geometric phases (again)people.roma2.infn.it/~cini/ts2016/ts2016-20.pdf · 2016-05-03 · the...

The geometric phase, or Pancharatnam-Berry phase is known in classical and quantum mechanics. Initially it has been studied in adiabatic processes, where it is simplest, but such a limitation to adiabatic transformations does not exist in general.

In many physical problems when the state of the system depends on parameters and cyclic transformations are done, one observes phenomena, which depend on the geometry and topology of the abstract parameter space.I have introduced this argument on part 11 of this course. . The physical effect in question can be a phase shift of a classical wave , a quantum phase shift, or an ordinary angle in the case of a classical mechanics system; there is a great variety of situations.

1

Below I discuss phenomena where the Hamiltonian has a parameter space R, and performs a closed cycle in it. Depending on the properties of the Hamiltonian in parameter space, the system does not return in the original state. More generally its ψ acquires a phase, which is related to the singularities of the Hamiltonian encircled by the loop in parameter space. To this end it is necessary that at least 2 parameters are varied and the return to the original point is not done by undoing the first steps in a self-retracing back-and-forth variation but is done by encircling singularities.

Geometric phases (again)

1

2

γ = ∇∫

( )

where is on instantaneous eigenket of H, and C a path in R space,is a topological phase, invariant under continuous deformations of C.

It vanishes in simply connected parameter space

n n R nC

n

C i a a dR

a

s where C can collapse to a point but in a multiply connected spaces it yields a good quantum number, which does not arise from any operator.

C

Professor Sir Michael Berry

Pancharatnam-Berry phase

2

Vector Potential Analogy

One naturally writes ( ) · , | . |n n n n R nCC A dR A i a aγ = = ⟨ ∇ ⟩∫

introducing a sort of vector potential (which depends on the H eigenstate n, however). The gauge invariance arises in the familiar way, that is, if we modify the basis with

( )[ ] [ ], ,i Rn n n n Ra R e a R A Aχ χ→ ⇒ → − ∇

then the extra term, being a gradient in R space, does not contribute to the integral over C. The Berry phase is real since

| 1 | 0 | | 0| . . 0 | is pure imaginary;

n n R n n R n n n R n

n R n n R n

a a a a a a a aa a c c a a

⟨ ⟩ = ⇔ ∇ ⟨ ⟩ = ⇔ ⟨∇ ⟩ + ⟨ ∇ ⟩ =⇔ ⟨ ∇ ⟩ + = ⇒ ⟨ ∇ ⟩

hence | | is real |Im, | .n n R n n n R nA i a a A a a= ⟨ ∇ ⟩ ⇒ ⟨ ∇= − ⟩

33

4

To avoid confusion with the electromagnetic field in real space one often calls An the Berry connection; calling ξα the parameters,

( ) ( ) ( ) .n n ni ξχ ξ ψ ξ ψ ξ= ∇

4

We prefer to work with a manifestly real and gauge independent integrand; going onwith the electromagnetic analogy, we introduce the field as well, such that

( ) · · .n n nS SC rot A ndS B ndSγ = ≡∫ ∫

In general, for R of any dimensions,

from | | , one defines the curvature tenso. The real point is that the Stokes theorem

extends to any dimen

r

si

ons.

n n R nn

n n

A i a aA Aµν µ ν ν µ

= ⟨ ∇ ⟩

Ω = ∂ ∂−

5

[ ]

[ ]

Im ( | ) Im ( | )

Im[ ( | ) ( | )].The last term vanishes, since it must be even and odd when .

( | ) Im[ ( | )] Im ( | | ) .

ni n n ijk j n k ni

ijk j n k n ijk n j k n

ni ijk j n k n ijk j k n n i

n

B a a a a

a a a aj i

B a a a a a a

B

ε

ε ε

ε ε

= − ∇ ∧ ∇ = − ∂

= − ∂ + ∂

= ∂ = − ∂ = − ∇ ∧ ∇

=

∂

∂

∂

−

∂

∂

[ ]Im ( | | ) .n na a∇ ∧ ∇

If the parameter space ,

Im an

is 3d,

| | dn n

n n R n

B rotA

A a a= −

=

⟨ ∇ ⟩

and, inserting a complete se

Im ( | | ) Im .

t,

n n n n m m nm n

B a a a a a a≠

= − ∇ ∧ ∇ = − ∇ ∧ ∇∑

We transform Imn n m m nm n

B a a a a≠

= − ∇ ∧ ∇∑

[ ] [ ] [ ][ ] [ ] [ ]

[ ] [ ] [ ] [ ] [ ] [ ]

Take of H R a R E R a [R]:

(H R a R ) (E R a [R])

( H R )a R H R a R ( E R )a [R] E R a [R]

R n n n

R n R n n

R n R n R n n n R n

∇ =

∇ = ∇

⇒ ∇ + ∇ = ∇ + ∇

6

Taking the scalar product with an orthogonal am

a H a a a E a a

a aa a divergence if degeneracy occurs along C.

