Diffusion–reaction approach to electronic relaxation in solution. An alternative simple derivation for two state model with a Dirac delta function coupling

Journal Pre-proof Diffusion-reaction approach to electronic relaxation in solution. An alternative simple derivation for two state model with a Dirac delta function coupling Swati Mudra, Aniruddha Chakraborty

PII: DOI: Reference:

S0378-4371(19)32104-1 https://doi.org/10.1016/j.physa.2019.123779 PHYSA 123779

To appear in:

Physica A

Received date : 11 March 2019 Please cite this article as: S. Mudra and A. Chakraborty, Diffusion-reaction approach to electronic relaxation in solution. An alternative simple derivation for two state model with a Dirac delta function coupling, Physica A (2019), doi: https://doi.org/10.1016/j.physa.2019.123779. This is a PDF file of an article that has undergone enhancements after acceptance, such as the addition of a cover page and metadata, and formatting for readability, but it is not yet the definitive version of record. This version will undergo additional copyediting, typesetting and review before it is published in its final form, but we are providing this version to give early visibility of the article. Please note that, during the production process, errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.

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Journal Pre-proof

Diffusion-reaction approach to electronic relaxation in solution. An alternative simple derivation for two state model with a Dirac delta function coupling Swati Mudra* and Aniruddha Chakraborty School of Basic Sciences, Indian Institute of Technology Mandi, Kamand, Himachal Pradesh, 175005, India (Dated: December 12, 2019)

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We give a very simple method for finding the exact analytical solution for the problem of electronic relaxation in solution, modeled by a particle undergoing diffusive motion under the influence of two potentials and the coupling between two potentials is assumed to be represented by Dirac delta function of arbitrary strength and position. Diffusive motion on both the potentials are described by Smoluchowski equation. Green’s function method has been used to solve the coupled equations. The solution requires expression of the Green’s function for the motion in isolated potentials. We apply our method to the case of coupled parabolic potentials.

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Electronic relaxation in solution is one of the very interesting example of barrierless reactions in solution [1, 2]. This is an interesting phenomenon from experimental as well as theoretical point of view [3–11]. A molecule in a ground state potential energy surface execute a walk in on that potential energy surface which may be assumed to be random, as it is immersed in a polar solvent. This molecule can be put on an electronically excited state potential energy surface by the absorption of radiation, the molecule executes a random walk again on that excited state potential energy surface. As it moves it may undergo non-radiative decay from certain specified regions of the excited state potential energy surface [3]. It may also undergo radiative decay from anywhere on that surface [3]. Most recent theoretical models assume motion on both the potential energy surface to be one dimensional and the relevant coordinate being denoted by x. In the following we will refer x as the position of a particle in both the potential energy surfaces and to the de-exciation of the molecule as the transfer of the particle of excited state potential energy surface to teh ground state potential energy surface [8]. In this paper we propose a very simple method for finding the exact analytical solution for this problem, earlier method was very complex for the same model [8]. As it is known that this two state model is more general than all the earlier models [9–17]. From the theoretical aspect, the problem is to calculate the probability that the molecule will still be in the electronically excited state, after time t. The probability Pe (x, t) that the particle may be found at x at time t on the excited state potential and the probability Pg (x, t) that the particle may be found at x at time t on the ground state potential, obeys the following modified Smoluchowski equation.

Here operator Li will be

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∂Pe (x, t) = Le Pe (x, t) − kr Pe (x, t) − k0 S(x)Pg (x, t), ∂t ∂Pg (x, t) = Lg Pg (x, t) − kr Pg (x, t) + k0 S(x)Pe (x, t). ∂t

Li = D

∂ dVi (x) ∂2 + . ∂x2 ∂x dx

(1)

(2)

