recurrence: random walks vs quantum walks€¦ · george pólya george pólya was puzzled by the...

RECURRENCE:

RANDOM WALKS vs

QUANTUM WALKS

MARTES CUÁNTICO 05/05/2015

RECURRENCE

RECURRENCE RETURN PROPERTIES≡

Example: the Ehrenfest model

Proposed by Tatiana and Paul Ehrenfest to reconcile reversibility of classical mechanics and irreversibility of thermodynamics

It is a simple model for the exchange of gas molecules between two isolated bodies

RECURRENCE





RECURRENCE


particles independently change container at rate N ∆t




RECURRENCE


N − k k





RECURRENCE


Even !!![N, 0]Any state has return probability R = 1

N − k k





RECURRENCE



N − k k


Differences come from the expected return time

τ [N−k,k] =2N

(

N

k

) ∆t




RECURRENCE



N − k k


Differences come from the expected return time

τ [N−k,k] =2N

(

N

k

) ∆t

millions of years ≈ 3×age of Universeτ[N,0] ≈ 40 000

secτ[N/2,N/2] ≈ 1.25× 10−11

For instance, if and sec N = 100 ∆t = 10−12

KEY RETURN PROPERTIES

Return probability

Expected return time

RECURRENCE


Return probability


RECURRENCE

chaos non-equilibriummicro-thermodynamics


Return probability


RECURRENCE

CLASSIFICATION OF STATES

Expected Return Time

< .

ReturnProbability

< 1 Transient

1 RecurrentPositive

Recurrent

∞∞



Return probability


RECURRENCE



< .

ReturnProbability

< 1 Transient

1 RecurrentPositive

Recurrent

∞∞

The Ehrenfest example shows that simple models (random, discrete) may help to uncover central aspects of recurrence avoiding unnecessary complications



Return probability


RECURRENCE



< .

ReturnProbability

< 1 Transient

1 RecurrentPositive

Recurrent

∞∞


RECURRENCEOLD RANDOM WALKS (RW)



Return probability


RECURRENCE



< .

ReturnProbability

< 1 Transient

1 RecurrentPositive

Recurrent

∞∞


RECURRENCEOLD RANDOM WALKS (RW)

NEW! QUANTUM WALKS (QW)


RANDOM WALKS (RW)

RANDOM WALKS formalization of a path consisting in successive random steps, widely used in physics, chemistry, biology, psychology, economy and computer science

≡

RANDOM WALKS (RW)


≡

George Pólya

George Pólya was puzzled by the fact that he recurrently run into a couple while strolling in a park

RANDOM WALKS (RW)


≡

George Pólya


He reduced the problem to the recurrence of a single walker and studied the return probability

RANDOM WALKS (RW)


≡

George Pólya



1D 2D 3D

RANDOM WALKS (RW)


≡

George Pólya



1D 2D 3D

R = 1 R = 1 R ≈ 0.34

RANDOM WALKS (RW)


≡

George Pólya



1D 2D 3D

R = 1 R = 1 R ≈ 0.34CRITICAL

DIMENSIOND= 3

⇒UNBIASED

D ≥ 3

D ≤ 2 RECURRENT

TRANSIENT

RANDOM WALKS (RW)


≡

George Pólya



1D 2D 3D

R = 1 R = 1 R ≈ 0.34

τ = ∞ τ = ∞

CRITICAL DIMENSION

D= 3

⇒UNBIASED

D ≥ 3

D ≤ 2 RECURRENT

TRANSIENT

RW RECURRENCE

STATES: elements of a countable set Ωi ∈ Ω

RANDOM EVOLUTION: given by a stochastic matrix P = (pij)

RW RECURRENCE


SIZE of

SIZE of

P

=

Ω


RW RECURRENCE

pij = Prob(i1 step−−−−→ j) ≥ 0


SIZE of

SIZE of

P

=

Ω


RW RECURRENCE



SIZE of

SIZE of

P

=

Ω

X

j

pij = Prob(i1 step−−−−→ Ω) = 1


RW RECURRENCE



SIZE of

SIZE of

P

=

Ω

X

j


Example: the Ehrenfest modelN − k k

≡ 0, 1, . . . , NΩ = [N, 0], [N − 1, 1], . . . , [0, N ]

pk,k+1 =

N − k

Npk,k−1 =

k

N= 1−

k

N


RW RECURRENCE



SIZE of

SIZE of

P

=

Ω

X

j



≡ 0, 1, . . . , NΩ = [N, 0], [N − 1, 1], . . . , [0, N ]

pk,k+1 =

N − k

Npk,k−1 =

k

N= 1−

k

N

10 2 3 N

1− 1/N 1− 2/N 1− 3/N1

1/N 2/N 3/N 4/N 1

1/N


RW RECURRENCE



SIZE of

SIZE of

P

=

Ω

X

j



≡ 0, 1, . . . , NΩ = [N, 0], [N − 1, 1], . . . , [0, N ]

pk,k+1 =

N − k

Npk,k−1 =

k

N= 1−

k

N

10 2 3 N

1− 1/N 1− 2/N 1− 3/N1

1/N 2/N 3/N 4/N 1

1/N

ji


RW RECURRENCE


Prob(in steps−−−−−→ ) =

X

ik

pii1pi1i2 · · · pin−1j j


SIZE of

SIZE of

P

=

Ω

X

j



≡ 0, 1, . . . , NΩ = [N, 0], [N − 1, 1], . . . , [0, N ]

pk,k+1 =

N − k

Npk,k−1 =

k

N= 1−

k

N

10 2 3 N

1− 1/N 1− 2/N 1− 3/N1

1/N 2/N 3/N 4/N 1

1/N

ji


RW RECURRENCE


= Pn

iProb(i

n steps−−−−−→ ) =

X

ik

pii1pi1i2 · · · pin−1j jj


SIZE of

SIZE of

P

=

Ω

X

j



≡ 0, 1, . . . , NΩ = [N, 0], [N − 1, 1], . . . , [0, N ]

pk,k+1 =

N − k

Npk,k−1 =

k

N= 1−

k

N

10 2 3 N

1− 1/N 1− 2/N 1− 3/N1

1/N 2/N 3/N 4/N 1

1/N

ji


iii

RW RECURRENCE


= Pn

iProb(i

n steps−−−−−→ ) =

X

ik

pii1pi1i2 · · · pin−1RETURN PROB.