E E

m R n m m R n n m R n

m R nm R n

m n

E

H

∇ + ∇ = ∇

∇∇ = →

−

Formula for the curvature

A nontrivial topology of parameter space is associated to the Berry phase, and degeneracies lead to singular lines or

surfaces, like wizard’s hat.

There must be a singularity!

where m,n indices refer to adiabatic eigenstates of H, and the term with m=n vanishes (vector product of a vector and itself).

6

7

Aharonov-Bohm effect

The electron(s) see no magnetic field. The phase difference between beams on either side of solenoid is

, magnetic flux in solenoid.

Dimensions: [ ] [ ] [ ] 1

qc

hcq

φ

φ

Φ∆ = Φ =

Φ = ⇒ ∆ =

8

Parameters : radius of solenoid,r B=

Berry connection : A

curvature : B

Berry phase : φ∆

singularity : solenoid

8

0B =

Molecular Aharonov-Bohm effect –recall exε

Longuet-Higgins H C, O¨ pik U, Pryce M H L and Sack R A 1958 Proc. R. Soc. A 244 1

The first discovery occurred early:

although systematic understanding occurred after the Berry paper.

H.Christopher Longuet-Higgins (1923-2004)In the BO approximation, the molecular wavefunction is factorized:

( ) ( )Assume , , where nuclear coordinates, electron coordinates,el nuclx xψ ξ ψ ξ ξΦ = = =

( ) ( ) ( ) ( )

Let component of momentum.

| , | | , | .

nuclear

el nucl el nucl

p i

p p x x p

αα

α α α

αξ

ψ ξ ψ ξ ψ ξ ψ ξ

∂= − =

∂

→ > > + > >

( )

( ) ( )

( ) ( ) ( ) ( )

The effective nuclear momentum acting on , averaged over electron degrees of freedom, is therefore :

,

, , .

nucl

nucl el

nucl el el nucl

x i

i x x

αα

α α

π ψ ξ

π ψ ξ ψ ξξ

ψ ξ ψ ξ ψ ξ ψ ξξ ξ

∂= − Φ

∂

∂ ∂= − +

∂ ∂

9

( ) ( ) ( )2, , , 0.el el elA dx x x dx xαα α

ψ ξ ψ ξ ψ ξξ ξ∂ ∂

= = =∂ ∂∫ ∫

If the electron wave functions can be taken real, the Berry connection vanishes since

The curvature (magnetic field of the Berry connection) also vanishes.

All OK? It depends on the irreps of electron wave functions and nuclear vibrations. In some cases like eXε molecules the electron wave functions cannot be taken real.

( ) ( ) ( ) ( ) ( )

( ) ( )

( ) ( )

, ,

means that there is an effective magnetic field for the nuclei, in this way:( )

, , vector potential: it has the form o

nucl nucl el el nucl

nucl nucl

el el

i x x

p eA

A x x

αα α

α α α

αα

π ψ ξ ψ ξ ψ ξ ψ ξ ψ ξξ ξ

π ψ ξ ψ ξ

ψ ξ ψ ξξ

∂ ∂= − +

∂ ∂

= −

∂=

∂

f a Berry connection.

10

11

To save the situation, the electron wave functions must be taken single-valued but complex, in the absence of a true magnetic field . The Berry connection then does not vanish. The gauge invariant curvature still vanishes, except at singularities, but the flux of the Berry connection through the surface bounded by curves in parameter space does not generally vanish. Again, this leads to the abovespecified observable consequences.

The trouble is that the wave functions change sign for an adiabaticrotation of the molecule by 2 π they fail to be single-valued, as in exε example.

Since the total wave function must be single valued, the nuclear wavefunction must also change sign. This is equivalent to a change of the boundary conditions. In turn, this hasobservable consequences, like rotovibrational levels withhalf integer rotational quantum numbers.

11

1212

Open path Pancharatnam phase

Then, open path geometric phase=discrete open path C with ‘equivalent’ initial and final points:

ψ ξ ψ ξγ ϕ ϕ ϕ ϕ

−

− − −= ∆ + ∆ + + ∆ + ∆

11

1,2 2,3 2, 1 1,

Fixing a gauge such that ( )=U ( ) we may definegauge invariant

Pancharatnam phase along open path.

n

n n n n

Two points in parameter space can sometimes correspond to symmetric situations.

