Theory of the kinetics of luminescence and its temperature dependence for Ag nanoclusters
dispersed in glass host.
Mikhail V. Shestakova, Liviu F. Chibotarua,b, Victor K. Tikhomirova, Vicente D. Rodriguezc, Jose
J. Velasquezc, Victor V. Moshchalkova
a
Institute for Nanoscale Physics and Chemistry, Katholieke Universiteit Leuven, Leuven,
Belgium
b
Chemistry Department, Katholieke Universiteit Leuven, Leuven, Belgium
c
Departamento de Física Fundamental y Experimental, Electrónica y Sistemas, Universidad de
La Laguna, Tenerife, Spain
S1
Contents
Introduction ................................................................................................................................... S3
Part 1 ............................................................................................................................................. S3
Part 2 ............................................................................................................................................. S5
Part 3 ............................................................................................................................................. S6
Part 4 ............................................................................................................................................. S8
References ................................................................................................................................... S10
S2
Introduction
In this supporting information file we find a solution of the system of the first order linear
homogeneous differential equations:
{
̇ = −(Г12 + 𝑘1 )𝑛1 (𝑡) + Г21 𝑛2 (𝑡)
𝑛1 (𝑡)
̇ = Г12 𝑛1 (𝑡) − (Г21 + 𝑘2 )𝑛2 (𝑡)
𝑛2 (𝑡)
(S1)
where n1(t) and n2(t) are time-dependent functions of concentrations of electrons at S1 and T2
excited energy levels in agreement with energy level diagram of Ag nanocluster shown in Fig. 2
̇ and 𝑛2 (𝑡)
̇ are the first time derivatives of time-dependent
in the main body of paper, 𝑛1 (𝑡)
functions of concentrations of electrons at S1 and T2 excited energy levels, k1 and k2 are the rates
of spontaneous emission from S1 and T2 the excited states to S0 the ground state levels which are
functions of emission wavelength, Γ12 and Γ21 are temperature-dependent rate functions of the
intersystem crossing process from S1 to T2 excited levels and from T2 to S1 excited levels, t is
independent variable which stands for time.
According to Ref.1 a general solution of the system of linear homogeneous differential equations
(S1) can be written using the time-dependent function:
𝑛(𝑡) = {
𝑛1 (𝑡) = 𝑐1 𝑒 −𝜆1 𝑡 + 𝑐2 𝑒 −𝜆2 𝑡
𝑛2 (𝑡) = 𝑐3 𝑒 −𝜆1 𝑡 + 𝑐4 𝑒 −𝜆2 𝑡
(S2)
where c1, c2, c3 and c4 are arbitrary coefficients, λ1 and λ2 are reciprocals of lifetimes τ1 and τ2, t
is independent variable which stands for time. Putting t = 0 in the general solution (S2) leads to
the system of initial values equations for the system of differential Eqs. (S1):
𝑛 (0) = 𝑐1 + 𝑐2
{ 1
𝑛2 (0) = 𝑐3 + 𝑐4
(S3)
Part 1
In Part 1 we find the unknown arbitrary coefficients c1, c2, c3 and c4 for the general solution (S2).
Let us find these coefficients according to a procedure described in Ref.1 We start with
substituting n1 = 𝑐1 𝑒 −𝜆1𝑡 and n2 = 𝑐3 𝑒 −𝜆1𝑡 from the general solution (S2) into the system of Eqs.
