Lecture notes on topological insulators

Ming-Che Chang
Department of Physics, National Taiwan Normal University, Taipei, Taiwan

(Dated: May 24, 2017)

I. ELECTROMAGNETIC RESPONSE OF WEYL (a) z (b)

SEMIMETAL

The Weyl nodes in a solid lead to several interesting χ=−1 χ=1

BE ε
effects. One is the anomalous Hall effect mentioned in
previous chapter. The others are, for exmaple, chiral kz
anomaly, and chiral magnetic effect. Before introducing B
these two effects, let’s first study the Landau levels in a E
Weyl semimetal. FIG. 2 (a) Applying a pair of electric and magnetic fields
along the two Weyl nodes. (b) The magnetic field generates
the chiral 0-th LLs in momentum space. The electric field
A. Landau levels in magnetic field pumps electrons from one node to another.

Consider a Weyl node with helicity τz , √

where ~ω ≡ v 2~eB, and σ± = (σx ± iσy )/2.
H = vτz σ · p. (1.1) We now solve
In the presence of a magnetic field B = B ẑ, the Hamil-
HΨn = εn Ψn , (1.8)
tonian becomes (Zyuzin and Burkov, 2012),    
1 0
H = vτz (σx πx + σy πy ) + vτz σz pz , (1.2) with Ψn = un |n − 1i + vn |ni, (1.9)
0 1
where π = ~i ∇ + eA, and √ √
and a|ni = n|n − 1i, a† |ni = n + 1|n + 1i. Then, for
~eB n = 0, one has
[πx , πy ] = . (1.3)
i
ετ0z = τz v~kz . (1.10)
ikz z
The state along z-direction is just a plane wave e .
Introduce the creation and annihilation operators, For n ≥ 1, one has
(
1
 √
a = √2~eB (πx − iπy ) v~k√z un + ~ω nvn = τz εn un ,
, (1.4) (1.11)
† 1
a = 2~eB (πx + iπy )
√ ~ω nun − v~kz vn = τz εn vn .

then To have non-trivial solutions, one needs

 √ 
[a, a† ] = 1. (1.5) det
− τz ε
v~kz √ ~ω n
= 0. (1.12)
~ω n −v~kz − τz ε
The Hamiltonian can be re-written as,
This gives
H = ~ωτz (σ+ a + σ− a† ) + vτz σz ~kz (1.6)
  p
v~kz ~ωa ετn±
z
= ±τz ~ω n + (vkz /ω)2 . (1.13)
= τz , (1.7)
~ωa† −v~kz
The energy dispersion of LLs along kz are shown in
Fig. 1. Notice that for Weyl nodes with opposite helici-
ties, the 0-th LLs slant toward opposite directions, par-
allel or anti-parallel to the magnetic field. There is no
τz=−1 τz=1 energy dispersion within the plane perpendicular to the
ε magnetic field.

kz
B. Chiral anomaly
FIG. 1 For Weyl nodes with opposite helicities, the 0-th Lan-
dau levels slant toward opposite directions. The figure is from Consider a pair of Weyl nodes separated in momentum
Hosur and Qi, 2013 space (see Fig. 2(a)). We first apply a magnetic field,
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preferably along the line connecting the two nodes. Then,

apply an additional electric field to push the electrons. If
E ⊥ B, then the electrons won’t move, since the LLs do µ+
not disperse along a direction perpendicular to B. That
is, only the component Ek parallel to B could transport µ−
the electrons.
The rate of electron pumping along the z-direction for
a node with chirality χ(= τz ) is given by (see Fig. 2(b)) FIG. 3 In a non-equilibrium state, the chemical potentials
near two nodes are different.
∆k
dQzχ 2π/Lz
z