ξ ξ−∃ = 11unitary U: ( ) U ( )UnH H

12

ξ1 ξ2

1313

( ) ( )1

Continuous version : , open curve

with equivalent end points .

nk k nk i fC

f i

i dk u u C

U

γ ξ ξ

ψ ξ ψ ξ−

= ∇ = →

=

∫

Single point Berry phase: n=2 equivalent points

13

ξ ξ−= 12 1

At the limit, one can get a Pancharatnam phase with 2 equivalent points: ( ) U ( )U.H H

( ) ( )( ) ( )

( ) ( ) ( ) ( )φψ ξ ψ ξ

φ ψ ξ ψ ξ ψ ξ ψ ξψ ξ ψ ξ

− ∆ −= ⇒ ∆ = − = −12 1 2 11 1 1 1

1 2

Im log argiije U U

Single point Pancharatnam phase: n=2 equivalent points

1414

One electron in a periodic potential with pbc.

How to define the mean position of a wavepacket?

( )ψ∞

−∞= ∫

2 with a Bloch wave does not exist

and does not mean anything!

x dx x x

We want a new definition. 14

Use of Open path Pancharatnam phase:

xa0-a 2a

n(x)

1515

Selloni et al. defined arg( ) Im log( ), where2 2a ax z zπ π

= =

πψ ψ ψ= = = ∈∫ 2

0

2| ( )| , reciprocal lattice.a iGx iGx

cellz dx e x e G

a

( )ψ δ= − +∑20 0

0Assume ψ perfectly localized at x in first cell and periodic,

( ) . A sensible definition should yield .m

x x x ma x

15

xa0-a 2a

n(x)

1616

One electron in a periodic potential with pbc: position of wavepacket

( )20Indeed, if ( ) , the definition

mx x x maψ δ= − +∑

00

2 ixiGx az e e

π

= =

= 0x x

ππψ= == ∫ 2

0

2| ( )| , yieldsarg(2

:),a iGxz dx e Gx x

aa z

Then, if wave packet is localized in length <<L, this works fine

For poor localization it fails, but then failure is acceptable.

02( ) xArg za

π=

16

1717

π

πψ ψ ψ

ψ ψ

= = = ∈∫ 2

0

2

2| ( )| , reciprocal lattice

= single-point Pancharatnam phase,phase difference between and .

The unitary transformation is U= .

This Unitary transformation applied to Blo

a iGx iGx

iGx

i xa

z dx e x e Ga

ze

e

ψ ψ

+

+ +

→ + →

= ⇔ =

( )

ch statesU : , ( ) ( )

shifts each k to equivalent k point: ( ) ( ) u ( ) ( )

ikx i k G xk k

iGxk k G k k G

k k G e u x e u x

x x x u x e

BZ

π= arg( ) ; has the form of a Pancharatnam phase.

2ax z z

17

( ) ( )( ) ( )

( ) ( ) ( )φψ ξ ψ ξ

φ ψ ξ ψ ξ ψ ξψ ξ ψ ξ

− ∆ = ∆ = =12 1 21 2 1

1 2

Recall: phase difference between and .iije U

Bloch oscillations and the Wannier–Stark ladder

Within the semiclassical approach, an electron in a given band is a wave packet centered at r, whose crystal momentum is q. Neglecting collisions, the following semiclassical picture in an external electric field E seems almost obvious:

( )1speed: ,

acceleration:

since ( ) ( ) is periodic, ( ( )) is periodic.

is periodic in time periodic orbit in real space!

n

n n n

qdrdt q

dq eEteE qdt

q q G q tdrdt

ε

ε ε ε

∂=

∂

= ⇒ =

⇒ = +

⇒

Wannier in 1960 predicted that quantization would lead to the so calledWannier –Stark ladders, with an oscillator-like discrete spectrum. He assumed

that adding G to q the wave function resumed the original value.

, = band-energy average, m=integer, integer4m m mW eEa m αε ε α = + + =

Frequency estimate of Bloch oscillations : , lattice constant.Periodic motion implies quantized energy.

B eEa aω = =

q

( )n qε

18( )1But the equation of motion was criticized by Zac.n qdrdt q

ε∂=

∂

1919

one uses a parametric Hamiltonian for periodic part u, k is the parameter2( ) ( ) ( ) ( )

2 k k kp k V x u x u x

mε

++ =

Bloch states in solids: Berry phase and Zak’s phase

We are allowed to study the cell as if it were isolated, but then—as emphasized in Berry’s original paper [1]—the interaction with the ‘rest of the Universe’ gives rise to a non-trivial phase which is observable: indeed, this is a Berry’s phase.

Since we use a k- dependent Hamiltonian for uk there is a Berry’s connection. K space is the parameter space R.

χ = ∇( )n nk k nkk i u u

γ = ∇ ≠∫ 0nk k nkCi dk u u

So, there is a Berry’s phase ante litteram if we can chooseC such that

How can we take contour C? What is the physical content?

