Fractional and Majorana Fermions

R. Jackiw (MIT)

A bit of physics history about Dirac and his equation:

Dirac was looking for a relativistic equation for electrons and

eventually arrived at the following first-order matrix equation

i γµ ∂µΨ(x) +mΨ(x) = 0

(α · p+ βm)Ψ = i ∂tΨ

α = γ0 γ, β = γ0, p = 1i ∇

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2

To expose properties of the Dirac equation, we make the

usual decomposition

Ψ = e−iEtψ ⇒ [α · p+ βm]ψ = Eψ

and discover that there exist positive energy solutions E > 0,

which can describe electrons.

But because “square roots” come in both signs, there exist

also negative energy solutions, E < 0. These need interpre-

tation because they cannot describe electrons, which carry

positive energy. After some hesitation, Dirac concluded that

the negative energy solutions correspond to anti-electrons,

viz. positrons, which soon were discovered.

In a further conceptual leap, Dirac posited that in the ground

state all negative energy levels are filled, but nevertheless the

charge of the ground state is zero. This means that the

Dirac equation is really a many particle equation, where the

particles populate the energy levels.

3

The Dirac equation is a beautiful equation, with a rich hidden

physical structure. It goes beyond a single particle interpreta-

tion and it predicts new (anti-)particles: the positrons. These

characteristics make it a beautiful equation for physics.

Also it is a beautiful equation for mathematics because by

using matrices, it succeeds in taking a square root of a second

order differential equation.

The mathematical and physical beauty of the Dirac equation

suggests to mathematicians and physicists that deformations

of the equation may also yield beautiful and interesting re-

sults.

4

Which deformations should we consider?

One alteration that can be made is to reconsider the equa-

tion in dimensions different from the three spatial and one

time dimension. If we take fewer dimensions we gain the

mathematical advantage of simplicity and also have the pos-

sibility of describing physical systems that are confined to

lower dimensions, for example to a line or to a plane. Such

configurations can occur in condensed matter physics. There

one encounters situations where the low energy dynamics is

well described by a matrix equation, linear in the momenta.

Also there may be a constant mass term which separates

the positive energy solutions from the negative energy ones

by a “gap.” Depending on the nature of the material, the

equation may describe excitations on a line, on the plane in

addition to those in the three-dimensional bulk.

5

Dirac Equations

(First-order matrix equations)

[α · p+ βm]ψ = Eψ

continuum solutions E > |m| and E < -|m|

“vacuum”:

Particle interpretation: E < 0 states filled (antiparticles)E > 0 states empty (particles)

Condensed matter: E < 0 states filled (valence band)interpretation: E > 0 states empty (conduction band)

m produces gap 2 |m|“vacuum” carries no net charge

One dimension, physics on a line, Dirac matrices realized with

2× 2 Pauli matrices

1-d : α = σ2, β = σ1, p = 1i

ddx

6

A further more profound deformation allows the mass term

m to depend on position.

What sort of dependence should we consider?

Surely a weak dependence will produce only insignificant changes

from the usual homogenous mass case; we are interested in a

significant dependence on position, which could significantly

alter the physical situation. Note that the gap 2 |m| dependsonly on the magnitude of m, and not on its sign. +m pro-

duces the same gap as -m. This suggest a deformation of

the mass term that interpolates between positive and nega-

tive values. In this way we are led to a Dirac equation in the

presence of a defect.

7

(mass term position-dependent, soliton)

m→ ϕ (r)

[α · p+ β ϕ (r)]ψ = Eψ

Find:

continuum solutions E > 0, E < 0

AND isolated, normalizable E = 0 solution

“mid-gap” state is found by explicit calculation

is guaranteed by index theorems

CENTRAL QUESTION: in “vacuum” is mid-gap state empty

or filled, what is its charge?

UNEXPECTED ANSWER: charge Q = ± 12.

