kinks and cusps in the transition dynamics of a bloch state

Kinks and Cusps in the Transition Dynamics of aBloch State

Jiang-min Zhang

In collaboration with Masudul Haque and Hua-Tong Yang

March 22, 2016

Outline

I A very simple problemI transition dynamics of a Bloch state

I Two scenariosI periodically driving case

I kinks in the perturbative regime

I sudden quench caseI cusps in the non-perturbative regime

I Open problems

JMZ and M. Haque, Nonsmooth and level-resolved dynamicsillustrated with the tight binding model, arXiv:1404.4280,ScienceOpen (2014).

JMZ and H. T. Yang, Cusps in the quenched dynamics of abloch state, arXiv:1601.03569.

The 1D tight binding modelThe Hamiltonian

HTBM = −N−1∑n=0

(a†nan+1 + a†n+1an).

The eigenvalues and eigenstates (Bloch states)

ε(q) = −2 cos q, q = 2πk/N,

|k〉 =1√N

N−1∑n=0

eiqn|n〉.

k = 0, 1, . . . , N − 1.

Scenario I: Local periodic driving

Initially, the particle is in some Bloch state |Ψ(0)〉 = |ki〉. Nowstart driving the system harmonically:

H(t) = HTBM + V (t),

V (t) = U sinωta†0a0.

Question: how does the survival probability

p(t) = |〈ki|Ψ(t)〉|2.

evolve?

Fermi golden rule fails

An example

Fermi golden rule breaks down and the problem has to be solvednon-perturbatively if the driving amplitude U is large.

Kinks!

A relatively weak driving (U small):

Three successiveBloch state |ki〉

Dotted line: Fermigolden rule

I Kinks appear periodically in time.

I Piece-wise linear!

I Initial level dependence (beyond some critical time)!

Essential features of the model

I Equally spaced spectrum (locally)

En+1 − En = ∆.

I Equal coupling

|〈n|V |i〉| = g =U

N.

I Small U means it is a perturbative effect.

Firsr-order perturbation theoryThe transition probability is

1− p(t) ' 4g2∑m∈Z

sin2[(Em − Ef )t/2]

(Em − Ef )2=

(4g2

∆2

)Wα(T ),

with the rescaled, dimensionless time T ≡ ∆t/2, and

Wα(T ) ≡ T 2∑m∈Z

sinc2[(m− α)T ]

with α = (Efinal − En)/(En+1 − En), En < Efinal < En+1.

A uniform sampling of the sinc2 x =(

sinxx

)2function!

Poisson summation formula I

+∞∑m=−∞

f(a+mT ) =1

T

+∞∑m=−∞

F

(2πm

T

)exp

(i2πma

T

).

Periodic sampling in real space⇐⇒ Periodic sampling in momentum space (simpler?)

Poisson summation formula II

For our specific problem, we sample the sinc2 function,

Wα(T ) ≡ T 2∑m∈Z

sinc2[(m− α)T ]

= T∑n∈Z

F2

(2πn

T

)exp (−i2πnα) ,

with

F1(q) ≡∫ +∞

−∞dxe−iqx

(sinx

x

),

F2(q) ≡∫ +∞

−∞dxe−iqx

(sinx

x

)2

,

Both F1 and F2 are of finite supports (Paley-Wiener theorem)!

Only a finite number of terms contribute to the summation!

A piece-wise linear function of time

I If 0 < T ≤ π,

Wα(T ) = πT.

Fermi golden rule! No α dependence!I If mπ < T ≤ (m+ 1)π,

Wα(T ) = πT

m∑n=−m

exp(inθ)−m∑n=1

2π2n cos(nθ).

The slope of this linear function is m-dependent and alsoα-dependent (θ = 2πα)!

I The period is 2π/∆, the so-called Heisenberg time.

Story I: Kinks were discovered long long ago

A two-level atom coupled to a multi-mode optical cavity

J. Parker and C. R. Stroud, Jr.,Phys. Rev. A 35, 4226 (1987).

H. Giessen, J. D. Berger, G.Mohs, P. Meystre, Phys. Rev. A53, 2816 (1996).

They just attribute the kinks to interference.

More robust kinks and an exactly solvable model

H = Eb|b〉〈b|+∑n∈Z

n∆|n〉〈n|+ g∑n∈Z

(|b〉〈n|+ |n〉〈b|)

ψ(0) = |b〉, 〈b|ψ(t)〉 = ?

Realization: a single two-level atom in a multi-mode optical cavity

G. C. Stey and R. W. Gibberd, Physica 60, 1 (1972).

M. Ligare and R. Oliveri, Am. J. Phys. 70, 58 (2002).

Scenario II: Sudden quenchInitially, the particle is in some Bloch state |Ψ(0)〉 = |ki〉. Nowturn on the potential at site n = 0 suddenly:

H(t > 0) = HTBM + Ua†0a0.

Question: how do the survival probability

Pi(t) = |〈+ki|Ψ(t)〉|2.

and the reflection probability

Pr(t) = |〈−ki|Ψ(t)〉|2.

evolve?

