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

2. 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, (1999)

3. 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 →

4. 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)

5. 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)

6. 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

7. 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 𝜕𝛾 𝜕 𝑘 + 𝑢 𝑘 𝑖 𝜕 𝑢 𝑘 𝜕 𝑘
𝑟 𝑐 = 𝜕 𝛾 𝑐 𝜕 𝑘 𝑐 + 𝑢 𝑘 𝑐 𝑖 𝜕 𝑢 𝑘 𝑐 𝜕 𝑘 𝑐

8. 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)

9. 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

10. 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

11. 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

12. 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

13. 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

14. 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

15. 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

16. 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

17. 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

18. 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

19. 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 − 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 )

20. 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

21. 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

22. 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

23. 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

24. 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

25. 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