To compute Bloch functions ( ) ( ),ikrnk nkr e u rψ =

J. Zak,Phys. Rev. Lett. 20, 1477 – Published 24 June 1968

19

2020

nkknkCuudki ∇= ∫γ

G

( ), ,

, ,

n,k

Natural gauge: ( ) ( ) ( ) ( ),

with same phase (periodic gauge) ( ) ( )(Thus, u is periodic in r and up to a phase in k)

i k G r ikrn n k G n k n

iGrn k n k G

k G e u r e u r k

u r e u r

ψ ψ++

+

+ = = =

⇒ =

C

Open path geometric phase

Zak’s phase

C connects k with k+G

(k, ) ( )ikrn nkr e u rψ =

20

2121

Berry phase of Slater determinant (Hartree-Fock or Kohn-Sham) in terms of Berry phases of orthogonal spin-orbitals

No general result for Pancharatnam phases, but taking the continuum limit the problem simplifies. One can show:

In the continuum limit the closed path Berry phase is the sum of

Individual spinorbital Berry phases- This is not granted in general for finite systems.

21

We can rewrite it as a Pancharatnam phase discretizing with M+1 points q0….qM

γ = ∇ ≠∑ ∫

Consider transporting over C a determinant with a number of bands: the Berry phase is

0bandsn

nk k nkCn

i dk u u

γ = ∇∑ ∫ is related to the bulk polarization of insulators!bandsn

nk k nkCn

i dk u u

22

( )G.Sundaram and Q.Niu in1 999 . . 59, 14 915 discovered that the seemingly obvious equations of mot

( )1, ,

not complet

ion

e.

n qdq dreEdt d

Phys Rev

tare

B

qε∂

= =∂

Actually,

( )

( )

( )

2Berry curvature of band, clearly L ,

so is a velocity.

is right but

( )1 , so there is a drift term.

.

q nq q nq

n

q i u u

dq qdt

dq eEdt

qdr dq qdt q dt

ε

Ω = ∇ ∧ ∇ =

∧ Ω

=

∂= − ∧ Ω

∂

' phase is obviously the Berry phase

u( ) u( ) qC

Zak s

i q q dqγ = ∇∫

They show that .4 2m mW eEa m α γε

π = + + −

23

More applications and derivations in Raffaele Resta, J. Phys.: Condens. Matter 12 (2000) R107–R143. and J. Phys.: Condens. Matter 14 (2002) R625–R656

23

24

In ordinary bulk materials these Stark oscillations cannot be seen, because collisions dephase the coherent motion of electrons on a time-scale which is much shorter than TB = 2π/ωB. Eventually they have been seen in semiconductorsuperlattices.

25

2626

Dipole moment of insulators

ρ ρ

=

= = Ψ Ψ

=

∫∑

0 0

1

For a molecule, one defines the dipole momentfunctional of the density

N

ii

d e rdV e R

R r

- +

Polar molecule

This definition does not apply to periodic solids!r takes outside Hilbert space of periodicfunctions; makes surfaces crucial

+ − + − + −+ − + − + −+ − + − + −

− + − + − +− + − + − +− + − + − +

The results of calculations in a slab geometry depend on how the slab is defined 26

2727

- - - -- + + + - + + +

-+

Periodic array of polar molecules: same pattern can be obtained e.g. by repeating

The polarization of metals is mainly due to screening at the surface, but polarization of piezoelectric insulators is much more a bulk (not surface) property.The dipole moment of the unit cell is not well defined.

- +

and the results of slab calculations depend on the way one terminates the lattice

27

Simple-minded definition of dipole averaged over a determinant

σ σσ

ρ

ψ ψ

=

−= =

∫

∑

31 ( ) .

Since it is a one-body operator, we may sum over spin-orbitals.2 ( , ) ( , ) , where .

The plane wave fac

el elcrystal

el n n cellnk

P d r r rV

eP k r r k r V NVV

σ σσ

−= ∑

tors cancel and2 ( , ) ( , ) . el n n

nk

eP u k r r u k rV

2828But result depends on origin, which is arbitrary!

To characterize a piezoelectric crystal we may use an arrangement in which the crystal is uniaxially strained in a shorted capacitor and the current in the external circuit is measured. E and H are negligible.

Amperometer

J

Indeed, the mechanical action varies the polarization P and produces a current.

dispacement current 4 but is negliglible 4 D E PJ D E P Jt t t

π π∂ ∂ ∂⇒ = − = + ⇒ = −

∂ ∂ ∂

Note:the phase of the wave function is involved in the current (J=0 if ψ is real); the mechanical action modulates P. The phenomenon is not a surface effect. We need a pressure gauge parameter.

but H is negligibleDH Jt

∂∇ ∧ = +

∂

29

3030

Introduce a deformation parameter λ such that the real solid

corresponds to λ=1

λ=0.5

λ=0.

the dipole is halved for

the dipole vanishes for

30

σ σσ

σ σσ

λ ρ λ λ ψ λ ψ λ

λ λ λ

−= ⇒ =

−⇒ =

∑∫

∑

31 2( ) ( , ) ( ) ( , , ) ( , , ) ,

2 ( ) ( , , ) ( , , ) .

