# Bogoliubov-Fermi surface with inversion symmetry and electron-electron
interactions: relativistic analogies and lattice theory

Igor F. Herbut and Julia M. Link  Department of Physics, Simon Fraser
University, Burnaby, British Columbia, Canada V5A 1S6

###### Abstract

We show that the general low-energy Bogoliubov-de Genness Hamiltonian in a
multiband superconductor with broken time reversal and preserved inversion
symmetry is a generator of a real four-dimensional representation of $SO(4)$.
In the particular representation such an effective Hamiltonian is a purely
imaginary matrix, and it is proportional to the antisymmetric tensor of a
fictitious electromagnetic field which one can define in the momentum space.
The quantum time evolution of the low-energy quasiparticle state becomes this
way closely related to the classical relativistic motion of a charged particle
in the presence of the Lorentz force that would be derived from such an
electromagnetic field configuration. The condition for the emergence of a
Bogoliubov-Fermi surface can then be understood as orthogonality of the
fictitious electric and magnetic fields, which would allow zero Lorentz force.
The corresponding zero-energy eigenstates are identified as the physical
timelike and the unphysical spacelike solutions of the Lorentz force equation.
We study the looming instability of the inversion-symmetric Bogoliubov-Fermi
surface in the presence of electron-electron interaction by formulating a
concrete interacting model on the Lieb lattice that features the requisite
$SO(4)$ kinetic energy term together with nearest-neighbor two-body repulsion.
The latter is shown to favor dynamical breaking of the inversion symmetry. The
inversion symmetry in our lattice model indeed becomes spontaneously broken at
zero temperature at infinitesimal repulsion, with the original Bogoliubov-
Fermi surface deformed and reduced in size. General features of this symmetry-
breaking phenomenon are discussed and a comparison with other works in the
literature is presented.

##  I Introduction

The appearance of the gap in the quasiparticle spectrum has been identified as
a key feature of the superconducting state of matter since the early days of
the field and the formulation of the foundational BCS theory of the
superconducting phenomenon. schrieffer  It has also been long known that the
gap may not extend everywhere on the Fermi surface, and that measure-zero
sections of the Fermi surface in the form of gapless points and or gapless
lines are also possible, and in fact common volovik ; sigrist . It came as a
surprise, however, when it was shown recently that in centrosymmetric
multiband superconductors with broken time-reversal symmetry, the outcome
could be none of the above options, but a new, and typically much smaller
surface in the momentum space, named the Bogoliubov-Fermi (BF)
surface.agterberg ; brydon ; yang  In contrast to the previous examples of the
BF surfaces, wilczek ; gubankova  here it is not a portion of the normal Fermi
surface that is being left ungapped, but the BF surface is better thought of
as a gapless point or a line inflated to a surface by the presence of other
bands. Of course, the presence of a BF surface in the quasiparticle spectrum
of a superconducting state in principle leaves distinct a signature on the
crucial low-temperature properties, such as the temperature dependence of the
penetration depth, of the specific heat, and of the thermal conductivity,
which would all reflect a finite density of states left. timm ; setty  Signs
of finite density of states in the superconducting state have been possibly
observed in $\text{U}_{1-x}\text{Th}_{x}\text{Be}_{13}$ stewart ; zieve ,
although the precise nature of the superconducting order there seems not yet
entirely clear.

The presence of inversion symmetry in centrosymmetric superconductors had been
assumed to be crucial for the appearance of the BF surface, as well as for its
protection by the $Z_{2}$ topological invariant, which requires the inversion
symmetry for its definition. agterberg ; bzdusek  However, examples of time-
reversal-broken multiband superconductors without inversion that nevertheless
featured BF surfaces emerged,volovik1 ; schnyder ; sim ; link1  and it has
been subsequently shown that this is a rather generic feature of
noncentrosymmetric superconductors as well. link2  Furthermore, the stability
of the inversion-symmetric BF surface has been questioned oh ; tamura ; as
will be discussed in this paper at length as well, the inversion symmetry
makes the BF surface everywhere doubly degenerate, and this degeneracy can be
removed by a manifest or a spontaneous breaking of inversion. It was shown,
for example,oh  that in presence of favorable effective electron-electron
interactions inversion symmetry at zero temperature becomes spontaneously
broken, and the BF surface then reduced or eliminated. Another example is an
inversion-reducing lattice distortion, which via electron-phonon coupling can
also cause the reduction of the BF surface in the quasiparticle spectrum. tim2
The net effect of these examples of dynamical breaking of inversion symmetry
is either a fully gapped quasiparticle spectrum, or a new non-degenerate BF
surface, of the type that exists in the noncentrosymmetric case. link2

In this paper we first revisit the formation of the BF surface in the
inversion-symmetric case and examine it from the point of view of the
effective low-energy quasiparticle Hamiltonian $H_{ef}$ in the superconductor
brydon ; link2 ; berg ; venderbos , previously derived for the
noncentrosymmetric superconductors in ref. link2 . The effective Hamiltonian
describes the two particle and two hole states that intersect the Fermi level
in the normal phase, intraband-coupled by the presence of the superconducting
order parameter, and then “renormalized” by the interband coupling to other
states that lie farther from the Fermi level. We show that $H_{ef}$ is in
certain preferred basis and at every momentum a four-dimensional imaginary
matrix, and as such it is a generator of real representation of the group of
four-dimensional rotations in Euclidean space, i. e. of the standard $SO(4)$.
The emergence of $SO(4)$ suggests possible analogies to classical relativity,
and indeed the time-dependent Schrödinger equation governed by such an
$H_{ef}$ is related to the covariant form of the classical second Newton law
in the presence of an “electromagnetic” Lorentz force in the momentum space.
rindler  Although the full analogy between the two time evolutions does not,
and as we explain, cannot exist, the BF surface can be understood as an
orthogonality condition between the fictitious momentum-dependent “electric”
and “magnetic” fields, which can be read off as the coefficients of $H_{ef}$
when expanded in terms of the generators of the $SO(4)$ Lie algebra. The
orthogonality condition allows the Lorentz force to vanish on the BF surface
provided that the velocity of the fictitious classical particle with the right
magnitude is orthogonal to both the “electric” and “magnetic” fields, which is
tantamount to finding the eigenstates with zero energy in the original quantum
problem. Interestingly, since the quantum problem has two orthogonal zero
modes at each momentum at the BF surface, whereas the analogous classical
Lorentz equation of motion can have only one physical solution, the second
quantum solution corresponds to the unphysical “spacelike” tachyonic solution
for the velocity four-vector. The latter has no physically acceptable
classical analog, but is nevertheless formally a solution of the Lorentz
equation, and as such it appears in the analogous quantum problem.

The relativistic analogy becomes particularly useful in studying the potential
interaction-induced instability of the inversion-symmetric BF surface. To this
purpose we formulate a single-particle model of spinless fermions hopping on
the Lieb lattice designed to fall into the topological class D bzdusek , i. e.
to anticommute only with an antiunitary operator “$\mathcal{A}$” with a
positive square, and violate time reversal symmetry. The operator
$\mathcal{A}$ can be thought of as representing the combined effects of
inversion and particle-hole transformations, and its anticommutation with
$H_{ef}$ is tied to the inversion symmetry of the full original Bogoliubov-de
Genness (BdG) quasiparticle Hamiltonian. Since the Lieb lattice has a four-
component unit cell our lattice single-particle Hamiltonian is then an $SO(4)$
generator, with a doubly degenerate manifold of zero-energy states, fully
equivalent to a BF surface in the superconducting problem. Having such a real-
space lattice model allows easy addition of two-body interaction terms of
one’s choice: we show that the simplest nearest-neighbor repulsion between the
fermions, for example, favors spontaneous breaking of inversion, that is a
dynamical generation of a single-particle term in the mean-field Hamiltonian
which, in contrast to $H_{ef}$, commutes with the operator $\mathcal{A}$. At
zero-temperature the combined effects of finite density of the zero-energy
states and the matrix structure of the dynamically generated term makes the BF
surface unstable at infinitesimal repulsion. The instability produces a
smaller, deformed, and non-degenerate BF surface.

The paper is organized as follows. In sec. II we discuss the multiband BdG
Hamiltonian as describing Cooper pairing between time-reversed states, for a
general time-reversal operator. The advantage of this representation is that
the existence of a nonunitary operator $\mathcal{A}$ that anticommutes with
the BdG Hamiltonian can be seen to be a universal feature tied to the general
commutativity of spatial symmetries such as inversion and the time reversal. A
critical discussion of the standard construction of the all-important operator
$\mathcal{A}$ is provided in Appendix A, and further support for the above
mentioned commutativity on the example of the standard Dirac Hamiltonian is
given in Appendix B. In sec. III we derive the low-energy effective
Hamiltonian by invoking the Schur complement, tantamount to integration over
bands with finite energy, and discuss its energy eigenvalues and the $SO(4)$
structure. The effective Hamiltonian in the canonical representation of
$SO(3)\times SO(3)\cong SO(4)$ and its relation to the inter- and intraband
pairing, as well as the transformation between the $SO(4)$ representation to
the canonical representation of the effective Hamiltonian, can be found in
Appendix C. The zero-energy eigenstates are computed in sec. IV, and the
analogy with the classical Lorentz force equation is expounded in sec. V. How
the preservation of time-reversal forbids the BF surface in this formulation
is explained in sec. VI. In sec. VII we define a hopping Hamiltonian on the
Lieb lattice that falls into the required topological class D and provides a
realization of a BF surface, and introduce nearest-neighbor repulsive
interactions. The mean-field theory of the BF surface instability is given in
sections VIII and IX. Conclusions and discussion are presented in sec. X.

##  II BdG Hamiltonian with inversion

The quantum-mechanical action for the Bogoliubov quasiparticles in the
superconducting state is given by:

$S=k_{B}T\sum_{\omega_{n},\textbf{p}}\Psi^{\dagger}(\omega_{n},\textbf{p})[-i\omega_{n}+H_{\rm{BdG}}(\textbf{p})]\Psi(\omega_{n},\textbf{p})\>,$
(1)

where the Nambu spinor is here defined as
${\Psi(\omega_{n},\textbf{p})=\big{(}\psi(\omega_{n},\textbf{p}),\mathcal{T}\psi(\omega_{n},\textbf{p})\big{)}^{\rm
T}}$, p is the momentum, $\omega_{n}=(2n+1)\pi k_{B}T$ is the Matsubara
frequency, and $T$ is the temperature. $\psi=(\psi_{1},\cdots,\psi_{N})$ is a
$N$-component Grassmann number describing $N$ eigenstates of the normal state
Hamiltonian H(p), and its time-reversed counterpart is
$\mathcal{T}\psi(\omega_{n},\textbf{p})=U\psi^{*}(-\omega_{n},-\textbf{p})$,
where $\mathcal{T}$ is the antiunitary time-reversal operator, with $U$ as its
unitary part. This way the BdG Hamiltonian becomes:

$\displaystyle H_{\rm BdG}(\textbf{p})$ $\displaystyle=$
$\displaystyle\begin{pmatrix}H(\textbf{p})-\mu&\Gamma(\textbf{p})\\\
\Gamma^{\dagger}(\textbf{p})&-\big{[}H(\textbf{p})-\mu\big{]}\end{pmatrix}\>.$
(2)

For simplicity, we assume first that the $N$-dimensional Hermitian Hamiltonian
$H(\textbf{p})$ is time-reversal-symmetric, so that

$U^{\dagger}H(\textbf{p})U=H^{*}(-\textbf{p}),$ (3)

or equivalently, in terms of the commutator, $[H(\textbf{p}),\mathcal{T}]=0$.
The off-diagonal (pairing) matrix needs to satisfy

