Symmetry and classification of Topological Orders


The terminology of symmetries is often invoked, especially in the field of topological classification (of non-interacting system), with conflicting meanings. Here I intend to summarize a self-consistent explanation to some well-known conclusion, e.g. particle-hole symmetry of Hubbard model, Chiral symmetry of SSH model …

main reference:

Chiu, C.-K., Teo, J. C. Y., Schnyder, A. P., & Ryu, S. (2016). Classification of topological quantum matter with symmetries. Reviews of Modern Physics, 88(3), 035005.


Any symmetry can be represented as an operator, which is either linear and unitary, or anti-linear and anti-unitary. Wigner’s theorem

Time-reversal symmetry

antilinear operator $A$ is a map in Hilbert space ${x, x \in \mathcal{H}}$, with following definitions:

\[\newcommand{\expect}[1]{\langle #1 \rangle} \begin{aligned} \text{antilinear: } & A (\alpha x + \beta y)= \bar{\alpha} \cdot Ax + \bar{\beta} \cdot Ay \\ \text{adjoint operator: } & \expect{Ax, y} = \overline{\expect{x, A^* y}} \\ \text{antiunitary: } & \expect{Ax, Ay} = \overline{\expect{x, y}} \Leftrightarrow A A^* = A^*A = I \end{aligned}\]

Time reversal is an antiunitary operator defined by:

\[\newcommand{\ti}{\mathscr{T}} \ti \psi_I \ti^{-1} = (U_T)_I^J \psi_J \label{eq:defTR}\]

for an antiunitary, $\ti^{-1}$ is equivalent with the adjoint operator $\ti^*$

As a symmetry, $\ti$ preserves the commutation relation and leaves $\hat{H}$ invariant:

\[\begin{aligned} & \ti \{\psi_I, \psi^\dagger_J\} \ti^{-1} = \{\psi_I, \psi^\dagger_J\}, \, \dots \\ & \ti \hat{H} \ti^{-1} = \hat{H} \end{aligned}\]

with observations:

  1. $\ti^2, \ti \psi_I \ti^{-1}, \dots$ are all linear operators, $(\ti \psi_I \ti^{-1})^\dagger = \ti \psi_I^\dagger \ti^{-1}$
  2. Matrix $U_T$ is unitary (can be deduced from the preserving of commutator)
  3. under Heisenberg representation, for a system with TRS, $\ti \hat{O}(t) \ti^{-1} = \hat{O}(-t)$

Kramer’s degeneracy comes when $\ti^2 = -1$, proof:

\[\braket{\phi | \ti \phi} = \overline{\expect{\ti\phi, \ti^2\phi}} = - \overline{\expect{\ti \phi, \phi}} = -\braket{\phi | \ti \phi} \Leftrightarrow \braket{\phi | \ti \phi} = 0\]

indicating $\ti\ket{\phi}$ and $\ket{\phi}$ are two distinct, degenerate state.

how this relates to topology :

when $\hat{H} = \psi^\dagger H \psi$ is single-particle Hamiltonian inside a symmetry class (defined such that matrix $H$ runs over an irreducible representation space), it can be shown that:

  • $(U_T^* U_T)^\dagger H (U_T^* U_T) = H$
  • Schur’s lemma gives $U_T^*U_T = \pm I \Rightarrow \ti^2 \psi_I \ti^{-2} = \pm \psi_I$
  • for any $N$-particle state, $\ti^2 = (\pm 1)^{\hat{N}}$

a common scenario with $\ti^2 = -1$ :

one common example is the spin-$\frac{1}{2}$ single particle case, where the Time reversal is chosen as (ordinary textbook notation)

\[i K \sigma_y = K \pmatrix{0 & 1 \\ -1 & 0}\]

which is a rather confusing notation. It may be understood by saying that this $\ti$ does not affect spatial coordinate, only reverse each individual on-site spin by setting

\[U_T = \pmatrix{0 & 1 \\ -1 & 0}\]

in the definition of TR, i.e.

\[\ti \psi_{\vec{r}, s} \ti^{-1} = (U_T)_{s s'} \psi{_{\vec{r}, s'}}\]

It is the only choice available if one requires $\ti^{2} = -1$ in a 2-dimensional Hilbert space.

particle-hole symmetry

particle-hole, or charge conjugation is a unitary transformation given by

\[\newcommand{\cg}{\mathscr{C}} \begin{aligned} & \cg \psi_I \cg^{-1} = (U_C^*)^J_I \psi_J^\dagger \\ & \cg \psi^\dagger_I \cg^{-1} = \psi_J (U_C)^J_I \\ \end{aligned}\]

with observations:

  • $\hat{N} = \sum \psi^\dagger_I \psi_I$, $\cg \hat{N} \cg^{-1} = N - \hat{N}$, with $N$ the total number of sites
  • it is natural to define total charge as $Q \equiv \hat{N} - N/2$, $\cg Q \cg^{-1} = -Q$

e.g. Hubbard model at half-filling

if a many body Hamiltonian, combining with the chemical potential term, is invariant under particle-hole transformation:

\[\cg (\hat{H} - \mu \hat{N}) \cg^{-1} = (\hat{H} - \mu \hat{N})\]

note that this can only be true for specific $\mu$’s.

