Exact analytical evaluation of time dependent transmission coefficient from the method of reactive flux for an inverted parabolic barrier

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by  Rajarshi Chakrabarti

Exact analytical evaluation of time dependent transmission coefficient from the
method of reactive flux for an inverted parabolic barrier
Rajarshi Chakrabarti
Department of Inorganic and Physical Chemistry,
7 Indian Institute of Science, Bangalore, 560012, India
0
0 In this paper we derive a general expression for the transmission coefficient using the method
2 of reactive flux for a particle coupled to a harmonic bath surmounting a one dimensional inverted
parabolic barrier. Unlike Kohen and Tannor [J. Chem. Phys. 103, 6013 (1995)] we use a normal
n
mode analysis where the unstable and the other modes have a complete physical meaning. Impor-
a
tantlyourapproachresultsaverygeneralexpressionforthetimedependenttransmissioncoefficient
J
notrestrictedtooverdampedlimit. Oncethespectraldensityfortheproblemisknownonecanuse
4
ourformulatoevaluatethetimedependenttransmission coefficient. Wehavedonethecalculations
1
withtimedependentfrictionusedbyXie[Phys. Rev. Lett93,180603(2004)]andalsotheoneused

by Kohen and Tannor [J. Chem. Phys. 103, 6013 (1995)]. Like the formula of Kohen and Tannor
]
h our formula also reproduces the results of transition state theory as well as the Kramers theory in
c thelimits t→0 and t→∞ respectively.
e
m
- I. INTRODUCTION constantkTST. Thusonewrites,k(t)=κ(t)kTST. Inthe
t limitt 0,κ(t) 1andoneobtainsk(0)=k . Here
a TST
→ →
t wefollowthemethodofreactivefluxtocalculatethetime
s
The dynamics of a particle surmounting a barrier is dependent rate constant for a particle bilinearly coupled
.
t
a an important and interesting problem in chemistry and to a harmonic bath [9, 10, 11]. The bilinear coupling
m physics. In chemistry, chemical reactions are most com- allowsus to get ananalyticalexpressionfor the time de-
mon examples where one studies the dynamics of bar- pendent transmission coefficient. Although the method
-
d riercrossingalongasuitablydefinedreactioncoordinate. ofreactivefluxandthehamiltonianwehaveusedarewell
n While in physics nucleation phenomena, electricaltrans- known in the literature, as far as our knowledge nobody
o
port etc. involve barrier crossing. A review by Ha¨nggi, has used such hamiltonian to calculate the time depen-
c
Talkner and Borkovec[1] gives a comprehensive account dent transmission coefficient analytically. The calcula-
[
on the subject of barrier crossing. The simplest possible tion is of interest as the transmission coefficient for the

4 approachtocalculatetherateconstantforsuchprocesses overdampedlimit of the model was obtained recently by
v
is givenby transitionstate theory [1, 2]. As the simplest simulation [12]. Our approach directly gives expression
1
version of the problem one can describe the rate pro- for the time dependent transmission coefficient in terms
9
0 cess along a single reaction co-ordinate. Then the one of the spectral density J(ω) and not restricted to over-
1 dimensional version of the transition state rate constant damped limit. A normal mode analysis of the coupled
0 k reads as, k = ω0 exp[ Eb ], where ω is the Hamiltonian was made by Pollak [13, 14, 15] and others
7 anTgSuTlarfrequencyTiSnTthe r2eπactan−twkBeTllandE is t0he bar- toanalyzethe Kramersproblem. Herewepointoutthat
b
0 rier height or the activation energy. The derivation of the approximation is very powerful and can be used to
/
t transition state rate constant is based on two assump- gettheexactexpressionforκ(t). Furtherwebelievethat
a
tions. Firstly an overall thermal equilibrium is assumed it can also be used to get the quantum expression for
m
andsecondlyfrictionaleffectisnottakenintoaccountby κ(t). An exact calculation of κ(t) using the phase space