el el el n ncrystalnk

el n nnk

eP d r r r P k r r k rV V

eP u k r r u k rV

3131

The current arises from a variation of pressure and of the pressure gauge parameter. Therefore set

Thus we focus on the derivative of dipole

λ λλ λ

∂ − ∂=

∂ ∂∑

2 ( , , ) ( , , )eln n

nk

P e u k r r u k rV

λ λ λ λλ λ

λ λλ

− ∂ ∂= +

∂ ∂

− ∂=

∂

∑ ∑

∑

2 ( ( , , ) ( , , ) ( , , ) ( , , ) )

2 2Re ( , , ) ( , , ) .

n n n nnk nk

n nnk

e u k r r u k r u k r r u k rV

e u k r r u k rV

and doing the derivative

3232

λ λλ λ

∂ − ∂=

∂ ∂∑

2 2Re ( , , ) ( , , ) .eln n

nk

P e u k r r u k rV

Thus, the object we want is the Bloch function contribution :

( )λ λ λ λ λλ λ∂ ∂

≡ =∂ ∂

2Re ( , , ) ( , , ) 2Re ( , , ) ( , , ) .nk n n n nB u k r r u k r u k r r u k r

( )

α αα

αα

λ

λ

λ λ λ λλ λ

λ λ

∂∂

∂∂

∂ ∂=

∂ ∂

=

∑

∑

ˆdoes not mix different k, since k is eigenvalue of , and a is not changed

however mixes different bands,and we expandover bands:it holds that

( , , ) ( , , ) ( , , ) ( , , )

Thus, 2Re ( , , )

a

n n

nk n

T

u k r u k r u k r u k r

B u k r r u αλ λ λλ∂

∂( , , ) ( , , ) ( , , )nk r u k r u k r

( )λλ

∂ −=

∂ ∑2elnk

nk

P e BV

3333

......but r is a tricky operator for a periodic solid. We must replace it.

( )λλ

∂ −=

∂ ∑2elnk

nk

P e BV

( ) α αα

λ λ λ λ λλ∂

=∂∑2Re ( , , ) ( , , ) ( , , ) ( , , )nk n nB u k r r u k r u k r u k r

ψ ψ ∂= =

∂( , ) ( , )

( Appendix: tedious but elementary consequence of Bloch's theorem)

m n m n m nr u r u u k r i u k rk

see

Recall: we met : it is Zac's phase (page 20)nk k nkCi dk u uγ = ∇∫

summarizing: 4 PJt

π ∂= −

∂

( )

( )

α αα

α αα

λ λ λ

ψ

λ λλ

λ λ λλ

ψ

∂=

∂

∂ ∂=

∂ ∂

∂= =

∂

∑

∑

using:

Embellish

2 Re ( , , ) ( , , ) ( , , ) ( , , )

2Re ( , , ) ( , ) ( , ,

( , ) ( , ) , and get

) ( , ,

m n m

nk n n

nk n

n

n

m n

B u k r r u k r u k r u k r

B u k r i u k r u k r

r u r u u k r i u k rk

u k rk

λ) .

34

α

∂∂∑Now, we wish to do and to this need to shiftend we :k

αα αλ ∂ ∂= −

∂ ∂∂∂

=

( , ) ( , )

where the blue stuff 0. N

( , , ) ( , ) ( , )

o

(

,

,

w

)n n nu k r i u k r i u ki u k r u rk rk

r uk

kk

( ) α αα

λ λ λ λ λλ

∂ ∂= −

∂ ∂∑2Re( ) ( , , ) ( , , ) ( , , ) ( , , ) ,

and we can sum over the complete set.

nk n nB i u k r u k r u k r u k rk

34( )λ λ λλ

∂ ∂= −

∂ ∂2Re( ) ( , , ) ( , , )nk n nB i u k r u k r

k

35

( )λ λ λλ

∂ ∂= −

∂ ∂2Re( ) ( , , ) ( , , )nk n nB i u k r u k r

k

( )λ λλ

∂ ∂=

∂ ∂2Im ( , ) ( , , )nk n nB u k r u k r

kthat is:

λ λλ λ

∂ − ∂ ∂=

∂ ∂ ∂∑2 2Im ( , , ) ( , , ) .

Next, we sh

Summary: we have shown tha

ow that this is a Berry phase by a model calculat

t

ion.

eln n

nk

P e u k r u k rV k

35

λλ λ

∂ − ∂ ∂=

∂ ∂ ∂∑2Model calculation of 2 Im ( , ) ( , , ) .eln n

nk

P e u k r u k rV k

π π π

π π ππ

π π

− − −

= =

=

∑

∑ ∫ ∫ ∫ ∫

∑

33

First, weassume that only the nth band contributes,andis inserted backat the end.