(S1):
−𝜆1 𝑐1 𝑒 −𝜆1 𝑡 = −(Г12 + 𝑘1 )𝑐1 𝑒 −𝜆1 𝑡 + Г21 𝑐3 𝑒 −𝜆1 𝑡
−𝜆1 𝑐3 𝑒 −𝜆1 𝑡 = Г12 𝑐1 𝑒 −𝜆1 𝑡 − (Г21 + 𝑘2 )𝑐3 𝑒 −𝜆1 𝑡
We divide both sides of the system of Eqs. (S4) by 𝑒 −𝜆1𝑡 :
−𝜆 𝑐 = −(Г12 + 𝑘1 )𝑐1 + Г21 𝑐3
{ 1 1
−𝜆1 𝑐3 = Г12 𝑐1 − (Г21 + 𝑘2 )𝑐3
{
S3
(S4)
(S5)
Adding the first and the second rows of the system of Eqs. (S5), we find:
−𝜆1 𝑐1 − 𝜆1 𝑐3 = −(Г12 + 𝑘1 )𝑐1 + Г21 𝑐3 + Г12 𝑐1 − (Г21 + 𝑘2 )𝑐3
(S6)
Eq. (S6) gives:
𝜆1 𝑐1 + 𝜆1 𝑐3 = 𝑘1 𝑐1 + 𝑘2 𝑐3
(S7)
Using the Eq. (S6) we find the expressions for the coefficients c1 and c3:
(𝜆 −𝑘 )
𝑐1 = (𝑘1 −𝜆2 ) 𝑐3
1
(S8)
1
(𝑘 −𝜆 )
𝑐3 = (𝜆1−𝑘1 ) 𝑐1
1
(S9)
2
By the analogy, we substitute n1 = 𝑐2 𝑒 −𝜆2𝑡 and n2 = 𝑐4 𝑒 −𝜆2𝑡 from the general solution (S2) into the
system of Eqs. (S1) and apply the same procedure for finding of relations between the
coefficients, we have:
(𝜆 −𝑘2 )
𝑐4
1 −𝜆2 )
(𝑘 −𝜆 )
𝑐4 = (𝜆1−𝑘2 ) 𝑐2
2
2
𝑐2 = (𝑘2
(S10)
(S11)
Substituting (S9) and (S11) into the second row of the system of initial values Eqs. (S3), we find:
{
𝑐1 = 𝑛1 (0) − 𝑐2
(𝑘 −𝜆1 )
𝑐1
1 −𝑘2 )
𝑛2 (0) = (𝜆1
(S12)
(𝑘 −𝜆2 )
𝑐2
2 −𝑘2 )
+ (𝜆1
Replacing the coefficient c1 in the second row of the system of Eqs. (S12) by the first row of the
system of Eqs. (S12), we get:
𝑐1 = 𝑛1 (0) − 𝑐2
(𝑘1 −𝜆1 )
(𝑘 −𝜆 )
{
(𝑛1 (0) − 𝑐2 ) + (𝜆1 2 ) 𝑐2
𝑛2 (0) = (𝜆
)
1 −𝑘2
(S13)
2 −𝑘2
From the second row of the system of Eqs. (S13), we have:
(𝜆2 −𝑘2 )(𝜆1 −𝑘2 )
(𝑘 −𝜆 )(𝜆 −𝑘 )
(𝜆 −𝑘 )(𝑘 −𝜆 )
𝑛 (0) = (𝜆1−𝑘1 )(𝜆2 −𝑘2 ) (𝑛1 (0) − 𝑐2 ) + (𝜆1 −𝑘2 )(𝜆1−𝑘2 ) 𝑐2
(𝜆1 −𝑘2 )(𝜆2 −𝑘2 ) 2
1
2
2
2
1
2
2
2
(𝜆2 −𝑘2 )(𝜆1 −𝑘2 )
(𝑘1 −𝜆1 )(𝜆2 −𝑘2 )
(𝑘1 −𝜆1)(𝜆2 −𝑘2 )
(𝜆1 −𝑘2 )(𝑘1 −𝜆2 )
𝑛 (0) − (𝜆
𝑛1 (0) = − (𝜆
𝑐2 + (𝜆
𝑐2
(𝜆1 −𝑘2 )(𝜆2 −𝑘2 ) 2
1 −𝑘2 )(𝜆2 −𝑘2 )
1 −𝑘2 )(𝜆2 −𝑘2 )
1 −𝑘2 )(𝜆2 −𝑘2 )
(S14)
(S15)
(𝜆2 − 𝑘2 )(𝜆1 − 𝑘2 )𝑛2 (0) + (𝜆1 − 𝑘1 )(𝜆2 − 𝑘2 )𝑛1 (0) = (𝜆1 𝑘1 + 𝜆2 𝑘2 − 𝜆2 𝑘1 − 𝜆1 𝑘2 )𝑐2 (S16)
Let us define c2 as equal to the left side of (S16):
𝑐2 =
(𝜆2 −𝑘2 )(𝜆1 −𝑘2 )𝑛2 (0)+(𝜆1 −𝑘1 )(𝜆2 −𝑘2 )𝑛1 (0)
(𝜆1 −𝑘1 )(𝜆2 −𝑘2 )
(𝜆1 −𝜆2 )(𝑘1 −𝑘2 )
(𝑘2 −𝜆2 )(𝜆1 −𝑘2 )
𝑛 (0)
(𝜆1 −𝜆2 )(𝑘1 −𝑘2 ) 1
− (𝜆
1 −𝜆2 )(𝑘1 −𝑘2 )
(𝜆 −𝑘 )(𝜆 −𝑘 )
(𝜆 −𝑘 )(𝜆 −𝑘 )
= (𝜆1 −𝜆1)(𝑘2 −𝑘2) 𝑛1 (0) + (𝜆2 −𝜆2)(𝑘1 −𝑘2 ) 𝑛2 (0) =
1
𝑛2 (0)
2
1
2
1
2
1
2
(S17)
Now we find the coefficient c1. Substitution Eq. (S17) into the first row of the system of Eqs.