in STEPSn


SIZE of

SIZE of

P

=

Ω

X

j



≡ 0, 1, . . . , NΩ = [N, 0], [N − 1, 1], . . . , [0, N ]

pk,k+1 =

N − k

Npk,k−1 =

k

N= 1−

k

N

10 2 3 N

1− 1/N 1− 2/N 1− 3/N1

1/N 2/N 3/N 4/N 1

1/N

ji


iii

RW RECURRENCE


= Pn

iProb(i

n steps−−−−−→ ) =

X

ik


in STEPSn


SIZE of

SIZE of

P

=

Ω

X

j



≡ 0, 1, . . . , NΩ = [N, 0], [N − 1, 1], . . . , [0, N ]

pk,k+1 =

N − k

Npk,k−1 =

k

N= 1−

k

N

10 2 3 N

1− 1/N 1− 2/N 1− 3/N1

1/N 2/N 3/N 4/N 1

1/N

i


iii

RW RECURRENCE


OVERCOUNTING!!!

= Pn

iProb(i

n steps−−−−−→ ) =

X

ik


in STEPSn

Prob(i → i) =X

n≥1

Prob(in steps−−−−−−→ i)


SIZE of

SIZE of

P

=

Ω

X

j


i


iii

RW RECURRENCE


PROBABILITY of RETURNING to

for the first time in the -th STEPn

i

= Pn

iProb(i

n steps−−−−−→ ) =

X

ik


in STEPSn

FIRST TIMEProb(i → i) =

X

n≥1



SIZE of

SIZE of

P

=

Ω

X

j


i


iii

RW RECURRENCE




i

= Pn

iProb(i

n steps−−−−−→ ) =

X

ik


in STEPSn

FIRST RETURN PROB.

in STEPSnFIRST TIME

Prob(in steps−−−−−−→ i) =

X

ik 6=i

pii1pii2 · · · pin−1i


X

n≥1



SIZE of

SIZE of

P

=

Ω

X

j


i


iii

RW RECURRENCE




i

= Pn

iProb(i

n steps−−−−−→ ) =

X

ik


in STEPSn

FIRST RETURN PROB.

in STEPSnFIRST TIME


X

ik 6=i



X

n≥1


ik 6=i


SIZE of

SIZE of

P

=

Ω

X

j


i


iii

RW RECURRENCE




i

= Pn

iProb(i

n steps−−−−−→ ) =

X

ik


in STEPSn

FIRST RETURN PROB.

in STEPSnFIRST TIME


X

ik 6=i



X

n≥1


ik 6=i


SIZE of

SIZE of

P

=

Ω

X

j


i

i

SIMPLE LOOP


iii

RW RECURRENCE




i

= Pn

iProb(i

n steps−−−−−→ ) =

X

ik


in STEPSn

FIRST RETURN PROB.

in STEPSnFIRST TIME


X

ik 6=i


p(n)i

=

q(n)i

=


X

n≥1


ik 6=i


SIZE of

SIZE of

P

=

Ω

X

j


i

i

SIMPLE LOOP


iii

RW RECURRENCE




i

= Pn

iProb(i

n steps−−−−−→ ) =

X

ik


in STEPSn

FIRST RETURN PROB.

in STEPSnFIRST TIME


X

ik 6=i


p(n)i

=

q(n)i

=

=

X

n≥1

q(n)iFIRST TIME

Prob(i → i) =X

n≥1


ik 6=i


SIZE of

SIZE of

P

=

Ω

X

j


i

i

SIMPLE LOOP


iii

RW RECURRENCE




i

= Pn

iProb(i

n steps−−−−−→ ) =

X

ik


in STEPSn

FIRST RETURN PROB.

in STEPSnFIRST TIME


X

ik 6=i


p(n)i

=

q(n)i

=

=

X

n≥1

q(n)iFIRST TIME

Ri = Prob(i → i) =X

n≥1

Prob(in steps−−−−−−→ i)RETURN

PROBABILITY

ik 6=i


SIZE of

SIZE of

P

=

Ω

X

j


i

i

SIMPLE LOOP


iii

RW RECURRENCE




i

= Pn

iProb(i

n steps−−−−−→ ) =

X

ik


in STEPSn

FIRST RETURN PROB.

in STEPSnFIRST TIME


X

ik 6=i


p(n)i

=

q(n)i

=

=

X

n≥1

q(n)iFIRST TIME


n≥1


EXPECTED RETURN TIME

RETURNPROBABILITY

ik 6=i


SIZE of

SIZE of

P

=

Ω

X

j


q(n)i∆t( )τi =

X

n≥1

n

i

i

SIMPLE LOOP


iii

RW RECURRENCE




i

= Pn

iProb(i

n steps−−−−−→ ) =

X

ik


in STEPSn

FIRST RETURN PROB.

in STEPSnFIRST TIME


X

ik 6=i


p(n)i

=

q(n)i

=

=

X

n≥1

q(n)iFIRST TIME


n≥1



RETURNPROBABILITY

CONVENTION

∆t = 1[ [

ik 6=i


SIZE of

SIZE of

P

=

Ω

X

j


q(n)i

τi =

X

n≥1

n

i

i

SIMPLE LOOP


iii

RW RECURRENCE


= Pn

iProb(i

n steps−−−−−→ ) =

X

ik


in STEPSn

FIRST RETURN PROB.

in STEPSnFIRST TIME


X

ik 6=i


p(n)i

=

q(n)i

=

=

X

n≥1

q(n)i

Ri =


RETURNPROBABILITY

CONVENTION

∆t = 1[ [

ik 6=i


SIZE of

SIZE of

P

=

Ω

X

j


q(n)i

τi =

X

n≥1

n

i

i

SIMPLE LOOP

RW RECURRENCE: GENERALITIES


RETURN PROB.

in STEPSnp(n)i

= Pn

iiP = (pij)


RETURN PROB.

in STEPSnp(n)i

= Pn

iiP = (pij)

q(n)i

FIRST RETURN PROB.

in STEPSn

?