- incorporating what is known as “no recrossing”assump- distributionfunctionformulationwascarriedoutinclas-
d
n tion. Subsequently,Kramers[1,2,3,4]theorytakesinto sicalregimebyKohenandTannor[16,17],Bao[18],and
o account of the friction in a phenomenological way. His morerecentlyinthe contextofsingleenzymekineticsby
c formulationwasbasedonMarkoviandynamicsofaBrow- ChowdhuryandCherayil[19]. Alsothecalculationofthe
:
v nian particle escaping from a metastable state. Further transmissioncoefficientinthequantumdomainiscarried
i extensionsofKramers’stheorywerecarriedoutbyGrote outbyRay’sgroup[20]. Ouranalysisintermsofnormal
X
andHynes [5] followedby Ha¨nggiandMojtabi [6]. They modeshastheadvantagethatitmakesthephysicalideas
r tookintoaccountofthenon-Markoviannatureofthedy- very clear. The paper is arranged as follows. In section
a
namics. Anotherapproachtocalculatethe rateconstant II we introduce our system plus harmonic bath hamilto-
isduetoChandler[7,8]knownasthemethodofreactive nian and the Generalized Langevin Equation (GLE). In
flux in chemical physics literature. In this approach the section III we mention the method of reactive flux and
thermalrateconstantiswrittenasacorrelationfunction usingourhamiltonianderiveananalyticalexpressionfor
which is nothing but the ensemble average over infinite the time dependent transmission coefficient, κ(t). Sec-
number of trajectories starting at the barrier top and tion IV deals with the calculation of κ(t) for Markovian
ending on the product side at time t. The rate constant and Non Markoviandynamics. Section V is conclusion.
at time t is written as a product of the time dependent
transmissioncoefficientκ(t)andthetransitionstaterate
2
II. GENERALIZED LANGEVIN EQUATION where U is an orthogonal matrix such that U D UT is
AND NORMAL MODE ANALYSIS diagonal. D is a dynamical matrix given by
We consider a particle coupled to a harmonic bath [9,
10] with the total hamiltonian ω2+ N c2j . . cj
 − b j=1Mmjωj2 −√Mmj 
D = P. . . 0 (8)
H = 2PM2 +V(Q)+21 jN=1 mPj2j +mjωj2(cid:18)Qj − mcjjωjQ(cid:19)2!  −√M.cjmj 0. 0. ω0j2 
X  
(1) The reaction co-ordinate q may then be written as
Here (P ,Q ) are the momenta and coordinates of the
j j
jth bath oscillatorwhose mass and frequency are m ,ω N
j j
respectively. (P,Q) is the momentum and coordinate of q = U0jηi
the system. c couples the bath oscillator to the system. j=0
j X
Now in terms of the mass weighted coordinates
Of the modes η , we take η to be the unstable mode.
j 0
Wedenotethefrequencyofη asλ . η hastheimaginary
j j 0
frequencyandwewriteitasλ =iΛ,whereΛisreal. As
q =√MQ,qj =√mjQj (2) η is a normal mode, its time0development is given by
j
our Hamiltonian becomes
sin(λ t)
η (t)=η (0)cos(λ t)+p (0) j
j 0 j ηj λ
j
p2 1 N c 2 with j =0,1,2,.....N.
H = +V(q)+ p2+ω2 q j q
2 2Xj=1 j j j − Mmj !  noNrmowalmthoedeHsatmheiltcoonuipalninwghisennowmriotrteenthienretearnmdsisogfivthene
 p (3) by
and then with this Hamiltonian one can derive what
is know as the Generalized Langevin Equation (GLE)
which reads as N p2 1
H = j + λ2η2 (9)
nor 2 2 j j
!
j=0
X
t
′ ′ ′ ′ Inthenextsectionweanalyticallyderivethe ratecon-
q¨(t)+V (q)+ dt γ(t t )q˙(t )=ζ(t) (4)
− stant in the normal mode description using the above
Z
0 Hamiltonian.
Onecanalsowritethetimedependentfrictioninterms
of the spectral density of the bath J(ω) as follows
III. METHOD OF REACTIVE FLUX AND THE
RATE CONSTANT
∞
N
Θ(t)2 J(ω)
γ(t)= dω cos(ωt) (5) In this section we briefly describe the method of reac-
M π ω
Xj=1Z0 tiveflux[7,8]tocalculatetherateconstant. Themethod
ofreactivefluxexpressesthethermalrateconstant(writ-
where ten in the mass-weightedcoordinates) as
J(ω)= π2 N mcj2jωjδ(ω−ωj) (6) k(t)= hq˙(t)δh(θqR(0(q))(θ0P))(iq(t))i = kknd((tt)) (10)
j=1
X where the angular bracket indicates the thermal av-
One can easily show that erage over the system as well as the bath degrees of
freedom. The top of the barrier is chosen as q(0) = 0.