$U^{\dagger}\Gamma(\textbf{p})U=-s\Gamma^{\rm T}(-\textbf{p}),$ (4)

where $s=\mathcal{T}^{2}=UU^{*}=\pm 1$. For real electrons the sign $s=-1$, of
course, but we keep the general sign $s$ nevertheless, to include fermions
with (effective) integer spin sim ; nandkishore  as well. As any other matrix,
the pairing matrix can also be written as
$\Gamma(\textbf{p})=\Gamma_{1}(\textbf{p})-i\Gamma_{2}(\textbf{p})$, where
$\Gamma_{1,2}$ are Hermitian. Then

$U^{\dagger}\Gamma_{1,2}(\textbf{p})U=-s\Gamma^{*}_{1,2}(-\textbf{p}),$ (5)

and for $s=-1$ ($s=1$) $\Gamma_{1,2}$ are simply even (odd) under time
reversal, and $[\Gamma_{1,2}(\textbf{p}),\mathcal{T}]=0$
($\\{\Gamma_{1,2}(\textbf{p}),\mathcal{T}\\}=0$, where $\\{,\\}$ is the
anticommutator). boettcher1 ; boettcher2

Let us now also assume the inversion symmetry, i. e. the existence of the
inversion operator $P$ with the effect:

$P^{\dagger}H(\textbf{p})P=H(-\textbf{p}),$ (6)
$P^{\dagger}\Gamma(\textbf{p})P=\Gamma(-\textbf{p}).$ (7)

The inversion transformation $\mathcal{P}$ in momentum representation is then
the combination of the operator $P$ and the momentum reversal
$\textbf{p}\rightarrow-\textbf{p}$. The inversion symmetry of the BdG
Hamiltonian means that $[O(\textbf{p}),\mathcal{P}]=0$, for $O=H$, and
$O=\Gamma$.

In contrast to the time reversal, the inversion operator is unitary, and
$P^{\dagger}P=1$. We also require that it is a physical observable, so that
$P^{\dagger}=P$ as well. This enforces that

$P^{2}=+1\>,$ (8)

so that the eigenvalues of the operator $P$ are $\pm 1$, i. e. the “parity” of
the eigenstates of $P$.

Finally, we postulate that, in general, inversion and time-reversal operations
commute:

$[\mathcal{P},\mathcal{T}]=0\>.$ (9)

The motivation is that inversion is an operation in real space, and as such
should have its action completely independent of the notion of time. The same
mutual commutation relation applies to any $SO(3)$ rotation and time reversal,
which can also be understood as the underlying reason for the antiunitarity of
the time-reversal operator. Additional arguments in support of this postulate
are given in Appendix B.

The BdG Hamiltonian can be rewritten as

$H_{\rm
BdG}(\textbf{p})=\sigma_{3}\otimes[H(\textbf{p})-\mu]+\sigma_{1}\otimes\Gamma_{1}(\textbf{p})+\sigma_{2}\otimes\Gamma_{2}(\textbf{p})\>,$
(10)

where $\sigma_{i}$, $i=1,2,3$ are the usual Pauli matrices. We observe that if
$\Gamma_{2}$ is finite, $[H_{\rm BdG},1\otimes\mathcal{T}]\neq 0$, if $s=-1$.
Similarly, when $s=1$, $[H_{\rm BdG},1\otimes\mathcal{T}]\neq 0$ for finite
$\Gamma_{1}$. When $s=1$ and $\Gamma_{1}=0$ the overall phase factor of $i$
can be gauged away, and the matrix $\Gamma$ again chosen to be Hermitian. It
is non-Hermiticity of the pairing matrix $\Gamma$ in either case that signals
the breaking of the time reversal in the superconducting state. $[H_{\rm
BdG}(\textbf{p}),1\otimes\mathcal{P}]=0$, on the other hand, and the BdG
Hamiltonian is even under inversion.

One can now construct a new antiunitary operator

$\mathcal{A}=\sigma_{k}\otimes(\mathcal{P}\mathcal{T})$ (11)

with $k=2$ for $s=-1$, and $k=1$ for $s=1$. Evidently,

$\\{H_{\rm BdG}(\textbf{p}),\mathcal{A}\\}=0,$ (12)

and the BdG Hamiltonian is odd under $\mathcal{A}$. By construction

$\mathcal{A}^{2}=(\sigma_{k}\sigma_{k}^{*})\otimes(\mathcal{P}^{2}\mathcal{T}^{2})=+1\>,$
(13)

where we used the fact that $\sigma_{k}\sigma_{k}^{*}=\mathcal{T}^{2}=s$, and
Eqs. (8) and (9). An equivalent antiunitary operator was constructed
before,agterberg  and it was responsible for the topological nontriviality of
the ensuing BF surface. The alternative construction is presented and
critically discussed in Appendix A. We see here that its existence is
guaranteed even when the inversion operator matrix $P$ is not diagonal, or a
real matrix in a given representation, and that it may be understood as a
consequence of basic postulates on the discrete symmetries involved. The
existence of an operator that anticommutes with the BdG Hamiltonian implies
that at fixed momentum the eigenstates of $H_{\rm BdG}(\textbf{p})$ come in
pairs of states with opposite signs of energy. Such an operator does not exist
when the system has no inversion symmetry in the normal phase link1 . $H_{\rm
BdG}(\textbf{p})$ with inversion and without time reversal therefore falls
into the topological class D. bzdusek

We have so far assumed that the time reversal symmetry may be violated only by
the off-diagonal pairing terms in $H_{\rm{BdG}}(\textbf{p})$ in Eq. (10). One
can, however, imagine it being broken, additionally or exclusively, by
diagonal terms in Eq. (10). In addition to the time-reversal-invariant part of
the normal state Hamiltonian $H(\textbf{p})$, this would require an addition
of a time-reversal-odd term to it: $H(\textbf{p})\rightarrow
H(\textbf{p})+H^{\prime}(\textbf{p})$, with

$U^{\dagger}H^{\prime}(\textbf{p})U=-(H^{\prime}(-\textbf{p}))^{*}.$ (14)

It is easy to see that the extra minus sign in the above expression relative
to Eq. (3) yields then an additional term in Eq. (10):

$1\otimes H^{\prime}(\textbf{p}).$ (15)

Assuming that $H^{\prime}(\textbf{p})$ is also even under inversion, it is odd
under the combined operation of time reversal and inversion, and the extra
term then evidently also anticommutes with the operator $\mathcal{A}$. With
this term included $H_{\rm{BdG}}(\textbf{p})$ in fact adopts its most general
form that exhibits this property.

An important observation can be made at this point: the fact that
$\mathcal{A}^{2}=+1$ implies that there exist a “real” basis in which the
unitary part of $\mathcal{A}$ is trivial, and $\mathcal{A}=K$, i. e. it is
just complex conjugation.herbutprb  In this basis therefore $H_{\rm
BdG}(\textbf{p})$ at every (real) momentum p is a purely imaginary matrix. Of
course, that also makes it antisymmetric, since it is Hermitian. Both of these
facts play a role in the rest of our discussion.

##  III  Effective Hamiltonian and emergence of $SO(4)$

Let us define the eigenvalues and the eigenstates of the normal state
Hamiltonian $H(\textbf{p})$, as $E_{i}(\textbf{p})$ and
$\phi_{i}(\textbf{p})$, $i=1,...N$. We may call the eigenstates with their
energy arbitrary close to the Fermi surface $\phi_{i}(\textbf{p})$ with
$i=1,...M$ “light”, and the remaining $N-M$ eigenstates “heavy”. When $s=-1$,
the Kramers theorem implies that $M$ is even, and when $s=1$, $M$ can be both
even or odd. Obviously, $M=2$, corresponding to the usual spin-1/2 fermions
such as electrons, would be of the greatest interest.

The spectrum of the Bogoliubov quasiparticles at a momentum p is given by the
solution of the equation for the real frequency $\omega$:

$\det(H_{\rm BdG}(\textbf{p})-\omega)=0\>.$ (16)

With the separation into light and heavy states at a given momentum near the
normal Fermi surface one can write the BdG Hamiltonian in the basis
$\\{(\phi_{i}(\textbf{p}),0)^{T},(0,\phi_{i}(\textbf{p}))^{T}\\}$, $i=1,...N$
as

$\displaystyle H_{\rm BdG}(\textbf{p})$ $\displaystyle=$
$\displaystyle\begin{pmatrix}H_{l}(\textbf{p})&H_{lh}(\textbf{p})\\\
H_{lh}^{\dagger}(\textbf{p})&H_{h}(\textbf{p})\end{pmatrix}\>.$ (17)

The block for the light particle and hole states $H_{l}(\textbf{p})$ is a
$2M$-dimensional matrix and describes the dispersion of the light particle and
hole states as well as the intraband pairing. The heavy modes are described by
the $2(N-M)$-dimensional matrix $H_{h}(\textbf{p})$ which denotes the energy
eigenstates of the heavy particle and holes and the intra- and interband
pairing only between the heavy modes. At last, the coupling between the light
and heavy states $H_{lh}(\textbf{p})$ is a $2M\times 2(N-M)$ matrix. (An
explicit expression of $H_{l,h,lh}$ for $M=2$ can be found in Appendix C.1).
The above determinant can now be rewritten as

$\det(H_{\rm BdG}(\textbf{p})-\omega)=\det(H_{h}(\textbf{p})-\omega)\det
L_{ef}(\omega,\textbf{p})\>,$ (18)

where the effective Lagrangian $L_{ef}$ is the Schur complement schur  of the
block matrix for the heavy modes:

$L_{ef}(\omega,\textbf{p})=H_{l}(\textbf{p})-\omega-
H_{lh}(\textbf{p})(H_{h}(\textbf{p})-\omega)^{-1}H_{lh}^{\dagger}(\textbf{p}).$
(19)

The first factor in Eq. (18) may also be understood as the fermionic partition
function for the heavy modes, and the second factor is therefore the residual
partition function for the light modes, renormalized by the integration over
the heavy modes link2 . $L_{ef}(\omega,\textbf{p})$ is well defined whenever
the heavy block is invertible, which is fulfilled for
$|\omega|<|E_{i}(\textbf{p})-\mu|$ for $i>M$. Under this condition the
eigenvalue equation in Eq. (16) reduces to $\det L_{ef}(\omega,\textbf{p})=0.$
In particular, $\omega=0$ is a solution only when

$\det H_{ef}(\textbf{p})=0,$ (20)

with $H_{ef}(\textbf{p})=L_{ef}(0,\textbf{p})$. We call $H_{ef}(\textbf{p})$
the effective Hamiltonian. brydon ; link2 ; venderbos  The same notion has
been used in the past in studies of stability of point nodes in two-
dimensional d-wave superconductors. berg  We emphasize, that only the
solutions for zero modes of $H_{ef}(\textbf{p})$ are exactly the same as those
for the original $H_{\rm BdG}(\textbf{p})$; the rest of their spectra differ.
This is, however, all that is needed to understand the emergence of the BF
surface, the dispersion of quasiparticles close to it, and even the
instability of the BF surface, as we show below.

