Lecture 1: Semiclassical wave-packet dynamics and the Boltzmann equation Dimi Culcer UNSW

Outline of this lecture Wave-packet dynamics Definition of a wave-packet, Lagrangian and equations of motion Berry curvature and Berry phase Combining wave-packet dynamics with the Boltzmann equation Intrinsic and extrinsic terms in transport Example: the longitudinal conductivity and the anomalous Hall effect Where does dissipation come from? When can transport be dissipationless? TODAY WILL BE MORE FORMAL Phys. Rev. B 59, 14915 (1999)

Philosophy of transport theory Topological materials funded because transport can be dissipationless Money is about power General picture of DC transport An electric field accelerates electrons Impurities, dislocations, phonons scatter them Random collisions give rise to electrical resistance This in turn gives rise to dissipation Semi-classical description of transport Electrons described by wave-packets These obey the semi-classical equations of motion The equations describe the motion of electrons between collisions The distribution of wave-packets in r- & k-space obeys the Boltzmann equation The effect of collisions is taken into account in the distribution function jHall ↑ E → jlong →

Justification of the semi-classical model Why do we need to use wave-packets? Charge carriers are particles which have a finite size We need a finite extent in real space This requires a finite extent in momentum space In transport we need position and momentum Especially over large scales over which transport is always diffusive Spatial gradients important even when applied fields are homogeneous Electric and magnetic fields are frequently inhomogeneous They depend on position Some problems are most easily solved semi-classically For example Landau-Zener tunnelling This is solved by WKB (rc, kc)

Assumptions of the semiclassical model Electrons described by wave-packets These are made of Bloch wave-functions 𝜓 𝑛 𝑘 = 𝑒 −𝑖 𝑘 ∙ 𝑟 𝑢 𝑛 𝑘 𝑢 𝑛 𝑘 has the periodicity of the lattice Electrons live in one band No transitions to other bands We can drop the subscript n Uncertainty principle satisfied Δ 𝑟 𝑐 Δ 𝑘 𝑐 ≥ℏ/2 This means the wave-packet stretches over a few unit cells External fields slowly varying over lattice length scale Wave-packet dynamics describe motion between collisions To introduce collisions we need the Boltzmann equation 𝜀 𝐹 𝜏/ℏ >> 1, equivalently 𝑘 𝐹 𝑙 >> 1, where 𝑙 is the mean free path (rc, kc)

Definition of the wave-packet Distribution function 𝑎 𝑘 , 𝑡 not specified It is assumed to be narrow compared to the Brillouin zone Written as 𝑎 𝑘 , 𝑡 = 𝑎 𝑘 , 𝑡 𝑒 −𝑖𝛾( 𝑘 , 𝑡) The wave packet is defined by 𝑤 = 𝑑 3 𝑘 𝑎 𝑘 , 𝑡 𝜓 𝑘 (𝑡) Normalisation implies 𝑑 3 𝑘 𝑎 𝑘 , 𝑡 2 =1 Do not forget the eigenstate normalisation 𝜓 𝑘 𝜓 𝑘 ′ = 𝛿( 𝑘 − 𝑘 ′ ) The distribution is never needed, whenever we have to integrate 𝑑 3 𝑘 𝑎 𝑘 , 𝑡 2 𝑠𝑜𝑚𝑒𝑡ℎ𝑖𝑛𝑔 we replace the something by its value at the centre of the wave-packet kc The mean wave vector 𝑑 3 𝑘 𝑎 𝑘 , 𝑡 2 𝑘 = 𝑘 𝑐

Expectation values The mean wave vector 𝑑 3 𝑘 𝑎 𝑘 , 𝑡 2 𝑘 = 𝑘 𝑐 The mean position is 𝑟 𝑐 = 𝑤 𝒓 𝑤 = 𝑑 3 𝑘 𝑑 3 𝑘′ 𝑎 ∗ 𝑘 , 𝑡 𝜓 𝑘 𝒓 𝑎 𝑘 ,′ 𝑡 𝜓 𝑘 ′ We need the matrix element of 𝒓 between Bloch states (see Callaway) 𝜓 𝑛 𝑘 𝒓 𝜓 𝑛′ 𝑘 ′ = 𝑖𝛿 𝑛𝑛′ 𝜕 𝜕 𝑘 δ 𝑘 − 𝑘 ′ + 𝑢 𝑛 𝑘 𝑖 𝜕 𝑢 𝑛 𝑘 𝜕 𝑘 δ 𝑘 − 𝑘 ′ This gives for the expectation value of 𝒓 𝑟 𝑐 = 𝑤 𝒓 𝑤 = 𝑑 3 𝑘 𝑎 𝑘 , 𝑡 2 𝜕𝛾 𝜕 𝑘 + 𝑢 𝑘 𝑖 𝜕 𝑢 𝑘 𝜕 𝑘 𝑟 𝑐 = 𝜕 𝛾 𝑐 𝜕 𝑘 𝑐 + 𝑢 𝑘 𝑐 𝑖 𝜕 𝑢 𝑘 𝑐 𝜕 𝑘 𝑐