θ (q(0)) is 1 if q(0)<1 and 0 otherwise. θ is just 1-θ ,
R p R
ζ(t )ζ(t ) =Mk Tγ(t t ) (7) q˙ is the velocity of the particle.
1 2 B 1 2
h i −
The numerator of Eq. (10) could be written as
We take the barrier top to be at q = 0 and introduce
ω2 = ∂2V . Forsmallamplitudemotionaroundthe
b ∂q2 ∂
q=0 k (t)= i exp(ir q˙(0))δ(q(0))θ (q(t))
(cid:16) (cid:17) N n − ∂r h 0 P i
barriertop,onecanusethenormalmodes,η = U q (cid:18) 0 (cid:19)r0=0
j ji i (11)
i=0
P
3
Expressing the step function as an integral over delta delta function the expression for k (t) becomes
n
function and using the Fourier integral representationof
∞ ∞ ∞
1 ∂
k (t)= dh( i) ds ds exp( is h) exp(ir q˙(0)+is q(t)+is q(0))
n (2π)2 Z0 − ∂r0−Z∞ 0−Z∞ 1 − 1 h 0 1 0 ir0=0
 
Theangularbracketindicatesthethermalaverageover exp( βH ) , and carrying out the integration over all
nor
−
all degrees of freedom of the problem and it is done by coordinates and momenta degrees of freedom. Writing
multiplying the quantities within the angular bracketby k (t) in terms of η leads to
n j
∞dh( i) ∂ ∞ ds ∞ ds exp( is h)exp( β p2η0 Λ2η2 )exp(ir U p )
 0 − ∂r0−∞ 0−∞ 1 − 1 − (cid:18) 2 − 2 0(cid:19) 0 00 η0 
kn(t)= exp((2−πβ)2Eb) R N ∞edxppR(ir∞0Ud0η0Rηe0x)pex(p(βis1p(cid:16)2jU+00λη2j0ηc2osh)(eΛxpt)(i+rpUη0(η0))sienxhΛp(Λ(ti)r(cid:17)U) p )  (12)
 j=1−∞ j−∞ j − 2 2 j 0 j0 j 0 j0 ηj 
 Q R R exp(is1 Uj0(cid:16)ηjcos(λjt)(cid:17)+pηj(0)sinλ(λjjt) r0=0
 (cid:16) (cid:17) 
where E is the barrier height. After one carries out the integrations the expression for k (t) can be written as
b n
N
2π 1 c˙(t)
k (t)= − exp( βE ) (13)
n jY=1(cid:18)βλj(cid:19)(cid:18)βΛ(cid:19) (c(t)2−c(0)2)21! − b
SimilarlythedenominatoroftheEq. (10)canbewrit- ten as
∞ ∞
N p2 λ02 N 2π
k (t)= dp dη exp( β j + j η2 )= (14)
d jY=0−Z∞ j−Z∞ j − 2 2 j! jY=0 βλ0j!
Then the time dependent rate constant becomes
N
λ0
j
k(t)= kn(t) = 1 j=0 −c˙(t) exp( βE )
kd(t) 2πΛ QN (c(t)2 c(0)2)21! − b
λ
j −
j=1
Q (15)
4
N λ0 Inthenextsectionwecalculatec(t)andfromthatκ(t)
j
Here we use the following identity [1], j=0 = ω0, for Markovianas well as for Non Markoviannoise.