τi =

X

n≥1

nq(n)i

RETURNPROBABILITY

Ri =

X

n≥1

q(n)i

RETURN PROB.

in STEPSnp(n)i

= Pn

iiP = (pij)

q(n)i

FIRST RETURN PROB.

in STEPSn

?



τi =

X

n≥1

nq(n)i

RETURNPROBABILITY

Ri =

X

n≥1

q(n)i

is RECURRENT ifi ∈ Ω Ri = 1

POSITIVE RECURRENT if τi < ∞

RETURN PROB.

in STEPSnp(n)i

= Pn

iiP = (pij)

q(n)i

FIRST RETURN PROB.

in STEPSn

?



τi =

X

n≥1

nq(n)i

RETURNPROBABILITY

Ri =

X

n≥1

q(n)i



RETURN PROB.

in STEPSnp(n)i

= Pn

iiP = (pij)

q(n)i

FIRST RETURN PROB.

in STEPSn

?

p(n)i

q(n)i≥



τi =

X

n≥1

nq(n)i

RETURNPROBABILITY

Ri =

X

n≥1

q(n)i



RETURN PROB.

in STEPSnp(n)i

= Pn

iiP = (pij)

q(n)i

FIRST RETURN PROB.

in STEPSn

?

τi can be ANY real number in [1,∞]

p(n)i

q(n)i≥

11



τi =

X

n≥1

nq(n)i

RETURNPROBABILITY

Ri =

X

n≥1

q(n)i



RETURN PROB.

in STEPSnp(n)i

= Pn

iiP = (pij)

q(n)i

FIRST RETURN PROB.

in STEPSn

?


p(n)i

q(n)i≥

T R

11



τi =

X

n≥1

nq(n)i

RETURNPROBABILITY

Ri =

X

n≥1

q(n)i



RETURN PROB.

in STEPSnp(n)i

= Pn

iiP = (pij)

q(n)i

FIRST RETURN PROB.

in STEPSn

?


FINITE systems may have TRANSIENT states

p(n)i

q(n)i≥

T R

Ω

SUBSET RECURRENCE

i

11



τi =

X

n≥1

nq(n)i

RETURNPROBABILITY

Ri =

X

n≥1

q(n)i



RETURN PROB.

in STEPSnp(n)i

= Pn

iiP = (pij)

q(n)i

FIRST RETURN PROB.

in STEPSn

?



p(n)i

q(n)i≥

T R

Ω

SUBSET RECURRENCE

Si

11



τi =

X

n≥1

nq(n)i

RETURNPROBABILITY

Ri =

X

n≥1

q(n)i



RETURN PROB.

in STEPSnp(n)i

= Pn

iiP = (pij)

q(n)i

FIRST RETURN PROB.

in STEPSn

?



p(n)i

q(n)i≥

Prob(i → S) Prob(i → i)≥ = Ri

T R

RECURRENCE & GENERATING FUNCTIONS

How to calculate these quantities?


How to calculate these quantities? P p(n)i

= Pn

ii

Pn



= Pn

ii

Pn

q(n)i

?



= Pn

ii

Pn

q(n)i

?

bpi(z) =X

n≥0

p(n)i

zn bqi(z) =

X

n≥1

q(n)i

zn

RETURN g.f. FIRST RETURN g.f.



= Pn

ii

Pn

q(n)i

?

bpi(z) =X

n≥0

p(n)i

zn bqi(z) =

X

n≥1

q(n)i

zn


bqi(z) = 1−1

bpi(z)RENEWAL EQUATION

RETURNPROBABILITY

=

X

n≥1

q(n)i

Ri


τi =

X

n≥1

nq(n)i



= Pn

ii

Pn

q(n)i

?

bpi(z) =X

n≥0

p(n)i

zn bqi(z) =

X

n≥1

q(n)i

zn


bqi(z) = 1−1


RETURNPROBABILITY

=

X

n≥1

q(n)i

Ri


τi =

X

n≥1

nq(n)i



= Pn

ii

Pn

q(n)i

?

bpi(z) =X

n≥0

p(n)i

zn bqi(z) =

X

n≥1

q(n)i

zn


bqi(z) = 1−1


= bqi(1)

=

dbqidz

∣∣∣∣z=1

RETURNPROBABILITY

=

X

n≥1

q(n)i

Ri


τi =

X

n≥1

nq(n)i



= Pn

ii

Pn

q(n)i

?

bpi(z) =X

n≥0

p(n)i

zn bqi(z) =

X

n≥1

q(n)i

zn


bqi(z) = 1−1


= bqi(1)

=

dbqidz

∣∣∣∣z=1

= 1−1

bpi(1)

= limz→1

1

(1− z)bpi(z)

RECURRENTi ∈ Ω Ri = 1⇔

POSITIVE RECURRENT ⇔ τi < ∞

RETURNPROBABILITY

=

X

n≥1

q(n)i

Ri


τi =

X

n≥1

nq(n)i



= Pn

ii

Pn

q(n)i

?

bpi(z) =X

n≥0

p(n)i

zn bqi(z) =

X

n≥1

q(n)i

zn


bqi(z) = 1−1


= bqi(1)

=

dbqidz

∣∣∣∣z=1

= 1−1

bpi(1)

= limz→1

1

(1− z)bpi(z)

RECURRENTi ∈ Ω Ri = 1⇔

POSITIVE RECURRENT ⇔ τi < ∞

RETURNPROBABILITY

=

X

n≥1

q(n)i

Ri


τi =

X

n≥1

nq(n)i



= Pn

ii

Pn

q(n)i

?