According to Eq. (19) the effective Hamiltonian is thus

$H_{ef}(\textbf{p})=H_{l}(\textbf{p})-H_{lh}(\textbf{p})H_{h}^{-1}(\textbf{p})H_{lh}^{\dagger}(\textbf{p})\>.$
(21)

The effective Hamiltonian computed in the standard (“canonical”)
representation where the diagonal terms of the two matrices
$H_{l,h}(\textbf{p})$ are the energy dispersions of the states and the off-
diagonal terms of the three matrices $H_{l,h,lh}(\textbf{p})$ are the intra-
and interband pairing between the different states can be found in Appendix C.
To understand its general structure, however, it is better to work in the real
basis. In the real basis $\mathcal{A}=K$, and thus all of the matrices
$H_{l}(\textbf{p})$, $H_{lh}(\textbf{p})$ and $H_{h}(\textbf{p})$ are
imaginary. Clearly, $H_{ef}(\textbf{p})$ is then imaginary as well. The
effective low-energy Hamiltonian inherits the antiunitary (anticommuting)
symmetry of the full BdG Hamiltonian, and therefore in general is a Hermitian
imaginary $2M$-dimensional matrix, i. e. a generator of the real
representation of $SO(2M)$ group of rotations. In the physically most
pertinent case of $M=2$, $H_{ef}(\textbf{p})$ is a generator of the $SO(4)$,
and in the real basis can be written as

$H_{ef}(\textbf{p})=\sum_{k=1}^{3}(a_{k}(\textbf{p})N_{k}+b_{k}(\textbf{p})J_{k})$
(22)

where $[N_{k}]_{\mu\nu}=-[N_{k}]_{\nu\mu}=-i\delta_{\mu 0}\delta_{\nu k}$, and
$[J_{k}]_{ij}=-i\epsilon_{ijk}$, $[J_{k}]_{0j}=[J_{k}]_{j0}=0$. Here the Greek
indices run from 0 to 3, and Latin indices from 1 to 3. We observe that in the
real basis the matrix elements of the effective Hamiltonian may be written as

$[H_{ef}(\textbf{p})]_{\mu\nu}=iF^{\mu\nu}(\textbf{p}),$ (23)

where $F^{\mu\nu}(\textbf{p})$ is the standard antisymmetric electromagnetic
tensor, with the “vector” coefficients
$\textbf{a}(\textbf{p})=(a_{1}(\textbf{p}),a_{2}(\textbf{p}),a_{3}(\textbf{p}))$
and
$\textbf{b}(\textbf{p})=(b_{1}(\textbf{p}),b_{2}(\textbf{p}),b_{3}(\textbf{p}))$
playing the role of momentum-dependent “electric” and “magnetic” fields. This
analogy will be deepened and will come in handy shortly when we discuss the
form of the zero-energy eigenstates of the effective Hamiltonian.

The six four-dimensional imaginary matrices $N_{k}$ and $J_{k}$ are chosen to
close the standard SO(4) Lie algebra in the following form:

$[J_{i},J_{j}]=i\epsilon_{ijk}J_{k},$ (24)
$[N_{i},J_{j}]=i\epsilon_{ijk}N_{k},$ (25)
$[N_{i},N_{j}]=i\epsilon_{ijk}J_{k}.$ (26)

Indeed, it is easily seen that the fully imaginary representation of the
generators $N_{k}$ and $J_{k}$ defined above is equivalent to the more
standard representation of real symmetric Lorentz boosts $K_{k}$, with
$[K_{k}]_{\mu\nu}=\delta_{\mu 0}\delta_{\nu k}$, and the same imaginary
generators of rotations $J_{k}$; explicitly $N_{k}=SK_{k}S^{\dagger}$, and
$J_{k}=SJ_{k}S^{\dagger}$, where

$S=e^{-i\frac{\pi}{4}G}$ (27)

and the matrix $G=\operatorname{diag}(1,-1,-1,-1)$.

By forming the symmetric and the antisymmetric linear combinations

$R_{i,\pm}=\frac{1}{2}(J_{i}\pm N_{i})\>,$ (28)

it readily follows that

$[R_{i,r},R_{j,r}]=i\epsilon_{ijk}R_{k,r}\>,$ (29)

for $r=\pm$, whereas

$[R_{i,+},R_{j,-}]=0\>.$ (30)

The Lie algebra of the generators of the $SO(4)$ is the same as the Lie
algebra of the generators of the $SO(3)\times SO(3)$, as is well known.georgi
The effective Hamiltonian can therefore be rewritten as

$H_{ef}(\textbf{p})=\sum_{k=1}^{3}\sum_{r=\pm}(a_{k}(\textbf{p})+rb_{k}(\textbf{p}))R_{k,r}.$
(31)

The four-dimensional matrices $N_{i}$ and $J_{i}$ form the irreducible
$(1/2,1/2)$ representation of the Lie algebra $SO(3)\times SO(3)$, where
$j=1/2$ refers to the spin-1/2 representation of $SO(3)$.georgi  The matrices
$R_{k,r}$ can thus be brought by a unitary transformation into
$1\otimes(\sigma_{k}/2)$ and $(\sigma_{k}/2)\otimes 1$. The explicit unitary
transformation that does so is provided in Appendix C.2. The spectrum of
$H_{ef}$ can then be readily discerned as

$E(\textbf{p})=\pm\frac{1}{2}(|\textbf{a}(\textbf{p})+\textbf{b}(\textbf{p})|\pm|\textbf{a}(\textbf{p})-\textbf{b}(\textbf{p})|)\>.$
(32)

In particular, it is evident that there are two zero eigenvalues at the
momenta at which

$\textbf{a}(\textbf{p})\cdot\textbf{b}(\textbf{p})=0\>.$ (33)

Since this is a single equation for three components of the momentum, the
solutions, when they exist, will form a surface in the momentum space.

Multiplying the four eigenvalues $E(\textbf{p})$ yields
$\det[H_{ef}(\textbf{p})]=(\textbf{a}(\textbf{p})\cdot\textbf{b}(\textbf{p}))^{2}$.
The last equation is therefore precisely the condition for vanishing of the
Pfaffian agterberg  of the effective Hamiltonian. The relation between our
electric and magnetic fields $\textbf{a}(\textbf{p})$ and
$\textbf{b}(\textbf{p})$ and the coefficients of the canonical representation
of the effective Hamiltonian, which describe the emergence of the BF surface
in terms of the “pseudomagnetic” field of Refs. agterberg ; brydon , can be
found in Appendix C (Eqs. (C26)-(C27)).

##  IV Zero modes at the BF surface

We will also need the explicit form of the eigenstates of $H_{ef}(\textbf{p})$
with zero energy, measured of course from the chemical potential. The
eigenvalue equation is then

$H_{ef}(\textbf{p})\Psi(\textbf{p})=0\>,$ (34)

where $\Psi=(v_{0},v_{1},v_{2},v_{3})^{\rm T}$, and the ubiquitous momentum
dependence of all variables suppressed for legibility. The eigenvalue equation
can then be compactly written in the vector notation

$\textbf{a}\cdot\textbf{v}=0\>,$ (35)
$v_{0}\textbf{a}+\textbf{b}\times\textbf{v}=0\>,$ (36)

with $\textbf{v}=(v_{1},v_{2},v_{3})$. Assume $v_{0}\neq 0$ first. Multiplying
Eq. (36) with b we get that

$v_{0}\textbf{b}\cdot\textbf{a}+\textbf{b}\cdot(\textbf{b}\times\textbf{v})=0.$
(37)

and therefore $\textbf{b}\cdot\textbf{a}=0$, as we already found. In this case
then $\textbf{v}\sim\textbf{b}\times\textbf{a}$. When normalized, the first
zero-energy solution may be taken to be

$\Psi_{t}=\frac{1}{\sqrt{1+\textbf{v}^{2}}}(1,\textbf{v})^{\rm T},$ (38)

where $\textbf{v}=(\textbf{b}\times\textbf{a})/\textbf{b}^{2}$. The second,
orthogonal, solution is then with $v_{0}=0$: in this case v needs to be
orthogonal to a and parallel to b, which again requires that the vectors a and
b are mutually orthogonal. In that case therefore $\textbf{v}\sim\textbf{b}$,
and the normalized zero-energy solution is

$\Psi_{s}=(0,\textbf{b}/|\textbf{b}|)^{\rm T}.$ (39)

Both solutions are manifestly real, and $\Psi_{t}^{\dagger}\Psi_{s}=0$. One
can rotate them into a pair of complex conjugate zero-energy solutions

$\Psi_{\pm}=\frac{1}{\sqrt{2}}(\Psi_{t}\pm i\Psi_{s})\>,$ (40)

which satisfy $\Psi_{+}=\mathcal{A}\Psi_{-}$, since $\mathcal{A}=K$ in the
real basis we are assuming.

We explain the motivation behind the labels “t” and “s” in the two basic zero-
energy solutions next.

##  V Relativistic analogy to Lorentz force equation

There exists an instructive analogy between our time-dependent Schrödinger
equation at low energies and the classical covariant second Newton law with
the Lorentz force for a charged particle in the electromagnetic field. The
time-dependent Schrödinger equation for the effective Hamiltonian is

$\frac{d}{dt}\Psi=F\Psi,$ (41)

once one recalls that $H_{ef}=iF$, with $F=F^{\mu\nu}$ as the real
antisymmetric electromagnetic tensor. Newton’s second law in the
electromagnetic field, on the other hand, in the covariant formulation takes
the form

$m\frac{d}{d\tau}V=FGV,$ (42)

where $G=G_{\mu\nu}=\operatorname{diag}(1,-1,-1,-1)$ is Minkowski’s metric
tensor, $V=V^{\mu}=\gamma(v)(c,\textbf{v})^{\rm T}$ is the velocity four-
vector, $c$ the velocity of light, v the velocity three-vector, and
$\gamma(v)=1/\sqrt{1-(v/c)^{2}}$. $\tau$ is the proper time, and $m$ the rest
mass of the particle. The velocity four-vector has the fixed positive norm
with respect to the Minkowski metric: rindler

$V^{\mu}V_{\mu}=V^{\rm T}GV=c^{2}.$ (43)

The presence of Minkowski’s metric tensor $G$ in the Lorentz equation, of
course, makes it decidedly not a Schrödinger equation; the Lorentz group in
not $SO(4)$ but $SO(1,3)$, which is not compact, and its finite-dimensional
representations are consequently not unitary.georgi  Multiplying both sides of
the Lorentz equation by the imaginary unit will fail to make the matrix $iFG$,
which appears in place of a Hamiltonian, Hermitian, for example. Nevertheless,
the solutions of the Schrödinger equation for which $F\Psi=0$ do have a
classical analog: they correspond to the four-velocity $V$ for which the
forces from the electrical and magnetic fields precisely cancel. Obviously
this is possible only at the points in space where the electric and the
magnetic fields are mutually orthogonal, and the unique three-velocity, of the
right magnitude and right direction, is orthogonal to both. Apart from our
normalization with respect to Euclidean and not Minkowski’s metric, the zero-
energy solution $\Psi_{t}$ is precisely such a four-vector, with the velocity
of light being simply unity. Index “t” in this solution was chosen to suggest
a “timelike” four-vector that would have positive Minkowski norm for
velocities below the velocity of light, as the physical velocity four-vector
by its definition has to be.

The second real solution we found, on the other hand, does not correspond to a
physical velocity in our analogy, since the form of $\Psi_{s}$ is “spacelike”,
i. e. with a negative Minkowski norm. As a physical solution for four-velocity
of the classical Lorentz equation it is thus unacceptable. But as a solution
of the Schrödinger equation it is perfectly regular, and it can be used in a
linear combination with the timelike solution to form a pair of complex-
conjugate zero modes. It is the fact that the positive-norm quantum state
$\Psi$ can be complex whereas the positive-norm four-velocity $V$ can only be
real that leads to an additional zero-mode in the quantum case, relative to
the closely related but not entirely equivalent Newton equation with the
Lorentz electromagnetic force.