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Here Vi (x) is the potential causing the drift of the particle on the i-th potential energy surface (for excited state potential energy surface i is e and for ground state potential energy surface i is g), the position dependent coupling term S(x) is assumed to be represented by a Dirac delta function, k0 is the rate of non-radiative decay and kr is the rate of radiative decay. D is the diffusion coefficient. Here decay in probability from excited state is represented by the term −k0 S(x)Pg (x, t) which incorporates the effect of ground state on excited state and +k0 S(x)Pe (x, t) term shows the increase in population is ground state due to decay of population in excited state. The molecule is initially on the ground state potential energy surface and as the solvent is at a temperature T , it’s distribution over the coordinate x can be assumed to be random. From this distribution it undergoes Franck-Condon excitation to the electronically excited potential energy surface. Therefore, x0 , the initial position of the particle on the excited stat potential energy surface is most likely Rto be random. We assume it to be given by Pe0 (x). So first we do Laplace transformation of Eq. ∞ (1), using Pi (x, s) = 0 Pi (x, t)e−st dt and Eq. (1) gives [s − Le + kr ]Pe (x, s) + k0 S(x)Pg (x, s) = Pe0 (x), [s − Lg + kr ]Pg (x, s) − k0 S(x)Pe (x, s) = 0,

(3)

Here we are assuming that all the molecules are in excited state at time t = 0+, just after electronic excitation, so initial probability distribution at the ground state is zero, i.e., Pg (x, 0) = 0. The initial probability distribution at

Journal Pre-proof 2 the electronically excited state is given by Pe0 (x0 ). The coupling between two states is taken as Dirac delta function at xc . Now Eq.(3) becomes [s − Le + kr ]Pe (x, s) + k0 δ(xc )Pg (xc , s) = Pe0 (x), [s − Lg + kr ]Pg (x, s) − k0 δ(xc )Pe (xc , s) = 0.

(4)

This equations can be re-written in the following form [s − Le + kr ]Pe (x, s) = Pe0 (x) − k0 δ(xc )Pg (xc , s), [s − Lg + kr ]Pg (x, s) = k0 δ(xc )Pe (xc , s).

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We can write the solution of above equations in terms of Green’s function Z ∞ Z ∞ Pe (x, s) = dx0 G0e (x, s; x0 )Pe0 (x0 ) − k0 Pg (xc , s) dx0 δ(xc )G0e (x, s; x0 ), −∞ −∞ Z ∞ dx0 δ(xc )G0g (x, s; x0 ), Pg (x, s) = k0 Pe (xc , s)

(5)

(6)

−∞

where



G0i (x, s; x0 ) = x [s − Li + kr ]−1 x0

(7)

−∞

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this Green’s function corresponds to the propagation of the particle placed initially at x0 under the influence of the uncoupled potential Vi (x). Now using the properties of Dirac delta function, the above equations can be simplified as Z ∞ Pe (x, s) = dx0 G0e (x, s; x0 )Pe0 (x0 ) − k0 Pg (xc , s)G0e (x, s; x0 ), (8) Pg (x, s) = k0 Pe (xc , s)G0g (x, s; xc ).

Now we can eliminate Pg (xc , s) from the top equation of Eq.(8) by substitution and we get Z ∞ Pe (x, s) = dx0 G0e (x, s; x0 )Pe0 (x0 ) − (k0 )2 Pe (xc , s)G0g (xc , s; xc )G0e (x, s; x0 ). −∞

(9)

Now the above equation has two unknowns, Pe (x, s) and Pe (xc , s), so we put x = xc , and so that we get an equation with one unknown only Z ∞ Pe (xc , s) = dx0 G0e (xc , s; x0 )Pe0 (x0 ) − (k0 )2 Pe (xc , s)G0g (xc , s; xc )G0e (xc , s; x0 ). (10)

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−∞

Now we solve the above equation for Pe (xc , s) to get R∞

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Pe (xc , s) =

−∞

dx0 G0e (xc , s; x0 )Pe0 (x0 )

1 + (k0 )2 G0g (xc , s; xc )G0e (xc , s; x0 )

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This solution, when substituted back into Eq. (9) we get " # Z ∞ k0 2 G0e (x, s; xc )G0g (xc , s; xc )G0e (xc , s; x0 ) 0 Pe (x, s) = dx0 Ge (x, s; x0 ) − Pe0 (x0 ). 1 + k0 2 G0e (xc , s; xc )G0g (xc , s; xc ) −∞