bpi(z) =X

n≥0

p(n)i

zn bqi(z) =

X

n≥1

q(n)i

zn


bqi(z) = 1−1


⇔ bpi(1) = ∞

⇔ limz→1

(1− z)bpi(z) > 0

= bqi(1)

=

dbqidz

∣∣∣∣z=1

= 1−1

bpi(1)

= limz→1

1

(1− z)bpi(z)

RECURRENCE & SPECTRUM


=

X

n≥0

Pn

iizn

P bpi(z)


=

X

n≥0

Pn

iizn

P bpi(z)

= (1− zP )−1

ii


P bpi(z) = (1− zP )−1

ii


P bpi(z)

τi = limz→1

1

(1− z)bpi(z)

RENEWAL Eq.Ri = 1−

1

bpi(1)= (1− zP )−1

ii


P bpi(z)

SPECTRAL SHORTCUT?τi = lim

z→1

1

(1− z)bpi(z)


1

bpi(1)= (1− zP )−1

ii


P bpi(z)

SPECTRAL SHORTCUT?τi = lim

z→1

1

(1− z)bpi(z)


1

bpi(1)= (1− zP )−1

ii ONLY

mattersz → 1


P bpi(z)

SPECTRAL SHORTCUT?

Under quite general conditions P becomes self-adjoint with kPk 1

τi = limz→1

1

(1− z)bpi(z)


1

bpi(1)= (1− zP )−1

ii ONLY

mattersz → 1

FINITEΩ


P bpi(z)

SPECTRAL SHORTCUT?

P FINITE matrix with spectrum in [−1, 1]


τi = limz→1

1

(1− z)bpi(z)


1

bpi(1)= (1− zP )−1

ii ONLY

mattersz → 1

FINITEΩ


P bpi(z)

SPECTRAL SHORTCUT?


Any vector has a spectral decomposition

v

X

k

vλkv =

eigenvector with EIGENVALUE λk


τi = limz→1

1

(1− z)bpi(z)


1

bpi(1)= (1− zP )−1

ii ONLY

mattersz → 1

If = (0, . . . , 0, 1, 0, . . . )v

i)

τi =1

kvλ=1k2then

vλ=1

veig. λ=1

FINITEΩ


P bpi(z)

SPECTRAL SHORTCUT?



v

X

k

vλkv =



τi = limz→1

1

(1− z)bpi(z)


1

bpi(1)= (1− zP )−1

ii ONLY

mattersz → 1

If = (0, . . . , 0, 1, 0, . . . )v

i)

τi =1

kvλ=1k2then

vλ=1

veig. λ=1

FINITEΩ


P bpi(z)

SPECTRAL SHORTCUT?



v

X

k

vλkv =


⇔ vλ=1 6= 0i ∈ ΩPOSITIVE

RECURRENT


τi = limz→1

1

(1− z)bpi(z)


1

bpi(1)= (1− zP )−1

ii ONLY

mattersz → 1

If = (0, . . . , 0, 1, 0, . . . )v

i)

τi =1

kvλ=1k2then

vλ=1

veig. λ=1

FINITEΩ


P bpi(z)

SPECTRAL SHORTCUT?



v

X

k

vλkv =


RECURRENT⇔⇔ vλ=1 6= 0i ∈ ΩPOSITIVE

RECURRENT


τi = limz→1

1

(1− z)bpi(z)


1

bpi(1)= (1− zP )−1

ii ONLY

mattersz → 1

If = (0, . . . , 0, 1, 0, . . . )v

i)

τi =1

kvλ=1k2then

vλ=1

veig. λ=1

FINITEΩ


IN IN

P bpi(z)

SPECTRAL SHORTCUT?



v

X

k

vλkv =



RECURRENT


τi = limz→1

1

(1− z)bpi(z)


1

bpi(1)= (1− zP )−1

ii ONLY

mattersz → 1

If = (0, . . . , 0, 1, 0, . . . )v

i)

τi =1

kvλ=1k2then

vλ=1

veig. λ=1

FINITEΩ


IN IN

P bpi(z)

SPECTRAL SHORTCUT?



v

X

k

vλkv =


contribution from CONTINUOUS SPEC.

dv(λ)+

Z


RECURRENT


τi = limz→1

1

(1− z)bpi(z)


1

bpi(1)= (1− zP )−1

ii ONLY

mattersz → 1

If = (0, . . . , 0, 1, 0, . . . )v

i)

τi =1

kvλ=1k2then

vλ=1

veig. λ=1

FINITEΩ


IN IN

P bpi(z)

SPECTRAL SHORTCUT?



v

X

k

vλkv =



dv(λ)+

Z

RECURRENT⇔


RECURRENT or dkv(λ)k2 = 1Z

1

1− λ


τi = limz→1

1

(1− z)bpi(z)


1

bpi(1)= (1− zP )−1

ii ONLY

mattersz → 1

If = (0, . . . , 0, 1, 0, . . . )v

i)

τi =1

kvλ=1k2then

vλ=1

veig. λ=1

FINITEΩ


IN INP FINITE matrix with spectrum in [−1, 1]


v

X

k

vλkv =



dv(λ)+

Z

RECURRENT⇔



1

1− λ


If = (0, . . . , 0, 1, 0, . . . )v

i)

τi =1

kvλ=1k2then

vλ=1

veig. λ=1

FINITEΩ




v

X

k

vλkv =



dv(λ)+

Z

RECURRENT⇔



1

1− λ

Recurrence ONLY depends on the spectral decomposition around λ = 1


If = (0, . . . , 0, 1, 0, . . . )v

i)

τi =1

kvλ=1k2then

vλ=1

veig. λ=1

FINITEΩ




v

X

k

vλkv =



dv(λ)+

Z

RECURRENT⇔



1

1− λ

Recurrence ONLY depends on the spectral decomposition around λ = 1

FINITE systems: RECURRENT POSITIVE RECURRENT⇒


RecurrenceQUANTUM WALKS (QW)

QUANTUM WALKS models for a quantum particle in discrete space-time≡



1940s Feynman: checkerboard model for path integral of free Dirac eq.Later related to D Ising model1

Feynman

1+1



1993 Aharonov et al:

quantum version of RW spreads out much faster

Aharonov


Feynman

1+1





Aharonov


Feynman

1+1

source: http://physik.uni-paderborn.de/?id=178571

http://physik.uni-paderborn.de/?id=178571





Aharonov

RW


Feynman

1+1







Aharonov

RW QW


Feynman

1+1




Why?


