Table Of ContentAdmissible memory kernels for random unitary qubit evolution
Filip A. Wudarski, Pawel Nalez˙yty, Gniewomir Sarbicki and Dariusz Chru´scin´ski
Institute of Physics, Faculty of Physics, Astronomy and Informatics
Nicolaus Copernicus University, Grudzia¸dzka 5/7, 87–100 Torun´, Poland
We analyze random unitary evolution of a qubit within memory kernel approach. We provide
sufficient conditions which guarantee that the corresponding memory kernel generates physically
legitimate quantum evolution. Interestingly, we are able to recover several well-known examples
and to generate new classes of nontrivial qubit evolution. Surprisingly, it turns out that a class of
quantumevolutions with memory kernelgenerated byourapproach gives rise tothevanishingof a
non-Markovianity measure based on the distinguishability of quantum states.
5
1 PACSnumbers: 03.65.Yz,03.65.Ta,42.50.Lc
0
2
t
I. INTRODUCTION the standard GKSL form (8), then Λ =T exp( L du)
r t 0 u
p defines the so-called divisible dynamical map [12, 13]
A Dynamics of open quantum systems plays an impor- whichisoftenconsideredasthe generalizationoRfMarko-
tant role in the analysis of various phenomena like dis- vianity (see [14] for a generalization of the notion of di-
2
1 sipation, decoherence and dephasing [1, 2]. The usual visibility). Thesecondapproachisbasedonthenonlocal
approach to the dynamics of an open quantum system Nakajima-Zwanzig equation [15] (see also [16])
] consists of applying the Born-Markov approximation [1]
h
whichleadstoalocalmasterequationfortheMarkovian
p t
- semigroup ρ˙t = Kt−τρτdτ, (4)
t
n Z0
ρ˙ =L[ρ ] , (1)
a t t
u
whereρ is the density matrixofthe investigatedsystem inwhichquantummemoryeffectsaretakenintoaccount
q t
[ andListhetime-independentgeneratorofthedynamical through the introduction of a memory kernel Kt. This
semigroup defined as follows means that the rate of change of the state ρt at time t
4 depends on its history (starting at t = 0). The Marko-
9v L[ρ]=−i[H,ρ]+ 12 [Vα,ρVα†]+[Vαρ,Vα†] . (2) Tvihaentmimaest-derepEeqn.den(1t)kiesrnreelobistauinsueadllwyhreenferKretd=to2aδs(tt)hLe.
7 α
X(cid:0) (cid:1) generator of the non-Markovianmaster equation. Equa-
1
5 Here H denotes the effective system Hamiltonian, and tion (4) applies to a variety of situations (see eg. [17]).
0 Vα represent noise operators [3, 4]. We call (2) the Becauseoftheconvolutionstructureof(4)thetime-local
. GKSL form (Gorini-Kossakowski-Sudarshan-Lindblad). approach is often called time-convolutionless [1, 18, 19].
1
0 The solution of (1) defines the Markoviansemigroup The structure and the properties of (4) were carefully
5 analyzed in [20–29]. In particular the generalization of
ρ =Λ [ρ]=etLρ , (3)
1 t t Markovian evolution to the so-called semi-Markov was
: investigated within the memory kernel approach by Bu-
v where ρ is an initial state. The dynamical map Λ =etL
t
i iscompletelypositiveandtrace-preserving(CPTP)[1,3– dini[21]andBreuerandVacchini[23](seealsodiscussion
X
in [28]).