For

Brilloui

simplici

n Zo

ty assume(2 )

2 2but neglecting dependence on

ne su

, ,

mmationn

a b c

x y zk BZ

a b c

x yk

V d k dk dk dk

k ka b

π

π−

∫c

z

c

dk

π

π

π π λλ λπ

−

∂ − ∂ ∂=

∂ ∂ ∂∫3

Then,

2 2 2 2 Im ( , ) ( , , )(2 )

cel

z n z n zz

c

P e V dk u k r u k rV a b k

36

36

π

π

π π λλ λπ

λ

−

∂ − ∂ ∂=

∂ ∂ ∂

= =

∫3

z

2 2 2Integrating 2Im ( , ) ( , , )(2 )

over λ with ( 0) 0, and writing k for k

cel

z n z n zz

c

el

P e V dk u k r u k rV a b k

P

π

π

π πλ λλπ

−

− ∂ ∂=

∂ ∂∫ ∫1

30

2 2 2 2 Im ( , ) ( , , ) .(2 )

c

el n n

c

e VP d dk u k r u k rV a b k

37

k

λ

cπ

−cπ

1

This is the change of P from no dipole to actual dipole, so it is the actual dipole. We want to transform this surface integral to the integral of a curl 37

*The identity 2 Im Im Im , with z ( , ) ( , , )n nz z z u k r u k rk

λλ

∂ ∂= − = ⇒

∂ ∂

λ λ λλ λ λ

∂ ∂ ∂ ∂ ∂ ∂= −

∂ ∂ ∂ ∂ ∂ ∂2Im ( , ) ( , , ) Im ( , ) ( , , ) Im ( , , ) ( , )n n n n n nu k r u k r u k r u k r u k r u k r

k k k

38

π

π

π λ λ λλ λ

−

=

− ∂ ∂ ∂ ∂− ∂ ∂ ∂ ∂

∫ ∫1

0

Putting all toget

Im ( , ) ( , , ) ( , , ) ( , )

her: el

c

n n n n

c

P

e d dk u k r u k r u k r u k rab k k

π

π

π πλ λλπ

−

− ∂ ∂=

∂ ∂∫ ∫1

30

2 2 2 2 Im ( , ) ( , , ) .(2 )

c

el n n

c

e VP d dk u k r u k rV a b k

The derivatives on the bra must be taken out the matrix elements38

λ λ λλ λ λ

λ λ λλ λ λ

λ λλ λ

λ

∂ ∂ ∂ ∂ ∂ ∂= −

∂ ∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂ ∂

= −∂ ∂ ∂ ∂ ∂ ∂

⇒

∂ ∂ ∂ ∂−

∂ ∂ ∂ ∂∂ ∂

=∂ ∂

( , ) ( , , ) ( , ) ( , , ) ( , ) ( , , )

( , , ) ( , ) ( , , ) ( , ) ( , , ) ( , )

( , ) ( , , ) ( , , ) ( , )

( , ) (

n n n n n n

n n n n n n

m n n m

m n

u k r u k r u k r u k r u k r u k rk k k

u k r u k r u k r u k r u k r u k rk k k

u k r u k r u k r u k rk k

u k r u kk

λ λλ∂ ∂

−∂ ∂

, , ) ( , , ) ( , )n mr u k r u k rk

π

π

λ λ λπ λ λ

−

− ∂ ∂ ∂ ∂⇒ = − ∂ ∂ ∂ ∂

∫ ∫1

0Im ( , ) ( , , ) ( , , ) ( , )

Now it is a curl.

c

el n n n n

c

eP d dk u k r u k r u k r u k rab k k

3939λ λ λλ

∂ ∂= ∂ ∂

( , ) ( , , ) ( , ) , ( , ) ( , , ) .k n m n nV V u k r u k r u k r u k rk

λ λ λλ

∂ ∂→ = ∂ ∂

( , ) ( , ) ( , , ) ( , ) , ( , ) ( , , ) .x y k n m n nV V V V u k r u k r u k r u k rk

∂ ∂− =

∂ ∂

For any vector , ( ) . In this case,y x zV V V rot Vx y

π

π

λ λ λπ λ λ

−

− ∂ ∂ ∂ ∂= − ∂ ∂ ∂ ∂

∫ ∫1

0Im ( , ) ( , , ) ( , , ) ( , )

c

el n n n n

c

eP d dk u k r u k r u k r u k rab k k

π

π

λ λ λπ

λ

−

− = ∇ ⊥

∇

∫ ∫

1

0Im ( , ) ( , , ) , ( , ) plane

is indeed the fluxof the curl of ( , ) ( , , ) .

c

el n n z

c

n n

eP d dk curl u k r u k r z kab

u k r u k r

k

λ

cπ

−cπ

1

40

λ λ= ∇

( , ) ( , , ) in ( , ) space, andn nV u k r u k r k

40

4141

k

λ

cπ

−cπ

1

λ γπ π− −

⇒ = ∇ =∫

Stokes theorem Im ( , ) ( , , ) . ( )el n n nC

e eP u k r u k r ds Cab ab

George Gabriel Stokes

ξ

ξ

γ ξ ξ ξ

λ

= ∇

∂ ∂∇ =

∂ ∂

∫

( ) ( ) ( ) ,

( , )

n n nC

C i u u d

k

γ Berry phaseof the periodic functions u ( , )along the

( ) is a

rectan

ctually a

gular circuit.

n

n

Ck r

41

.