Lagrangian Wave-packet dynamics found from the Lagrangian ℒ= 𝑤 𝑖ℏ 𝑑 𝑑𝑡 − 𝐻 𝑤 Justification for this Lagrangian ~ proposed by Dirac The Euler-Lagrange equation for 𝑤 reproduces the Schrodinger equation Here we evaluate the Lagrangian as a function of 𝑟 𝑐 , 𝑟 𝑐 , 𝑘 𝑐 , 𝑘 𝑐 , 𝑡 Hamiltonian assumed to have the form 𝐻 ≈ 𝐻 𝑐 +𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑡𝑒𝑟𝑚𝑠 The gradient terms are needed in a magnetic field, but we will not discuss those here Hence from now on I will drop them 𝐻 𝑐 ( 𝑟 𝑐 , t) 𝜓 𝑘 ( 𝑟 𝑐 , t) ≈ 𝜀 𝑐 ( 𝑟 𝑐 , 𝑘 , t) 𝜓 𝑘 ( 𝑟 𝑐 , t)

Evaluating the Lagrangian First evaluate the time-derivative term 𝑤 𝑖ℏ 𝑑 𝑑𝑡 𝑤 Total derivative 𝑑 𝑑𝑡 = 𝜕 𝜕𝑡 + 𝑟 ∙ 𝜕 𝜕 𝑟 + 𝑘 ∙ 𝜕 𝜕 𝑘 and these will become 𝑟 𝑐 , 𝑘 𝑐 The distribution gives 𝜕𝑎 𝜕𝑡 = 𝜕 𝑎 𝜕𝑡 𝑒 −𝑖𝛾 −𝑖 𝜕𝛾 𝜕𝑡 𝑎 𝑒 −𝑖𝛾 Normalisation 𝑑 3 𝑘 𝑎 2 =1 → 2 𝑑 3 𝑘 𝑎 𝑑 𝑑𝑡 𝑎 =0 However we have a contribution from the phase →ℏ 𝜕 𝛾 𝑐 𝜕𝑡 We can rewrite this as 𝜕 𝛾 𝑐 𝜕𝑡 = 𝑑 𝛾 𝑐 𝑑𝑡 − 𝑘 𝑐 ∙ 𝜕 𝛾 𝑐 𝜕 𝑘 𝑐 and later throw away total derivatives Total derivatives only contribute a constant to the action 𝑆= ℒ 𝑑𝑡 Now we can replace 𝜕 𝛾 𝑐 𝜕 𝑘 𝑐 = 𝑟 𝑐 − 𝑢 𝑘 𝑐 𝑖 𝜕 𝑢 𝑘 𝑐 𝜕 𝑘 𝑐 This gives us finally 𝑤 𝑖ℏ 𝑑 𝑑𝑡 𝑤 = 𝑑 𝛾 𝑐 𝑑𝑡 − 𝑘 𝑐 ∙ 𝑟 𝑐 + 𝑘 𝑐 ∙ 𝑢 𝑘 𝑐 𝑖 𝜕 𝑢 𝑘 𝑐 𝜕 𝑘 𝑐 + 𝑟 𝑐 ∙ 𝑢 𝑘 𝑐 𝑖 𝜕 𝑢 𝑘 𝑐 𝜕 𝑟 𝑐 + 𝑢 𝑘 𝑐 𝑖 𝜕 𝑢 𝑘 𝑐 𝜕𝑡