8

Fractional Charge (Analytic Derivation)

Vacuum charge density:

ρ(r) =

0∫

−∞dE ρE (r) ρE = ψ

†EψE

renormalized charge in soliton background

Q =

∫

dr

0−∫

−∞dE

(

ρsE(r)− ρ0E(r))

Evaluation simple in the presence of

an energy reflection symmetry:

ρ-E = ρE (charge conjugation)

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Fractional Charge Calculation

Completeness:∞∫

−∞dE ψ

†E(r)ψE(r

′) = δ(r− r′)

⇒∞∫

−∞dE[ρsE(r)− ρ0E (r)] = 0

Conjugation (ρE = ρ-E) and zero mode ⇒0−∫

−∞dE

(

2ρsE(r)− 2ρ0E (r))

+ ψ†E=0 (r)ψE=0(r) = 0

0−∫

−∞dE

(

ρsE(r)− ρ0E(r))

= −1

2ψ†E=0(r)ψE=0(r)

Q = −1

2Any dimension!

Empty mid-gap state: Q = −12

Filled mid-gap state: Q = +12

Eigenvalue, not expectation value!

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Fractional Charge

(Second Quantized Description)

Expansion of quantum Fermi field in presence of defect &

zero mode

Ψ =∑

E>0

(bE ψsE + d

†Eψ

∗sE ) + aψE=0

Ψ† =∑

E>0

(b†E ψ

s∗E + dE ψ

sE) + a†ψE=0

a†a+ aa† = 1

How to realize on states:

a | + >=| − >, a† | + >= 0,

a | − >= 0, a† | − >=| + >

11

Q =1

2

∫

d r(Ψ†Ψ−ΨΨ†)

=1

2

∑

E>0

(b†E bE + dE d

†E − bE b

†E − d

†E dE) +

1

2(a† a− a a†)

=∑

E>0

(b†E bE − d

†E dE) + a† a− 1

2

Q | − >= − 1

2| − >,

Q | + >= +1

2| + >

eigenvalue !

12

One Dimensional Example (Polyacetylene)

Dirac equation with varying mass [σ2p+ σ1ϕ(x)]ψ = Eψ

E ≶ 0

0 − d

dx

ddx 0

+

0 ϕ(x)

ϕ(x) 0

ψuE

ψlE

= E

ψuE

ψlE

E = 0

0 − d

dx + ϕ(x)

ddx + ϕ(x) 0

ψu0

ψl0

= 0

ψu,l0 = N exp ∓ ∫ x dx′ ϕ(x′)

Normalizable provided ϕ(x) has kink profile.

Rebbi & RJ, PRD 13, 3398 (76)

Su, Schrieffer & Heeger, PRL 42, 1698 (79)

[high conductivity in polymers]

13

Peierls’ Instability in Polyacetylene

O

B

A

Energetics of Polyacetylene Phonon Field

0

Energy density V (φ), as a function of a

constant phonon field φ. The symmetric

stationary point, φ = 0, is unstable. Sta-

ble vacua are at φ = + |φ0|, (A) andφ =

−|φ0|, (B).

Profiles of Phonon Field

A

S

S

B

The two constant fields, ± | φ0 |, cor-

respond to the two vacua (A and B).

The kink-soliton fields, ±φs, interpolate

between the vacua and represent domain

walls.

14

Polyacetylene realization of (1− d) “Dirac” equation

Kinetic term: linearization at Fermi level p

Potential term:

Peierls’ instability

energy profile seen

by phonon field

0m

B A

unstable

m

0

V (ϕ)

ϕ0

configurations

of phonon

field ϕS

S

mϕ(x)

+|m|

-|m|

ϕ(x)

15

Fractional Charge Schematic

B

A

2S

S S

Two soliton state carries one fewer link

relative to no-soliton vacuum A.

Separate solitons to ∞ ⇒split quantum numbers of link ⇒fermion number fractionalization!

16

(2− d) ⇒ Fractional charge in quantum Hall effect inspired by soli-

tons result, but different mechanism.