QN ωb
Λ λj
j=1
and this simplifies our k(t) Q
IV. CALCULATION OF c(t) AND κ(t)
1 ω c˙(t)
k(t)= 0 − exp( βE ) (16)
2πωb (c(t)2 c(0)2)21! − b Calculation of κ(t) involves the calculation of c(t).
−
First we describe a general derivation of c(t) then we
where, c(t) = N U02jcos(λjt) . As usual k(t) [16] is evaluate κ(t) for different cases.
λ2
j=0 j !
defined by P
A. Calculation of c(t)
k(t)=κ(t)k(0) (17)
whWicehsihsonwotihninthgebAuptptehnedtixratnhsaittiokn(0s)ta=teω2πr0aetxepc(o−nβsEtabn)t, c(t) is defined by c(t) = jN=0Uj20cλos2j(λjt). One proceeds
kTST. In the Appendix we also show how one gets by writing N U2 δ(ω2 λ2)P= 1Im N Uj20 . To
Kramers rate constant in the t limit. j0 − j −π (ω2+iη−λ2)
→∞ j=1 j=1 j
Usingtheaboveexpressionfork(0)ourtimedependent
transmission coefficient, κ(t) becomes evaluate NP Uj20 here we adoptPnotations similar
(ω2+iη−λ2)
j=1 j
to quantuPm mechanics and also use the fact that D is a
1 c˙(t)
κ(t)= − (18) matrix with eigen values λ2,j =1,2,.....N.
ωb (c(t)2 c(0)2)12! j
−
N U2 N 1 N 1
j0 = 0 j j 0 = 0 j j 0 (19)
ω2+iη λ2 h | i ω2+iη λ2 h | i h | i(ω2+iη D)h | i
j=1 − j j=1 − j j=1 −
X X X
(cid:0) (cid:1) (cid:0) (cid:1)
Then by introducing j j in between the sum and
| ih |
N
using the resolution of identity j j =I.
| ih |
j=1
P
N
1 1
U2δ(ω2 λ2)= 0 0 = Im G(ω2+iη) (20)
j0 − j h |(ω2+iη D)| i −π
j=1 −
X (cid:0) (cid:1)
Usingpartitioningtechniqueonecanevaluate(G) to N
00 As 1Im G(ω2) = U2δ(ω2 λ2)=ρ(ω2). We
get −π 00 j0 − j
j=0
(cid:0) ∞(cid:1) P
G(ω2+iη) can write c(t)=2 dω cos(ωt)ρ(ω2).
00 −1 0 ω
= ω2+ω2 (cid:0) 1 N (cid:1) c2jω2 (21) R
b − M j=1mjωj2(ω2+iη−ωj2)!
P
Then using the definition of J(ω) Eq.(6) it becomes
B. Calculationof κ(t) with γ(t)=γ2H(2H−1)|t|2H−2
G(ω2+iη)
= ω2+ω2(cid:0) 2ω2 ∞ (cid:1)J00(ω˜)dω˜ −1 (22) First we perform the calculation with the time depen-
b − Mπ ω˜(ω2+iη−ω˜2) dent friction γ(t) used by Xie [21], which is
(cid:18) 0 (cid:19)
R
5
a inverse Fourier transform of Eq. (5) and write J(ω) in
terms of γ(t).