Evaluating the Lagrangian The Hamiltonian term is very simple since we ignore the gradients 𝑤 𝐻 𝑤 ≈ 𝑤 𝐻 𝑐 𝑤 = 𝜀 𝑐 The general expression for the Lagrangian ℒ=− 𝜀 𝑐 − 𝑑 𝛾 𝑐 𝑑𝑡 − 𝑘 𝑐 ∙ 𝑟 𝑐 + 𝑘 𝑐 ∙ 𝑢 𝑘 𝑐 𝑖 𝜕 𝑢 𝑘 𝑐 𝜕 𝑘 𝑐 + 𝑟 𝑐 ∙ 𝑢 𝑘 𝑐 𝑖 𝜕 𝑢 𝑘 𝑐 𝜕 𝑟 𝑐 + 𝑢 𝑘 𝑐 𝑖 𝜕 𝑢 𝑘 𝑐 𝜕𝑡 Some simplifications 𝑑 𝛾 𝑐 𝑑𝑡 − 𝑘 𝑐 ∙ 𝑟 𝑐 = 𝑑 𝛾 𝑐 𝑑𝑡 − 𝑑 𝑑𝑡 𝑘 𝑐 ∙ 𝑟 𝑐 + 𝑟 𝑐 ∙ 𝑘 𝑐 Now we can throw away the total derivatives Moreover, the last 3 terms on the RHS of the Lagrangian are just 𝑢 𝑘 𝑐 𝑖 𝑑 𝑢 𝑘 𝑐 𝑑𝑡 The final Lagrangian is ℒ=− 𝜀 𝑐 + 𝑟 𝑐 ∙ 𝑘 𝑐 + 𝑢 𝑘 𝑐 𝑖 𝑑 𝑢 𝑘 𝑐 𝑑𝑡

Electromagnetic fields These go in through the scalar and vector potentials The Hamiltonian has the form 𝐻 → 𝐻 𝑘 + 𝑒 𝐴 𝑟 , 𝑡 ℏ −𝑒𝑉 𝑟 , 𝑡 The local Hamiltonian 𝐻 𝑐 = 𝐻 𝑘 + 𝑒 𝐴 𝑟 𝑐 , 𝑡 ℏ −𝑒𝑉 𝑟 𝑐 ,𝑡 The eigenstates have the form 𝑢 𝑘 → 𝑢 𝑘 +𝑒 𝐴 The eigen-energy has the form 𝜀 𝑐 = 𝜀 0 ( 𝑘 )− 𝑒𝑉 𝑟 𝑐 ,𝑡 The Lagrangian can be greatly simplified Now all the time- and space-dependence comes from kc ℒ=− 𝜀 0 +𝑒𝑉 𝑟 𝑐 ,𝑡 + 𝑟 𝑐 ∙ 𝑘 𝑐 −𝑒 𝑟 𝑐 ∙ 𝐴 + 𝑘 𝑐 ∙ 𝑢 𝑘 𝑐 𝑖 𝜕 𝑢 𝑘 𝑐 𝜕 𝑘 𝑐

Semiclassical equations of motion Using the Euler-Lagrange equations 𝑑 𝑑𝑡 𝜕ℒ 𝜕 𝑟 𝑐 = 𝜕ℒ 𝜕 𝑟 𝑐 𝑑 𝑑𝑡 𝜕ℒ 𝜕 𝑘 𝑐 = 𝜕ℒ 𝜕 𝑘 𝑐 We obtain the SEMICLASSICAL EQUATIONS OF MOTION These describe the motion of the centre of the wave-packet ℏ 𝑟 𝑐 = 𝜕𝜀 𝜕 𝑘 𝑐 − 𝑘 𝑐 × Ω ℏ 𝑘 𝑐 =− 𝑒 𝐸 + 𝑟 𝑐 × 𝐵 Ω =𝑖 𝜕𝑢 𝜕 𝑘 𝑐 × 𝜕𝑢 𝜕 𝑘 𝑐

