1. Lecture 3: Topological insulators
Dimi Culcer
UNSW

2. Outline of this lecture
3D Topological Insulators
What is a topological insulator?
𝑘 ∙ 𝑝 theory for Bi2Se3
Surface states of Bi2Se3
Topological Insulators films
Thin layer of Bi2Se3 and interlayer tunnelling
Quantum anomalous Hall effect in 3D and 2D TIs
Extended state picture
Edge state picture of 2D QAHE
Is transport really dissipationless?

3. What is a topological insulator?
A fancy name for a schizophrenic material
All that matters is spin-orbit coupling and time reversal
2D topological insulators
Insulating surface, conducting edges – chiral edge states with definite spin orientation
3D topological insulators
Insulating bulk, conducting surfaces – chiral surface states with definite spin orientation
Many kinds of insulators
Band insulator – energy gap ≫ room temperature
Anderson insulator – large disorder concentration
Mott insulator – strong electron-electron interactions
Kondo insulator – localized electrons hybridize with conduction electrons – gap
All of these can be topological insulators if spin-orbit strong enough
All the materials in this talk are band insulators
Also topological superconductors
Quasiparticles – Cooper pairs

4. What is a topological insulator?
The first topological insulator was the quantum Hall effect (QHE)
QHE is a 2D topological insulator
No bulk conduction (except at special points), only edge states
Edge states travel in one direction only
They cannot back-scatter – have to go across the sample
Hall conductivity quantised
𝜎 𝑥𝑦 = 𝑛 𝑒 2 ℎ
n is a topological invariant – Chern number
This is the integral of the Berry curvature over the Brillouin zone
n counts the number of Landau levels below the Fermi energy ~ like the filling factor
QHE breaks time-reversal because of the magnetic field
The current generation of TIs is time-reversal invariant
C.L. Kane & E.J. Mele, Physical Review Letters 95 (2005)
M.Z. Hasan & C.L. Kane, Reviews of Modern Physics 82 (2010) 3045.
X.-L. Qi & S.-C. Zhang, Reviews of Modern Physics 83 (2011) 1057.
X.-L. Qi, T.L. Hughes & S.-C. Zhang, Physical Review B 78 (2008)

5. How do we identify a TI? Kane & Mele found a topological invariant
Called Z2 invariant
Related to the matrix elements of the time-reversal operator
Sandwich this operator between all pairs of bands in the crystal
Need the whole band structure – difficult calculation
Z2 invariant counts the number of surface states
0 or even is trivial
1 or odd is non-trivial – odd number of Dirac cones
Theorem says fermions come in pairs
The pair is on the on opposite surface
In practice in a TI slab all surfaces have TI states
Especially important when looking at Hall transport
Bulk conduction
Surface
states
Bulk valence

6. Most common TI – Bi2Se3 Equivalent layers
Se1 equivalent to Se1’
Bi1 equivalent to Bi1’
The system has inversion symmetry
Se2 site plays the role of an inversion centre
Under spatial inversion
Bi1 → Bi1’
Se1 → Se1’
Eigenstates have definite parity
Bi2Se3 is slightly non-Bravais
5 atoms per unit cell
Zhang et al, Nature Physics 5, 438 (2009)

7. Bulk Bi2Se3 Start from the atomic energy levels
Interested in energy eigenvalues around Γ point
What is the effect of crystal field splitting and SO?
States near Fermi surface come from p-orbitals
Both in Bi (6s26p3) and Se (4s24p4)
Crystal field removes degeneracy of p-orbitals
Only the pz orbitals are relevant – low-lying
This comes from the point-group symmetry of the crystal
The lattice has inversion symmetry
pz orbitals from Bi, Se near FS have opposite parities
Energy-gap between these two controlled by SO
Increasing SO may cause a level-inversion
See figure
In the crystal this corresponds to a band inversion
Picture from Shen (2nd edition, 2017)
Spin-orbit controls the energy splitting between the orbitals.
Increasing the spin-orbit will result in a level inversion.