17

Graphene Realization of (2-d) Dirac Equation

Graphene hexagonal lattice

Sublattices

Kinetic term:

Linearization at Fermi level

2 “Dirac points” per sublattice (conduction and valence bands

meet)

⇒ 4× 4 Dirac Hamiltonian in 2 spatial dimensions

Wallace,PR 71, 662 (47); Semenoff,PRL53, 2449 (84);

Gaim & Novoselov, Nobel Prize (10)

Potential term: ? Speculate: “Kekule’ distortion”

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Dirac Hamiltonian

H = ψ†α · pψ︸ ︷︷ ︸

+ψ†β [ϕre − i ϕim γ5]︸ ︷︷ ︸ψ (4×4)

��

��

��

���✠

ϕ ≡ ϕre+ i ϕim❍❍❍❍❍❍❍❍❍❍❍❍❍❍❍❥

α =(

σ 00 −σ

), β =

(0 II 0

), γ5 =

(I 00 −I

), p = 1

i∇

(All vectors are 2-dimensional)

linearization Kekule distortion

at Fermi level ϕ constant: ϕ0 ⇒ mass gap

ϕ soliton/vortex: ϕs = |ϕ(r)|eiθ, ϕ(0) = 0, |ϕ(∞)| = ϕ0

⇒ mid-gap state

Conclusion : Q = ±12 (existence proof!)

Hou, Chamon & Mudry, PRL 98, 186809 (07) [cond-mat/0609740]

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Majorana Equation

electrically charged particles:

particle is different from anti-particle

created by complex field

electrically neutral particles:

particle can be identified with its anti-particle

created by real fielde.g. neutral pion (S = 0)

photon (S = 1)graviton (S = 2)

⟩

all bosons

Majorana fermion = neutral fermion

Majorana Matrix Equation

(α · p+ βm)Ψ = i ∂∂tΨ

Ψ real (neutral excitations)

p = 1i ∇ imaginary

α real = α∗β imaginary = −β∗

}

Majorana Representation

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Majorana Representation

α1M =

(

0 σ1

σ1 0

)

α2M =

(

I 00 −I

)

α3M =

(

0 σ3

σ3 0

)

βM =

(

0 σ2

σ2 0

)

Ψ∗M = ΨM

Majorana in arbitrary representation

Cα∗C−1 = α, Cβ∗C−1 = −β CΨ∗ = Ψ

e.g. Weyl α =

(

σ 00 −σ

)

β =

(

0 II 0

)

C =

(

0 −iσ2

iσ2 0

)

(

σ · p m

m −σ · p

) (

ψχ

)

= i∂

∂t

(

ψχ

)

CΨ∗ = Ψ ⇒ χ = iσ2ψ∗

σ · pψ+ iσ2mψ∗ = i∂

∂tψ (2× 2)

ψ mixes with ψ∗

NB C = I in Majorana representation

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Majorana Equation (2 component)

σ · pψ+ iσ2mψ∗ = i∂

∂tψ

NB. Dirac mass term: preserves quantum numbers (charge, particle number)

Majorana mass term: does not preserve any quantum numbers

⇒ no distinction between particle and anti-particlesince there are no conserved quantities to tellthem apart, particle is its own anti-particle

Dirac field operator

Ψ =∑

E>0

(

aE e−iEtΨE + b†E e

iEtCΨ∗E

)

Majorana field operator

Ψ =∑

E>0

(

aE e−iEtΨE + a†E e

iEtCΨ∗E

)

anti-particle operators (b, b†) have disappeared

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Are there Majorana fermions in Nature?

neutrinos?

– recent development in neutrino physics

experimental observation of neutrino oscillations ⇒• neutrinos have mass (< 0.1eV)

• lepton number is not conserved separately for each

flavor.