Figure 1: Hoppings on the Lieb lattice as defined by the Hamiltonian in Eq.
(53). The fermions hop with amplitude $+i$ along the direction of the arrow,
and with the amplitude $-i$ in the direction opposing the arrow on a link
between two sites. Pink lines connect nearest neighbors and correspond to
hopping of magnitude $t$, and yellow lines connect next-nearest neighbors with
hopping of magnitude $\chi$. The two-body repulsion is between the fermions
residing on the nearest-neighboring sites. Panel (a) shows the full three
dimensional picture of the Lieb lattice, while panel (b) shows the projection
of the Lieb lattice in the $xy$-plane.

##  VI Time reversal preserved

When the $H_{\rm BdG}$ preserves not only inversion but time reversal symmetry
as well, there cannot be a Bogoliubov-Fermi surface of zero modes. The
elimination of the heavy modes will in this case produce an effective
Hamiltonian which will commute with an operator that represents the combined
operation of $\mathcal{P}\mathcal{T}$, i. e. with an antiunitary operator with
a square of $-1$. At the level of $H_{ef}$ let us call this operator
$\mathcal{B}=WK$, with a unitary representation-dependent matrix $W$, which is
four-dimensional if we focus on the physically most urgent case of $M=2$.
Operation $\mathcal{B}$ leaves the momentum invariant. To recognize the matrix
$W$ it is useful to write the explicit form of the matrices $R_{k,\pm}$ in our
representation:

$R_{1,+}=1\otimes\frac{\sigma_{2}}{2}\>,$ (44)
$R_{2,+}=\sigma_{2}\otimes\frac{\sigma_{3}}{2}\>,$ (45)
$R_{3,+}=\sigma_{2}\otimes\frac{\sigma_{1}}{2}\>,$ (46)

and similarly for $R_{k,-}$:

$R_{1,-}=-\frac{\sigma_{3}}{2}\otimes\sigma_{2}\>,$ (47)
$R_{2,-}=-\frac{\sigma_{2}}{2}\otimes 1\>,$ (48)
$R_{3,-}=-\frac{\sigma_{1}}{2}\otimes\sigma_{2}\>.$ (49)

Consider now $W=\sigma_{2}\otimes X$, with $X$ a Pauli matrix. The matrix $X$
only needs to be real, so that $\mathcal{B}^{2}=-1$ as required. Direct
inspection then gives that for any such $X$ the operator $\mathcal{B}$ would
commute with two, and anticommute with the remaining four out of six matrices
$R_{i,\pm}$. Furthermore, the three of the latter four matrices are either all
$R_{i,+}$, or all $R_{i,-}$. For example, for $X=\sigma_{1}$, $\mathcal{B}$
commutes only with $R_{1,+}$ and $R_{2,+}$. This means that when time reversal
symmetry is present, first, it must be that

$\textbf{a}(\textbf{p})+r\textbf{b}(\textbf{p})\equiv 0\>,$ (50)

for either $r=+1$ or $r=-1$. In the relativistic analogy this means that the
electric and magnetic fields are either parallel or antiparallel everywhere,
and therefore the Lorentz force can never vanish, unless both fields vanish.
Second, since for one of the components we also have that

$a_{k}(\textbf{p})-rb_{k}(\textbf{p})=0\>,$ (51)

there are only two finite terms in the representation in Eq. (31). For the
specific choice in the example above the spectrum would therefore be

$E(\textbf{p})=\pm 2(a_{1}(\textbf{p})^{2}+a_{2}(\textbf{p})^{2})^{1/2}.$ (52)

In general therefore $E(\textbf{p})=0$ leads to two conditions to be satisfied
for three components of the momenta, i. e. a line in the momentum space.
boettcher2

The algebra involved in the above argument becomes particularly transparent in
the canonical representation of the generators $R_{k,\pm}$ (Appendix C).

##  VII Lattice Hamiltonian and interactions

We now define a lattice single-particle Hamiltonian which provides a minimal
realization of the above $H_{ef}(\textbf{p})$ in Eq. (22) for spin-1/2
electrons. The only requirement is that it is a four-dimensional matrix
Hamiltonian that admits an antiunitary operator with positive square that
anticommutes with it.

With this in mind we consider the Lieb lattice in three dimensions: the unit
cell consists of four sites, one that is at the sites of the primitive cubic
lattice at positions $\textbf{R}=\sum_{i}n_{i}\textbf{e}_{i}$ with $n_{i}$ as
integers, $\textbf{e}_{i}\cdot\textbf{e}_{j}=\delta_{ij}$, and the other three
which are at the centers of the three links in orthogonal directions that
connect the sites of the cubic lattice at positions
$\textbf{R}+(\textbf{e}_{i}/2)$, with $i=1,2,3$. The Hamiltonian is then
defined as:

$\displaystyle
H_{0}=-it\sum_{\textbf{R},k=1,2,3}c^{\dagger}(\textbf{R})c(\textbf{R}\pm\frac{\textbf{e}_{k}}{2})+\big{[}i\chi\sum_{\textbf{R},s=\pm
1}(s)c^{\dagger}(\textbf{R}+\frac{\textbf{e}_{1}}{2})[c(\textbf{R}+\textbf{e}_{1}+s\frac{\textbf{e}_{2}}{2})$
(53) $\displaystyle-c(\textbf{R}+s\frac{\textbf{e}_{2}}{2})]+(1\rightarrow
2,2\rightarrow 3)+(2\rightarrow 3,3\rightarrow 1)\big{]}+herm.conj.$

with parameters $t$ and $\chi$ real, so that the hoppings are all purely
imaginary. $c^{\dagger}(\textbf{R})$ is the usual fermionic creation operator
on site R. (See Figure 1.) The phases of the hopping terms are chosen so that
in momentum space the Hamiltonian becomes

$H_{0}=\sum_{\textbf{p}}\Psi^{\dagger}(\textbf{p})H_{ef}(\textbf{p})\Psi(\textbf{p}),$
(54)

with

$\Psi(\textbf{R})=\big{[}c(\textbf{R}),c(\textbf{R}+\frac{\textbf{e}_{1}}{2}),c(\textbf{R}+\frac{\textbf{e}_{2}}{2}),c(\textbf{R}+\frac{\textbf{e}_{3}}{2})\big{]}^{\rm
T},$ (55)

and $H_{ef}(\textbf{p})$ precisely as in Eq. (22), with

$a_{i}(\textbf{p})=2t\cos(\frac{p_{i}}{2}),$ (56)

and

$b_{i}(\textbf{p})=4\chi\sin(\frac{p_{j}}{2})\sin(\frac{p_{k}}{2}),$ (57)

with $i\neq j$, $i\neq k$, $j\neq k$ in the last equation. The BF surface is
now determined by the equation

$\big{[}\prod_{i=1,2,3}\sin\big{(}\frac{p_{i}}{2}\big{)}\big{]}\sum_{i=1,2,3}\cot\big{(}\frac{p_{i}}{2}\big{)}=0,$
(58)

which is independent of the hopping parameters $t$ and $\chi$ as long as they
are both finite. The BF surface is depicted in Fig. 2. Note that whereas the
three axes belong to the BF surface, poles of the cotangents remove the planes
$p_{i}=0$ from it.

Figure 2: BF surface of zero-energy states of the Hamiltonian $H_{0}$ in Eq.
(53) in the first Brillouin zone.

One may now also define the two-body interaction term as

$H_{int}=V\sum_{\textbf{R},i}n(\textbf{R})n(\textbf{R}\pm\frac{\textbf{e}_{i}}{2}),$
(59)

with $n(\textbf{R})=c^{\dagger}(\textbf{R})c(\textbf{R})$ as the usual
particle number operator, which describes repulsion between nearest neighbors
on the Lieb lattice ($V>0$). The full interacting lattice model is then

$H=H_{0}+H_{int}.$ (60)

We assume half filling, which corresponds to the spectral symmetry of the BdG
Hamiltonian between positive and negative states. Besides possessing
translational symmetry, the Hamiltonian remains invariant under $2\pi/3$
rotations around the $(1,1,1)$ diagonal and under inversion around any site R.

##  VIII Mean field theory

To study the effects of two-body interactions we first rewrite the interaction
Hamiltonian as

$H_{int}=\frac{V}{4}\sum_{\textbf{R},i}\big{[}\big{(}n(\textbf{R})+n(\textbf{R}\pm\frac{\textbf{e}_{i}}{2})\big{)}^{2}-\big{(}n(\textbf{R})-n(\textbf{R}\pm\frac{\textbf{e}_{i}}{2})\big{)}^{2}\big{]}.$
(61)

It may then be decoupled with two Hartree variables (in the sense of Hubbard -
Stratonovich transformation)

$\displaystyle
H_{int}=\frac{1}{V}\sum_{\textbf{R},i}\big{[}\zeta^{2}(\textbf{R},\textbf{R}\pm\frac{\textbf{e}_{i}}{2})-\mu^{2}(\textbf{R},\textbf{R}\pm\frac{\textbf{e}_{i}}{2})\big{]}$
(62)
$\displaystyle+\sum_{\textbf{R},i}\\{\zeta(\textbf{R},\textbf{R}\pm\frac{\textbf{e}_{i}}{2})\big{[}n(\textbf{R})-n(\textbf{R}\pm\frac{\textbf{e}_{i}}{2})\big{]}$
$\displaystyle+\mu(\textbf{R},\textbf{R}\pm\frac{\textbf{e}_{i}}{2})\big{[}n(\textbf{R})+n(\textbf{R}\pm\frac{\textbf{e}_{i}}{2})\big{]}\\}.$

Anticipating the energetically preferable uniform mean-field configuration, we
take

$\zeta(\textbf{R},\textbf{R}\pm\textbf{e}_{i}/2)=\langle
n(\textbf{R})-n(\textbf{R}\pm\textbf{e}_{i}/2)\rangle=\zeta,$ (63)

and

$\mu(\textbf{R},\textbf{R}\pm\textbf{e}_{i}/2)=\langle
n(\textbf{R})+n(\textbf{R}\pm\textbf{e}_{i}/2)\rangle=\mu,$ (64)

and both constant. The mean-field interaction term then becomes

$\displaystyle H_{int,mf}$ $\displaystyle=$
$\displaystyle\frac{6N}{V}(\zeta^{2}-\mu^{2})$ $\displaystyle+$ $\displaystyle
2\sum_{\textbf{R},i}\big{[}(\zeta+\mu)n(\textbf{R})+(\mu-\zeta)n(\textbf{R}+\frac{\textbf{e}_{i}}{2})\big{]}\>,$

with $N$ as the number of primitive lattice sites. In the momentum space the
full mean-field Hamiltonian $H_{mf}=H_{0}+H_{int,mf}$ can therefore be
arranged into

$\displaystyle H_{mf}=\sum_{\textbf{p}}$
$\displaystyle\big{\\{}\Psi^{\dagger}(\textbf{p})[H_{ef}(\textbf{p})+u\mathbb{1}+vG]\Psi(\textbf{p})$
$\displaystyle+$ $\displaystyle\frac{v^{2}-u^{2}}{2V}\big{\\}},$

with the matrix $G$ as the previously encountered Minkowski metric matrix, and
the two new Hubbard-Stratonovich variables being $u/4=\mu+(\zeta/2)$ and
$v/4=\zeta+(\mu/2)$.