Why?

Simple models for quantum dynamics


Why?


The widespread use of RW in randomized algorithms motivated the use of QW in quantum algorithms exponential speedup over any classical algorithm!!!


Why?

Step towards information processing machines using quantum mechanical laws (Feynman ’82 proposal) quantum simulators, quantum computers




Why?



Quantum biology quantum coherence in photosynthetic energy transfer



Why?



Experimental realizations

Atoms in optical lattices

Trapped ions

Wave guide arrays

Optical fibres



STATES: (up to phases) unit vectors of a Hilbert spaceψ (H , h · | · i)

QW RECURRENCE


QW RECURRENCE

ψ1 step−−−−→ UψEVOLUTION: given at every step by a unitary operator U


QW RECURRENCE

Example: D coined QW1

Generalization of checkerboard model: particle with spin on a lattice x ∈ Zs ∈ ↑, ↓

spanned by states H = Hspace ⊗ Hspin |xi |si

U = S(1space ⊗ C) unitary step



QW RECURRENCE





C =

a b

c d

∈ U(2) spin rotation (‘coin flip’)



QW RECURRENCE





C =

a b

c d


conditional shiftS =

X

x∈Z

|x− 1ihx| |#ih#|+ |x+ 1ihx| |"ih"|



QW RECURRENCE





C =

a b

c d


conditional shiftS =

X

x∈Z

|x− 1ihx| |#ih#|+ |x+ 1ihx| |"ih"|

ψ1 step−−−−→ Uψ

PROBABILITYAMPLITUDE

MEASUREMENT: Probability of measuring the state when the system is in state

φ

ψ

Probψ(φ) = |hφ|ψi|2

EVOLUTION: given at every step by a unitary operator U


QW RECURRENCE




φ

ψ



= Probψ(Unψ) = |hψ|Unψi|2Prob(ψ

n steps−−−−−→ ψ) RETURN PROB.

in STEPSn


QW RECURRENCE




φ

ψ



FIRST TIME



in STEPSn

Prob(ψ → ψ) =X

n≥1

Prob(ψn steps−−−−−→ ψ)

RETURNPROBABILITY


QW RECURRENCE




φ

ψ



FIRST RETURN PROB.

in STEPSnFIRST TIME



in STEPSn

Prob(ψ → ψ) =X

n≥1


RETURNPROBABILITY


QW RECURRENCE




φ

ψ



PROBLEM: the quantum measurement to check the return after every step collapses the state altering irreversibly the “natural” evolution

FIRST RETURN PROB.

in STEPSnFIRST TIME



in STEPSn

Prob(ψ → ψ) =X

n≥1


RETURNPROBABILITY


QW RECURRENCE




φ

ψ



PROBLEM: the quantum measurement to check the return after every step collapses the state altering irreversibly the “natural” evolution

We will take the collapse as an intrinsic ingredient of monitored quantum recurrence.

This is in QM spirit, which gives to measurements a role absent in classical physics

FIRST RETURN PROB.

in STEPSnFIRST TIME



in STEPSn

Prob(ψ → ψ) =X

n≥1


RETURNPROBABILITY

MEASUREMENT & COLLAPSE

ψ

φ⊥

φ

ORTHOGONAL PROJECTION onto φ

Qφ = I − Pφ ORTHOGONAL PROJECTION onto φ⊥

Pφ = |φihφ|


ψ

φ⊥

φ

Pφψ



Pφ = |φihφ|

Qφψ

PROBABILITY OF FINDING φ


ψ

φ⊥

φ

Pφψ

PROBABILITY OF NOT FINDING φ



Pφ = |φihφ| 1 = kψk2 = kPφψk2+ kQφψk

2

=

Probψ(φ)

=

Probψ(φ⊥)

Qφψ



ψ

φ⊥

φ

Pφψ





2

=

Probψ(φ)

=

Probψ(φ⊥)

Two possible results when measuring at state φ ψ

Qφψ



ψ

φ⊥

φ

Pφψ





2

=

Probψ(φ)

=

Probψ(φ⊥)


φ is found:Pφψ

k · k= φψ

COLLAPSE−−−−−−−−→

Qφψ


Qφψ

kQφψk


ψ

φ⊥

φ

Pφψ





2

=

Probψ(φ)

=

Probψ(φ⊥)


φ is found:Pφψ

k · k= φψ


φ is NOT found:Qφψ

k · kψ


Qφψ

MONITORED RECURRENCE How to calculate first return probabilities?

