wave vector $-\pmb{k}$., i.e. $ \ket{-\pmb{k},n} = \sum_{m \in \text{occ}} \mathcal{B}^*_{nm}(\pmb{k}) \ket{\pmb{k},m}, $ ==I think the indices are wrong in the paper== where $\mathcal{B}_{mn}$ is the [[Sewing Matrix|sewing matrix]] of $\hat{\Theta}_S$ and generally $m\neq n$ due to degeneracy. Notice that, other than at TRIM (i.e. for generic $\pmb{k}$), $\hat{\Theta}_S\ket{\pmb{k}, n}$ and $\ket{\pmb{k}, n}$ have different superlattice translation eigenvalues and are thus orthogonal. I.e., these are two different states. We have also already shown that these two states are degenerate since they have the same energy as shown above. Moreover, since $\hat{\Theta}_S\ket{\pmb{k}, n}$ is proportional to another state at $-\pmb{k}$, we conclude the following: Recall that, at [[Time-Reversal Invariant Momenta|TRIM]], when $2 \pmb{k}_{inv} \cdot \pmb{D} = (2n+1) \pi$, i.e. $\exp(- 2 i \pmb{k}_{inv} \cdot \pmb{D}) = -1$, $\ket{\pmb{k}_{inv}, n}$ and $\hat{\Theta}_S\ket{\pmb{k}_{inv}, n}$ are *not* two different states. Thus, at these TRIM, the bands are generically (i.e., if no other symmetries exist) non-degenerate. These TRIM will henceforth be called the *A-TRIM*. On the other hand, at [[Time-Reversal Invariant Momenta|TRIM]] where $2 \pmb{k}_{inv} \cdot \pmb{D} = 2n \pi$, i.e. $\exp(- 2 i \pmb{k}_{inv} \cdot \pmb{D}) = +1$, $\ket{\pmb{k}_{inv}, n}$ and $\hat{\Theta}_S\ket{\pmb{k}_{inv}, n}$ are two different states. At these TRIM, the bands are doubly degenerate. These TRIM will henceforth be called the *B-TRIM*. > ### [[Kramers Theorem|Kramers' Degeneracy]] in [[Antiferromagnetic Topological Insulator|AFTI]] > At all non-TRIM in AFTI, the eigenstate at $\pmb{k}$ must be degenerate with another eigenstate at $-\pmb{k}$. At the *B-TRIM*, the bands are doubly degenerate. At the *A-TRIM*, the bands are generically degenerate. Since $2 \pmb{D}$ is a magnetic superlattice vector, it can be written as: $ \begin{align} 2 \pmb{D} = \sum_i f_i \pmb{a}_i^M \implies \pmb{D} = \sum_i f_i/2 \pmb{a}_i^M = \sum_{i=1,\ldots,d} x_i \pmb{a}_i^M/2 + \pmb{L}^M \end{align} $ where the $f_i$ are integers and the $x_i$ are either 0 or 1 with at least one of them being $1$. Meanwhile, the TRIM can be written as: $ \pmb{k}_{inv} = \sum_{i=1,\ldots,d} y_i \pmb{b}_i^M/2 $ where $y_i$ are either 0 or 1. Using the fact that $\pmb{a}_i^M \cdot \pmb{b}_j^M = 2 \pi \delta_{ij}$. We have $ \begin{align} -2 \pmb{k}_{inv} \cdot \pmb{D} &= -2\sum_{i = 1,\ldots,d} \sum_{j = 1,\ldots,d} x_i y_j \pmb{a}_i^M \cdot \pmb{b}_j^M/4 -\sum_{i = 1,\ldots,d} y_i \pmb{L}^M \cdot \pmb{b_i}^M, \\ &= -\pi \sum_{i,\ldots,d} x_i y_i - 2 \pi \sum_{i,\ldots,d} y_i f_i = (2 n + 1 )\pi, \quad \text{(At A-TRIM)}, \\ &= \sum_{i,\ldots,d} x_i y_i + 2\sum_{i,\ldots,d} y_i f_i = (2 n + 1), \quad \text{(At A-TRIM)}, \\ \end{align} $ Noting that the second term is an even integer (since $f_i$ are integers), we get the following result: $ \sum_{i=1,\ldots,d} x_i y_i \in \text{odd} $ > ### [[Time-Reversal Invariant Momenta|TRIM]] Classification in [[Antiferromagnetic Topological Insulator|AFTI]] > The TRIM are classified as follows: > > | TRIM | $2 \pmb{k}\cdot D$ | $\exp(-2 i \pmb{k}\cdot D)$ | $\sum_{i=1,\ldots,d} x_i y_i$ | > |------|-|-|-| > | A | $(2n + 1) \pi $| $- 1$ | odd | > | B | $2n\pi $| $- 1$ | even |