Let us define the two “critical” eigenvalues of the $H_{ef}$ which vanish at
the BF surface as $\pm\xi(\textbf{p})$, with

$\xi(\textbf{p})=\frac{1}{2}(|\textbf{a}(\textbf{p})+\textbf{b}(\textbf{p})|-|\textbf{a}(\textbf{p})-\textbf{b}(\textbf{p})|)\>,$
(67)

and the remaining two “massive” eigenvalues which are finite everywhere as
$\pm m(\textbf{p})$ with

$m(\textbf{p})=\frac{1}{2}(|\textbf{a}(\textbf{p})+\textbf{b}(\textbf{p})|+|\textbf{a}(\textbf{p})-\textbf{b}(\textbf{p})|).$
(68)

There exists a unitary transformation $U_{ef}(\textbf{p})$ that diagonalizes
$H_{ef}(\textbf{p})$, so that

$\displaystyle
U_{ef}(\textbf{p})H_{ef}(\textbf{p})U_{ef}^{\dagger}(\textbf{p})$
$\displaystyle=$ $\displaystyle\begin{pmatrix}m(\textbf{p})\sigma_{3}&0\\\
0&\xi(\textbf{p})\sigma_{3}\end{pmatrix}\>.$ (69)

The two-component fermions that correspond to the massive and critical states
are then given by
$U_{ef}(\textbf{p})\Psi(\textbf{p})=(\Psi_{m}(\textbf{p}),\Psi_{\xi}(\textbf{p}))^{T}$.
The mean-field Hamiltonian in terms of the critical and massive fermions now
becomes

$\displaystyle
H_{mf}=\sum_{\textbf{p}}[\Psi^{\dagger}_{m}(\textbf{p})(m(\textbf{p})\sigma_{3}+u\mathbb{1}+vX_{m}(\textbf{p}))\Psi_{m}(\textbf{p})$
(70)
$\displaystyle+\frac{v^{2}-u^{2}}{2V}+\Psi^{\dagger}_{\xi}(\textbf{p})(\xi(\textbf{p})\sigma_{3}+u\mathbb{1}+vX_{\xi}(\textbf{p}))\Psi_{\xi}(\textbf{p})$
$\displaystyle+v(\Psi_{m}^{\dagger}(\textbf{p})X_{m\xi}(\textbf{p})\Psi_{\xi}(\textbf{p})+\Psi^{\dagger}_{\xi}(\textbf{p})X_{m\xi}^{\dagger}(\textbf{p})\Psi_{m}(\textbf{p}))],$

where the two-dimensional matrices $X$ are defined by

$\displaystyle U_{ef}(\textbf{p})GU_{ef}^{\dagger}(\textbf{p})$
$\displaystyle=$
$\displaystyle\begin{pmatrix}X_{m}(\textbf{p})&X_{m\xi}(\textbf{p})\\\
X_{m\xi}^{\dagger}(\textbf{p})&X_{\xi}(\textbf{p})\end{pmatrix}\>.$ (71)

The imaginary-time mean-field quantum mechanical action at finite temperatures
is then

$\displaystyle
S=\int_{0}^{\beta}d\tau[\sum_{\textbf{p},r=m,\xi}\Psi_{r}^{\dagger}(\textbf{p},\tau)\partial_{\tau}\Psi_{r}(\textbf{p},\tau)+H_{mf}]$
(72)

($\beta=1/k_{B}T$) in terms of the usual Grassmann variables for the massive
and critical fermions. negele  Minimization of the free energy, which is the
logarithm of the usual path integral over Grassmann and Hubbard-Stratonovich
variables, determines the saddle-point values of $u$ and $v$, which then equal
their expectation values in the ground state:
$u=\langle\sum\Psi^{\dagger}(\textbf{p})\Psi(\textbf{p})\rangle$ is the shift
in the chemical potential, and
$v=\langle\sum\Psi^{\dagger}(\textbf{p})G\Psi(\textbf{p})\rangle$ is the
“staggered” chemical potential herbut2006 ; hjr , i. e. the imbalance between
the average occupations of sites on the corners R and sites on the links
$\textbf{R}\pm\textbf{e}_{i}/2$. If either $u$ or $v$ is finite the inversion
symmetry is broken, since $H_{mf}$ would acquire real terms and so cease to
anticommute with the operator $\mathcal{A}$.

We now integrate over fermions to get the remaining action $S$ in terms of the
variables $u$ and $v$ only, and expand in powers of both variables to examine
the stability of the inversion-symmetric BF surface. The integration over the
massive fermions, of course, can only produce infrared-finite terms in the
expansion of such $S$ in powers of $u$ and $v$.herbutbook  In particular, the
terms $\sim uv$ and $\sim u^{2}$ produced by this integration vanish exactly
at $T=0$. The same absence of $\sim uv$ and $\sim u^{2}$ terms is also found
in the integration over the more important critical modes, as we explain
below.

The integration over the critical modes yields the following term in the
action $S$, quadratic in $u$ and $v$:

$\frac{k_{B}T}{2}\sum_{\omega_{n},\textbf{p}}\operatorname{Tr}\big{[}\frac{i\omega_{n}+\xi\sigma_{3}}{\omega_{n}^{2}+\xi^{2}}(u+vX_{\xi})\frac{i\omega_{n}+\xi\sigma_{3}}{\omega_{n}^{2}+\xi^{2}}(u+vX_{\xi})\big{]},$
(73)

where $\xi=\xi(\textbf{p})$. This can be rearranged into

$\displaystyle\frac{k_{B}T}{2}\sum_{\omega_{n},\textbf{p}}$
$\displaystyle\operatorname{Tr}$
$\displaystyle\big{[}\frac{-\omega_{n}^{2}+\xi^{2}}{(\omega_{n}^{2}+\xi^{2})^{2}}u(u+2vX_{\xi})$
$\displaystyle+$ $\displaystyle
v^{2}\frac{i\omega_{n}+\xi\sigma_{3}}{\omega_{n}^{2}+\xi^{2}}X_{\xi}\frac{i\omega_{n}+\xi\sigma_{3}}{\omega_{n}^{2}+\xi^{2}}X_{\xi}\big{]}\>.$

The first term ($\sim u^{2}$ and $\sim uv$) vanishes at $T=0$ due to the exact
property of the integral over frequencies

$\int_{-\infty}^{\infty}d\omega\frac{\xi^{2}-\omega^{2}}{(\omega^{2}+\xi^{2})^{2}}=0\>,$
(75)

whereas it would be finite at $T\neq 0$. It cannot therefore produce a $T=0$
instability of the BF surface at infinitesimal coupling by itself. The
remaining second term ($\sim v^{2}$), on the other hand, upon expanding

$X_{\xi}(\textbf{p})=\sum_{\mu=0}^{3}g_{\mu}(\textbf{p})\sigma_{\mu}$ (76)

becomes

$v^{2}k_{B}T\sum_{\omega_{n},\textbf{p}}\frac{\xi^{2}(g_{0}^{2}(\textbf{p})+g_{3}^{2}(\textbf{p})-g_{1}^{2}(\textbf{p})-g_{2}^{2}(\textbf{p}))-\omega_{n}^{2}g_{\mu}^{2}(\textbf{p})}{(\omega_{n}^{2}+\xi^{2}(\textbf{p}))^{2}}.$
(77)

At $T=0$, using Eq. (75), the last expression can be written as

$\displaystyle-v^{2}\int_{-\infty}^{+\infty}\frac{d\omega}{2\pi}\sum_{\textbf{p}}\frac{g_{1}^{2}(\textbf{p})+g_{2}^{2}(\textbf{p})}{\omega^{2}+\xi^{2}(\textbf{p})}=$
(78)
$\displaystyle-v^{2}\int_{0}^{\Omega}\frac{d\xi}{|\xi|}\mathcal{N}(\xi)\>,$

where

$\mathcal{N}(\xi)=\sum_{\textbf{p}}\delta(\xi-\xi(\textbf{p}))\big{(}g_{1}^{2}(\textbf{p})+g_{2}^{2}(\textbf{p})\big{)}$
(79)

and $\Omega$ is a UV cutoff. The integral is logarithmically divergent if
$\mathcal{N}(0)$ is finite, i. e. if the expansion coefficients
$g_{1,2}(\textbf{p})$ of $X_{\xi}(\textbf{p})$ have finite support on the BF
surface. The sign of the integral implies that the coefficient of the
quadratic term $\sim v^{2}$ is in that case always negative, which signals the
instability of the inversion-symmetric BF surface at $T=0$. Computing the
energy of the ground state with a finite uniform $v$ and minimizing it then
yields the characteristic form when $V\mathcal{N}(0)\rightarrow 0$:

$v=\Omega e^{-1/(2V\mathcal{N}(0))}.$ (80)

The critical temperature below which $v\neq 0$ exhibits the same essential
singularity in the interaction $V$, common to all weak-coupling instabilities.

Since the integration over the fermions at $T=0$ does not contribute to the
coefficients of $\sim u^{2}$ and $\sim uv$ terms in the action, the saddle-
point value of $u$ vanishes at $T=0$.

Finally, it is easy to show that although the integration over massive states
modifies the propagator for the critical fermions to the order of $v^{2}$,
this does not alter the log-divergent coefficient of the quadratic term above.

##  IX  Fate of BF surface

The lesson of the previous section is that the stability of the BF surface
depends only on whether the matrix $G$ that couples the light fermions to the
order parameter $v$, once projected onto the critical states, has finite off-
diagonal elements for the momenta at the BF surface. For momenta at the BF
surface $\xi(\textbf{p})=0$, and the matrix $X_{\xi}$ is then by definition

$\displaystyle X_{\xi=0}(\textbf{p})$ $\displaystyle=$
$\displaystyle\begin{pmatrix}\Psi_{+}^{\dagger}(\textbf{p})G\Psi_{+}(\textbf{p})&\Psi_{+}^{\dagger}(\textbf{p})G\Psi_{-}(\textbf{p})\\\
\Psi_{-}^{\dagger}(\textbf{p})G\Psi_{+}(\textbf{p})&\Psi_{-}^{\dagger}(\textbf{p})G\Psi_{-}(\textbf{p})\end{pmatrix}\>,$
(81)

with the states $\Psi_{\pm}(\textbf{p})$ given by Eq. (40). This readily
yields $g_{k}(\textbf{p})=0$ for $k=2,3$ and

$g_{0}(\textbf{p})=-\frac{\textbf{a}^{2}(\textbf{p})}{\textbf{a}^{2}(\textbf{p})+\textbf{b}^{2}(\textbf{p})},$
(82)
$g_{1}(\textbf{p})=\frac{\textbf{b}^{2}(\textbf{p})}{\textbf{a}^{2}(\textbf{p})+\textbf{b}^{2}(\textbf{p})}.$
(83)

$g_{1}(\textbf{p})$ is finite everywhere on the BF surface, except at the
three coordinate axis. The integral in Eq. (78) is then indeed logarithmically
divergent, and the inversion-symmetric BF surface is unstable at $T=0$ and
infinitesimal repulsive nearest-neighbor interaction $V$.