1.00
nt (t)0.99
e
effici0.98 ∞
o
n c0.97 J(ω)=Mω dtγ(t)cos(ωt) (24)
sio Z
mis0.96 0
s
n
a0.95
Tr Using the definition of γ(t) we get
0.94
0.93
∞
1 2 3 4 5 6 7 8 9 10 2ω2 J(ω˜)dω˜
t =γΓ(2H +1)Λ2−2H
Mπ ω˜(ω2+iη ω˜2)
. Z −
0
FIG.1: Timedependenttransmissioncoefficient,κ(t)against
time,forH = 1 (dashedline)andH = 3 (solidline),param- First we consider the case ω2 = Λ2 which means we
2 4 −
eters used are ωb=1,γ =0.1 consider the unstable mode of the problem. Then
1.0 G(−Λ2) 00 = −Λ2+ωb2−γΓ(2H +1)Λ2−2H −1
oefficient (t) 000...468 moOd(cid:0)uersnfoerxtwt(cid:1)haisckhisG(cid:0)t0o0fi−ndΛ2outbtlohwespoulpe,s,i.ie.e..tthheesuonl(cid:1)ustta(io2bn5les)
C of the equation
on 0.2 (cid:0) (cid:1)
si
s
ansmi 0.0 −Λ2+ωb2−γΓ(2H+1)Λ2−2H =0 (26)
Tr -0.2
-0.4 On the other hand, to find ρ(ω2) we have to consider a
situation where ω2 is positive. In other words the stable
-0.6
10 20 30 40 modes. Now the Evaluation of ρ(ω2) involves the inte-
t ∞
gral 2ω2 J(ω˜)dω˜ . Using the identity 1 =
Mπ ω˜(ω2+iη−ω˜2) (ω2+iη−ω˜2)
0
P 1 R iπδ ω2 ω˜2 ,whereP 1 istheprin-
FIG.2: Timedependenttransmissioncoefficient,κ(t)against ω2−ω˜2 − − ω2−ω˜2
time for Non-Markovian friction, γ(t)=γhexp(−ht). Three cip(cid:16)al value(cid:17)of (cid:0)1 we(cid:1)find (cid:16) (cid:17)
different regimes, caging regime (solid line) with parameters, ω2−ω˜2
γ = 150,h = 0.01,ωb = 1, intermediate regime (dotted line) (cid:16) (cid:17)
with parameters γ = 5,h = 1,ωb = 1, and nonadiabatic
∞
regime (dashed line) with parameters γ =50,h=0.01,ωb = 2ω2 J(ω˜)dω˜
1. = exp(iHπ)Γ(2H+2)ω2−2Hγ
Mπ ω˜(ω2+iη ω˜2) −
Z −
0
(27)
Thus
γ(t)=γ2H(2H −1)|t|2H−2 (23) G(ω2) = ω2+ω2+exp(iHπ)Γ(2H +2)ω2−2Hγ −1
00 b
with, 1 < H < 1, thus the thermal fluctuations are de- (28)
2 (cid:0) (cid:1) (cid:0) (cid:1)
scribed by fractional Gaussian noise [21]. Now we take and
ρ(ω2)= 1Im G(ω2)
−π 00
= 1 γΓ(2H+1)ω2−(cid:0)2Hsin(H(cid:1)π) (29)
π (ω2+ω2+γΓ(2H+1)ω2−2Hcos(Hπ))2+γ2sin2(Hπ)Γ(2H+1)2ω4−4H
b
“ ”
6
1. The Case H = 1 case Eq. (26) becomes, Λ2 + ω2 γΛ = 0 and has
2 − b −
two roots. But only one of them is physically accept-
ainnFdEiqwrs.itt(hw7)tehbciesocntohsmiedenesrodiHseel-t=naofi12us,enfccotorirorwnehlcaiotcirhornetlhafuetenγdc(,ttir)oen=add2ienγfigδn(etd) afrbolme. ETqh.e r(2o9o)t iosn,eΛgHe=ts21 ρ=(ω−2)γ2H+=q21 =(cid:0)γ2(cid:1)π2ω+2γω2b+2.ω(ωγS2i+mωi2la)2rly.
b
ζ(t )ζ(t ) =2Mγk Tδ(t t ) (30) ThenΛH=12 andρ(ω2)H=21 areusedtoc“alculatec(t)an”d
1 2 B 1 2 taking t 0 limit one gets c(0). In this case we could
h i −
→
do these analytically. The time dependent transmission
Thenthe dynamicsis describedbyusualLangevinequa-
coefficient in this case is given by
tion rather than a GLE. Thus the dynamics is then
Markovian i.e. with no memory. For the Markovian
1+et(f−γ)
κ(t)= −
1 4eft+2et(f−γ)+2e2t((cid:0)f−γ) ( γ )((cid:1)f fe2t(f−γ)+2γ( 1+eft))
− − 2ωb − −
q(cid:0) (cid:1)
where f =γ+(4ω 2+γ)1/2. C. Calculation of κ(t) with γ(t)=γhexp(−ht)
b
κ(t) is then calculated using the above expression and
plotted against time (Fig. 1).