QW RECURRENCE

UNITARY STEP

U


QW RECURRENCE

Dynamics perturbed by measurements≡

MEASUREMENT

Return to ?ψ

UNITARY STEP

U


QW RECURRENCE


MEASUREMENT

Return to ?ψ

UNITARY STEP

U


QW RECURRENCE

YES END


MEASUREMENT

Return to ?ψ

UNITARY STEP

U


QW RECURRENCE

YES END

projection onto ψ⊥

NO UQψ


MEASUREMENT

Return to ?ψ

UNITARY STEP

U


QW RECURRENCE

YES END

=eU


NO UQψ


MEASUREMENT

Return to ?ψ

UNITARY STEP

U


QW RECURRENCE

YES END

=eU


NO UQψ


NO return to prior to -th step meansnψ

NO ψψ

step 1−−−−−→ eUψ

MEASUREMENT

Return to ?ψ

UNITARY STEP

U


QW RECURRENCE

YES END

=eU


NO UQψ



NO ψψ


NO ψ

step 2−−−−−→ eU2ψ

MEASUREMENT

Return to ?ψ

UNITARY STEP

U


QW RECURRENCE

YES END

=eU


NO UQψ



NO ψψ


NO ψ


NO ψ NO ψ

step 3−−−−−→ · · ·

step n−1−−−−−−−→ eUn−1ψ

MEASUREMENT

Return to ?ψ

UNITARY STEP

U


QW RECURRENCE

YES END

=eU


NO UQψ



NO ψψ


NO ψ


step n

−−−−−→ U eUn−1ψNO ψ NO ψ

step 3−−−−−→ · · ·

step n−1−−−−−−−→ eUn−1ψ

MEASUREMENT

Return to ?ψ

UNITARY STEP

U

FIRST RETURN

PROBABILITY

in STEPSn

= |hψ|U eUn−1ψi|2FIRST TIME



QW RECURRENCE

YES END

=eU


NO UQψ



NO ψψ


NO ψ


step n

−−−−−→ U eUn−1ψNO ψ NO ψ

step 3−−−−−→ · · ·

step n−1−−−−−−−→ eUn−1ψ

QW RECURRENCE

RETURN

AMPLITUDESµ(n)ψ = hψ|Unψi = |µ

(n)ψ |2Prob(ψ

n steps−−−−−→ ψ)

QW RECURRENCE


eU = QψU

FIRST RETURN

AMPLITUDESa(n)ψ = hψ|U eUn−1ψi

FIRST TIMEProb(ψ

n steps−−−−−→ ψ)= |a

(n)ψ |2

RETURN


(n)ψ |2Prob(ψ


QW RECURRENCE


eU = QψU

FIRST RETURN


FIRST TIMEProb(ψ


(n)ψ |2

RETURN


(n)ψ |2Prob(ψ


QW RECURRENCE

RETURNPROBABILITY

Rψ = Prob(ψ → ψ) =X

n≥1

|a(n)ψ |2


τψ =

X

n≥1

n |a(n)ψ |2


eU = QψU

FIRST RETURN


FIRST TIMEProb(ψ


(n)ψ |2

RETURN


(n)ψ |2Prob(ψ


QW RECURRENCE

is RECURRENT if ψ Rψ = 1

POSITIVE RECURRENT if τψ < ∞

RETURNPROBABILITY

Rψ = Prob(ψ → ψ) =X

n≥1

|a(n)ψ |2


τψ =

X

n≥1

n |a(n)ψ |2


-1 10

-1 10

a

b

c

d

b

a

c

d

C =

a b

c d

∈ U(2)

COIN FLIPSITE

R|xi|si =2

π|c|4

(1 + 2|a|2)|ac|+ (1− 4|a|2) arcsin |c|

|a|2 + |c|2 = 1

|c|0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0


-1 10

-1 10

a

b

c

d

b

a

c

d

C =

a b

c d

∈ U(2)

COIN FLIPSITE

R|xi|si =2

π|c|4

(1 + 2|a|2)|ac|+ (1− 4|a|2) arcsin |c|

|a|2 + |c|2 = 1

|c|0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0


-1 10

-1 10

a

b

c

d

b

a

c

d

C =

a b

c d

∈ U(2)

COIN FLIPSITE

D ≥ 3

D ≤ 2 RECURRENT

TRANSIENT

UNBIASED RW

CRITICAL DIMENSION

D= 3

R|xi|si =2

π|c|4

(1 + 2|a|2)|ac|+ (1− 4|a|2) arcsin |c|

|a|2 + |c|2 = 1

|c|0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0


-1 10

-1 10

a

b

c

d

b

a

c

d

C =

a b

c d

∈ U(2)

COIN FLIPSITE

UNBIASED COINED QW

|a|2 = |b|2 = |c|2 = |d|2 =1

2

D ≥ 3

D ≤ 2 RECURRENT

TRANSIENT

UNBIASED RW

CRITICAL DIMENSION

D= 3

Every state is TRANSIENT

already for !!! D= 1

Example: cyclic shift

Consider a system with Hilbert state space and unitary step H = span|0i, |1i, |2i

2 1

0

1 1

1

U = |1ih0|+ |2ih1|+ |0ih2|



Prob(ψ2 steps−−−−−→ ψ) =

1

4

Suppose the system initially in the state ψ =1p2(|1i − |2i)

µ(2)ψ = hψ|U2ψi = −

1

2

2 1

0

1 1

1

U = |1ih0|+ |2ih1|+ |0ih2|




1

4


µ(2)ψ = hψ|U2ψi = −

1

2


9

16FIRST TIMEa(2)ψ = hψ|U eUψi = −

3

4

2 1

0

1 1

1

U = |1ih0|+ |2ih1|+ |0ih2|


<



1

4


µ(2)ψ = hψ|U2ψi = −

1

2


9


3

4

2 1

0

1 1

1

U = |1ih0|+ |2ih1|+ |0ih2|


QUANTUM PARADOX

FIRST return probabilities can be greater than return probabilities!!!

<



1

4


µ(2)ψ = hψ|U2ψi = −

1

2


9


3

4

2 1

0

1 1

1

U = |1ih0|+ |2ih1|+ |0ih2|


RETURN

AMPLITUDESµ(n)ψ = hψ|Unψi FIRST RETURN



RETURN



bµψ(z) =X

n≥0

µ(n)ψ z

nRETURN g.f. baψ(z) =

X

n≥1

a(n)ψ z

nFIRST RETURN g.f.


RETURN



bµψ(z) =X

n≥0

µ(n)ψ z


X

n≥1

a(n)ψ z

nFIRST RETURN g.f.

QUANTUM RENEWAL EQUATION

baψ(z) = 1−1

bµψ(z)


RETURN



bµψ(z) =X

n≥0

µ(n)ψ z


X

n≥1

a(n)ψ z

nFIRST RETURN g.f.


baψ(z) = 1−1

bµψ(z)

For amplitudes

instead of

probabilities!!!

