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6 of insulators

6.1. Gauge theory (E&M) for Chern insulators In an crystal, we have nucleons, electrons, and they couple together through gauge fields (E and B). If we don’t consider the spin degree of freedom (ignoring magnetic ordering and magnetization, etc), the only degrees of freedom are the motion of charge and the gauge field. In an insulator, because charged particles cannot move, the only useful degrees of freedom are the gauge fields. So let’s consider the gauge field in an insulator.

6.1.1. Hamiltonian for E and B fields E and B field carries . In vacuum, the energy (Hamiltonian) is

1 B2 Ø 2 H = „ r e0 E + (6.1) 2 m0 In an insulator with time-reversal , the energy of gauge fields must be a function of E and B. For weak fields (small E and B) we can do a power series expansion:

1 Ø 1 H = „ r eij Ei E j + Bi B j + higher order terms (6.2) 2 mij

We cannot have any other quadratic terms, like E ÿB, because it violates the time-reversal symmetry (B is odd under time-reversal but E is even).  Higher order terms are in principle allowed, and they are responsible for non-linear E&M responses. These nonlinear effects are negligible if E and B are weak enough. So we will ignore them here. The bottom line: the energy is still E2 + B2. The coefficients can be renoramlized, but the gauge theory in an insulator is qualitatively the same as in vacuum. Q: What will happen if we break the T-reversal symmetry? A: Additional terms are allowed by symmetry, e.g. E ÿ B if the lattice also breaks the inversion symmetry. If the Hamiltonian contains this term, it means that we will induce magnetization by applying E field (and induce electric dipoles by applying a B field). However, this is a “small” effect, which we will not consider. In addition to E ÿ B, the system can get another term which is the Chern-Simon’s gauge term.

6.1.2. the Chern-Simon’s gauge theory in 2+1D For simplicity, we use the theorists’ unit: e=m = everything = 1. The Cherm-Simon’s theory is easier to handle using Lagrangian, instead of the Hamiltonian. The Lagrangian is k Ø m n l LCS = „ r emnl A ∑ A (6.3) 4 p Here, k is the coupling constant; m, n and l run over t, and y. Ax, Ay is the vector potential and At is the electric potential f. etxy =exyt =eytx= 1 and etyx =eyxt =exty =-1 and e=0 for any other subindcies. The is

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k Ø m n l SCS = „ tL= „ r „ t emnl A ∑ A (6.4) 4 p The coupling constant k is an integer for topological reasons, which will be discussed below.   Gauge invariance: Let’s consider a gauge transformation:

Am Ø Am ' = Am +∑m c (6.5)

k k Ø m m n l l Ø m m n l SCS Ø SCS ' = „ r „ t emnl A +∑ c ∑ A +∑ c = „ r „ t emnl A +∑ c ∑ A = 4 p 4 p (6.6) k k k Ø m n l Ø m n l Ø m n l „ r „ t emnl A ∑ A + „ r „ t emnl ∑ c∑ A = SCS + „ r „ t emnl ∑ c∑ A = SCS + surface term 4 p   4 p     4 p   Here, the last term is an integral of a total deriviative, which is just a surface integral, according to the Stokes’ theorem. If we don’t worry about the surface term (assuming the system has no boundary), the action is gauge . So this is a valid gauge theory (if there is no boundary). The surface term will be consider later. In fact, the surface term of the Chern-Simon’s gauge theory violates the gauge symmetry. But luckily, the edge metallic state (chiral state) that we have at the edge also violates the gauge symmetry through anomaly. And these two effects cancels each other, so the whole system is gauge invariant. For the whole insulator:

1 Ø 1 S = SCS + „ r eij Ei E j - Bi B j + higher order terms (6.7) 2 mij

If the field A varies very slowly in space and time, ∑A is small. The CS term contains only one derivative, but E2 and B2 term contain two derivatives (E and B both take the form of ∑ A). Two derivatives means it is much smaller than the CS term, if A changes very slowly. There- fore, the CS term gives the dominate contribution (so we can ignore the E2 and B2 terms). Electric Current: dS d k Ø m n l ji r0, t0 = = „ r „ t emnl A r, t ∑ A r, t = dAi r0, t0 dAi r0, t0 4 p k d Am r, t k d Al r, t Ø n l Ø m n   „ r „ t emnl ∑ A r, t + „ r „ temnl A r,t ∑ 4 p dA r , t 4 p dA r , t   i 0 0  i 0 0 (6.8) k Ø   k Ø   = „ r „ t e d d r - r d t - t ∑n Al r, t + „ r „ t e Am r, t ∑n d d r - r d t - t  mnl m,i 0 0   mnl  l,i  0 0 4 p   4 p   k k k k Ø n l n m m n Ø = einl ∑ A r0, t0 - emni ∑ A r, t = eimn ∑ A r0, t0 =- ei μ E j 4 p  4 p    2 p   2 p       Ø k m n k Ø Q: Why e ∑ A r0, t0 =- ei μ E j? 2 p imn   2 p     A: let’s do i = x as an example: k   m n jx r0, t0 = exmn ∑ A r0, t0 = 2 p (6.9) k k k k k t y y t t y y y exty ∑ A r0, t0 + exyt ∑ A r0, t0 = -∑ A r0, t0 +∑ f r0, t0 = ∑t A r0, t0 +∑y f r0, t0 =- Ey  2 p   2 p  2 p 2 p 2 p

k Ø Ø Since, ji r0, t0 =- ei μ E j, the Hall conductivity is: 2 p              k sxy = (6.10)  2 p  Above, we set e and Ñ to 1 in the calculations above. Here we add them back Phys620.nb 97

k e2 e2 sxy = = k (6.11) 2 p Ñ h

Because k is an integer for the CS gauge theory, sxy is quantized. Q: What happens if we have a metal? A: For metals, the charge degrees of freedom must also be considered. So we have both gauge fields and charge motions. The situation is much more complicated and in general, we cannot get a clear answer of sxy.

6.1.3. and magnetic field in a quantum Hall system The Cherm-Simon’s gauge theory k Ø m n l SCS = „ r „ t emnl A ∑ A (6.12) 4 p Charge density:  dS d k Ø m n l r r0, t0 = = „ r „ t emnl A r, t ∑ A r, t = dA0 r0, t0 dA0 r0, t0 4 p k d Am r, t k d Al r, t Ø n l Ø m n   „ r „ t emnl ∑ A r, t + „ r „ temnl A r, t ∑ 4 p dA r , t 4 p dA r , t   0 0 0  0 0 0 (6.13) k Ø   k Ø   = „ r „ t e d d r - r d t - t ∑n Al r, t + „ r „ t e Am r, t ∑n d d r - r d t - t  mnl m,0 0 0   mnl  l,0  0 0 4 p   4 p   k k k k Ø k n l n m m n = e0 nl ∑ A r0, t0 - emn0 ∑ A r, t = e0 mn ∑ A r0, t0 = “μA = B 4 p  4 p    2 p   2 p   2 p     Bottom line: charge density is propotional to magnetic field       k r r0, t0 = B (6.14) 2 p Integral on both sides:   k „ r r r0, t0 = „ rB (6.15) D 2 p D So we find that electric charge is proportional to the magnetic flux     k q = FB (6.16) 2 p Flux attachment:

† Electrons in an IQH system carries magnetic flux FB = 2 p k (the Chern-Simon's gauge theory attaches a magnetic flux FB = 2 p k to each of the electrons). † Magnetic flux in an IQH system carries charge q=k FB 2 p   Why? Q: Faraday’s law+Hall effect

Let’s add an magnetic flux in an IQH system, by increasing the magnetic field in some region D. From time ti to t f , we increase the B field from B r, ti to B r, t f . If B t f > B ti , we add some magnetic flux into the system

DFB = „ rBr, t f - „ rBr, ti (6.17)   D     D   Increasing magnetic flux will create some electric field (Faraday’s law)       „FB E ÿ„l =- (6.18) „ t