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Survival probability for excited state can be calculated by using the following formula Z ∞ Pe (t) = dxP e(x, t),

(11)

(12)

(13)

−∞

Now it is possible to calculate the Laplace Transform Pe (s) directly from Pe (x, t) and also Pe (s) can be calculated from Pe (x, s) by using the following formula Z ∞ Pe (s) = dxP˜ (x, s). (14) −∞

Journal Pre-proof 3 Now we write an explicit formula for Pe (s)   Z ∞ 1 Pe (s) = 1 − [1 + k0 2 G0g (xc , s; xc )G0e (xc , s + kr |xc )]−1 k0 2 G0g (xc , s; xc ) dx0 G0e (xc , s + kr |x0 )P (x0 , 0) . s + kr −∞ (15) In driving the above equation we have used the following property of Green’s function Z ∞ dx0 G0e (x, s|x0 ) = 1/s. (16) −∞

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Now we can calculate the average and long time rate constants by using the analytical expression of Pe (s). So kI−1 = Pe (0) and kL = negative of the pole of Pe (s), which is close to the origin. From Eq. (15), we get the rate constant kI as given below   Z ∞ 1 1 − [1 + k0 2 G0g (xc , s; xc )G0e (xc , kr |xc )]−1 k0 2 G0g (xc , s; xc ) dx0 G0e (xc , kr |x0 )P (x0 , 0) . (17) kI−1 = kr −∞ Here from the expression, it can be seen that kI depends on the initial probability distribution P (x, 0). The long
−1 time rate kL = − pole of [1 + k0 2 G0g (xc , s; xc )G0e (xc , s + kr |xc )][s + kr ] , on the negative s - axis and closest to the origin, which is independent of the initial distribution P (x0 , 0). The G0i (x, s; x0 ) can be found out by using the following equation [11]:
s − Li G0i (x, s; x0 ) = δ(x − x0 ). (18)

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Green’s function for this equation can be found using standard method [18] G0i (x, s; x0 ) = F (z, s; z0 )/(s + kr )

(19)

here for a parabolic potential Green’s function will be

2

F (z, s; z0 ) = Dν (−z< )Dν (z> )e(z0 −z

2

)/4

1/2

Γ(1 − ν)[B/(2πD)]

.

(20)

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In the above equation zj = xj (A/B)1/2 , ν = sB and Γ(ν) represents the gamma function. Also, z< = min(z, z0 ) and z> = max(z, z0 ). Dν are called parabolic cylinder functions. For further calculations, we will assume that the initial distribution Pe0 (x0 ) is represented by δ(x − x0 ). So, we get ! k0 2 G0g (xc , s; xc )F (zs , kr |z0 ) −1 −1 kI = (kr ) 1− . (21) kr + k0 2 G0g (xc , s; xc )F (zs , kr |zs ) Again

kL = kr − [values of s f or which s + k0 2 G0g (xc , s; xc )F (zs , s|zs ) = 0].

(22)

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Here, we assume kr → 0 and by using the properties of parabolic cylinder function Dv (z), we find that in this limit, F (zs , kr |z0 ) and F (zs , kr |zs ) → exp(−zs2 /2)/[B/(2πD)]1/2 , so that k0 2 G0g (xc , s; xc )F (zs , kr |z0 )/[kr + k0 2 G0g (xc , s; xc )F (zs , kr |zs )] → 1.

Hence

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kI−1

∂ =− ∂kr

"

k0 2 G0g (xc , s; xc )F (zs , kr |z0 )

kr + k0 2 G0g (xc , s; xc )F (zs , kr |zs )

#

(23)

.

(24)

kr →0

Now here we take z0 < zs , the initial position of the particle is at left side of sink. Then

kI−1

=e

zs 2 /2

1/2 /{k0 G0g (xc , s; xc )[B/(2πD)] } 2

"

∂ + ∂kr

"

2

2

e[(z0 −zs )/4] Dv (−z0 ) Dv (−zs )

##

. v=0

(25)

Journal Pre-proof 4 After simplification kI−1 = ezs

2

/2

1/2

/{k0 2 G0g (xc , s; xc )[B/(2πD)]

}+

Z

zs

dze(z

2

/2)

z0

h

 √ i 1 + erf (z/ 2 (π/2)1/2 /B.