⇒ they could be Majorana fermions

Hypothetical Majorana fermions:

• supersymmetry – supersymmetric partners of

photon, neutral Higgs boson, etc. are necessarily

Majorana fermions

• cosmology – dark matter candidates

23

Majorana fermions in superconductor in contact with a topologicalinsulator

superconductor

proximity effects ⇒ Cooper pairs

tunnel through to the surface of TI

topological insulator

Hamiltonian density for the model:

H = ψ∗ (σ · 1i∇− µ) ψ+

1

2(△ψ∗ i σ2ψ∗ + h.c.)

ψ =

(

ψ↑

ψ↓

)

,σ = (σ1, σ2),

µ is chemical potential, △ is the order parameter

△ may be constant:△ = △0j

or take vortex profile: △(r) = v(r)eiθ, v(0) = 0, v(∞) = △0.

Equation of motion: i ∂t ψ = (σ · p− µ) ψ+△ i σ2ψ∗

In the absence of µ, and with constant △ , the above system is a (2+1)-dimensional version of the (3+1)-dimensional, two component Majoranaequation!

⇒ governs chargeless spin 12fermions with Majorana mass |△|.

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Zero Mode

In the presence of a single vortex order parameter ∆(r) = v(r)eiθ thereexists a zero-energy (static) isolated mode(Fu & Kane, PRL 100, 096407 (08); Rossi & RJ NPB 190, 681 (81)

)

ψ0 = N

(

J0(µr) exp {−iπ/4− V (r)}J1(µr) exp {i(θ+ π/4)− V (r)}

)

N real constant, V ′(r) = v(r)

Majorana field expansion:Ψ = ................ + aΨ0

E 6=0 modes

where zero mode operator a satisfies

{a, a†} = 1, a† = a ⇒ a2 = 1/2

25

[Chamon, Nishida, Pi, Santos & RJ; PRB 81, 224515 (10)]

(i) Two 1-dimensional realizations: take vacuum state to be eigenstateof a, with possible eigenvalue ±1/

√2.

a |0±〉 = ± 1√2

|0±〉

There are two ground states |0+〉 and |0−〉. Two towers of states

are constructed by repeated application of a†E. No operator connectsthe two towers.

Fermion parity is broken because a is a fermionic operator. Like inspontaneous breaking, a vacuum |0+〉 or |0−〉 must be chosen, andno tunneling connects to the other ground state.

(ii) One 2-dimensional realization: vacuum doubly degenerate |1〉 , |2〉,and a connects the two vacua.

a |1〉 = 1√2

|2〉

a |2〉 = 1√2

|1〉

Two towers of states are constructed by repeated application of a†E.a connects the towers. Fermion parity is preserved.

We shall assume that fermion parity is preserved, and adopt secondpossibility

26

• Curious fact in (1-d)

total L for scalar kink ⊕ fermions

L = 12∂µΦ ∂µΦ+ µ

22Φ2 − λ

8

2Φ4 + iΨ γµ ∂µΨ− gΦΨΨ

L possesses SUSY for g = λ,Ψ Majorana

Center anomaly in SUSY algebra ⇒ fermion parity can be absent.

[Losev, Shifman & Vainshtein, PLB 522, 327 (01)

Any relevance for condensed matter?

Semenoff & Sodano, EJTP 10, 57 (08)]

27

Multiple Vortices

With N vortices, governed by operators a1, a2, . . . , aN that satisfy

{ai, aj} = 2 δij (Clifford algebra)

one can show that one needs

N = 2N

2 states for even N

and N = 2N+1

2 states for odd N

N = 1 N = 2 σ1 or σ2 (not σ3) (2× 2)

N = 2 N = 2 σ1 and σ2 (not σ3) (2× 2)

N = 3 N = 4 α1 α2 α3 or β (not diagonal) (4× 4)

N = 4 N = 4 α1 α2 α3 and β (not diagonal) (4× 4)

etc.

Clifford algebra, with a restriction: use for ai Pauli, Dirac, . . . , matrices

excluding diagonal one since it would correspond to diagonalizing a mode

operator and would produce fermion parity violation [Pi & RJ, PRB 85,

033102 (12)]

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