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If we have E, it will creates Hall current perpendicular to E (Hall effect). Because here E is in the circlar direction, the current j is along the radials direction, which means that charges are flowing away from (or towards) the place where we insert the magnetic flux

j =sxy E (6.19)

Total current flowing away from the region D:

Ø Ø Ø Ø „FB I = j μ„ r =sxy E ÿ„ l =-sxy (6.20) „ t Total charge accumulated due to this current flow:   „FB Dq = q t f - q ti =- „ tI t =sxy „ t =sxy FB t f -FB ti =sxy DFB (6.21) „ t

k 2 If we create an flux FB in a IQH system, some charge will accumulate at this flux: q =sxy FB. Because sxy = (in the unit of e Ñ), we find              2 p that k  q = FB (6.22) 2 p This result fully agrees with the prediction from the Chern-Simon’s gauge theory.

6.1.4. Why k is quantized? (a not-so-rigorous proof) (For more a rigorous proof, see of Many-body Systems, Xiao G, Oxford 2004). Integral over the whole system: k „ r r r0, t0 = „ rB (6.23) D 2 p D Total charge is proportional to the total magnetic flux     B FB Ne = k „ r = k = kNm (6.24) D 2 p 2 p

For a closed , Nm is the monopole charge, which is quantized to integer values (as we proved early on). Ne is the total number of electrons, which is always an integer. If we increase the magnetic field by a little bit such that Nm turns into Nm + 1, the charge of the system need to change from Ne to Ne + k. Because charge is always an integer, k must be an integer.

If we increase the charge by 1, Ne Ø Ne + 1, for general k (if k is not ≤1), Nm will increase by a fractional value 1 k in order to satisfies the relation: Ne = kNm. This is impossible, because Nm can not take fractional value. This implies that if we have a IQHE with Ne electrons, by adding one electron, the system will no longer be in an IQHE state. This is reasonable because the quantum Hall state is an insulator. For an insulator, if we add one more electron, it becomes a metal. (In reallity, due to the existence of impurities, adding one electr on will not change an insulator into an metal, but for an idea model without impurities, one electron is enough to destroy an insulator).

6.1.5. Chern-Simon’s theory for an open manifold (with edges)

Am Ø Am ' = Am +∑m c (6.25)

k Ø m n l SCS Ø SCS ' = SCS + „ r „ t emnl ∑ c∑ A = SCS + surface term (6.26) 4 p It is not gauge invariant near the boundar. This is because we have (metallic) chiral edge states near the boundary, which also violate the gauge symmetry (known as gauge anomaly), and the violation of the gauge theory there introduce an extra term to the action, which cancles the gauge dependent term here.

6.1.6. Gauge anomaly (NOT required) Ref: Quantum theory of fields, Weinberg, VOL 2, Chapter 22, Anomalies A stand approximation: ignore high energy states, because it cost too much energy to excite the system into such state. This approximation is NOT always valid and could be problematic. Phys620.nb 99

Example (statistical physics): at 1K, what is the probability of finding a hydrogen atom in its ground state?

Energy levels of a hydrogen atom: E =-13.6 eV n2

Boltzmann distribution P ∂ exp -En T . Low energy state have a larger probably. For very high energy states, the probability is much smaller.

Ground state energy of a hydrogen atom: E1 =-13.6 eV

First excited states: E2 =-3.4 eV   Temperature T = 1 K º 10-4 eV.

T << E2 - E1, so let’s ignore the high energy states beyond n > 2, only focusing on the lowest to states.

59 064 exp E1 T 10 Pn=1 = º = 1 (6.27) 59 064 14 766 exp E1 T + exp E2 T 10 + 10 This answer is WRONG! One  cannot ignore high energy states. Their probably is small, but their number is huge.