5]. TheBorn-Markovapproximationassumesweakinter-
r
a action and a separation of time scales between the sys- Inapresentarticlewestudyrandomunitaryevolution
temanditsenvironment. Suchapproachworksperfectly ofaqubitwithinthememorykernelapproach. Inpartic-
well for many quantum optical systems [6–8]. When the ularweaddressthefollowingproblem: whatisthestruc-
aboveassumptionisnolongervalidthedescriptionbased ture of the correspondingmemory kernel K which leads
t
on (1) is not satisfactory. Recent technological progress to the legitimate CPTP dynamical map Λ . The article
t
andmodernlaboratorytechniquescallforamorerefined has the following structure: in Section II we recall basic
approach which takes into account memory effects com- factsaboutrandomunitaryevolutionsandinSectionIII
pletely neglected in the description based on Markovian we formulatethe sufficientconditionfor K to guarantee
t
semigroups. In recent years we observed an intense re- legitimatephysicalevolutions. InSectionIVweexamine
search activity in the field of non-Markovian quantum the issue of Markovianity. Surprisingly,it turns out that
evolution (see the recent review [9], a collection of arti- a subclass of quantum evolutions with memory kernel
cles in [10] and a recent comparative analysis in [11]). generated by our approach gives rise to the vanishing of
There are basically two approaches which generalize a non-Markovianity measure based on the distinguisha-
the standard Markovian master equation (1): time-local bility of quantum states [31]. Section V illustrates our
approach replaces L by a time-dependent generator L . approach with several examples. Final conclusions are
t
Interestingly,ifforalltthetime-dependentgeneratorhas collected in Section VI.
2
II. RANDOM UNITARY QUBIT EVOLUTION III. CONSTRUCTION OF LEGITIMATE
MEMORY KERNELS
AquantumchannelE :B(H)→B(H)iscalledrandom
unitary [32] if its Kraus representation is given by In this article we analyze the nonlocal description
based on the following memory kernel equation
E[X]= p U XU† , (5)
k k k
t
k
X Λ˙ = K Λ dτ , (15)
where U is a collection of unitary operators and {p } t t−τ τ
k k
stands for a probability distribution. In this article we Z0
considera randomunitary dynamicalmapΛ defined by
t with
3
3
Λ [ρ]= p (t)σ ρσ , (6)
t α α α K [ρ]= k (t)(σ ρσ −ρ), (16)
t i i i
α=0
X i=1
where σ are Pauli matrices with σ = I [33]. Initial X
α 0 2 where k (t) (i = 1,2,3) represent nontrivial memory ef-
conditionΛ =1limpliesp (0)=δ . Recentlyatime- i
t=0 α α0 fects. Note, that equation (15) considerably simplifies
local description based on the following master equation
after performing the Laplace transform
was analyzed [34, 35]
1
Λ˙ =L Λ , (7) Λ˜ = , (17)
t t t s s−K˜
s
where L is a time-local generator defined by
t
whereΛ˜ := ∞e−stΛ dtandsimilarlyforK˜ . Theques-
3 s 0 t s
tionweaddressis: whataretheconditionsfork (t)which
L [ρ]= γ (t)(σ ρσ −ρ), (8) i
t k k k R
guarantee that the solution Λ provides a legitimate dy-
t
k=1
X namical map?
with time-dependent decoherence rates γ (t). One asks
k Denoting by κ (t) the eigenvalues of K ,
α t
the following question: what are the conditions for γ (t)
k
which guarantee that the solution Λt = exp( 0tLτdτ) Kt[σα]=κα(t)σα , (18)
provides a legitimate dynamical map? Note, that the
R equation(15) gives rise to the following set of equations:
solution defines a random unitary evolution with p (t)
α
given by t
λ˙ (t)= κ (t−τ)λ (τ)dτ, i=1,2,3. (19)
i i i
3
1 Z0
p (t)= H λ (t) , (9)
α αβ β
4 Note, that κ (t) = 0 and hence λ (t) = 1 = const. In
0 0
βX=0 termsoftheLaplacetransformsλ˜ (s)andκ˜ (s)onefinds
