Consider a non-interacting superconducting Hamiltonian in an arbitrary dimension. This is most conveniently expressed in terms of Majorana modes, which are defined as $$\gamma_{2n-1} = c_n + c_n^\dagger \quad \textrm{ and } \quad\gamma_{2n} = -i \left( c_n - i c_n^\dagger \right)$$ for every complex fermionic mode $c_n$. For convenience we label $n = 1, 2, \cdots N$ where $N$ is our number of sites, but I am not presuming a one-dimensional structure (i.e. I do not enforce a notion of locality with respect to this labeling). A generic Hamiltonian is then written as $$ H = i \sum_{n,m} \gamma_n A_{nm} \gamma_m $$ where $A \in \mathbb R^{2N}\times \mathbb R^{2N}$ is anti-symmetric: $A^T = - A$.

Requiring a gap above the ground state is equivalent to demanding that $\det A\neq 0$. (Indeed: the positive eigenvalues of $A$ tell us the single-particle excitation energies.) But in that case the sign of the Pfaffian of $A$ is a well-defined quantity (i.e. $\frac{\textrm{pf} A}{\sqrt{\det A}}$).

This seemingly gives a topological $\mathbb Z_2$ invariant, independent of dimensionality! But this must be wrong, since the classification of non-interacting topological insulators/superconductors tells us this class of Hamiltonians (`class D') only has a $\mathbb Z_2$ invariant when the dimension of space is $d = 1 \mod 8$.


1 Answer 1


[Disclaimer: I find the conclusion a bit surprising, so perhaps it is provisional, but I do not see any other way out.]

As Kitaev (cf. equation (19) of his seminal paper) pointed out, the above $\mathbb Z_2$ invariant in fact simply equals the fermionic parity (= parity of the number of fermions) of the ground state, i.e. $P|\psi\rangle = \pm |\psi\rangle$. This is consistent with what we know of the Kitaev/Majorana chain: with closed boundary conditions, the fermionic parity of the ground state is $-1$.

This makes it clear that the above $\mathbb Z_2$ invariant is indeed well-defined in any dimension! The only way out must be: in e.g. dimensions $d=3,4,5,7$ (for which the `D class' only has a trivial phase according to the classification table, cf Table 3 of the paper by Ryu et al.) it must be impossible to write down a Hamiltonian for which the ground state has an odd number of fermions. (There would have to be some conditions on this to make it precise: e.g. 'for an even number of sites', 'with a well-defined thermodynamic limit', ...)

In summary: the $\mathbb Z_2$ invariant is indeed well-defined, but one cannot always find an example which realizes a non-trival value.

  • $\begingroup$ Consider the much easier example of the complex classes of the table: one can always define a winding number, but it is non-zero only if the BZ is of odd dimension. $\endgroup$
    – PPR
    Commented Aug 2, 2017 at 16:51
  • $\begingroup$ For sure you can find systems of odd parity in any dimension: simply stack Kitaev chains on top of each other in a suitable way. This could hint at a resolution: Not all invariants manifest themselves in the same way on the boundary. In particular the groups in the periodic table are strong invariants, associated to top dim cycles in the BZ (in non-interacting, translation invariant physics). There are many more weak invariants associated to lower dim ones. $\endgroup$ Commented Jul 25, 2019 at 16:17

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