= (−e)χ (1.14) C. Chiral magnetic effect

dt ∆t
k̇z
= −eχ , ~k̇z = −eEz (1.15) Suppose that under a pair of E, B fields, the system
2π/Lz is maintained in a steady state with different chemical
Ek Lz potentials near the two nodes (see Fig. 3). Assume µ+ >
= e2 χ . (1.16)
h µ− , and an electric field moves positive charges Q from
left to right. The displacement of charges costs an energy,
Furthermore, each LL has a huge degeneracy (see p. 245
of Kittel, 2005), Q
δE = (µ+ − µ− ). (1.22)
e
φtot Asamp B To balance the energy, the rate of work (per unit volume)
D= = , (1.17)
φ0 h/e done by the applied electric field should be
∂(ρ/e)
which is the ratio between total magnetic flux (through j·E = (µ+ − µ− ) (1.23)
the sample) and flux quantum φ0 = h/e. Asamp is the ∂t
2
projected area of the sample perpendicular to B. There- e
= 2µ5 2 E · B, (1.24)
fore, the total charges transported via the 0-th LL are, h
where µ5 ≡ (µ+ − µ− )/2 is the chiral chemical potential.
dQ3D
χ AB dQzχ Choose E k B, and let E → 0, then one has
= (1.18)
dt h/e dt
e2
e3 j = 2µ5 B. (1.25)
= χ 2 ALz BEk . (1.19) h2
h
This is called the chiral magnetic effect (CME). The
For the chiral charge density, we have (Hosur and Qi, current vanishes when µ5 = 0, so a non-equilibrium state
2013; Nielsen and Ninomiya, 1983), is required. The CME has not been directly verified in
experiments yet. For a critical review of the CME (in the
∂ρχ e3 context of high-energy physics), see Fukushima, 2013.
= χ 2 E · B. (1.20) Some remarks on the symmetries of the Hall effect and
∂t h
the chiral magnetic effect:
Note: It is possible to get this result using a semiclassical jy = σH E x , jz = αz Bz
analysis without discrete Landau levels (Son and Spivak,
2013). SI − − − +
This is essentially the same as the equation for the chi- TR − + − −
ral anomaly in particle physics (Adler, Bell, and Jackiw, The equations above show the symmetries of j, E, and
1969), B under SI and TR. It follows that the Hall conductivity
σH needs be even under SI, and odd under TR. That is,
e3 1 µνρλ Hall effect requires the breaking of TRS.
∂µ j5µ = − ε Fµν Fρλ , (1.21)
h2 4 On the other hand, the CME coefficient α needs be
odd under SI, and even under TR. That is, the CME
where jµ5 is the chiral current density, and j50 = ρ+ − ρ− . requires the breaking of SIS.
Note that the derivation here requires no field quantiza-
tion. Furthermore, in the context of particle physics, 1.
Since there is no lattice in vacuum, there is no node dou- D. Negative magnetoresistance
bling there. 2. The Dirac sea of vacuum is not bounded
from below, so the chiral charges are supplied from an Since a magnetic field tends to restrict the motion of
infinite reservoir, not from the other node (which does electrons, the magnetoresistance (MR) is usually posi-
not exist). tive. That is, the resistance increases with the magnetic
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field. An exception is the disordered medium with weak When E k B, the current, and thus the longitudinal con-
localization. In this case, the localization is due to the ductivity, has a part proportional to B 2 . That is, we’ll
phase coherence of electrons. A magnetic field breaks the have a negative MR.
phase coherence and delocalizes the electrons. Furthermore, because of the E · B factor, when E ro-
In Weyl semimetal, the charge pumping due to the tates away from B, the current should reduce with the
chiral anomaly also would result in negative magnetore- angle. Such a locking of the maximum current to the
sistance. This is explained as follows. After allowing direction of the magnetic field is a signature of the chiral
for the relaxation due to inter-node scatterings, the rate anomaly in Weyl semimetals (Xiong et al., 2015).
equation for chiral charges becomes,

∂ρχ e3 ρχ References
= χ 2E · B − , (1.26)
∂t h τv
Fukushima, K., 2013, Views of the Chiral Magnetic Effect
where τv is the inter-node scattering time. In steady (Springer Berlin Heidelberg, Berlin, Heidelberg), pp. 241–
state, ∂ρχ /∂t = 0, and one has 259.
Hosur, P., and X. Qi, 2013, Comptes Rendus Physique
e3 14(910), 857 .
ρ± = ± E · Bτv (1.27) Kittel, C., 2005, Introduction to Solid State Physics (John
h2
Wiley & Sons, Inc.), 8th edition.
→ µ5 ∝ E · Bτv . (1.28)
Nielsen, H. B., and M. Ninomiya, 1983, Phys. Lett. B 130(6),
389.
Because of the chiral magnetic effect, a non-zero chiral
Son, D. T., and B. Z. Spivak, 2013, Phys. Rev. B 88, 104412.
chemical potential leads to Xiong, J., S. K. Kushwaha, T. Liang, J. W. Krizan,
M. Hirschberger, W. Wang, R. J. Cava, and N. P. Ong,
e2 2015, Science 350(6259), 413.
j = 2µ5 B (1.29)
h2 Zyuzin, A. A., and A. A. Burkov, 2012, Phys. Rev. B 86,
∝ (E · B)Bτv . (1.30) 115133.