i i
where H is the Hadamard matrix
αβ
1
1 1 1 1 λ˜i(s)= . (20)
s−κ˜ (s)
1 1 −1 −1 i
H = , (10)
1 −1 1 −1 In terms of λ˜ (s) conditions (13)–(14) may be equiva-
i
1 −1 −1 1 lently reformulated as follows:
and λβ(t) are time-dependent eigenvalues of Λt 1 +λ˜ (s)+λ˜ (s)+λ˜ (s) is CM , (21)
1 1 2
Λ [σ ]=λ (t)σ , (11) s
t α α α
which read as follows: Λ (t)=1 and and
0
1
λ1(t) = exp(−2[Γ2(t)+Γ3(t)]), +λ˜3(s)−λ˜1(s)−λ˜2(s) is CM ,
s
λ (t) = exp(−2[Γ (t)+Γ (t)]), (12)
2 1 3 1
λ3(t) = exp(−2[Γ1(t)+Γ2(t)]), s +λ˜2(s)−λ˜1(s)−λ˜3(s) is CM , (22)
t 1
with Γk(t) = 0 γk(τ)dτ. Now, the map (6) is CP iff +λ˜1(s)−λ˜3(s)−λ˜2(s) is CM ,
p (t) ≥ 0, which is equivalent to the following set of s
α
R
conditions for λs [34, 35]: where CM stands for a completely monotone function
1+λ (t)+λ (t)+λ (t)≥0 , (13) [37], i.e. a smooth function f : [0,∞) → R satisfying
1 2 3
the condition
and
dn
λ (t)+λ (t) ≤ 1+λ (t), (−1)n f(s)≥0, s≥0, n=0,1,2,... (23)
1 2 3 dsn
λ (t)+λ (t) ≤ 1+λ (t), (14)
3 1 2 The equivalence of (14) and (22) results fromthe follow-
λ2(t)+λ3(t) ≤ 1+λ1(t). ing
3
Theorem 1 (Bernstein’s Theorem) A function and
f: [0,∞) → R is completely monotone on [0,∞) if and
1 1 1
only if it is a Laplace transform of a finite non-negative + ≥ ,
a a a
Borel measure µ on [0,∞), i.e. f is of the form 1 2 3
1 1 1
+ ≥ , (30)
∞
a a a
f(s)= e−stdµ(t). (24) 2 3 1
1 1 1
Z0 + ≥ ,
a a a
Note that the initial condition p (0) = 1 and p (0) = 0 3 1 2
0 k
for k =1,2,3 is equivalent to λ (0)=1 due to define a legitimate memory kernel
k
Theorem 2 (Initial Value Theorem) Let f˜(s) be the 1 3
Laplace transform of f(t). Then the following relation is K˜s[ρ]= κ˜α(s)σktr[σkρ] , (31)
2
true: k=1
X
limf(t)= lim sf˜(s) (25) i.e. the corresponding λ˜k(s) satisfy (21)–(22) and (26).
t→0 s→∞
Proof: note that formula (28) implies
it is equivalent to
1 1
lim sλ˜k(s)=1 , (26) λ˜k(s)= s 1− a W(s) , (32)
s→∞ (cid:18) k (cid:19)
for k =1,2,3. This way we have proved and hence
Theorem 3 The map Λ˜s represented by the following 1 +λ˜3(s)−λ˜1(s)−λ˜2(s)
s
spectral decomposition
1 1 1 1 1
= + − , (33)
3
1 sW(s) a a a
Λ˜ [ρ]= λ˜ (s)σ tr[σ ρ] , (27) (cid:18) 1 2 3(cid:19)
s α α α
2
α=0 which proves that 1 +λ˜ (s)−λ˜ (s)−λ˜ (s) is CM due
X s 3 1 2
gwiittihmλa˜t0e(sm)a=p1Λ/s,ifdaenfidneosnltyheifLcaopnladciteiotnrasn(s2f1o)r,m(2o2f)aalned- rtoemthaienifnagctctohnadtiti1soWns1(s()14is).CM. Similarly one proves th(cid:3)e
t
(26) are satisfied. Note,thatsince 1sW1(s) isCM,hence,due tothe Bern-
stein theorem, it is the Laplace transform of a positive
Itisworthemphasisingthattherearefewanalyticaltools function. Hence
for dealing with CM functions, which is due to the fact
1
that an infinite set of conditions (23) must be verified. W(s)= , (34)
Nevertheless, we found an important class of CM func- f˜(s)
tions giving rise to CPTP dynamics with a straightfor-
where f˜(s) is the Laplace transform of f(t) satisfying
ward interpretation. To present them, let us first ob-
t
serve that CM functions have the following two proper- f(τ)dτ ≥0 for all t≥0. One finds
0
ties, which will not be proved:
R −sf˜(s)
κ˜ (s)= . (35)
Property 1 Let f and g be arbitrary completely mono- k a −f˜(s)
k
tone functions. Then
Note, that condition (29) implies
1. f ·g is CM,
1 1 1 t
2. αf +βg is CM for any α,β >0, + + f(τ)dτ ≤4 . (36)
a a a
Property 2 If s ≥0 then 1 is CM. (cid:18) 1 2 3(cid:19)Z0
0 s+s0 Hencetosummarize: ourclassischaracterizedbyasingle
We are now ready to prove our main result: function f(t) and three numbers a1,a2,a3 >0 such that
t
F(t)= f(τ)dτ ≥0 and conditions (30) and (36) hold.