59 064 59 064 exp -E 1 T    10 10 Pn=1 = º = = 0 (6.28) 59 064 14 766 exp -En T 10 + 10 + …1 + 1 + 1 + 1 + … ¶ One cannot ignore the high energy states here. If one ignore the high energy states, the probably is 1. But if one include the high energy states, one get 0. In other words, the high energy states changed the probably by “-1”, which is an “anomaly”.     Sometimes, high energy states (ultraviolet) can change the low energy behavior (infrared) very strongly. Such a contribution from the high energy states are often called an “anomaly”. Anomaly here means that one cannot see it in the low energy theory. So in the low energy theory, it seems to be something comes from nowhere. But in really, there is nothing abnormal here. Just one cannot ignore the high energy physics. In gauge theory, there is such an effect, which was first noticed in a high-energy experiments: pion decay pion decay: p 0 Æ 2 g Pions are bosons made by quark and anti-quark pairs. Quarks are fermions, in 3+1D, we have left moving and right moving quarks. In the effective low energy (when we ignore the of fermions), the number of left and right moving quarks are separately conserved. This is one important conservation law in QED and QCD. It is not an exact conservation law, because fermions have some mass. But it is a very good approximate conservation law. This conservation law suppresses the pion decay rate very strongly. If we keep this (approximate) conservation law in mind, the decay rate is 1.9μ1013 s-1(theory). In experiments, the decay rate is 1.19μ1016, which is 1000 time larger. The problem is the conservation law. At low energy, (if we ignore all the high energy states), there is such an conservation law (particle numbers in the left and right moving sectors are conserved separately). But the high energy state cannot be ignored in this system and destroys this conservation law, which increases the decay rate of p0 by 3 order of magnitudes, which is known as the . A baby version of the chiral anomaly, Let’s consider a 1D condensed system, electrons hopping on a lattice. The dispersion is a e=-2 t cos ka. If we only care about states close to the Fermi energy, we can expand the dispersoin around +kF and -kF. So the low energy effective theory looks like † † H = vF kL L - vF kR R (6.29) k k k k k k

We have two species of fermions, L and R. Their dispersions are ≤vF k (left and right moving). The left moving mode is located at -kF, while the right moving mode is at +kF. The particle numbers for left moving particles is

† NL = L L (6.30) k k k The particle numbers for right moving particles is

 † NR = R R (6.31) k k k

If we consider the effective low-energy theory, NL and NR are both conserved quantities  NL, H = NR, H = 0 (6.32)

This is NOT true! Because if we cannot ignore high energy states here. If we taken into account the high energy states, we find that the left and right moving branches  are connected at energy far away from the chemical potential (near the band bottom). So in reality, we only have one 100 Phys620.nb

conservation law, which is the total fermion number NL + NR. NL - NR is NOT conserved. We only have one . This is the reason why one gets a wrong answer in the p decay rate, if one assumes that there are two conserved quantities. The two separate conservation laws for NL and NR are fake. It is because one ignored high energy states, which should not be ignored in this case. In high energy, this problem is addressed in a slightly different way, using gauge theory. Because gauge field couples to the phase of fermions and phase is related to fermion numbers, it is reasonable to use gauge theory to describe the system. In high energy, the way people describe this effect is to couple the left moving particles with a gauge field, and right moving particles with another gauge field. For left moving particles, one finds that the gauge theory is NOT gauge invariant. When we take into account high energy states, we get an extra term in the theory, which is not gauge invariant: e 1 c∑m An Similarly, the right moving theory is NOT gauge invariant either, and the gauge anomaly mnl 4 p takes the same form but with opposite sign - 1 c“A0 . To keep gauge invariance, one must have one left moving channel and one right 4 p moving channel and their gauge anomaly cancels each other. So at the end of the day, only the total number is conserved, and one cannot write down a theory which conserved the number for both left and right moving particles. For IQHE, the edge state only have one branch (left modes only). So, the high energy theory tells us that we will have a gauge anomaly and we cannot keep the gauge theory. This gauge anomaly is precisely the extra piece we get in the CS gauge theory near the boundary. And they cancel each other, keep the whole system gauge invariant.