Berry curvature How do we evaluate the Berry curvature? Ω =𝑖 𝜕𝑢 𝜕 𝑘 𝑐 × 𝜕𝑢 𝜕 𝑘 𝑐 How do we evaluate the Berry curvature? Component by component Ω 𝑖 = 𝜀 𝑖𝑗𝑙 𝜕𝑢 𝜕 𝑘 𝑐,𝑗 𝜕𝑢 𝑘 𝑐,𝑙 For example Ω 𝑥 = 𝜀 𝑥𝑦𝑧 𝜕𝑢 𝜕 𝑘 𝑐,𝑦 𝜕𝑢 𝑘 𝑐,𝑧 + 𝜀 𝑥𝑧𝑦 𝜕𝑢 𝜕 𝑘 𝑐,𝑧 𝜕𝑢 𝑘 𝑐,𝑦 = 𝜕𝑢 𝜕 𝑘 𝑐,𝑦 𝜕𝑢 𝑘 𝑐,𝑧 − 𝜕𝑢 𝜕 𝑘 𝑐,𝑧 𝜕𝑢 𝑘 𝑐,𝑦 What is the Berry phase? Going back to Berry’s original paper Proc. R. Soc. London Ser. A 392, 45 (1984) Particle moving adiabatically along some loop C Γ 𝐶 = 𝑑 𝑘 𝑐 ∙ 𝑢 𝑘 𝑐 𝑖 𝜕 𝑢 𝑘 𝑐 𝜕 𝑘 𝑐 𝐶 Using Stokes’ theorem we can express the line integral as a surface integral Γ 𝐶 = 𝑑 𝑆 𝑘 ∙ 𝛻 × 𝑢 𝑘 𝑐 𝑖 𝜕 𝑢 𝑘 𝑐 𝜕 𝑘 𝑐 = 𝑑 𝑆 𝑘 ∙ Ω Sk is any surface in k-space whose boundary is the loop C See also J. J. Sakurai Modern Quantum Mechanics for the derivation of the Berry phase Wherever we have topological materials we have 𝜴

When is the Berry curvature non-zero? LHS, RHS must have same properties under time reversal (and inversion) Time-reversal invariance implies Ω − 𝑘 𝑐 =− Ω 𝑘 𝑐 There is also a geometric argument – Berry phase depends on path C Γ 𝐶 = 𝑑 𝑘 𝑐 ∙ 𝑢 𝑘 𝑐 𝑖 𝜕 𝑢 𝑘 𝑐 𝜕 𝑘 𝑐 𝐶 Under time-reversal, both C and 𝑘 𝑐 are reversed, as is 𝜕 𝜕 𝑘 𝑐 The Berry phase changes sign under time-reversal and Ω must do the same Note the Berry phase is also gauge-dependent There is a gauge in which it is zero This corresponds to parallel transport ℏ 𝑟 𝑐 = 𝜕𝜀 𝜕 𝑘 𝑐 − 𝑘 𝑐 × Ω

Berry curvature Suppose time-reversal is preserved Ω 𝑘 𝑐 in principle does not have to be zero However time-reversal implies Kramers degeneracy 𝜀 𝑐 𝑘 𝑐 = 𝜀 𝑐 − 𝑘 𝑐 Spatial inversion symmetry may or may not be broken If the state at 𝑘 𝑐 is occupied then the state at − 𝑘 𝑐 is also occupied Recall time-reversal also implies Ω − 𝑘 𝑐 =− Ω 𝑘 𝑐 So the integral of Ω over all filled states vanishes Any effect that depends purely on Ω must vanish Meaning of curvature correction 𝑘 changing adiabatically Aharonov-Bohm effect = example of Berry phase Known before it was called Berry phase

Boltzmann equation The wave-packets follow a distribution 𝑓( 𝑟 𝑐 , 𝑘 𝑐 , 𝑡) Electric field cause electrons to drift, defects cause scattering The distribution will change as a result of 𝐸 and scattering In general 𝑓( 𝑟 𝑐 , 𝑘 𝑐 , 𝑡) obeys the Boltzmann equation 𝜕𝑓 𝜕𝑡 + 𝑟 𝑐 ∙ 𝜕𝑓 𝜕 𝑟 𝑐 + 𝑘 𝑐 ∙ 𝜕𝑓 𝜕 𝑘 𝑐 + ℐ 𝑓 =0 Here ℐ 𝑓 is the collision integral – simplify to relaxation time approx. ℐ 𝑓 = 𝑓− 𝑓 𝑒𝑞 𝜏 𝑓 𝑒𝑞 is the equilibrium distribution – in general Fermi-Dirac 𝜕 𝑓 𝑒𝑞 𝜕𝑡 + 𝑟 𝑐 ∙ 𝜕 𝑓 𝑒𝑞 𝜕 𝑟 𝑐 + 𝑘 𝑐 ∙ 𝜕 𝑓 𝑒𝑞 𝜕 𝑘 𝑐 + ℐ 𝑓 𝑒𝑞 =0 Collisions seek to return the system to equilibrium