Theorem 4 Let W(s) be a function such that 1 1 is 0
sW(s) One finds for p (t):
α
CM. Then the functions R
1 1 1 1
s p (t) = + − F(t) ,
κ˜k(s)=−akW(s)−1, k =1,2,3, (28) 1 4(cid:18)a2 a3 a1(cid:19)
1 1 1 1
with a1,a2,a3 >0 such that p2(t) = 4 a + a − a F(t) , (37)
(cid:18) 3 1 2(cid:19)
1 1 1 1 1 1 1 1 1
4− + + is CM , (29) p (t) = + − F(t) ,
3
s W(s) a a a 4 a a a
(cid:18) (cid:20) 1 2 3(cid:21)(cid:19) (cid:18) 1 2 3(cid:19)
4
andp (t)=1−p (t)−p (t)−p (t). Inparticular,taking where x :=tr[ρσ ] and the relaxationtimes are defined
0 1 2 3 k k
a =a =a and a =∞ one finds via
1 2 3
−sf˜(s) T = 1 , T = 1 , T = 1 . (48)
κ˜1(s)=κ˜2(s)= a−f˜(s) , κ˜3(s)=0 , (38) 1 γ2+γ3 2 γ3+γ1 3 γ1+γ2
Itiswellknown[5]thatcomplete positivityisequivalent
and hence
to the following set of conditions upon T :
k
1 sf˜(s)
k˜ (s)=k˜ (s)=0 , k˜ (s)= , (39) 1 1 1
1 2 3 2a−f˜(s) + ≥ ,
T T T
1 2 3
1 1 1
gives rise to the legitimate memory kernel + ≥ , (49)
T T T
2 3 1
K [ρ]=k (t)(σ ρσ −ρ), (40) 1 1 1
t 3 3 3 + ≥ ,
T T T
3 1 2
with arbitrary f(t) and a > 0 satisfying additional con-
dition It is therefore clear that condition (30) is an analogue
of (49). Note that condition (30) means that there exist
t
b ,b ,b >0 such that
0≤F(t):= f(τ)dτ ≤2a, (41) 1 2 3
Z0 1 1 1 1
= + ,
for all t≥0. The corresponding solution reads 2a b b
1 2 3
1 1 1 1
1 = + , (50)
p0(t) = 1− 2aF(t) , 2a2 b3 b1
1 1 1 1
p1(t) = p2(t)=0 , (42) = + .
2a b b
1 3 1 2
p (t) = F(t) .
3
2a Now,ittermsofb ,b ,b ourresultmaybereformulated
1 2 3
as follows
This approach resembles very much the semi-Markov
construction [23, 28]: for any f(t) ≥ 0 satisfying
∞f(t)dt≤1 the memory kernel (40) with Corollary 1 For any b1,b2,b3 >0 and the function f(t)
0 satisfying
R k˜ (s)= sf˜(s) , (43) t 1 1 1 −1
3 1−f˜(s) 0≤F(t):= f(τ)dτ ≤ + + , (51)
b b b
Z0 (cid:18) 1 2 3(cid:19)
gives rise to CPTP evolution. In this case one finds
and
1
p (t) = [1+λ (t)] , lim f˜(s)=0, (52)
0 1
2 s→∞
p (t) = p (t)=0 , (44)
1 2 the memory kernel defined by
1
p (t) = [1−λ (t)] ,
3 2 1 sf˜(s)
κ˜ (s)=− , (53)
where k ak−f˜(s)
f˜(s)+1 defines legitimate quantum evolution. Moreover one has
λ˜ (s)=λ˜ (s)= . (45)
1 2 f˜(s)−1
1
p (t)= F(t) , (54)
k
b
It is therefore clear that our approach goes beyond the k
semi-Markov construction.
and p (1)=1−p (t)−p (t)−p (t).