Cusps!

Subtle structures in the overall picture of Rabi oscillation

Apparently non-perturbative

More examples

0 1000 20000

0.2

0.4

0.6

0.8

1

t

Pi&

Pr

(a)

0 1000 20000

0.2

0.4

0.6

0.8

1

t

(b)

0 1000 20000

0.2

0.4

0.6

0.8

1

t

(c)

(a) N = 401, ki = 80, U = 1.5;(b) N = 401, ki = 100, U = 2, especially regular;(c) N = 401, ki = 100, U = 12.

A linearized model

Key numerical observation: Only those Bloch states with energyε(q) ' ε(qi) participate significantly in the dynamics.

−π π q

ε

qi−qi

Two groups of Bloch states

I Right going: |Rn〉 = |+ ki + n〉;I Left going: |Ln〉 = | − ki − n〉, −∞ ≤ n ≤ +∞.

The dispersion curve at around q = ±ki can be linearized.

Similar ideas in the Luttinger liquid theory.

Taking into account the parity symmetryA new basis (even and odd states)

|A±n 〉 =1√2

(|Rn〉 ± |Ln〉).

In the yet to be diagonalized subspace of {|A+n 〉}, the matrix

elements of H0 = HTBM and H1 = Ua†0a0 are

〈A+n |H0|A+

n 〉 = n∆, 〈A+n1|H1|A+

n2〉 = 2g, ∀ n1 & n2.

Initially, |Ψ(0)〉 = |R0〉 = 1√2(|A−0 〉+ |A+

0 〉). Some time later,

|Ψ(t)〉 =1√2|A−0 〉+

1√2

∞∑n=−∞

ψn(t)|A+n 〉,

since |A−0 〉 is already an eigenstate of H. The quantity wanted isψ0(t), in terms of which the probabilities Pi and Pr are

Pi =1

4|1 + ψ0|2 , Pr =

1

4|1− ψ0|2 .

Solution strategyThe Schrodinger equation for the ψ’s is

i∂

∂tψn = n∆ψn + 2g

∞∑m=−∞

ψm. (1)

Note that the term in the summation is independent of n.Therefore, we define the collective, auxiliary quantity

S(t) =

∞∑m=−∞

ψm(t). (2)

The solution of (1) is then

ψn(t) = e−in∆tδn,0 − i2g∫ t

0dτe−in∆(t−τ)S(τ). (3)

Plugging this into (2), we get an integral equation of S,

S(t) = 1− i2g∫ t

0dτ

( ∞∑n=−∞

e−in∆(t−τ)

)S(τ). (4)

Solution strategy continuedWe have (again by the Possion summation formula)

+∞∑n=−∞

e−in∆(t−τ) = T

+∞∑n=−∞

δ(t− τ − nT ),

where the period T ≡ 2π/∆ is the Heisenberg time. The integralequation is then easily solve. We have

S(t) =1

1 + igT,

which is a constant, for 0 < t < T . In turn, we get

ψ0(t) = 1− i2gt

1 + igT=

1− i2g(t− T/2)

1 + igT,

which is linear in t. We note that as t→ T−,

ψ0(t)→ 1− igT1 + igT

= e−iθ.

After one period, ψ0 returns to its initial value, except for a phaseaccumulated. The same evolution then repeats!

Trajectory of ψ0—A ball in a circular billiard

Irrational rotation

θRe(ψ0)

Im(ψ0)

(1, 0)

(0, 1)

Survival probabilityPi(t) = 1

4 |1 + ψ0|2

Reflection probabilityPr(t) = 1

4 |1− ψ0|2

0 1000 20000

0.2

0.4

0.6

0.8

1

tPi&

Pr

(a)

0 1000 20000

0.2

0.4

0.6

0.8

1

t

(b)

0 1000 20000

0.2

0.4

0.6

0.8

1

t

(c)

I The overall picture is Rabi oscillation

I Instead of piece-wise linear, now it ispiece-wise quadratic

I When the cusps show up, Pi + Pr = 1.

An open problem (in the real space)

How is the picture in the real space? (So far, the discussion isconfined to the momentum space)

Why sharp steps? Can you predict the values of the steps?

Story II: Dynamical phase transitionTransverse field Ising model

I an exactly solvable many-body systemI completely different mechanismI no physical understanding (as in our case)

M. Heyl, A. Polkovnikov, S. Kehrein, Dynamical quantumphase transitions in the transverse-field Ising model, Phys.Rev. Lett. 110, 135704 (2013).

Summary

I Transition dynamics of a Bloch stateI perturbative kinks in the weak periodical driving scenario

I a crystal-clear derivation of the Fermi golden rule

I non-perturbative cusps in the sudden quench scenarioI reminiscent of the so-called “dynamical phase transition”

I Lesson: simple models are not easy!I especially when it comes to dynamics

I Many open problemsI only one is shownI go ahead if you are interested

JMZ and M. Haque, arXiv:1404.4280, ScienceOpen (2014).

JMZ and H. T. Yang, arXiv:1601.03569.