Boltzmann equation Out of equilibrium but 𝐸 not too large 𝑓= 𝑓 𝑒𝑞 +𝑔 At small 𝐸 solve 𝑔 to linear order in 𝐸 Assume homogeneous system so no 𝑟 𝑐 – dependence Assume no magnetic field so 𝑘 𝑐 =− 𝑒 𝐸 ℏ and steady state so 𝑓 has no explicit t-dependence 𝑔= 𝑒 𝐸 𝜏 ℏ ∙ 𝜕 𝑓 𝑒𝑞 𝜕 𝑘 𝑐 For a more general proof see M. Marder Condensed Matter Physics At 𝑇=0, 𝑓 𝑒𝑞 𝑘 𝑐 =Θ 𝜀 𝐹 − 𝜀 𝑐 → 𝜕 𝑓 𝑒𝑞 𝜕 𝑘 𝑐 =−δ 𝜀 𝐹 − 𝜀 𝑐 𝜕 𝜀 𝑐 𝜕 𝑘 𝑐 𝑔=− 𝑒 𝐸 𝜏 ℏ ∙ 𝜕 𝜀 𝑐 𝜕 𝑘 𝑐 δ 𝜀 𝐹 − 𝜀 𝑐 So 𝑔 is responsible for a Fermi-surface effect

Boltzmann equation Conductivity – multiply by (−𝑒 𝑟 𝑐 ) 𝑟 𝑐 = 1 ℏ 𝜕 𝜀 𝑐 𝜕 𝑘 𝑐 + 𝑒 𝐸 × Ω ℏ 2 𝑓= 𝑓 𝑒𝑞 − 𝑒 𝐸 𝜏 ℏ ∙ 𝜕 𝜀 𝑐 𝜕 𝑘 𝑐 δ 𝜀 𝐹 − 𝜀 𝑐 The term independent of 𝐸 integrates to zero over phase-space There are two terms linear in 𝐸 The extrinsic term containing 1 ℏ 𝜕 𝜀 𝑐 𝜕 𝑘 𝑐 − 𝑒 𝐸 𝜏 ℏ ∙ 𝜕 𝜀 𝑐 𝜕 𝑘 𝑐 δ 𝜀 𝐹 − 𝜀 𝑐 This depends on scattering through 𝜏 The intrinsic term containing 𝑒 𝐸 × Ω ℏ 2 𝑓 𝑒𝑞 This is INDEPENDENT of scattering – no 𝜏 present

Boltzmann equation The extrinsic term leads to a longitudinal conductivity − 𝑒 ℏ 𝜕 𝜀 𝑐 𝜕 𝑘 𝑐 − 𝑒 𝐸 𝜏 ℏ ∙ 𝜕 𝜀 𝑐 𝜕 𝑘 𝑐 δ 𝜀 𝐹 − 𝜀 𝑐 Consider a boring band 𝜀 𝑐 = ℏ 2 𝑘 2 2𝑚 and take 𝐸 ∥ 𝑥 Integrate over all 𝑘 𝑐 including density of states and 2 for spin 2 𝑑 3 𝑘 𝑐 2𝜋 3 − 𝑒 ℏ 𝜕 𝜀 𝑐 𝜕 𝑘 𝑐 − 𝑒 𝐸 𝜏 ℏ ∙ 𝜕 𝜀 𝑐 𝜕 𝑘 𝑐 δ 𝜀 𝐹 − 𝜀 𝑐 = 𝑛 𝑒 2 𝐸𝜏 𝑚 This is the well-known Drude conductivity Note 𝜏 breaks time reversal as required by the relationship 𝐽 =𝜎 𝐸 Note 𝜎 is a tensor ( 𝐽 𝑖 = 𝜎 𝑖𝑗 𝐸 𝑗 )

Boltzmann equation The intrinsic term has the form 𝑓 𝑒𝑞 𝑒 𝐸 × Ω ℏ 2 There is no 𝜏 but Ω breaks time-reversal symmetry here It contains a cross product so it contributes to the Hall conductivity This is not the ordinary Hall conductivity It contributes but typically overwhelmed by Lorentz-force term ℏ 𝑘 𝑐 =−𝑒( 𝑟 𝑐 × 𝐵 ) It is significant when the system already breaks time reversal in equilibrium For example there is a magnetisation 𝑀 present 𝑀 by itself does not give a Lorentz force (because 𝑀 comes from the spin) Hence the Berry-curvature term is important and can be dominant Hall effect with no external magnetic field ≡ anomalous Hall effect We will calculate this explicitly for a specific model in Lecture 2