0 1 2 3
Let us recall that Markoviansemigroup generated by
Letusobservethatitisveryhard,ingeneral,toinvert
3
1 formula (35) to the time domain. Now, we provide a
L[ρ]= γ [σ ρσ −ρ], (46)
k k k
2 familyofW(s)whichenablesonetoeasilycomputeκ (t)
i
k=1
X and have the memory kernel in time domain.
the corresponding Bloch equation reads
Theorem 5 Let W(s) be a polynomial
2
x˙ (t)=− x (t) , (47)
k Tk k W(s)=(s+z1)...(s+zn), (55)
5
with z >0. If a ,a ,a satisfy (30) and where the integral is evaluated over the region where
i 1 2 2
d||Λ [ρ − ρ ]|| > 0. Now, it has been proved [34]
n dt t 1 2 tr
1 1 1 1 thatforrandomunitaryqubitevolutionifalleigenvalues
z ≥ + + , (56)
i
4 a a a λ (t)≥0, then (61) is equivalent to
i=1 (cid:18) 1 2 3(cid:19) k
Y
then κ (t) defined via (28) define a legitimate memory d
i λ (t)≤0 ; k =1,2,3. (63)
k
kernel. dt
Proof: It is clear that it is enough to prove (21). Proposition 1 For a1,a2,a3 satisfying (30) and
W(s)= 1 , where f˜(s) is CM and
Lemma 1 One has the following decomposition f˜(s)
1 =A 1 − n ij−=11zj , (57) tf(τ)dτ ≤amin, amin =min{a1,a2,a3}, (64)
s ni=1(s+zi) s i=1 ijQ=1(s+zj)! Z0
X
thecorrespondingmemorykernelgivesrisetothedynam-
whereQ Q
ical map Λ such that N [Λ ]=0.
t BLP t
1
A= . (58)
n z Proof: Let us observe that condition (64) implies (29).
i=1 i
Indeed, from (64) one has
For the proof see the AppeQndix. Now we show that con-
dition (21) holds. According to (57) one has 1 1 1 t
+ + f(τ)dτ ≤3, (65)
a a a
1 +λ˜ (s)+λ˜ (s)+λ˜ (s) (cid:18) 1 2 3(cid:19)Z0
1 2 3
s
and hence the condition (29) follows. Now, observe that
1 1 1 1 1
= 4− + +
s a a a W(s) 1 t
(cid:18) (cid:20) 1 2 3(cid:21) (cid:19) λ (t)=1− f(τ)dτ ≥0,
k
= 1 4− 1 + 1 + 1 1 (59) ak Z0
s a a a n z
(cid:18) (cid:20) 1 2 3(cid:21) i=1 i(cid:19) due to (64). Hence itis sufficient to showthat dλ (t)≤
+ (cid:20)a11 + a12 + a13(cid:21) ni=11ziQj=n1 jiQ=1ij(=−s11+zizi). 0a.ndIthiesnccleeatrakdditnλgki(ntt)o≤ac0coifuanntd(2o0n)lyitiifs1eq−usivλ˜akl(edsnt)tkitsoCthMe
X requirement that −κ˜ (s)λ˜ (s) is CM. One has therefore
A second term in (59) is CQM due to thQe fact that it is a k k
sumofCMfunctions. Hence,ifcondition(56)issatisfied f˜(s)
then (21) holds. (cid:3) −κ˜k(s)λ˜k(s)= , (66)
a
k
Note, that
which ends the proof since f˜(s) is CM and a >0 (cid:3)
s 1 s k
κ˜ (s)=− =− ,
i
akW(s)−1 ak(s−s1)...(s−sm) Remark 2 It W(s) = (s+z1)...(s+zn) with zk > 0
(60)
and a ,a ,a satisfying (30) together with
1 2 3
where {s ,...,s } are the roots of the polynomial
1 m
(akW(s)−1). Itisthereforeclearthatformula(60)may n 1
be easily inverted to the time domain. zi ≥ , k =1,2,3, (67)
a
k
i=1
Y
Remark 1 Note, that W(s) defined in (55) implies that
1 is CM and hence 1 1 is CM as well. then the corresponding dynamical map Λt satisfies
W(s) sW(s)
N [Λ ]=0.