8. Bulk Bi2Se3
Here z // (111)
Near Fermi surface the relevant energy levels are pz orbitals
There are called 𝑝 1𝑧 + ↑ , 𝑝 1𝑧 + ↓ , 𝑝 2𝑧 − ↑ , 𝑝 2𝑧 − ↓
Here +/- represents the parity, ↑/↓ represents spin
Using these states we can construct the effective 𝑘 ∙ 𝑝 Hamiltonian
𝐻= 𝜀 𝑝 + 𝑀− 𝐵 ∥ 𝑝 ∥ 2 − 𝐵 ⊥ 𝑝 𝑧 2 𝕀 2×2 ℏ 𝑣 ∥ 𝜎 ∙ 𝑝 ∥ + 𝑣 ⊥ 𝜎 𝑧 𝑝 𝑧 𝑣 ∥ 𝜎 ∙ 𝑝 ∥ + 𝑣 ⊥ 𝜎 𝑧 𝑝 𝑧 𝜀 𝑝 − 𝑀− 𝐵 ∥ 𝑝 ∥ 2 − 𝐵 ⊥ 𝑝 𝑧 2 𝕀 2×2
Very much like the Dirac equation
Here the energy 𝜀 𝑝 =𝐶 − 𝐷 ∥ 𝑝 ∥ 2 − 𝐷 ⊥ 𝑝 𝑧 2
This breaks the particle-hole symmetry of the system
p-linear term due to the different parities of the two sets of basis functions
The model is time-reversal invariant as expected

9. Surface states of Bi2Se3
Surface states can be derived from bulk Hamiltonian
Take 𝑘 ∙ 𝑝 Hamiltonian from previous slide
Truncate the system at 𝑧=0
𝑝 𝑥 , 𝑝 𝑦 are still good quantum numbers because translationally invariant in xy-plane
First set 𝑝 𝑧 →−𝑖ℏ𝜕/𝜕𝑧 and 𝑝 𝑥 , 𝑝 𝑦 →0
Find the solution to the equation 𝐻 𝑧 𝜓 𝑧 =𝜀𝜓 𝑧 where
𝐻(𝑧)= 𝐶+ ℏ 2 𝐷 ⊥ 𝜕 2 𝜕 𝑧 2 + 𝑀+ ℏ 2 𝐵 ⊥ 𝜕 2 𝜕 𝑧 2 𝕀 2×2 −𝑖ℏ𝑣 ⊥ 𝜎 𝑧 𝜕 𝜕𝑧 −𝑖ℏ𝑣 ⊥ 𝜎 𝑧 𝜕 𝜕𝑧 𝐶+ ℏ 2 𝐷 ⊥ 𝜕 2 𝜕 𝑧 2 − 𝑀+ ℏ 2 𝐵 ⊥ 𝜕 2 𝜕 𝑧 2 𝕀 2×2
Here the 𝜀 𝑝 term breaks particle-hole symmetry between conduction & valence bands
For 𝐷 ⊥ > 𝐵 ⊥ the band gap closes, the system is no longer an insulator
To have a surface state solution we need to focus on 𝐷 ⊥ < 𝐵 ⊥

10. Surface states of Bi2Se3
We can decouple rows 1&3 from rows 2&4 in H(z)
We take two trial solutions 𝜓 1 = 𝑒 𝜆𝑧 𝑎 𝑏 and 𝜓 2 = 𝑒 𝜆𝑧 𝑎 𝑏 2
Now we have two independent sets of equations
Let 𝐵 ± = 𝐵 ⊥ ± 𝐷 ⊥
𝑀+ 𝐵 + 𝜆 2 −𝑖 𝑣 ⊥ 𝜆 −𝑖 𝑣 ⊥ 𝜆 −𝑀+ 𝐵 − 𝜆 𝑎 1 𝑏 1 =𝜀 𝑎 1 𝑏 1
𝑀+ 𝐵 + 𝜆 2 𝑖 𝑣 ⊥ 𝜆 𝑖 𝑣 ⊥ 𝜆 −𝑀+ 𝐵 − 𝜆 𝑎 2 𝑏 2 =𝜀 𝑎 2 𝑏 2