Dissipation Dissipation is related to irreversibility This in turn is related to entropy production Let’s look at the first law of thermodynamics (s = entropy density) 𝑑𝑢=𝑇𝑑𝑠+ 𝜇𝑑𝑁+𝑉𝑑𝜌→𝑑𝑠= 𝑑𝑢 𝑇 − 𝜇 𝑇 𝑑𝑁− 𝑉 𝑇 𝑑𝜌= 𝑖 𝑉 𝑖 𝑑 𝜌 𝑖 The 𝜌 𝑖 are generalised densities, 𝑉 𝑖 are generalised potentials 𝐽 𝑖 are generalised current densities leading to continuity equations 𝜕 𝜌 𝑖 𝜕𝑡 + 𝛻 ∙ 𝐽 𝑖 =0 For the entropy the continuity equation has a source term 𝜕𝑠 𝜕𝑡 + 𝛻 ∙ 𝐽 𝑠 =Σ where Σ = local entropy production rate

Dissipation Express the entropy current in terms of the others similar to 1st law 𝐽 𝑠 = 𝑖 𝑉 𝑖 𝐽 𝑖 From the 1st law we have after some simplifications Σ= 𝑖 ( 𝛻 𝑉 𝑖 )∙ 𝐽 𝑖 Compare this result to The standard classical mechanics result that 𝑃𝑜𝑤𝑒𝑟= 𝐹 ∙ 𝑣 Note for dissipation the force must have a component parallel to the velocity Satisfied for longitudinal conductivity since 𝐹 , 𝑣 are both parallel to 𝐸 The standard electrodynamics result that 𝑃𝑜𝑤𝑒𝑟= 𝐼 2 𝑅= 𝑉 2 /𝑅 Note that 𝛻 𝑉= 𝐸 and 𝐽 =𝜎 𝐸 so entropy production ∝ 𝐸 2 𝜏 Meaning that scattering is directly involved in dissipation

Dissipation The Hall conductivity does not give dissipation Simplest argument 𝐹 ∙ 𝑣 =0 Because 𝐹 is parallel to 𝐸 Whereas 𝑣 is perpendicular to 𝐸 Onsager relations on the symmetry of kinetic coefficients For the conductivity tensor this takes the general form in a magnetic field 𝜎 𝑖𝑗 𝐵 = 𝜎 𝑗𝑖 − 𝐵 For the longitudinal component 𝜎 𝑖𝑖 𝐵 = 𝜎 𝑖𝑖 − 𝐵 so can only depend on magnitude of 𝐵 Meaning that the 𝐵 -dependent terms in 𝜎 𝑖𝑖 𝐵 cannot break time-reversal However in the Hall conductivity this restriction does not exist In the ordinary Hall effect it is the 𝐵 -term in 𝜎 𝑥𝑦 that breaks time-reversal NOT the scattering time 𝜏 Whatever breaks time-reversal there does not have to lead to dissipation

Dissipationlessness To get a system that is dissipationless Arrange so that 𝜎 𝑥𝑥 =0 This way the dissipative term is eliminated Keep only 𝜎 𝑥𝑦 𝜎 𝑥𝑦 does not lead to dissipation We do not want a magnetic field because that costs a lot of energy To generate a magnetic field you need a current – lots of dissipation The earlier analysis suggests we want an anomalous Hall effect So we want a magnetisation and a system that exhibits No longitudinal transport But a finite anomalous Hall effect stemming from a finite Berry curvature So far this is only possible in topological materials We will derive the anomalous Hall effect in a dissipationless system explicitly in Lecture 3

Summary Transport of electrons in crystals Dissipation in DC transport Electric field drives electrons Impurities, phonons scatter them – resistance Collisions taken into account in the Boltzmann equation Drift of electrons between collisions described by wave-packet dynamics Semi-classical equations contain a non-trivial correction due to the Berry curvature This correction gives rise to a Hall conductivity Dissipation in DC transport Associated with entropy production It comes from the longitudinal part of the conductivity The Hall conductivity is NOT associated with dissipation In order to get dissipationless transport Make the longitudinal conductivity zero Keep only the Hall conductivity To date this is ONLY possible in topological materials – following lectures