4242

ξγ ξ ξ ξ= ∇∫

( ) ( ) ( )n n nC

C i u u d

( ),

Along vertical tracks, , hence commutes with e and one can

replace periodic part by full Bloch function:

( ) ( ) ( ) ( ) .

Recall Zac phase and Natural gauge:( )

ikx

n n n n

i k G rn n

d dd d

u u d d

k G e u

ξ

ξ ξ

λ λ

ξ ξ ξ ψ ξ ψ ξ ξ

ψ +

∇ =

∇ = ∇

+ =

∫ ∫

,( ) ( ) ( ),vertical tracks areintegrated in oppositesenseand cancel.

is given by the difference of the horizontal tracks, which are Zac's phases.

ikrk G n k n

el

r e u r k

P

ψ+ = =

⇒

γπ−

=Dipole = ( )el neP Cab

k

λ

cπ

−cπ

1

42

4343

nkknkCuudki ∇= ∫γ

G

C

Recall the open path geometric phase

(Zak’s phase)

C connects k with k+G

( )

λ

π =

=

∂=

∂

=

∑ ∫

∑

31

1

For =0 there is no dipole. General formula:

2 (King-Smith and Vanderbilt,1993)2

sum over bands

b

b

n

nq nqn c

n

n

ieP dq u uq

43

44

45

Mathematical digression: Topological space

A topological space is (X,N(x)) where X represents a set of points x and N(x) is a neighborhood topology of x:neighborhood topology of x means that

If a topological space is such that two different points have at least two distinct neighbourhoods than it is aHausdorff space.

∈ ∈ ⇒ ∈ and n(x) (x) n(x)that is a point belongs to all its neighbourhoodsThe intersection of two neighbourhoods is a neighbourhoodn(x) is a neighbourhood of all points of some neighbourhood of x

x X N x

46

Mathematical digression: Fiber bundle

If E and F are spaces, the product ExF is a trivial fiber bundle.One may take E=base space and F=fiber (fibre in UK English).More generally, a fiber bundle is a structure that locally is like ExF whileglobally can be more complex.

The moebius strip is a bundle which is locally the product of a circle times a segment, but is nontrivial because it has one face.One can also make a cylinder which is a trivial fiber bundle.

Fiber bundles obtained from two circles

Cats always fall on their legs. They manage to control the rotation angle while conserving angular momentum

Falling cats

47

48

Falling cat model as a problem in fiber bundle theory with base spaceM parametrized by α,β and the fibre is the rotation group SO(3) or SO(2) parametrized (in 2d) by θ The fibre specifies the orientation of the cat.

The shapespace isparamertizedby α, β

angle between 1 and 2angle between 2 and 3

αβ

==

angle between 1 and x axisdefines orientation in a fixed frameθ =

48

Masses 1, 2 and 3 are taken equal to m.3

21

x

y

θα

β

Change in shape rotation because of conservation of angular momentum

1 1 1 1 2 2 2 2 3 3 3 3

z component of angular momentum of the cat( )

can be written in terms of angleszL m x y y x x y y x x y y x= − + − + −

( ) ( )1 1cos y sinx R Rθ θ= =

( ) ( )2 2cos y sinx R Rθ α θ α= + = +

( ) ( ) ( ) ( )3 3cos + cos y sin + sinx R R R Rθ α θ α β θ α θ α β= + + + = + + +49

0 (4 2cos ) (3 2cos ) (1 cos ) 0,

3 2cos 1 cos4 2cos 4 2cos

zdL d d dd A d A d

A A

α β

α β

β θ β α β βθ α β

β ββ β

= ⇔ + + + + + =⇒ = +

+ += =

+ +

50

zLθ

∂∂

zLα

∂∂

zLβ

∂∂

Impose dL 0z z zz

L L Ld d dθ α βθ α β

∂ ∂ ∂= + + =

∂ ∂ ∂

51

A closedcircuit in shape spaceleading to net rotation

2π

α

1

β

2π

3

21

x

y

θα

β

( ) angle of rotation of cat in fixed frameA d A dα βθ α β∆ = + =∫

,3 2cos 1 cos4 2cos 4 2cos

d A d A d

A A

α β

α β

θ α β

β ββ β

= +

+ += =

+ +

52

A closedcircuit in shape spaceleading to net rotation

0 02 2

0 02 2

The rotation of the cat during the motion shown in theright panel

( )