=

X

n≥1

|a(n)ψ |2Rψ

=

X

n≥1

n|a(n)ψ |2

RETURNPROBABILITY

EXPECTED RETURN TIME τψ


RETURN



bµψ(z) =X

n≥0

µ(n)ψ z


X

n≥1

a(n)ψ z

nFIRST RETURN g.f.


baψ(z) = 1−1

bµψ(z)

For amplitudes

instead of

probabilities!!!

=

X

n≥1

|a(n)ψ |2Rψ

=

X

n≥1

n|a(n)ψ |2

RETURNPROBABILITY

EXPECTED RETURN TIME τψ


RETURN



bµψ(z) =X

n≥0

µ(n)ψ z


X

n≥1

a(n)ψ z

nFIRST RETURN g.f.


baψ(z) = 1−1

bµψ(z)

=

Z2π

0

|baψ(eiθ)|2dθ

2π

=

Z2π

0

baψ(eiθ) ∂θbaψ(eiθ)dθ

2πi

For amplitudes

instead of

probabilities!!!


U bµψ(z) =X

n≥0

hψ|Unψi zn


U bµψ(z) =X

n≥0

hψ|Unψi zn

= hψ|(1− zU)−1ψi


U bµψ(z) = hψ|(1− zU)−1ψi


U bµψ(z) = hψ|(1− zU)−1ψiRENEWAL Eq.

QUANTUMRψ =

Z2π

0

|baψ(eiθ)|2dθ

2π

=

Z2π

0


2πiτψ


SPECTRAL SHORTCUT?


QUANTUMRψ =

Z2π

0

|baψ(eiθ)|2dθ

2π

=

Z2π

0


2πiτψ


SPECTRAL SHORTCUT?


QUANTUMRψ =

Z2π

0

|baψ(eiθ)|2dθ

2π

=

Z2π

0


2πiτψ

U unitary ⇒spectrum in

unit circle

eiθ


SPECTRAL SHORTCUT?


QUANTUMRψ =

Z2π

0

|baψ(eiθ)|2dθ

2π

=

Z2π

0


2πiτψ

EVERY point matterseiθ


unit circle

eiθ


SPECTRAL SHORTCUT?


QUANTUMRψ =

Z2π

0

|baψ(eiθ)|2dθ

2π

=

Z2π

0


2πiτψ



unit circle

eiθ

Any vector has a spectral decompositionψ

⇒

ψ =

X

k

ψλk+

Zdψ(eiθ)



dimH = ∞


SPECTRAL SHORTCUT?


QUANTUMRψ =

Z2π

0

|baψ(eiθ)|2dθ

2π

=

Z2π

0


2πiτψ



unit circle

eiθ


⇒

ψ =

X

k

ψλk+ +

Zdψsc(e

iθ)

Zw(θ) dθ



dimH = ∞

ABSOLUTELY CONTINUOUS SINGULAR


SPECTRAL SHORTCUT?


QUANTUMRψ =

Z2π

0

|baψ(eiθ)|2dθ

2π

=

Z2π

0


2πiτψ



unit circle

eiθ


⇒

ψ =

X

k

ψλk+ +

Zdψsc(e

iθ)

Zw(θ) dθ



dimH = ∞





unit circle

eiθ


⇒

ψ =

X

k

ψλk+ +

Zdψsc(e

iθ)

Zw(θ) dθ



dimH = ∞





unit circle

eiθ


⇒

ψ =

X

k

ψλk+ +

Zdψsc(e

iθ)

Zw(θ) dθ



ψ ONLY SINGULAR part RECURRENT ⇔

⇔POSITIVE RECURRENT ONLY finite EIGENVECTORS

dimH = ∞





unit circle

eiθ


⇒

ψ =

X

k

ψλk+ +

Zdψsc(e

iθ)

Zw(θ) dθ




⇔POSITIVE RECURRENT ONLY finite EIGENVECTORS

τψ = number of EIGENVECTORS

dimH = ∞



⇔POSITIVE RECURRENT ONLY finite EIGENVECTORS τψ = number of EIGENVECTORS




Recurrence depends an ALL the spectral decomposition





FINITE DIMENSIONAL systems only have POSITIVE RECURRENT states


QUANTIZATION of EXPECTED RETURN TIME!!!









θα

eiθ e

iα

baψ(eiθ)

=∆α

2πτψ

WINDING NUMBER of baψ(eiθ)



EXPECTED RETURN TIME: Topological meaning INTEGER





θα

eiθ e

iα

baψ(eiθ)

=∆α

2πτψ

WINDING NUMBER of baψ(eiθ)



Even more interesting than the return properties of a state are the return properties of a subspace : starting at , what is the probability of returning to ? ψ ∈ V VV ⊂ H

SUBESPACE RECURRENCE


-1 10

-1 10

Example: site recurrence in D coined QW1



-1 10

-1 10


|0i |"iψ =

−→

Prob???

span|0i |"i, |0i |#iV =



-1 10

-1 10


|0i |"iψ =

−→

Prob???

span|0i |"i, |0i |#iV =


orthogonal projection onto=Pψ ψ

orthogonal projection onto=Qψ ψ⊥

orthogonal projection onto= VPV

orthogonal projection onto= V⊥QV

STATE RECURRENCEProb(ψ → ψ) SUBSPACE RECURRENCEProb(ψ → V )








The generating functions become matrix functions acting on V








first return g.f. SCALAR

baψ(z) baV (z)first -return g.f.