(S13) gives:
S4
(𝜆 −𝜆 )(𝑘 −𝑘 )
(𝜆 −𝑘 )(𝜆 −𝑘 )
(𝜆 −𝑘 )(𝜆 −𝑘 )
𝑐1 = (𝜆1 −𝜆2 )(𝑘1 −𝑘2 ) 𝑛1 (0) − (𝜆1 −𝜆1)(𝑘2 −𝑘2) 𝑛1 (0) − (𝜆2 −𝜆2)(𝑘1 −𝑘2) 𝑛2 (0) =
1
2
1
2
1
(𝜆1 𝑘1 −𝜆1 𝑘2 −𝜆2 𝑘1 +𝜆2 𝑘2 )
𝑛1 (0)
(𝜆1 −𝜆2 )(𝑘1 −𝑘2 )
(𝜆1 𝑘1 −𝑘1 𝑘2 −𝜆1 𝜆2 +𝜆2 𝑘2 )
𝑛1 (0)
(𝜆1 −𝜆2 )(𝑘1 −𝑘2 )
−
−
2
1
2
1
(𝜆1 𝜆2 −𝜆1 𝑘2 −𝜆2 𝑘1 +𝑘1 𝑘2 )
(𝜆1 −𝜆2 )(𝑘1 −𝑘2 )
(𝜆2 −𝑘2 )(𝜆1 −𝑘2 )
𝑛 (0)
(𝜆1 −𝜆2 )(𝑘1 −𝑘2 ) 2
2
1
2
(𝜆2 −𝑘2 )(𝜆1 −𝑘2 )
𝑛1 (0) − (𝜆
1 −𝜆2 )(𝑘1 −𝑘2 )
(𝜆 −𝑘 )(𝑘 −𝜆 )
𝑛2 (0) =
(𝜆 −𝑘 )(𝜆 −𝑘 )
= (𝜆1 −𝜆2)(𝑘1−𝑘2 ) 𝑛1 (0) − (𝜆2 −𝜆2)(𝑘1 −𝑘2 ) 𝑛2 (0)
1
2
1
2
1
2
1
2
(S18)
Now we find the coefficients c3 and c4. Replacing the coefficient c1 in the Eq. (S9) by the Eq.
(S18), we find:
(𝑘 −𝜆1 )
𝑐1
1 −𝑘2 )
𝑐3 = (𝜆1
(𝑘 −𝜆1 ) (𝜆1 −𝑘2 )(𝑘1 −𝜆2 )
(
𝑛1 (0)
1 −𝑘2 ) (𝜆1 −𝜆2 )(𝑘1 −𝑘2 )
= (𝜆1
(𝜆 −𝑘2 )(𝜆1 −𝑘2 )
𝑛2 (0))
1 −𝜆2 )(𝑘1 −𝑘2 )
− (𝜆2
(𝑘 −𝜆1 )(𝑘1 −𝜆2 )
(𝜆 −𝑘 )(𝑘 −𝜆 )
𝑛1 (0) − (𝜆2 2)(𝑘1 1 ) 𝑛2 (0)
1 −𝜆2 )(𝑘1 −𝑘2 )
1 −𝜆2
1 −𝑘2
= (𝜆1
(S19)
Replacing the coefficient c2 in the Eq. (S11) by the Eq. (S17), we find:
(k −λ2 )
c2
2 −k2 )
c4 = (λ1
(k −λ2 ) (𝜆1 −𝑘1 )(𝜆2 −𝑘2 )
(𝜆 −𝑘 )(𝜆 −𝑘 )
(
𝑛1 (0) + (𝜆2 2)(𝑘1 2) 𝑛2 (0))
2 −k2 ) (𝜆1 −𝜆2 )(𝑘1 −𝑘2 )
1 −𝜆2
1 −𝑘2
= (λ1
(𝜆 −𝑘1 )(k1 −λ2 )
(λ −k )(𝜆 −𝑘 )
𝑛1 (0) − (𝜆2 1)(𝑘1 2 ) 𝑛2 (0)
1 −𝜆2 )(𝑘1 −𝑘2 )
1 −𝜆2
1 −𝑘2
= (𝜆1
(S20)
Substitution the Eqs. (S17), (S18), (S19) and (S20) into the general solution (S2) gives the final