Inthissectionwecarryoutthecalculationofκ(t)with
atimedependentfrictionγ(t)=γhexp( ht). Thesame
−
2. The Case H = 3 friction is used by Kohen and Tannor [16, 17]. First we
4 calculate J(ω) with this friction as follows
WhenH = 1 thenthenoiseisnolongerdeltafunction
6 2 ∞
correlated, i.e. not a white noise with Gaussian distri- γMh2ω
bution. This regime is know as Non Markovian regime J(ω)=Mγhω exp( ht)cos(ωt)dt=
− (h2+ω2)
which has memory. We have similarly evaluated κ(t) for Z
0
H = 3 and plotted against time in Fig. 2. But in this
4
case the integrations are done numerically since analyti-
cal evaluation of integrals were not possible. Then the following integral is evaluated as follows
∞ ∞
2ω2 J(ω˜)dω˜ 2ω2h2 dω˜ γhω(ω ih)
= = −
Mπ (α2+ω˜2)(ω2+iη ω˜2) π (α2+ω˜2)(ω2+iη ω˜2) (h2+ω2)
Z − Z −
0 0
We haveused the identity, 1 =P 1 and
(ω2+iη−ω˜2) ω2−ω˜2 −
iπδ ω2 ω˜2 ,whereP 1 istheprincip(cid:16)alvalu(cid:17)eof
− ω2−ω˜2
(cid:0)1 . Th(cid:1)us (cid:16) (cid:17)
ω2−ω˜2
(cid:16) (cid:17)
γhω(ω ih) −1
G(ω2) = ω2+ω2 −
00 b − (h2+ω2)
(cid:18) (cid:19)
(cid:0) (cid:1)
1 ωγh2 h2+ω2
ρ(ω2)= Im G(ω2) =
−π 00 π ((h2+ω2)(h2+ω(cid:0)2) γhω(cid:1)2)2+ω2γ2h4
(cid:0) (cid:1) b −
(cid:16) (cid:17)
7
One should notice that
ωγh2 h2+ω2 γω
hl→im∞π ((h2+ω2)(h2+ω(cid:0)2) γhω(cid:1)2)2+ω2γ2h4 = π((ω2+ωb2)+γ2ω2) =ρ(ω2)H=12
b −
(cid:16) (cid:17)
Thus in this limit the result obtained is identical with method of reactive flux [7, 8] to calculate the time de-
the frictionγ(t)=γ2H(2H 1) t2H−2 with H = 1. In pendent transmission coefficient, κ(t). Analytically we
− | | 2
other words the Markoviandynamics is recovered. derive a general formula for κ(t) Eq. (18). For the
To calculate κ(t) we proceed as in the previous case. Markovian case with no memory (H = 1) for the fric-
2
First considering the case ω2 = Λ2, one gets tion γ(t) = γ2H(2H 1) t2H−2 with H = 1 which is
− − | | 2
sameas withh= for the frictionγ(t)=γhexp( ht),
∞
∞ −
2ω2 J(ω˜)dω˜ γhΛ κ(t) is calculated analytically. Whereas in the case with
=
Mπ (α2+ω˜2)(ω2+iη ω˜2) (h+Λ) memory (H = 3) and with any finite value of h the cal-
4
Z0 − culations were done numerically. In all the cases a plot
of κ(t) vs t starts from 1 at t = 0 and reaches a plateau
Thus
in the long time limit. As expected our formula for the
time dependent rate constant, k(t) becomes equal to the
γhΛ −1 transition state rate constant when one takes the limit
G(ω2) = Λ2+ω2 t 0 [16, 17]. Similarly Kramers rate constant is ob-
00 − b − (h+Λ) →
(cid:18) (cid:19) tained by taking t [16, 17]. In future we would like
(cid:0) (cid:1) →∞
to extend our formulation to quantum domain.