7.5 degrees

A d A d

A d A d A d A d

α β

π π

π πα β α β

θ α β

α β α β

∆ = +

= − − − −

= −

∫

∫ ∫ ∫ ∫

2π

α

1

β

2π

52

3

21

x

y

θα

β

53

2

0

Selloni et al. found how to compute the mean value .

arg( ),2

We shall also need to compute off-diagon

2| ( ) | , .

al .

a x

m

G

n

iz dx e x Ga

xax z

rπ

ψ

ψ

ψ

π= = =∫

Appendix: Coordinate and Momentum matrix elements between Bloch functions

[ , ] [ , ] ( ( ) ( ))m n n m m n m ni iH r p H r E k E k r pm m

ψ ψ ψ ψ ψ ψ= − ⇒ = − = −

so since , m n we must work out( ( ) ( ))

the momentum matrix elements.

m n m nn m

ir pm E k E k

ψ ψ ψ ψ= − ≠−

( )One finds ( , ) ( , ) and off-diagonal nn n

E kpk r k rm k

ψ ψ ∂=

∂

( , ) ( , ) ( ( ) ( )) ( , for) ( , ) , ψ ψ ∂≠= −

∂m n n m m npk r k r E k E k u k r u k r m

kn

m

54

2 2( )[ ( )] ( , ) ( )] ( , ) [ ( )] ( , ) ( ) ( , )2 2

ikr ikrn n n n n n

p p kV r e u k r E k e u k r V r u k r E k u k rm m

++ = ⇒ + =

2 2

so ( ) ( , ) ( ) ( , ), where ( ) ( ) . .2 2n n np kH k u k r E k u k r H k V r k pm m m

= = + + +

We need the equation for the periodic function un(k,r)

From the previous result ( ( ) ( ))

one finds ( , ) ( , ) for .

n m m n m n

m n m n

iE k E k r pm

r u k r i u k r nk

m

ψ ψ ψ ψ

ψ ψ

− = −

≠∂

=∂

Proof

55

2( )Apply on Schrodinger equation [ ( )] ( , ) ( ) ( , ) :2 n n n

p k V r u k r E k u k rk m∂ +

+ =∂

( ) ( , )( ) ( , ) ( ) ( , ) ( , ) ( )n nn n n n

E k u k rp k u k r H k u k r u k r E km k k k

∂ ∂+ ∂+ = +

∂ ∂ ∂

( , )( )(

but red terms van

Le

( ) ( , ) ( , )

( ) ( , ) ( , ) ,

( , ) real ( , ) ( , ) 0

t us take matrix elements, multiplying b

, ) ( , )

(

y . On diagonal, m

ish

si

)

=n

nce

,+

∂=

∂

∂+

∂∂

+∂

∂⇒ =

∂

m

n n n n

n n nn

n n n

n E k u k r u k rk

E k u k r u k rk

u k r u

u k rp

k r u

ku k

k

r u k rm

rk

E kk

Thus , we are left with:( )( )(

on diag

, )

a

( , )

on

.

l

nn n

E kp ku k r u k rm k

∂+=

∂

( )In terms of ( , ) we may write ( , ) ( , ) .nn n n

E kpk r k r k rm k

ψ ψ ψ ∂=

∂

( )( , ) ( ) ( , ) (

here red stuff 0 by orthogonality, a

, )

Off diagonal fo

n

r

d oneis le

(

) ( ,

ft wi

,( ) ;)

th

m n m n nnEpu k r H k u k r u k r E k uk kk

m

rm k k

nk ∂

=

∂+ =

++

∂ ∂∂

∂

≠

( , ) ( ) ( , ) ( , ) ( ) ( , )m m n m n npu k r E k u k r u k r E k u k rm k k

∂ ∂+ =

∂ ∂

56

that is,

( , ) ( , ) ( ( ) ( )) ( , ) ( , ) .m n n m m npu k r u k r E k E k u k r u k rm k

∂= −

∂

F

for .

rom the previous result ( ( ) ( ))

one finds ( , ) ( , )

n m m n m n

m n m n

iE

m

k E k r pm

r u k r i u k rk

n

ψ ψ ψ ψ

ψ ψ

− = −

∂∂

≠=

We may also write

( , ) ( , ) ( ( ) ( )) ( , for) ( , ) , ψ ψ ∂≠= −

∂m n n m m npk r k r E k E k u k r u k r m

kn

msince the extra term generated by differentiating the plane wave is killed by

orthogonality.

( ) ( , )( ) ( , ) ( ) ( , ) ( , ) ( )n nn n n n

E k u k rp k u k r H k u k r u k r E km k k k

∂ ∂+ ∂+ = +

∂ ∂ ∂

56