solution of the system of differential Eqs. (S1) is:
(𝜆 −𝑘2 )(𝑘1 −𝜆2 )
𝑛1 (0)
1 −𝜆2 )(𝑘1 −𝑘2 )
𝑛1 = ((𝜆1
(𝑘1−𝜆1 )(𝑘1 −𝜆2 )
𝑛2 = ((
𝜆1 −𝜆2 )(𝑘1 −𝑘2 )
(𝜆 −𝑘2 )(𝜆1 −𝑘2 )
𝑛2 (0)) 𝑒 −𝜆1𝑡
1 −𝜆2 )(𝑘1 −𝑘2 )
− (𝜆2
𝑛 (0) −
1
(𝜆2 −𝑘2 )(𝑘1−𝜆1)
𝑛 (0)) 𝑒
(𝜆1 −𝜆2 )(𝑘1 −𝑘2) 2
(𝜆 −𝑘1 )(𝜆2 −𝑘2 )
(𝑘 −𝜆 )(𝜆 −𝑘 )
𝑛1 (0) − (𝜆2 2)(𝑘1 2) 𝑛2 (0)
1 −𝜆2 )(𝑘1 −𝑘2 )
1 −𝜆2
1 −𝑘2
+ ((𝜆1
−𝜆1 𝑡
(𝜆1−𝑘1 )(k1 −λ2)
+ ((
𝜆1 −𝜆2 )(𝑘1 −𝑘2 )
𝑛1 (0) −
) 𝑒 −𝜆2 𝑡
(S21)
(λ2 −k1)(𝜆1−𝑘2)
𝑛 (0)) 𝑒 −𝜆2𝑡
(𝜆1−𝜆2)(𝑘1−𝑘2 ) 2
(S22)
Part 2
In Part 2 we check out the solution (S21) and (S22) for the system of differential Eqs. (S1).
Addition the first and the second rows of the system of differential Eqs. (S1) gives:
𝑛̇ 1 (𝑡) + 𝑛̇ 2 (𝑡) = −𝑘1 𝑛1 (𝑡) − 𝑘2 𝑛2 (𝑡)
(S23)
We replace n1(t) and n2(t) in the differential Eq. (S23) by the Eqs. (S21) and (S22):
(𝜆 −𝑘2 )(𝑘1 −𝜆2 )
(𝜆 −𝑘 )(𝜆 −𝑘 )
𝑛1 (0) − (𝜆2 2)(𝑘1 2) 𝑛2 (0)) 𝑒 −𝜆1 𝑡
1 −𝜆2 )(𝑘1 −𝑘2 )
1 −𝜆2
1 −𝑘2
− 𝜆2 ((𝜆1
(𝑘 −𝜆1 )(𝑘1 −𝜆2 )
(𝜆 −𝑘 )(𝑘 −𝜆 )
𝑛1 (0) − (𝜆2 2)(𝑘1 1 ) 𝑛2 (0)) 𝑒 −𝜆1 𝑡
1 −𝜆2 )(𝑘1 −𝑘2 )
1 −𝜆2
1 −𝑘2
− 𝜆2 ((𝜆1
−𝜆1 ((𝜆1
−𝜆1 ((𝜆1
(𝜆 −𝑘2 )(𝑘1 −𝜆2 )
𝑛1 (0)
1 −𝜆2 )(𝑘1 −𝑘2 )
(𝜆 −𝑘1 )(𝜆2 −𝑘2 )
(𝑘 −𝜆 )(𝜆 −𝑘 )
𝑛1 (0) − (𝜆2 2)(𝑘1 2) 𝑛2 (0)
1 −𝜆2 )(𝑘1 −𝑘2 )
1 −𝜆2
1 −𝑘2
(𝜆 −𝑘1 )(k1 −λ2 )
𝑛1 (0)
1 −𝜆2 )(𝑘1 −𝑘2 )
(𝜆 −𝑘2 )(𝜆1 −𝑘2 )
𝑛2 (0)) 𝑒 −𝜆1𝑡