Hencetofindoutthepolesonehastosolvethesolution
of the equation
γhΛ
Λ2+ω2 =0 (31)
− b − (h+Λ) VI. ACKNOWLEDGEMENTS
To explore the non-Markovian dynamics we choose
The author is grateful to Prof. K. L. Sebastian for his
three set of parameters corresponding to three different
useful comments and encouragements and thanks Prof.
regimesofnon-MarkoviandynamicsfollowingKohenand
B. J. Cherayil for his comments on the manuscript.
Tannor [16, 17]. These regimes are called non-adiabatic,
The author also acknowledges Council of Scientific and
caging and intermediate according to Kohen and Tan-
Industrial Research (CSIR), India for financial support.
nor [16, 17]. The time dependent coefficients in these
regimes are calculated numerically and plotted against
time. We observe similar kind of oscillatory behavior of
κ(t)in the cagingregimeasobtainedbythem. Similarly
in the intermediate regime the plot looks similar with a
nonmonotonic decay (Fig. 2).
VII. APPENDIX : TRANSITION STATE RATE
AND KRAMERS RATE FROM k(t)
V. CONCLUSIONS
The paper shows an elegant way of combining the Firstwe takethe limit t 0. ThenfromEq. (16) one
→
traditional system plus reservoir model [9, 10] and the writes
8
k(0)=limt→0k(0+t)= 21πωω0b limt→0 (c(0+−t)c˙2(−0+c(t0))2)21 exp(−βEb)=
21πωω0b limt→0 c(0)+tc˙(−0)c˙+(0t)22−c¨c¨((00))t2−c(0)2 12 exp(−βEb)
„“ ” «
= 21πωω0b limt→0 (c(0)2−c(0−)2c¨+(0c)(t0)c¨(0)t2)21 exp(−βEb)= 21πωω0b √−c(c¨0()0c¨)(0)exp(−βEb)= 21πωω0b √−1c(0)exp(−βEb)
Where we have used the facts, c˙(0)=0 and c¨(0)= 1 N
. This is because c˙(t) = 2∞dωρ(ω2)sin(ωt) maki−ng −j=0Uj20 = −1. Now one can also show, −c(0) = ω1b2
− asPfollows
0
∞
R
c˙(0) = 0. Similarly c¨(t) = 2 dωρ(ω2)sin(ωt)ω =
−
0
R
c(0)= N Uj20 = N U λ−2U = N 0 j λ−2 j 0
j=0 λ2j j=0 j0 j 0j j=0h | i j h | i
P P P
N N
=−limω2→0 h0|jihj|(ω21−D)|jihj |0i=−limω2→0 h0|(ω21−D)|0i=−limω2→0 G(ω2) 00
j=0 j=0
P P (cid:0) (cid:1)
From Eq. (22) one can see that limω2→0 G(ω2) 00 = Similarly taking t → ∞ one gets the Kramers rate
1 = c(0). Then k(0) is nothing but the transition constant as follows
ωb2 − (cid:0) (cid:1)
state rate constant, k .
TST
ω
k(0)= 0 exp( βE )=k (32)
2π − b TST
2
k( )= 1 ω0 UΛ00 sinh(Λt) exp( βE )= 1 ω0 (Λtanh(Λt)) exp( βE )
∞ 2πωb s„UΛ0220 cosh(Λt)«2t→∞ − b 2πωb t→∞ − b
 
Then Now if one replaces Λ by ΛH=12 = −γ2 + γ2 2+ωb2
1 ω (H = 1 means white noise) one gets q
k( )=Λ 0 exp( βE ) 2 (cid:0) (cid:1)
∞ 2πω − b
b
k( )= γ + γ 2+ω2 1 ω0 exp( βE )
∞ −2 2 b 2πωb − b
(cid:18) q (cid:19)
(cid:0) (cid:1)
Which is nothing but the usual Kramers rate. Also in
the overdamped limit i.e. when γ >>ω2 one gets
b
9
γ γ 1 ω2 ω ω ω
k( )= + 1+ 4 b 0 exp( βE )= 0 b exp( βE )
∞ −2 2 2 γ2 2πω − b 2πγ − b
(cid:18) (cid:18) (cid:18) (cid:19)(cid:19)(cid:19) b
which is Kramers rate expression in overdamped
regime.
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