BLP t
Remark 3 It was shown [14, 36] that BLP condition
IV. CHECKING FOR NON-MARKOVIANITY
(61) is equivalent to so-called P-divisibility. This means
that
Let us recall that according to [31] the evolution rep-
resented by Λt is non-Markovianif the condition Λt =Vt,sΛs, (68)
d
||Λt[ρ1−ρ2]||tr ≤0 , (61) and for any t>s the propagator Vt,s is positive (but not
dt necessarily completely positive).
is violated for some initial states ρ and ρ . One defines
1 2
Interestingly, our construction provides a class of legiti-
[31] a well-known non-Markovianity measure
mate random unitary qubit evolutions generated by the
d nontrivial memory kernel but still satisfying BLP condi-
N [Λ ]= sup ||Λ [ρ −ρ ]|| dt , (62)
BLP t dt t 1 2 tr tion (61), (cf. also [30]). It is clear that to violate (61)
ρ1,ρ2Z
6
one needs a more refined construction such that 1 is with z >0. One finds
W(s)
not CM but 1 1 is already CM. It deserves further s
sW(s) κ˜ (s)=− , (74)
analysis. k a (s+z)−1
k
Consider now the question of CP-divisibility which is
fullycontrolledbythelocaldecoherenceratesin(8). One and the inverse Laplace transform gives
may easily compute them in terms of f(t):
κk(t)=−1 δ(t)− z− 1 e−[z−a1k]t . (75)
z a
(cid:18) (cid:20) k(cid:21) (cid:19)
f(t) −1 1 1
γ (t) = + + ,
1 Note, that if a =1/z, then the dynamics is purely local.
4 a −F(t) a −F(t) a −F(t) k
(cid:18) 1 2 3 (cid:19) One easily finds
f(t) 1 1 1
γ (t) = − + ,
2 4 a −F(t) a −F(t) a −F(t) 1
(cid:18) 1 2 3 (cid:19) λk(t)=1− (1−e−zt), (76)
f(t) 1 1 1 zak
γ (t) = + − .
3
4 a −F(t) a −F(t) a −F(t)
(cid:18) 1 2 3 (cid:19) and finally the solution for pk(t) is defined by (37) with
The dynamical map Λt is CP-divisible iff γk(t) ≥ 0 for F(t)= 1(1−e−zt) . (77)
k =1,2,3. Let us assume that
z
a ≤a ≤a . (69) Note, that condition (56) implies the following relation
1 2 3
between z and a ,a ,a :
1 2 3
Proposition 2 If a ,a ,a and f(t) ≥ 0 satisfy condi-
1 2 3
1 1 1
tions (29) and (30) the corresponding memory kernel
4z ≥ + + , (78)
a a a
1 2 3
sf˜(s)
κ˜ (s)=− , (70) which guarantees that p (t) ≥ 0. In the symmetric case
k a −f˜(s) 0
k a =a = a =a one finds p (t) =p (t)= p (t)=: p(t)
1 2 3 1 2 3
with
leads to a CP-divisible dynamical map iff
1
p(t)= [1−e−zt], (79)
F(t)≤a1− (a2−a1)(a3−a1). (71) 4za
Proof: Due to (69) it ispsufficient to show that γ1(t)≥0 and p0(t) = 1 − 3p(t) with 4za ≥ 3. One finds that
which, for f(t)≥0, is equivalent to asymptotically
−1 1 1 3
+ + ≥0. (72) p0(t)→1− . (80)
a −F(t) a −F(t) a −F(t) 4za
1 2 3
Note that for za > 1 one has asymptotically p (∞) <
Let us assume that F(t) < a , which means, that γ (t) 0
1 1 1/4. This property cannot be reproduced within the local
is not singular. Inequality (72) is satisfied iff
approach with regular generators L . Indeed, it follows
t
from (9) (see also [34] for more details) that
F(t)∈(−∞,F ]∪[F ,+∞)
− +
1
with p (t)= [1+λ (t)+λ (t)+λ (t)], (81)
0 1 2 3
4
F± =a1± (a2−a1)(a3−a1). and hence, using (14), one finds
p
Now, taking into account that F(t) < a one finally 1
1
proves (71). (cid:3) p0(t)≥ 4. (82)
This Proposition shows that positivity of the function
This example shows that local and memory kernel ap-
f(t) is not sufficient for CP-divisibility. One needs an
proaches may lead to essentially different evolutions.
extra condition (71) which involves not only f(t) but
{a ,a ,a } as well.