11. Surface states of Bi2Se3
Focus on the first equation – for a non-trivial solution need
det 𝑀+ 𝐵 + 𝜆 2 −𝐸 −𝑖 𝑣 ⊥ 𝜆 −𝑖 𝑣 ⊥ 𝜆 −𝑀+ 𝐵 − 𝜆 2 −𝐸 =0
This gives 4 roots, ± 𝜆 1 ,± 𝜆 2 since it is a quartic in 𝜆
Impose Dirichlet boundary conditions 𝜓 𝑧 =0 at 𝑧=0 and 𝑧=−∞
For 𝑀 𝐵 ⊥ >0 the wave-function takes the form 𝜓 1 = (𝑒 𝜆 1 𝑧 − 𝑒 𝜆 2 𝑧 ) 𝑎 𝑏 1 0
Similarly from the second equationn we find 𝜓 2 = (𝑒 𝜆 1 𝑧 − 𝑒 𝜆 2 𝑧 ) 0 − 𝑎 𝑏 1
For the exact coefficients, exponents, normalisation factors – look up Shen

12. Surface states of Bi2Se3 Effective Hamiltonian for surface states
Use 𝜓 1 , 𝜓 2 as the basis states
Add the 𝑝 𝑥 , 𝑝 𝑦 terms in the Hamiltonian
Work out the matrix elements
𝜓 1 𝐻 𝜓 𝜓 1 𝐻 𝜓 𝜓 2 𝐻 𝜓 𝜓 2 𝐻 𝜓 2
This yields the effective Hamiltonian
𝐻 𝑒𝑓𝑓 = 𝐸 0 − 𝐷 ∥ 𝑝 ∥ 2 + 𝑣 𝑒𝑓𝑓 ( 𝜎 𝑥 𝑝 𝑦 − 𝜎 𝑦 𝑝 𝑥 )
Where 𝑣 𝑒𝑓𝑓 = 𝑣 ∥ 1− 𝐷 ⊥ 2 𝐵 ⊥ 2
The quadratic term does not affect the dispersion much at any transport densities
It is however important when we discuss tunnelling later on
Note this is exactly Rashba even though the prefactor has a different origin
However, here there is only ONE Fermi surface and NO spin precession
Bulk conduction
Surface
states
Bulk valence

13. Helical versus chiral ky kx Spin-momentum locking HgTe quantum well
Most of the time we can forget about the quadratic term
𝐻 𝑒𝑓𝑓 = ℏ𝑣 𝑒𝑓𝑓 ( 𝜎 𝑥 𝑘 𝑦 − 𝜎 𝑦 𝑘 𝑥 )
The momentum sets the direction of the spin
Direction of spin ⊥ momentum
We say the system is chiral
HgTe quantum well
𝐻 𝑒𝑓𝑓 = ℏ𝑣 𝑒𝑓𝑓 ( 𝜎 𝑥 𝑘 𝑥 + 𝜎 𝑦 𝑘 𝑦 )
Spin still locked to the momentum but ∥ momentum
This system is helical
You can rotate one H into the other
𝑘 𝑥 → 𝑘 𝑦 , 𝑘 𝑦 → −𝑘 𝑥
The physics is always the same
Graphene similar to HgTe QWs but characterised by pseudospin helicity
Same Hamiltonian except 𝜎 is not a real spin but a pseudospin (2 sub-lattices) ky kx