We may now examine the resulting low-energy spectrum of the quasiparticles in
the inversion-symmetry-broken state with $v>0$ and $u=0$. It is given by the
two-dimensional mean-field Hamiltonian for the critical fermions near the BF
surface

$H_{mf,\xi}=\xi(\textbf{p})\sigma_{3}+vg_{1}(\textbf{p})\sigma_{1}+vg_{0}(\textbf{p}),$
(84)

with $g_{k}(\textbf{p})$, $k=0,1$ given above, and $\xi(\textbf{p})$ as in Eq.
(67). Near the BF surface one can approximate

$\xi(\textbf{p})=\frac{\textbf{a}(\textbf{p})\cdot\textbf{b}(\textbf{p})}{[\textbf{a}(\textbf{p})^{2}+\textbf{b}(\textbf{p})^{2}]^{1/2}}\\{1+\frac{[\textbf{a}(\textbf{p})\cdot\textbf{b}(\textbf{p})]^{2}}{2[\textbf{a}(\textbf{p})^{2}+\textbf{b}(\textbf{p})^{2}]^{2}}+...\\}.$
(85)

The spectrum of $H_{mf,\xi}$ is therefore

$E(\textbf{p})=\pm[\xi^{2}(\textbf{p})+v^{2}g_{1}^{2}(\textbf{p})]^{1/2}+vg_{0}(\textbf{p})\>.$
(86)

In particular, the location in the momentum space of the zero modes of the new
spectrum is in general given by the solution of

$\xi^{2}(\textbf{p})=v^{2}[g_{0}^{2}(\textbf{p})-g_{i}^{2}(\textbf{p})]\>,$
(87)

which in the present case and with the order parameter $v$ small reduces to
the simple condition

$(\textbf{a}(\textbf{p})\cdot\textbf{b}(\textbf{p}))^{2}=v^{2}(\textbf{a}(\textbf{p})^{2}-\textbf{b}(\textbf{p})^{2}).$
(88)

The left-hand side of the last equation vanishes at the original BF surface.
The parts of the original BF surface where the right-hand side
($RHS=v^{2}(\textbf{a}^{2}-\textbf{b}^{2})$) of the equation is positive will
thus split into two wings of the new surfaces of zero modes, which merge at
the intersection of the original BF surface and the surface given by the zero
value of the right-hand side of the equation ($RHS=0$). So if such an
intersection of the two surfaces exists, a part of the original BF surface
will become gapped, and its complement will effectively remain gapless, i. e.
transform into a new surface. If there is no such intersection of the two
surfaces, on the other hand, the original BF surface is either completely
gapped out (if $RHS<0$ everywhere on it), or split into two new separate
nearby surfaces (if $RHS>0$ everywhere on it).

In our lattice model, since $\textbf{a}\sim t$ vanishes in the corners of the
Brillouin zone, and $\textbf{b}\sim\chi$ vanishes at the three axis, the
surface $RHS=0$ always intersects the original BF surface, and thus gaps out
only a part of it. The size of the remaining surface when $v\neq 0$ depends on
the ratio $\chi/t$: when $\chi/t\rightarrow 0$, the gapped part vanishes,
whereas as $\chi/t\rightarrow\infty$ only the parts of the BF surface around
the axis survive, and the gap is finite almost everywhere. A typical result is
depicted in Fig. 3.

Figure 3: New BF surface (yellow) after the inversion is spontaneously broken.
Parts of the original BF surface (blue) outside of the new BF surface are
gapped out, whereas those inside are split into the new surface, as described
in the text. The values of the parameters are here chosen to be $t=1$,
$\chi=0.5$, and $v=0.6$.

##  X Summary and discussion

We have discussed the formation of the BF surface in the multiband
superconductors with inversion symmetry by pointing out the analogy with
classical relativity, furnished by the $SO(4)$-generator form of the low-
energy Hamiltonian which ensues when the time reversal is broken, either in
the superconducting or the normal phase, or in both. In this analogy the zero-
energy solutions of the BdG Hamiltonian correspond to four-velocities for
which classical Lorentz force in fictitious corresponding electric and
magnetic fields vanishes, and the BF surface is linked to the orthogonality of
the electric and magnetic fields. The latter condition is found to be
tantamount to vanishing of the Pfaffian of the low-energy Hamiltonian. The
relativistic analogy suggested a simple single-particle lattice model which
falls into the class D, that is, which yields a hopping Hamiltonian that
anticommutes with an antiunitary operator of a positive square, the latter
encoding the joint particle-hole and inversion symmetries of the
superconducting state. We then added a two-body repulsive term between nearest
neighbors on the lattice, to find that the inversion symmetry becomes
spontaneously broken at $T=0$ at infinitesimal such interaction. The BF
surface of the noninteracting lattice model deforms and reduces in size as a
result, but does not completely disappear.

The relativistic analogy offers maybe the simplest way to understand why a BF
surface arises when the time reversal is broken: since the effective
Hamiltonian is a four-dimensional $SO(4)$ generator which belongs to the
$(1/2,1/2)$ representation equivalent to standard boosts and rotations in the
Minkowski space, the quasiparticle spectrum is a linear combination of two
familiar spectra of spin-1/2 particles (Eq. (32)). As such it yields a single
zero-energy condition on the three momentum components, which when satisfied
leads to a surface in the momentum space. The preservation of the time
reversal prevents the condition to be fulfilled, and leads to two equations on
momenta with zero energy, i. e. a line.

Following the same mode-elimination procedure of Ref. link2,  for the present
inversion-symmetric case, outlined also here in Appendix C, one finds that at
weak Cooper paring BF surfaces will inevitably form around those points in the
momentum space where the intraband pairing between the light states happens to
vanish. The size of the BF surface is then $\sim\Delta^{2}/E_{0}$, where
$\Delta$ is the overall norm of the multi-component pairing order parameter,
and $E_{0}$ the energy gap to the first higher energy level in the normal
state, and thus typically small in the weak-coupling limit. In precise analogy
to the case without inversion,link2  increasing the pairing order parameter
initially inflates the BF surfaces, but only up to a point, beyond which it
begins to reduce them, until they disappear via an example of a Lifshitz
transition. lifshitz

It was pointed out oh ; tamura  that the inversion symmetry is in danger of
being spontaneously broken by residual interaction effects, and the
concomitant BF surface further reduced or gapped out. This ensues, however,
only if the effective residual interactions between the low-energy
quasiparticles with momenta near the BF surface are attractive in the
particular inversion-symmetry-breaking channel, which seems difficult to
ascertain without a specific model in mind. To that purpose we proposed a
lattice model which is motivated by the phenomenon of the BF surface in the
inversion-symmetric and time-reversal-broken multiband superconductor; the
only requirement on it is that it falls into the class D bzdusek , as dictated
by the symmetries of the superconducting problem under consideration. The
model then features spinless fermions hopping on the three-dimensional Lieb
lattice and repelling each other when found on nearest-neighboring sites. We
show that this model indeed exhibits a surface of Weyl points, which spans the
entire Brillouin zone, and serves therefore as a magnified version of a BF
surface. Infinitesimal nearest-neighbor interaction leads however to
spontaneous dynamical breaking of the D-class condition in the mean-field
Hamiltonian, which should be interpreted as breaking of inversion in the
superconducting problem. The non-interacting BF surface is found to be
deformed and reduced by this mechanism, with its final size dependent on the
model parameters.

The dynamical inversion symmetry breaking in the present lattice model is
interesting from the point of view of the theory of quantum phase transitions
in fermionic systems. At the level of the model, it is not really, as usual, a
symmetry (a commuting linear operator) that becomes broken, but an
“antisymmetry” (an anticommuting, and even antiunitary operator) that does so.
Other modifications of our lattice model with different two-body interaction
terms, or disorder, may lead to further insights into this new phenomenon.

##  XI Acknowledgments

We are grateful to Igor Boettcher and Carsten Timm for useful comments on the
manuscript. JML is supported by DFG grant No. LI 3628/1-1, and IFH by the
NSERC of Canada.

##  Appendix A CP symmetry of the BdG Hamiltonian

Let us redefine the quantum-mechanical action for the Bogoliubov
quasiparticles in the superconducting state:

$S=k_{B}T\sum_{\omega_{n},\textbf{p}}\Phi^{\dagger}(\omega_{n},\textbf{p})[-i\omega_{n}+H_{\rm{BdG}}(\textbf{p})]\Phi(\omega_{n},\textbf{p}),$
(89)

where the Nambu spinor is now simply
${\Phi(\omega_{n},\textbf{p})=\big{(}\psi(\omega_{n},\textbf{p}),\psi^{*}(-\omega_{n},-\textbf{p})\big{)}^{\rm
T}}$, without the unitary part of the time reversal in the lower, hole
component. In this representation the BdG Hamiltonian assumes the standard
form: agterberg

$\displaystyle H_{\rm BdG}(\textbf{p})$ $\displaystyle=$
$\displaystyle\begin{pmatrix}H(\textbf{p})-\mu&\Delta(\textbf{p})\\\
\Delta^{\dagger}(\textbf{p})&-\big{[}H^{\rm
T}(-\textbf{p})-\mu\big{]}\end{pmatrix}\>,$ (90)

related to our form in an obvious way. The pairing matrix needs to satisfy

$\Delta^{\rm T}(-\textbf{p})=-\Delta(\textbf{p}).$ (91)

It is straightforward to check that the BdG Hamiltonian in this representation
possesses the particle-hole symmetry (by construction) in the following form:

$(\sigma_{1}\otimes 1_{N\times N})H_{\rm BdG}^{\rm
T}(-\textbf{p})(\sigma_{1}\otimes 1_{N\times N})=-H_{\rm BdG}(\textbf{p}).$
(92)

We now additionally assume that there is an inversion symmetry, so:

$P^{\dagger}H(-\textbf{p})P=H(\textbf{p})$ (93)
$P^{\dagger}\Delta(-\textbf{p})P=\Delta(\textbf{p}).$ (94)

For the BdG Hamiltonian this implies that

$(1\otimes P^{\dagger})H_{\rm BdG}(-\textbf{p})(1\otimes P)=H_{\rm
BdG}(\textbf{p})$ (95)

Recognizing that transposing the (Hermitian) BdG Hamiltonian is the same as
complex-conjugating it, one discerns the antiunitary operator
$\mathcal{A}^{\prime}$

$\mathcal{A}^{\prime}=(\sigma_{1}\otimes\mathcal{P})K$ (96)

which has the desired effect of anticommuting with the BdG Hamiltonian at
fixed momentum, i. e.

$(\mathcal{A}^{\prime})^{-1}H_{\rm
BdG}(\textbf{p})\mathcal{A}^{\prime}=-H_{\rm BdG}(\textbf{p})\>.$ (97)

The square of this operator is now

$(\mathcal{A}^{\prime})^{2}=(\sigma_{1})^{2}\otimes(PP^{*})=PP^{*}.$ (98)

When $P$ is simply a unit matrix this is $+1$, but when it is not, even if one
assumes the usual Hermiticity of $P$, i.e. that $P=P^{\dagger}$, the square of
$\mathcal{A}^{\prime}$ depends on whether the matrix for $P$ in the given
representation is real or imaginary. A simple example of the Dirac Hamiltonian
with imaginary Hermitian $P$ is provided in the next appendix. Of course, one
can always from the outset work in the eigenbasis of $P$ itself, in which it
is a real diagonal matrix, and in which consequently $PP^{*}=P^{2}=+1$. The
antiunitary operator that anticommutes with the BdG Hamiltonian therefore
always exists. Another way to see that is to construct the BdG Hamiltonian by
defining the hole component of the Nambu spinor as a time-reversed particle
component, as done in the body of the paper. Then the fact that
$\mathcal{A}^{2}=+1$ simply reflects the fundamental commutation relation
between spatial transformations such as inversion and time reversal. More on
this is next.

##  Appendix B Commutation between inversion and time reversal

Let us provide an argument as to why inversion and time reversal operations
need to be assumed to be commuting in general on the familiar example of the
Dirac Hamiltonian. First, modulo an overall sign, there is a unique four-
dimensional representation of five-dimensional Clifford algebra, which can
always be chosen so that three of the matrices are real ($\alpha_{i}$,
$i=1,2,3$), and two imaginary ($\beta_{i}$, $i=1,2$).herbutprb  We may choose
all five matrices to be Hermitian, and to be squaring to unity. These are
simple generalizations of the known properties of the Pauli matrices. Consider
then a massless inversion-symmetric Dirac Hamiltonian, which is the sum of two
Weyl Hamiltonians of opposite chirality. It can be written, for example, as

$H_{W}(\textbf{p})=\sum_{i=1}^{3}p_{i}\alpha_{i}.$ (99)

There is not one, but two options for the matrix part of the inversion
operation $\mathcal{P}$ at this stage: $P_{1}=\beta_{1}$, or
$P_{2}=\beta_{2}$. Both have the desired effect on the massless inversion-
symmetric Dirac Hamiltonian:

$P_{i}^{\dagger}H_{W}(-\textbf{p})P_{i}=H_{W}(\textbf{p}),$ (100)

and both are Hermitian and unitary matrices.