1 2 3 Example 2 Consider now the same polynomial W(s)=
s+z but let z =2c>0. Moreover
V. EXAMPLES 1 1
a =a = , a = . (83)
1 2 3
c 2c
Example 1 Consider the simplest case with a polyno-
One finds
mial of degree one
sc
κ˜ (s)=κ˜ (s)=− , κ˜ (s)=−2c,
W(s)=s+z, (73) 1 2 3
s+c
7
and hence Example 4 Let
κ1(t)=κ2(t)=−cδ(t)+c2e−ct , κ3(t)=−2cδ(t). W(s)=s2+ω2. (92)
Finally, one finds the following formula for the memory Note that 1 1 is CM since
sW(s)
kernel
c 1 1 1 1 ω
K [ρ] = δ(t)[σ ρσ +σ ρσ −2ρ] = ,
t 2 1 1 2 2 sW(s) ωs s2+ω2
(cid:18) (cid:19)
c2
− e−ct[σ ρσ −ρ]. (84) t
3 3 is theLaplace transform of sin(ωτ)dτ which is positive
2 0
for all t≥0. Condition (29) implies
One has R
1 1 1
λ1(t)=λ2(t)= 21 1+e−2ct , λ3(t)=e−2ct. a1 + a2 + a3 ≤2ω2 . (93)
Interestingly,thisevolut(cid:0)ionreprod(cid:1)ucestime-localdescrip- Thecorrespondingeigenvalues ofthememorykernelread
tion with
1 1
c c κ (t)=− cos ω2− t , (94)
γ (t)=γ (t)= , γ (t)=− tanh(ct). (85) i a a
1 2 2 3 2 i (cid:18)r i (cid:19)
asdiscussedin[35]. Itwasshown[36]thatΛt isaconvex for ω2 ≥1/ai, and
combination of two Markovian semigroups Λ(1) and Λ(2)
t t 1 1
generated by κ (t)=− cosh −ω2 t , (95)
i
a a
c i (cid:18)r i (cid:19)
L [ρ]= [σ ρσ −ρ] ; k =1,2, (86)
k 2 k k for ω2 <1/ai. Moreover one finds
that is, 1
λ (t)=1+ [cos(ωt)−1] , (96)
1 k akω2
Λ = etL1 +etL2 . (87)
t
2 and hence
(cid:0) (cid:1)
This simple example shows that a convex combination
1 1 1 1
of Markovian semigroups leads to a quantum evolution p (t) = + − [1−cos(ωt)] ,
1 4ω2 a a a
displaying essential memory effects. (cid:18) 2 3 1(cid:19)
1 1 1 1
p (t) = + − [1−cos(ωt)] , (97)
Example 3 Consider now a polynomial of degree two 2 4ω2 a a a
(cid:18) 3 1 2(cid:19)
W(s)=(s+c )(s+c ), (88) 1 1 1 1
1 2 p (t) = + − [1−cos(ωt)] ,
3 4ω2 a a a
(cid:18) 1 2 3(cid:19)
with c >c >0. Our construction gives rise to a legiti-
2 1
mate memory kernel if condition (30) holds and togetherwithp (t)=1−p (t)−p (t)−p (t). Inparticular
0 1 2 3
taking
1 1 1
4c c ≥ + + . (89)
1 2 a a a 1
1 2 3 a =a = , a =∞, (98)
1 2 ω2 3
One finds
one finds
1 s
κ˜ (s) = −
k ak(s+c1)(s+c2)− a1k κ1(t)=κ2(t)=−ω2 , κ3(t)=0 , (99)
1 s
= − , (90) and hence
a (s+s )(s+s )
k 1 2
ω2
with k1(t)=k2(t)=0 , κ3(t)= , (100)
2
1
s +s =c +c , s s =c c − . which proves that the constant (time independent)
1 2 1 2 1 2 1 2
a
k
k
Hence the solution has the form (37) with the function K [ρ]= (σ ρσ −ρ), (101)
t 3 3
2
F(t) given by
provides a legitimate memory kernel for arbitrary k =
1 1 1
F(t)= [1−e−c1t]− [1−e−c2t] . (91) ω2 > 0. Moreover one finds for the local decoherence
c2−c1 (cid:18)c1 c2 (cid:19) rates
8
CM. We stress that Theorem4 providesonly a sufficient
condition and further analysis is needed to cover physi-
ωsin(ωt) −1
γ (t)= callyinterestingcaseswhichdonotfittheassumptionsof
1 4 a1ω2−1+cos(ωt) the Theorem. Interestingly, it turns out that the quan-