MATRIX

V











MATRIX

V

LOOP in V

bψ(θ) := baV (eiθ)ψ











MATRIX

V

LOOP in V


Rψ(V ) = Prob(ψ → V ) -RETURNPROBABILITYV











MATRIX

V

LOOP in V

bψ(θ) := baV (eiθ)ψ=

Z2π

0

k bψ(θ)k2 dθ

2πRψ(V ) = Prob(ψ → V ) -RETURN

PROBABILITYV











MATRIX

V

LOOP in V

bψ(θ) := baV (eiθ)ψ=

Z2π

0

k bψ(θ)k2 dθ

2πRψ(V ) = Prob(ψ → V ) -RETURN

PROBABILITYV

τψ(V ) =

Z2π

0

h bψ(θ)|∂θ bψ(θ)idθ

2πi EXPECTED -RETURN TIMEV


1/ dimV

V

STATE RECURRENCE Expected return time was a “topological integer”

SUBESPACE CURRENCE

1/ dimV

V

τψ(V ) =

Z2π

0


2πi

EXPECTED

-RETURN

TIME

V


SUBESPACE CURRENCE

BERRY PHASE

of the loop

EXPECTED -RETURN TIME = geometrical phase NOT QUANTIZEDV


1/ dimV

V

τψ(V ) =

Z2π

0


2πi

EXPECTED

-RETURN

TIME

V


SUBESPACE CURRENCE

BERRY PHASE

of the loop



1/ dimV

V

Averaging over we find again “topological integers”ψ ∈ Vτψ(V )

τψ(V ) =

Z2π

0


2πi

EXPECTED

-RETURN

TIME

V


SUBESPACE CURRENCE

BERRY PHASE

of the loop



1/ dimV

V


τψ(V ) =

Z2π

0


2πi

EXPECTED

-RETURN

TIME

V


SUBESPACE CURRENCE

hτψ(V )iψ∈V =N

dimV

N =

WINDING NUMBER

of detbaV (eiθ)

θ

eiθ

α

eiα

detbaV (eiθ)

BERRY PHASE

of the loop



1/ dimV

V


τψ(V ) =

Z2π

0


2πi

EXPECTED

-RETURN

TIME

V


SUBESPACE CURRENCE

hτψ(V )iψ∈V =N

dimV

N =

WINDING NUMBER

of detbaV (eiθ)

θ

eiθ

α

eiα

detbaV (eiθ)

QUANTIZATION of MEAN EXPECTED -RETURN TIME!!!

Topological meaning INTEGER MULTIPLE of

V

1/ dimV

-1 10

-1 10

a

b

c

d

b

a

c

d

−→

Prob???

span|0i |"i, |0i |#iV =

ψ = |0i |si


|c|0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0

-1 10

-1 10

a

b

c

d

b

a

c

d

−→

Prob???

span|0i |"i, |0i |#iV =

ψ = |0i |si

|a|2 + |c|2 = 1

R|xi|si(V ) =2

π|c|2

|ac|+ (1− 2|a|2) arcsin |a|

SITE RECURRENCE


|c|0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0

0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0

-1 10

-1 10

a

b

c

d

b

a

c

d

−→

Prob???

span|0i |"i, |0i |#iV =

ψ = |0i |si

R|xi|si =2

π|c|4

(1 + 2|a|2)|ac|+ (1− 4|a|2) arcsin |c|

STATE RECURRENCE

|a|2 + |c|2 = 1

R|xi|si(V ) =2

π|c|2

|ac|+ (1− 2|a|2) arcsin |a|

SITE RECURRENCE


|c|0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0

0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0

-1 10

-1 10

a

b

c

d

b

a

c

d

−→

Prob???

span|0i |"i, |0i |#iV =

ψ = |0i |si

R|xi|si =2

π|c|4

(1 + 2|a|2)|ac|+ (1− 4|a|2) arcsin |c|

STATE RECURRENCE

|a|2 + |c|2 = 1

R|xi|si(V ) =2

π|c|2

|ac|+ (1− 2|a|2) arcsin |a|

SITE RECURRENCE


As intuition suggests, . Is this a general fact?Prob(ψ → V ) ≥ Prob(ψ → ψ)

Example: combined shifts

U =

X

x∈Z

|x+ 1ihx|+ |#ih"|+ |"ih#|

-1 101 1

1

1

Consider a Hilbert space spanned by with unitary step|xix∈Z [ |"i, |#i


U =

X

x∈Z

|x+ 1ihx|+ |#ih"|+ |"ih#|

-1 101 1

1

1


ψ = α|0i+ |"ip

2+ β|#iFor and we obtainV = span|0i+ |"i, |#i

Rψ = Prob(ψ → ψ) =1− 1

2|α|2

1 + 1

2|α|2

0.0 0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0

|α|

Rψ(V ) = Prob(ψ → V ) = 3

4− 1

4|α|2


QUANTUM PARADOX

Return probability to a subspace can be smaller than to the state!!!

U =

X

x∈Z

|x+ 1ihx|+ |#ih"|+ |"ih#|

-1 101 1

1

1


ψ = α|0i+ |"ip

2+ β|#iFor and we obtainV = span|0i+ |"i, |#i

Rψ = Prob(ψ → ψ) =1− 1

2|α|2

1 + 1

2|α|2

0.0 0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0

|α|

Rψ(V ) = Prob(ψ → V ) = 3

4− 1

4|α|2

RW vs QW

Random Walks Quantum Walks

Spectral shortcut NOT always applicable

Stochastic Self-adjoint Spectrum on [-1,1]

Spectral shortcut always applicable

Unitary Spectrum on unit circle

ONLY spectrum around 1 matters for recurrence ALL the spectrum matters for recurrence

Recurrence Singular part Recurrence ONLY singular part

eigenvector Positive recurrence

with eigenvalue 1

ONLY finite eigenvectors Positive recurrence

with ANY eigenvalue

Finite system ONLY recurrent states ONLY positive

Finite-dim system recurrent states

Expected return time is NOT quantized Expected return time is quantized

First return probabilities

are NOT greater than return probabilities

First return probabilities

can be greater than return probabilities

Return probability to a subset

is NOT smaller than to the initial state

Return probability to a subspace

can be smaller than to the initial state

⇔;:

⇔⇔

; ⇒

? ⇒ ⇒

∃

RECURRENCE COLLABORATORS

Reinhard WernerLeibniz U Hannover

Albert WernerFreie U Berlin

Alberto GrünbaumUC Berkeley

Jon WilkeningUC Berkeley

Jean BourgainIAS Princeton

recurrence: random walks vs quantum walks€¦ · george pólya george pólya was puzzled by the...

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