expectation of particle number:

\[\newcommand{\Tr}{\text{Tr}} \begin{aligned} \expect{\hat{N}} &= \frac{\Tr(\hat{N} e^{-\beta (\hat{H} - \mu \hat{N})})}{\Tr(e^{-\beta (\hat{H} - \mu \hat{N})})} \\ &= \frac{\Tr(\cg \hat{N} \cg^{-1} \cg e^{-\beta (\hat{H} - \mu \hat{N})} \cg^{-1})}{\Tr(e^{-\beta (\hat{H} - \mu \hat{N})})} \\ &= N - \expect{\hat{N}} \\ &\Rightarrow \expect{\hat{N}} = N / 2 \end{aligned}\]

i.e. at this certain chemical potential level, half of the total orbitals are “filled”, regardless of temperature $\beta$. While the exact particle number expectation corresponding to generic $\mu$ may not be able to calculate.

Explicitly write down the PH transformation of Hubbard model:

\[\newcommand{\greenF}[2]{c^\dagger_{#1} c_{#2}} \hat{H} = \sum_{i,j; \sigma} t_{ij} \greenF{i \sigma}{j\sigma} + U \sum n_{i\uparrow} n_{i\downarrow}\]

one may expect a simple form of $\cg$:

\[\cg c_{i\sigma} \cg^{-1} \propto c^\dagger_{i\sigma} = \alpha_{i\sigma} c^\dagger_{i\sigma}\]

where $\alpha$ has absolute value 1.

\[\begin{align} \cg \hat{H} \cg^{-1} &= \sum t_{ij} \alpha^*_{i\sigma} \alpha_{j\sigma} (-c^\dagger_{j\sigma} c_{i\sigma}) + U \sum (1 - n_{i\uparrow}) (1-n_{i\downarrow}) \nonumber \\ &= \hat{H} + \frac{N}{2}U - U \hat{N} \\ \cg (- \mu \hat{N} ) \cg ^{-1} &= -\mu N + \mu \hat{N} \end{align}\]

PH is then satisfied when $\mu = U/2$, plus one additional constraint that

\[t_{ji} = t^*_{ij} = - t_{ij} \alpha^*_{i\sigma} \alpha_{j\sigma}, \quad \forall \; \text{neighboring} \; ij\]

under the special case of real $t$, bipartite lattice, $\alpha$ can be chosen as $\pm 1$ on each sub lattice to meet the requirements.

Chiral symmetry

CS often refers to the case when both TRS and PHS are broken, while their combination is satisfied.

\[\newcommand{\cl}{\mathscr{I}} \begin{aligned} & \cl = \ti \cdot \cg \\ \end{aligned}\]

A well-known example for chiral symmetry is the SSH model (on a bipartite lattice, orbits in each unit cell are labeled by $a, b$)

\[\hat{H} = \sum (t a^\dagger_jb_j + t' b^\dagger_j a_{j+1} + H.C.)\]


using the translational symmetry:

\[\hat{H} = \sum_k \pmatrix{a^\dagger_k & b^\dagger_k} \pmatrix{0 & t + t'^* e^{ik}\\ t^* + t' e^{-ik} & 0} \pmatrix{a_k \\ b_k} \label{eq:diagink}\]

the property $h_k = \vec{h} \cdot \vec{\sigma}$, while $h_z = 0$ is often claimed to be Chiral symmetry.

Here for spinless fermion, TR:

\[\ti \psi_I \ti^{-1} = \psi_I\]

PH transformation is exactly the same as stated in above section:

\[\begin{aligned} \cg a_j \cg^{-1} &= + a^\dagger_j \\ \cg b_j \cg^{-1} &= - b^\dagger_j \\ \end{aligned}\]

In the general case of complex $t, t’$, TRS and PHS are all broken, but their combination is satisfied.

Chiral symmetry then forbids the exists of terms like $\sum_{k} u(k) a^\dagger_k a_k$.

A more general argument may be given, for non-interacting system

\[\cl \psi_I \cl^{-1} = (U^*_S)_I^J \psi^\dagger_J \Rightarrow U_S^\dagger H U_S = - H\]

(without loss of generality, assuming $\Tr(H) = 0$ by adjusting the zero of energy)

Chiral symmetry then gives rise to a symmetric spectrum:

\[H \ket{\phi} = E \ket{\phi} \Rightarrow H (U_S \ket{\phi}) = - E (U_S \ket{\phi})\]

In a basis where $U_S$ is diagonal, i.e., let $U^2_S = 1$ from the schur’s lemma:

\[\text{basis} = \{\ket{\phi} \pm U_S \ket{\phi}\}\]

corresponding to $\pm 1$ sector of operator $U_S$, Hamiltonian is thus block-off-diagonal

\[H = \pmatrix{0 & D \\ D^\dagger & 0}\]

to be continued …

This work is licensed under a Attribution-NonCommercial 4.0 International license. Attribution-NonCommercial 4.0 International

ᐕ) 赞 0