(cid:16)
1 1 tum evolution with a memory kernel generated by our
+ + ,
a ω2−1+cos(ωt) a ω2−1+cos(ωt) approach gives rise to vanishing non-Markovianity mea-
2 3
(cid:17) sure based on the distinguishability of quantum states
andsimilarlyforγ2(t)andγ3(t). Note,thatiffor somek [31]. We also have shown when the corresponding dy-
one has akω2 <1 then local decoherence rates are singu- namicalmapis CP-divisible. Itshowsthatthe evolution
lar and hence in this case the non-local approach is more satisfying nonlocal master equation does not necessarily
suitable. lead to a non-Markovian evolution. It would be also in-
teresting to analyze the relation between semi-Markov
evolutionandthe one governedby our approachin more
VI. CONCLUSIONS detail.
We analyzed random unitary evolution of a qubit
withinmemorykernelapproach. Ourmainresultformu- Acknowledgements
lated in Theorem 4 allows to construct legitimate mem-
orykernelsleadingtoCPTPdynamicalmaps. Thepower
We thank anonymous referee for valuable comments
of this method is based on the fact that 1) it allows to and Karol Z˙yczkowski for pointing out [33]. This article
reconstructwellknownexamplesoflegitimatequbitevo-
was partially supported by the National Science Center
lution,2)thestructureofpolynomialsW (s)enablesone
k project DEC- 2011/03/B/ST2/00136.
to perform the inverse Laplace transform and to find a
formula for the kernel in the time domain. The math-
ematical analysis heavily uses the notion of completely
Appendix: proof of Lemma 1
monotone functions. These functions are not commonly
used in theoretical physics and knowledge of their prop-
erties is rather limited. There are no known effective Let us observe that (57) may be represented in the
methods allowing to check whether a given function is following form
n (s+z )−s n (s+z )+z n (s+z )+... n−2z (s+z )+ n−1z
1 i=1 i i=2 i 1 i=3 i j=1 j n j=1 j
=A , (102)
s ni=1(s+zi) Q (cid:16)Q s Qni=1(s+zi) Q Q (cid:17)
Q Q
therefore, to prove the Lemma it suffices to show that
n n n n n−2 n−1
z = (s+z )−s (s+z )+z (s+z )+... z (s+z )+ z . (103)
i i i 1 i j n j
i=1 i=1 i=2 i=3 j=1 j=1
Y Y Y Y Y Y
Wewillprovethisbyinduction. Forn=1itisclearthat while RHS reads
LHS=RHS=z . We assume that (103) is true for n and
1
prove it is also true for (n+1). LHS may be written as
n
LHS= z ·z , (104)
i n+1
i=1
Y
9
n n n
RHS = (s+z )(s+z )−s (s+z )(s+z )+z (s+z )(s+z )+...+
i n+1 i n+1 1 i n+1
iY=1 (cid:16)iY=2 iY=3
n−2 n−1 n−1
+ z (s+z )(s+z )+ z (s+z )+ z ·z =
j n n+1 j n+1 j n
jY=1 jY=1 jY=1 (cid:17)
n n n n−2 n−1
= (s+z ) (s+z )−s (s+z )+z (s+z )+... z (s+z )+ z −
n+1 i i 1 i j n j
i=1 i=2 i=3 j=1 j=1
Y Y Y Y Y
n−1 n n n
− s z z =s z +z z −s nz , (105)
j n i n+1 i i
j=1 i=1 i=1 i=1
Y Y Y Y
which proves that RHS=LHS. (cid:3)
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