1 −𝜆2 )(𝑘1 −𝑘2 )
+ ((𝜆1
(𝑘 −𝜆1 )(𝑘1 −𝜆2 )
(𝜆 −𝑘 )(𝑘 −𝜆 )
𝑛1 (0) − (𝜆2 2)(𝑘1 1 ) 𝑛2 (0)) 𝑒 −𝜆1 𝑡
1 −𝜆2 )(𝑘1 −𝑘2 )
1 −𝜆2
1 −𝑘2
+ ((𝜆1
−𝑘1 (((𝜆1
−𝑘2 (((𝜆1
− (𝜆2
) 𝑒 −𝜆2𝑡 −
(λ −k1 )(𝜆1 −𝑘2 )
𝑛2 (0)) 𝑒 −𝜆2𝑡
1 −𝜆2 )(𝑘1 −𝑘2 )
− (𝜆2
(𝜆 −𝑘1 )(𝜆2 −𝑘2 )
𝑛1 (0)
1 −𝜆2 )(𝑘1 −𝑘2 )
− (𝜆2
(𝑘 −𝜆2 )(𝜆1 −𝑘2 )
𝑛2 (0)
1 −𝜆2 )(𝑘1 −𝑘2 )
(𝜆 −𝑘1 )(k1 −λ2 )
𝑛1 (0)
1 −𝜆2 )(𝑘1 −𝑘2 )
− (𝜆2
=
) 𝑒 −𝜆2 𝑡 ) −
(λ −k1 )(𝜆1 −𝑘2 )
𝑛2 (0)) 𝑒 −𝜆2 𝑡 )
1 −𝜆2 )(𝑘1 −𝑘2 )
(S24)
The Eq. (S24) is to be true for all values of t, then the coefficients of 𝑒 −𝜆1 𝑡 and 𝑒 −𝜆2 𝑡 must
vanish. Now we consider the equation for the coefficients of 𝑒 −𝜆1 𝑡 which follows from the Eq.
(S24):
S5
(𝜆 −𝑘2 )(𝑘1 −𝜆2 )
(𝜆 −𝑘 )(𝜆 −𝑘 )
𝑛1 (0) − (𝜆2 2)(𝑘1 2) 𝑛2 (0)) 𝑒 −𝜆1 𝑡
1 −𝜆2 )(𝑘1 −𝑘2 )
1 −𝜆2
1 −𝑘2
− 𝜆1 ((𝜆1
(𝑘 −𝜆1 )(𝑘1 −𝜆2 )
(𝜆 −𝑘 )(𝑘 −𝜆 )
𝑛1 (0) − (𝜆2 2)(𝑘1 1 ) 𝑛2 (0)) 𝑒 −𝜆1 𝑡
1 −𝜆2 )(𝑘1 −𝑘2 )
1 −𝜆2
1 −𝑘2
(𝜆 −𝑘2 )(𝑘1 −𝜆2 )
𝑛1 (0)
1 −𝜆2 )(𝑘1 −𝑘2 )
− 𝑘2 ((𝜆1
−𝜆1 ((𝜆1
−𝑘1 ((𝜆1
(𝜆 −𝑘2 )(𝜆1 −𝑘2 )
𝑛2 (0)) 𝑒 −𝜆1𝑡
1 −𝜆2 )(𝑘1 −𝑘2 )
(𝑘 −𝜆1 )(𝑘1 −𝜆2 )
𝑛1 (0)
1 −𝜆2 )(𝑘1 −𝑘2 )
− (𝜆2
=
(𝜆 −𝑘2 )(𝑘1 −𝜆1 )
𝑛2 (0)) 𝑒 −𝜆1𝑡
1 −𝜆2 )(𝑘1 −𝑘2 )
(S25)
− (𝜆2
Division of the Eq. (S25) by 𝑒 −𝜆1 𝑡 gives:
(𝜆 −𝑘2 )(𝑘1 −𝜆2 )
(𝜆 −𝑘 )(𝜆 −𝑘 )
𝑛1 (0) − (𝜆2 2)(𝑘1 2) 𝑛2 (0))
1 −𝜆2 )(𝑘1 −𝑘2 )
1 −𝜆2
1 −𝑘2
−𝜆1 ((𝜆1
(𝜆 −𝑘2 )(𝑘1 −𝜆2 )
𝑛1 (0)
1 −𝜆2 )(𝑘1 −𝑘2 )
−𝑘1 ((𝜆1
(𝜆 −𝑘2 )(𝜆1 −𝑘2 )