(26)

The long-term rate constant kL is determined by the value of s, which satisfy s + k0 F (zs , s|zs ) = 0. This equation can be written as an equation for ν(= −s/B) 1/2

ν = Dν (−zc )Dν (zc )Γ(1 − ν)(2)2 k0 2 G0g (xc , s; xc )[B/(2πD)]

.

(27)

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√ 2 For integer values of ν, Dν (z) = 2−ν/2 e−z /4 Hν (z/ 2),where Hν are Hermite polynomials. Γ(1 − ν) has poles at 1/2 ν = 1, 2, ..... Our interest is in ν ∈ [0, 1], as kL = Bν for kr = 0. If k0 2 G0g (xc , s; xc )[B/(2πD)]  1, or zc  1 then ν  1 and one can arrive 1/2

and hence kL = e−zs

2

/2

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ν = D0 (−zs )D0 (zs )k0 2 G0g (xc , s; xc )[B/(2πD)]

1/2

k0 2 G0g (xc , s; xc )[B/(2πD)]

.

(28)

(29)

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J. T. Hynes, Ann. Rev. Phys. Chem. 36, 573 (1985). G. R. Fleming and P. G. Wolynes, Phys. Today 43(5), 36 (1990). G. W. Robinson and R. P. Forsch, J. Chem. Phys. 38, 1187 (1963). D. Ben-Amotz and C. B. Harris, J. Chem. Phys. 86, 4856 (1987). D. Ben-Amotz and C. B. Harris, J. Chem. Phys. 86, 5433 (1987). D. Ben-Amotz, R. Jeanloz, and C. B. Harris, J. Chem. Phys. 86, 6119 (1987). B. Bagchi and G. R. Fleming, J. Phys. Chem. 94, 9 (1989). A. Chakraborty , J. Chem. Phys. 139 , 094101 (2013) B. Bagchi and G. R. Fleming and D. W. Oxtoby, J. Chem. Phys. 78, 7375 (1983). B. Bagchi, J. Chem. Phys. 87, 5393 (1987). K. L. Sebastian, Phys. Rev. A 46, R1732 (1992). A. Patra, K.A. Acharya and A. Samanta, J. Phys. Chem. B, 119, 11063 (2015). A. Patra, A. Samanta and S.K. Ghosh, Phys. Rev. E, 83, 026104 (2011). K. Gupta, A. Patra, K.Dhole, A.K. Samanta and S.K. Ghosh, J. Phys. Chem. Lett.,8, 4545 (2017) R. A. Marcus and N. Sutin, Biochem. Biophys. Acta 811, 265 (1985). H. Sumi and R. A. Marcus, J. Chem. Phys. 84, 4894 (1986). J. T. Hynes, J. Phys. Chem. 90, 3701 (1986). R. Courant and D. Hilbert, Methods of Mathematical Physics (Wiley Eastern, New Delhi, 1975), Vol. 1, p. 351.

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[1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18]

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In this limit, the rate constant kL exhibits Arrhenius type activation. In this paper we have given a very simple method to solve two state model for electronic relaxation problem in solution. The same problem was solved buy us earlier by using a very complex method [8]. From the expression of kI and kL , we can see that probability distribution in ground state G0g (xc , s; xc ) will affect the dynamics of excited state. The solution is expressed in terms of Green’s function of both the uncoupled states, so our solution is quite general and applicable to any set of two potentials. Here we use two parabolic potentials as an example to illustrate our method of solution.

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Highlights

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Develop a very simple new analytical technique to solve two states with point coupling model of electronic relaxation problem. Exact result for survival probability; a generalized solution has been reported. Exact results for rate constants in which the contribution of ground state distribution can be seen.

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Diffusion-reaction approach to electronic relaxation in solution. An alternative simple derivation for two state model with a Dirac delta function coupling