Likewise, there are two options for the time reversal operator:
$T_{1}=\beta_{1}K$, and $T_{2}=\beta_{2}K$. The time reversal operation
$\mathcal{T}$ in the momentum space is then given by the combined action of
$T_{k}$ and the momentum reversal $\textbf{p}\rightarrow-\textbf{p}$. Since
$\beta_{1,2}$ are imaginary, we have

$[\mathcal{P}_{i},\mathcal{T}_{j}]=0$ (101)

only if $i\neq j$, otherwise the two operations anticommute instead of
commuting. Let us chose then one anticommuting pair, say $\mathcal{P}_{1}$ and
$\mathcal{T}_{1}$. Is this a sensible choice? Add a relativistic mass term to
the massless Dirac Hamiltonian, and consider

$H_{D}(\textbf{p})=H_{W}(\textbf{p})+m\beta_{k},$ (102)

with $k=1$ or $k=2$. The mass $m$ is real. These are the only two options for
the mass term, since there are no further four dimensional matrices that would
anticommute with all three matrices $\alpha_{i}$. If we chose $k=1$, $H_{D}$
is symmetric under inversion operation $\mathcal{P}_{1}$, but the mass term
violates time reversal $\mathcal{T}_{1}$. If we had chosen $k=2$, then the
mass term would respect the time reversal $\mathcal{T}_{1}$, but violate the
inversion $\mathcal{P}_{1}$. Obviously if we would choose the second
anticommuting pair $\mathcal{P}_{2}$ and $\mathcal{T}_{2}$ it would be the
other way around. Still, either choice of the mass term would violate one of
the two discrete symmetries, if we allowed them to anticommute with each
other.

So the very existence of massive relativistic fermions in the world which is
both inversion-symmetric and time-reversal-symmetric implies that these two
symmetries must be assumed to be commuting. Then the mass term uniquely
selects the corresponding operators: if $k=1$, the required pair is
$\mathcal{P}_{1}$ and $\mathcal{T}_{2}$.

We may also note, in relation to the previous appendix, that in the above
representation, in spite of $P_{k}^{2}=1$, $P_{k}P_{k}^{*}=-1$, for both
$k=1,2$.

##  Appendix C The effective Hamiltonian in the canonical representation of
$SO(3)\times SO(3)$

Figure 4: (a) The energy bands of light particle and hole states (red) and the
heavy particle and hole states. Each energy band is doubly degenerate. (b) The
energy dispersion of the BdG quasiparticles with broken TR symmetry in the
special direction where the intraband coupling between the light states
vanishes and $f_{1,2}\propto\mathcal{O}(\Delta^{3})$. The interband pairing
induces a shift in momentum and energy with
$f_{3}\propto\mathcal{O}(\Delta^{2})$ and
$f_{0}\propto\mathcal{O}(\Delta^{2})$, respectively. Meanwhile $f_{1,2}$
introduces a gap between the critical- and the massive energy band. The two
critical energy bands are intersecting the Fermi level at $p_{1}$ and $p_{2}$
in that special direction. If one deviates from this special direction, the
two points $p_{1}$ and $p_{2}$ will approach each other until they merge. This
way a closed BF surface nucleates. (c) The energy dispersion of the BdG
quasiparticles with preserved TR symmetry again in the special direction. We
see that $f_{0}=0$, i.e. there is no shift in energy of the energy bands
induced by the interband pairing but only a shift in momentum. This will lead
to either point or line nodes.

In this appendix, we consider systems in the normal state with $M=2$ and
$\mathcal{T}^{2}=-1$, i.e. the energy band $E_{i}(\textbf{p})$ is doubly
degenerated due to the inversion symmetry and has the eigenstates
$\phi_{+,i}(\textbf{p})$ and $\phi_{-,i}(\textbf{p})$. The emergence of the BF
surface in such a system will be explained in terms of shifts in momentum and
in energy of the critical and massive energy bands due to inter- and intraband
pairing. To this end, the effective Hamiltonian is written in the canonical
representation of $SO(3)\times SO(3)$ and has the form

$\displaystyle H_{ef}$ $\displaystyle=$ $\displaystyle
f_{0}(\textbf{p})\>\sigma_{3}\otimes\mathbb{1}_{2\times
2}+\sum_{j=1}^{3}[f_{j}(\textbf{p})\mathbb{1}_{2\times
2}\otimes\sigma_{j}]\>,$ (103)

where the function $f_{0}(\textbf{p})$ is defined as

$f_{0}(\textbf{p})=\sqrt{Z_{1}(\textbf{p})^{2}+Z_{2}(\textbf{p})^{2}+Z_{3}(\textbf{p})^{2}}$
(104)

with $Z_{i}(\textbf{p})$ being the coefficient of
$Z_{i}(\textbf{p})\sigma_{i}\otimes\mathbb{1}_{2\times 2}$. The function
$f_{0}(\textbf{p})$ acts as a “pseudomagnetic field” responsible for the
emergence of the BF agterberg ; brydon . How $f_{0}(\textbf{p})$ is related to
the electric field $\textbf{a}(\textbf{p})$ and the magnetic field
$\textbf{b}(\textbf{p})$ in the body of paper is shown in the second part of
this appendix.

In the normal state with the pairing matrix $\Gamma(\textbf{p})=0$ and the
superconducting gap being zero, i.e. $\Delta=0$, the effective Hamiltonian is
only proportional to $f_{3}(\textbf{p})=E_{1}(\textbf{p})-\mu$ and describes
the two particle and two hole states of the light mode which arises due to the
inversion symmetry, see Fig. 4.

However, in the superconducting state with broken time-reversal (TR) symmetry,
i.e. $\Gamma=\Gamma_{1}-{\rm i}\Gamma_{2}$ with $\Gamma_{1,2}$ being finite,
the term $f_{3}(\textbf{p})-\big{[}E_{1}(\textbf{p})-\mu\big{]}$ introduces a
shift in the momentum of the energy band of order $\mathcal{O}(\Delta^{2})$
due to interband pairing, $f_{0}(\textbf{p})$ introduces a shift in the energy
of order $\mathcal{O}(\Delta^{2})$, while
$f_{1,2}(\textbf{p})\propto\mathcal{O}(\Delta)+\mathcal{O}(\Delta^{3})$ which
introduces a gap between the light particle and hole states. Whenever the
leading order term of $f_{1,2}(\textbf{p})$, which describes the intraband
pairing between the light particle and light hole state, vanishes in a certain
direction and $f_{1,2}(\textbf{p})\propto\mathcal{O}(\Delta^{3})$, the shift
in momentum and energy of the energy bands leads to two points $p_{1}$ and
$p_{2}$ along that special direction, where the energy bands of the
quasiparticles are zero, see Fig. 4. If one deviates from this special
direction, the two points will come closer to each other and merge at one
point due to continuity. This will lead to a closed BF surface. In the case of
a superconducting state with preserved TR symmetry, $f_{0}(\textbf{p})\equiv
0$, which means that no shift in the energy occurs. There is only a shift in
the momentum of the energy bands introduced by
$f_{3}(\textbf{p})-\big{[}E_{1}(\textbf{p})-\mu\big{]}$. This leads in general
to a line of gapless nodes.

###  C.1 The relation between the effective Hamiltonan and inter- and
intraband pairing

Next, we want to relate the intra- and interband pairing of the different
energy bands to the functions $f_{\alpha}(\textbf{p})$ with
$\alpha=\\{1,2,3,4\\}$ which shift the light states in energy and momentum
($f_{0,3}$) and open up a gap between the critical and massive energy bands
($f_{1,2}$).

To derive the effective Hamiltonian, we employ Eq. (21) where the effective
Hamiltonian is (again) given by

$H_{ef}(\textbf{p})=H_{l}(\textbf{p})-H_{lh}(\textbf{p})H_{h}^{-1}(\textbf{p})H_{lh}^{\dagger}(\textbf{p})\>.$
(105)

For the doubly degenerated energy bands, the matrix describing the light
states $H_{l}=H_{11}^{(0)}$, while $H_{ii}^{(0)}$ with $i\geq 2$ denote the
intraband pairing between the heavy states. The matrices $H_{ii}^{(0)}$
describing the energy dispersion and the intraband pairing between one energy
band are thus given by

$H_{ii}^{(0)}=\begin{pmatrix}E_{i}(\textbf{p})-\mu&R_{i}^{(0)}&0&0\\\
\bar{R}_{i}^{(0)}&-E_{i}(\textbf{p})+\mu&0&0\\\
0&0&E_{i}(\textbf{p})-\mu&R_{i}^{(0)}\\\
0&0&\bar{R}_{i}^{(0)}&-E_{i}(\textbf{p})+\mu\end{pmatrix}\>,$ (106)

with
$R_{i}^{(0)}=\phi_{\pm,i}^{\dagger}(\textbf{p})\Gamma\phi_{\pm,i}(\textbf{p})$.
The form of the matrices is determined by the inversion and TR symmetry of the
normal state Hamiltonian and the pairing matrix $\Gamma_{1,2}$ which is
defined by the operator $D=\mathcal{P}\cdot\mathcal{T}=P\cdot UK$ with
$D^{2}=-1$. The unitary part of $D$ is defined as $\tilde{\mathcal{U}}=P\cdot
U$. A consequence of this property is the fact that the eigenstates transform
as

$\displaystyle\phi_{-,i}(\textbf{p})$ $\displaystyle=$
$\displaystyle+D\phi_{+,i}(\textbf{p})$ (107)
$\displaystyle\phi_{+,i}(\textbf{p})$ $\displaystyle=$
$\displaystyle-D\phi_{-,i}(\textbf{p})\>,$ (108)

while the pairing term transforms as

$\tilde{\mathcal{U}}^{\dagger}\>\Gamma(\textbf{p})\>\tilde{\mathcal{U}}=\Gamma(\textbf{p})^{\rm
T}\>.$ (109)

The elements $(1,4),(2,3),(3,2)$, and $(4,1)$ of $H_{ii}^{(0)}$ are zero,
since these matrix elements describe the coupling between the Kramers pairs
and

$\displaystyle\phi_{+,i}^{\dagger}\Gamma\phi_{-,i}=\big{(}-D\phi_{-,i}\big{)}^{\dagger}\Gamma
D\phi_{+,i}=-\phi_{+,i}^{\dagger}\Gamma\phi_{-,i}=0\>.$ (110)

The matrices $H_{ij}^{(0)}$ with $i\neq j$ describe the coupling between the
light state and the $j$th heavy state in the case of $i=1$ and $j\neq 1$ and
between the $i$th and $j$th heavy state. They are defined as