14. No back-scattering ky kx To flip the momentum you must flip the spin
Ordinary Coulomb scattering cannot do this
Need time-reversal breaking – magnetic impurities
Eigenstates of Rashba Hamiltonian
In 𝑘 ∙ 𝑝 language 𝑢 𝑘 = 𝑒 −𝑖𝜃 ±1
One represents electrons, the other holes
Fermi’s Golden Rule
𝑆𝑐𝑎𝑡𝑡𝑒𝑟𝑖𝑛𝑔 𝑟𝑎𝑡𝑒= 2𝜋 ℏ 𝑘′ 𝜓 𝑘 𝑈 𝑑𝑖𝑠𝑜𝑟𝑑𝑒𝑟 𝜓 𝑘′ 𝛿 𝜀 𝑘 − 𝜀 𝑘′
Substitute Rashba eigenstates (for electrons)
You get a factor of 1+ cos 𝜃− 𝜃 ′
Vanishes when 𝜃 ′ =− 𝜃
i.e. no back-scattering
This provides topological protection against localisation
Which is coherent back-scattering ky kx

15. Spin-polarized current
𝐻 𝑠𝑜 =𝛼 𝜎 ∙ 𝑘 × 𝑧
Current operator proportional to spin
Charge current = spin polarization
Spin polarization exists throughout surface
Not in bulk because Bi2Se3 has inversion symmetry
This is a signature of surface transport
Smoking gun for TI behavior
Detection – Dan Ralph group, Nature 511, 449 (2014)
Conducting surface

Insulating bulk

16. Spin-polarized currentky ky kx kx
No E
E // x

17. Thin film of Bi2Se3 – non-magnetic
Formally it is exactly like a single surface
Except the system is ow truncated at 𝑧= 𝑧 𝑡𝑜𝑝 and 𝑧 𝑏𝑜𝑡𝑡𝑜𝑚
Results in tunnelling between top & bottom layers
Because of wave function overlap
Important in ultra-thin films < 5 nm
Now the effective Hamiltonian is 4 × 4
𝐻 𝑇𝐼𝑇𝐹 = 𝑅𝑎𝑠ℎ𝑏𝑎 𝑡 0 0 𝑡 𝑡 0 0 𝑡 −𝑅𝑎𝑠ℎ𝑏𝑎

18. Thin film of Bi2Se3 – non-magnetic
The energy eigenvalues are 2-fold degenerate
This is because time-reversal symmetry is still preserved
We have Kramers degeneracy
𝜀 ± = ± 𝑣 𝑒𝑓𝑓 2 𝑝 2 + 𝑡 2
But now the eigenstates contain spins from both layers
When tunnelling is strong we can no longer think of individual surfaces
The asymptotic limits are easy to understand
Expand the eigen-energies in the small parameter
Small t, like 2 TI surfaces
Large t, like ordinary parabolic semiconductor

19. Thin film of Bi2Se3 – magnetic
We need to add a magnetisation 𝑀 ∥ 𝑧
Same form for top and bottom
Let M have units of energy
𝐻 𝑇𝐼𝑇𝐹 = 𝑀 𝑖𝑣 𝑒𝑓𝑓 𝑝 − −𝑖𝑣 𝑒𝑓𝑓 𝑝 + −𝑀 𝑡 𝑡 𝑡 𝑡 𝑀 −𝑖𝑣 𝑒𝑓𝑓 𝑝 − 𝑖𝑣 𝑒𝑓𝑓 𝑝 + −𝑀
Rotate this to get rid of the off-diagonal blocks
𝜓 1 = 𝜓 1 + 𝜓 3
𝜓 2 = 𝜓 2 − 𝜓 4
𝜓 3 = 𝜓 1 − 𝜓 3
𝜓 4 = 𝜓 2 + 𝜓 4
Liu et al, arXiv:

20. Thin film of Bi2Se3 – magnetic
After rotation to the new basis
𝐻 𝑇𝐼𝑇𝐹 = 𝑀+𝑡 𝑖𝑣 𝑒𝑓𝑓 𝑝 − − 𝑖𝑣 𝑒𝑓𝑓 𝑝 + −𝑀−𝑡 −𝑀+𝑡 − 𝑖𝑣 𝑒𝑓𝑓 𝑝 + 𝑖𝑣 𝑒𝑓𝑓 𝑝 − 𝑀−𝑡
The eigenvalues are no longer degenerate because breaking TR
Top block 𝜀 ± = ± 𝑣 𝑒𝑓𝑓 2 𝑝 2 + 𝑀+𝑡 2
Bottom block 𝜀 ± = ± 𝑣 𝑒𝑓𝑓 2 𝑝 2 + 𝑀−𝑡 2
Notice this is just 2 copies of Rashba + magnetisation
Instead of M we have 𝑀±𝑡 some effective magnetisation
The two blocks do not represent the two surfaces
But the structure of the Hamiltonian is the same
Liu et al, arXiv:

21. Surface QAHE in Bi2Se3
TI thin/ultra-thin film Hamiltonian ~ 2 blocks
In both cases it is Rashba + effective magnetisation
The magnetisation opens a gap in the spectrum
We can place the Fermi energy in this gap
For now assume everything is happening at 𝑇=0
Valence band full, conduction band empty
There is definitely no 𝜎 𝑥𝑥 because there is no Fermi surface
Consider the contribution to the AHE due to one block
We already calculated this in Lecture 2
The Berry curvature is exactly the same
This time we only have one band – the valence band
We can extend the k-integral from −∞ to +∞
This is cheating but doing it formally correctly takes a lot of work – need full 3D model
One block gives 𝜎 𝑥𝑦 = 𝑒 2 2ℎ
Surface
states εF
Surface
states

22. Surface QAHE in Bi2Se3 The other block contributes 𝜎 𝑥𝑦 = 𝑒 2 2ℎ
These add up because the normal to the other surface is flipped, total 𝜎 𝑥𝑦 = 𝑒 2 ℎ
Easiest to understand for 𝑡=0 when we can think of independent surfaces
All surfaces are topological and the Hall current flows around the sample
This is the famous quantised anomalous Hall effect, the Chern number is 1
Notice we get the same result regardless of tunnelling
Because the structure of the Hamiltonians is the same
t appears as a renormalisation of the magnetisation
But we do not care because we are extending k from −∞ to +∞
The magnitude of the effective magnetisation becomes irrelevant
In the two cases the mechanism for current flow must be different
When t is large the electrons live on both surfaces – picture below not valid
Cross-sectional view
Electric field into the page

23. Edge states of Bi2Se3 film
See Shen for edge states – same demonstration as for surface states
For the edge state to be well-defined
We need non-magnetic tunnelling term t to be significant
So the film needs to be 3 nm thick and certainly less than 5 nm
Some sort of edge states exist even if t negligible
But very complicated and poorly understood
We understand the maths
We don’t fully understand the physics
E →
Top view

24. Is transport really dissipationless?
Not completely dissipationless for two reasons
The system is coupled to leads
Otherwise you cannot pass a current through it
So the system talks to the outside world
The leads are metallic and interact with phonons etc
Also, there is a potential difference across the sample
We can think of it as transport through an island
There is no dissipation in the island itself
But there is dissipation in the leads and ACROSS the island
Jared Cole is currently working on this
There is a top gate – needed to switch a transistor on/off
This is like a parallel-plate capacitor (conductor – insulator – conductor)
Actually most of the energy dissipated inn transistors comes from the gate
There is also the issue of room temperature operation
Bi2Se3 bulk gap 0.3 eV compared to Si 1.1 eV

25. Summary 3D Topological Insulators 2D Topological Insulators
Bi2Se3 surface states follow Rashba-like dispersion
Spin-orbit is everything: no kinetic-energy term
Each surface exhibits a quantum anomalous Hall effect adding up to 𝑒 2 ℎ
2D Topological Insulators
Thin layer of Bi2Se3
QAHE in 2D TIs can be explained by edge states
Only one state on each edge → transport without scattering like in QHE → dissipationless
Not truly dissipationless
Because of leads and top gate
But can still save a lot of energy
Thanks to Haizhou Lu for many explanations