$H_{ij}^{(0)}=\begin{pmatrix}0&C_{ij}^{(0)}&0&B_{ij}^{(0)}\\\
\bar{A}_{ij}^{(0)}&0&\bar{D}_{ij}^{(0)}&0\\\ 0&-D_{ij}^{(0)}&0&A_{ij}^{(0)}\\\
-\bar{B}_{ij}^{(0)}&0&\bar{C}_{ij}^{(0)}&0\end{pmatrix}\>,$ (111)

where the coefficients are given by

$\displaystyle A_{ij}^{(0)}$ $\displaystyle=$
$\displaystyle\phi_{-,i}^{\dagger}(\textbf{p})\Gamma\phi_{-,j}(\textbf{p})=\phi_{+,j}^{\dagger}(\textbf{p})\Gamma\phi_{+,i}(\textbf{p})\>,$
(112) $\displaystyle B_{ij}^{(0)}$ $\displaystyle=$
$\displaystyle\phi_{+,i}^{\dagger}(\textbf{p})\Gamma\phi_{-,j}(\textbf{p})\>,$
(113) $\displaystyle C_{ij}^{(0)}$ $\displaystyle=$
$\displaystyle\phi_{+,i}^{\dagger}(\textbf{p})\Gamma\phi_{+,j}(\textbf{p})\>,$
(114) $\displaystyle\bar{D}_{ij}^{(0)}$ $\displaystyle=$
$\displaystyle\phi_{+,i}^{\dagger}(\textbf{p})\Gamma^{\dagger}\phi_{-,j}(\textbf{p})\>.$
(115)

Note that the diagonal blocks $H_{ii}^{(0)}$ are Hermitian matrices whereas
the off-diagonal blocks $H_{ij}^{(0)}$ in general are not. To obtain a
physical intuition for how the functions $f_{\alpha}$ are related to the
inter- and intraband pairing, we consider the result of second-order
perturbation theory. Since the matrix blocks belonging to $H_{lh}$ are in
first order of $\Delta$, we neglect all intra- and interband coupling between
the heavy states, i.e. we set $\Gamma=0$ in all $H_{ij}^{(0)}$ with $i>2$,
which yields

$H_{ef}=H_{11}^{(0)}-\sum_{k=2}^{N}H_{1k}^{(0)}\big{(}H_{kk,\Gamma=0}^{(0)}\big{)}^{-1}H_{1k}^{(0)\dagger}+\mathcal{O}(\Delta^{3})\>.$
(116)

The BF surface emerges when the leading order of the intraband pairing between
the light particle and light hole state is vanishing in one special direction,
which is described in the effective Hamiltonian by

$f_{1}(\textbf{p})-if_{2}(\textbf{p})=\phi_{+,1}^{\dagger}(\textbf{p})\Gamma\phi_{+,1}(\textbf{p})+\mathcal{O}(\Delta^{3})\>.$
(117)

This can also be rewritten in terms of $\Gamma=\Gamma_{1}-{\rm i}\Gamma_{2}$
as

$\displaystyle f_{1}$ $\displaystyle=$
$\displaystyle\phi_{+,1}^{\dagger}(\textbf{p})\Gamma_{1}\phi_{+,1}(\textbf{p})+\mathcal{O}(\Delta^{3})\>,$
(118) $\displaystyle f_{2}$ $\displaystyle=$
$\displaystyle\phi_{+,1}^{\dagger}(\textbf{p})\Gamma_{2}\phi_{+,1}(\textbf{p})+\mathcal{O}(\Delta^{3})\>.$
(119)

The interband pairing between the light state and the heavy states shifts the
energy band crossing of the light particle and light hole state in momentum
and is given by

$f_{3}(\textbf{p})=E_{1}(\textbf{p})-\mu+\sum_{k=2}^{N}\frac{1}{2[E_{k}(\textbf{p})-\mu]}\big{[}|\phi_{+,1}^{\dagger}(\textbf{p})\Gamma\phi_{-,k}(\textbf{p})|^{2}+|\phi_{+,1}^{\dagger}(\textbf{p})\Gamma\phi_{+,k}(\textbf{p})|^{2}+|\phi_{+,1}^{\dagger}(\textbf{p})\Gamma^{\dagger}\phi_{-,k}(\textbf{p})|^{2}+|\phi_{+,1}^{\dagger}(\textbf{p})\Gamma^{\dagger}\phi_{+,k}(\textbf{p})|^{2}\big{]}\>,$
(120)

The function $f_{0}(\textbf{p})$ which introduces a shift in energy of the
energy bands and is thus responsible for the nucleation of the BF surface, is
defined as $f_{0}=\sqrt{Z_{1}^{2}+Z_{2}^{2}+Z_{3}^{2}}$. The functions
$Z_{1}(\textbf{p})$ and $Z_{2}(\textbf{p})$ are defined as the interband
pairing between the light states and the heavy states and are only finite when
TR symmetry is broken (i.e. $\Gamma_{2}$ is finite), as can be seen in

$Z_{1}(\textbf{p})-iZ_{2}(\textbf{p})=\sum_{k=2}^{N}\frac{1}{E_{k}(\textbf{p})-\mu}\big{[}\big{(}\phi_{+,1}^{\dagger}(\textbf{p})\Gamma^{\dagger}\phi_{+,k}(\textbf{p})\big{)}\big{(}\phi_{+,1}\Gamma\phi_{-,k}(\textbf{p})\big{)}-\big{(}\phi_{+,1}^{\dagger}\Gamma^{\dagger}\phi_{-,k}(\textbf{p})\big{)}\big{(}\phi_{+,1}^{\dagger}(\textbf{p})\Gamma\phi_{+,k}(\textbf{p})\big{)}\big{]}\>,$
(121)

or also with $\Gamma=\Gamma_{1}-{\rm i}\Gamma_{2}$

$\displaystyle Z_{1}$ $\displaystyle=$ $\displaystyle 0$ (122) $\displaystyle
Z_{2}$ $\displaystyle=$
$\displaystyle\sum_{k=2}^{N}\frac{2}{E_{k}(\textbf{p})-\mu}[(\phi_{+,1}^{\dagger}(\textbf{p})\Gamma_{2}\phi_{+,k}(\textbf{p}))(\phi_{+,1}^{\dagger}(\textbf{p})\Gamma_{1}\phi_{-,k}(\textbf{p}))-(\phi_{+,1}^{\dagger}(\textbf{p})\Gamma_{1}\phi_{+,k}(\textbf{p}))(\phi_{+,1}^{\dagger}(\textbf{p})\Gamma_{2}\phi_{-,k}(\textbf{p}))]\>.$
(123)

The same is true for $Z_{3}$ which is given by

$Z_{3}(\textbf{p})=\sum_{k=2}^{N}\frac{1}{2(E_{k}(\textbf{p})-\mu)}\big{[}|\phi_{+,1}^{\dagger}(\textbf{p})\Gamma\phi_{-,k}(\textbf{p})|^{2}+|\phi_{+,1}^{\dagger}(\textbf{p})\Gamma\phi_{+,k}(\textbf{p})|^{2}-|\phi_{+,1}^{\dagger}(\textbf{p})\Gamma^{\dagger}\phi_{-,k}(\textbf{p})|^{2}-|\phi_{+,1}^{\dagger}(\textbf{p})\Gamma^{\dagger}\phi_{+,k}(\textbf{p})|^{2}\big{]}\>.$
(124)

In Eqs. (121)-(124), we see explicitly that $f_{0}(\textbf{p})\equiv 0$ for a
TR-preserved superconducting state with $\Gamma_{2}=0$. This implies that the
interband pairing induces no shift in energy of the critical and massive
energy bands. The only shift induced by the interband pairing is in the
momentum of the light particle and hole states which exhibits only line or
point nodes, as can be seen in Fig. 4.

###  C.2 The effective Hamiltonian in the $SO(4)$ representation and in the
canonical $SO(3)\times SO(3)$ representation

In this section, we want to relate the $SO(4)$ representation of the effective
Hamiltonian to the canonical $SO(3)\times SO(3)$ representation.

Although the $SO(4)$ commutation relations guarantee the existence of the
unitary transformation that would bring the matrices in Eqs. (44)-(49) into
the standard form, we nevertheless provide it here, for completeness:

$\displaystyle\mathcal{U}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\begin{pmatrix}1&-i&i&1\\\ -1&-i&-i&1\\\
-i&-1&-1&i\\\ -i&1&-1&-i\end{pmatrix}\>.$ (125)

Then

$\mathcal{U}R_{+}\mathcal{U}^{\dagger}=\frac{1}{2}(1\otimes\sigma_{3},1\otimes\sigma_{1},1\otimes\sigma_{2}),$
(126) $\mathcal{U}R_{-}\mathcal{U}^{\dagger}=\frac{1}{2}(\sigma_{2}\otimes
1,\sigma_{3}\otimes 1,\sigma_{1}\otimes 1),$ (127)

which are cyclic permutations of the canonical form $1\otimes\sigma_{k}/2$,
$k=1,2,3$, and $\sigma_{k}/2\otimes 1$, respectively.

Hence, we can relate the coefficients $a_{i}(\textbf{p})$ and
$b_{i}(\textbf{p})$ of the $SO(4)$ representation to their canonical counter
part $f_{i}(\textbf{p})$ and $Z_{i}(\textbf{p})$ in the following way:

$\displaystyle a_{1}(\textbf{p})$
$\displaystyle=\frac{1}{2}(f_{3}(\textbf{p})+Z_{2}(\textbf{p})),\>b_{1}(\textbf{p})$
$\displaystyle=\frac{1}{2}(f_{3}(\textbf{p})-Z_{2}(\textbf{p}))$ (128)
$\displaystyle a_{2}(\textbf{p})$
$\displaystyle=\frac{1}{2}(f_{1}(\textbf{p})+Z_{3}(\textbf{p})),\>b_{2}(\textbf{p})$
$\displaystyle=\frac{1}{2}(f_{1}(\textbf{p})-Z_{3}(\textbf{p}))$ (129)
$\displaystyle a_{3}(\textbf{p})$
$\displaystyle=\frac{1}{2}(f_{2}(\textbf{p})+Z_{1}(\textbf{p})),\>b_{3}(\textbf{p})$
$\displaystyle=\frac{1}{2}(f_{2}(\textbf{p})-Z_{1}(\textbf{p}))\>.$ (130)

The condition for the emergence of the BF surface (see Eq. (33)) can now be
expressed by the coefficients $f_{i}$ and $Z_{i}$ as

$\boldsymbol{a}(\textbf{p})\cdot\boldsymbol{b}(\textbf{p})=\sum_{i=1}^{3}[f_{i}^{2}(\textbf{p})-Z_{i}^{2}(\textbf{p})]=\sum_{i=1}^{3}f_{i}^{2}(\textbf{p})-f_{0}^{2}(\textbf{p})=0\>,$
(131)

which is the same condition as Eq.(10) of Ref. link2, , and corresponds to the
condition that the Pfaffian of the effective Hamiltonian has to vanish. Or in
other words: the condition for the emergence of the BF surface in the $SO(4)$
representation is the orthogonality of the electric and magnetic fields
$\textbf{a}(\textbf{p})$ and $\textbf{b}(\textbf{p})$, whereas in the
canonical representation the condition for the emergence of the BF surface
translates into the fact that the interband coupling, which introduces the
shift in momentum of the critical and massive energy bands as well as the gap
between the critical and massive bands, has to be as large as the shift in
energy of the bands induced by the “pseudomagnetic” field $f_{0}$.

For a superconducting state with preserved TR symmetry, $Z_{i}(\textbf{p})=0$
(as proven above) which yields that the electric and magnetic fields are
parallel with $\textbf{a}(\textbf{p})=\textbf{b}(\textbf{p})$ (as demanded in
Eq. (50)). Furthermore, we see that the component
$a_{3}(\textbf{p})=b_{3}(\textbf{p})=0$ vanishes when TR symmetry is
preserved, i.e. $\Gamma_{2}$ is zero, which is in agreement with Eq. (51).

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