𝑛2 (0)) −
1 −𝜆2 )(𝑘1 −𝑘2 )
− (𝜆2
(𝑘 −𝜆1 )(𝑘1 −𝜆2 )
𝑛1 (0)
1 −𝜆2 )(𝑘1 −𝑘2 )
− (𝜆2
(𝜆 −𝑘2 )(𝑘1 −𝜆1 )
𝑛2 (0))
1 −𝜆2 )(𝑘1 −𝑘2 )
(𝑘 −𝜆1 )(𝑘1 −𝜆2 )
𝑛1 (0)
1 −𝜆2 )(𝑘1 −𝑘2 )
− (𝜆2
− 𝜆1 ((𝜆1
𝑘2 ((𝜆1
=
(𝜆 −𝑘2 )(𝑘1 −𝜆1 )
𝑛2 (0))
1 −𝜆2 )(𝑘1 −𝑘2 )
(S26)
Eq. (S26) gives:
(𝜆 −𝑘2 )(𝑘1 −𝜆2 )
1 −𝜆2 )(𝑘1 −𝑘2 )
(𝜆2 −𝑘2 )(𝜆1 −𝑘2 )
(−𝜆1 (𝜆1
𝑘1 (𝜆
1 −𝜆2 )(𝑘1 −𝑘2 )
(𝑘 −𝜆1 )(𝑘1 −𝜆2 )
1 −𝜆2 )(𝑘1 −𝑘2 )
(𝜆2 −𝑘2 )(𝑘1 −𝜆1 )
− 𝜆1 (𝜆1
(𝜆 −𝑘2 )(𝑘1 −𝜆2 )
1 −𝜆2 )(𝑘1 −𝑘2 )
+ 𝑘1 (𝜆1
(𝑘 −𝜆1 )(𝑘1 −𝜆2 )
) 𝑛1 (0)
1 −𝜆2 )(𝑘1 −𝑘2 )
+ 𝑘2 (𝜆1
(𝜆 −𝑘2 )(𝜆1 −𝑘2 )
1 −𝜆2 )(𝑘1 −𝑘2 )
= (−𝜆1 (𝜆2
(𝜆 −𝑘2 )(𝑘1 −𝜆1 )
1 −𝜆2 )(𝑘1 −𝑘2 )
− 𝜆1 (𝜆2
(S27)
) 𝑛2 (0)
+ 𝑘2 (𝜆
+
1 −𝜆2 )(𝑘1 −𝑘2 )
Eq. (S27) becomes:
(
(𝜆1 −𝑘2 )(𝑘1 −𝜆2 )(𝑘1 −𝜆1 )
(𝜆1 −𝜆2 )(𝑘1 −𝑘2 )
−
(𝜆1 −𝑘2 )(𝑘1 −𝜆2 )(𝑘1 −𝜆1 )
(𝜆2 −𝑘2 )(𝜆1 −𝑘2 )(𝑘1 −𝜆1 )
(𝜆1 −𝜆2 )(𝑘1 −𝑘2 )
(𝜆1 −𝜆2 )(𝑘1 −𝑘2 )
) 𝑛1 (0) = (
−
(𝜆2 −𝑘2 )(𝜆1 −𝑘2 )(𝑘1 −𝜆1 )
(S28)
) 𝑛2 (0)
(𝜆1 −𝜆2 )(𝑘1 −𝑘2 )
Now we consider the equation for the coefficients of 𝑒 −𝜆2 𝑡 which follows from the Eq. (S28):
(𝜆 −𝑘1 )(𝜆2 −𝑘2 )
𝑛1 (0)
1 −𝜆2 )(𝑘1 −𝑘2 )
− (𝜆2
(𝜆 −𝑘1 )(𝜆2 −𝑘2 )
𝑛1 (0)
1 −𝜆2 )(𝑘1 −𝑘2 )
− (𝜆2
−𝜆2 ((𝜆1
−𝑘1 ((𝜆1
(𝑘 −𝜆2 )(𝜆1 −𝑘2 )
𝑛2 (0)
1 −𝜆2 )(𝑘1 −𝑘2 )
) 𝑒 −𝜆2 𝑡 − 𝜆2 ((𝜆1
(𝜆 −𝑘1 )(k1 −λ2 )
(λ −k )(𝜆 −𝑘 )
𝑛1 (0) − (𝜆2 1)(𝑘1 2) 𝑛2 (0)) 𝑒 −𝜆2 𝑡
1 −𝜆2 )(𝑘1 −𝑘2 )
1 −𝜆2
1 −𝑘2
(𝑘 −𝜆2 )(𝜆1 −𝑘2 )
𝑛2 (0)
1 −𝜆2 )(𝑘1 −𝑘2 )
) 𝑒 −𝜆2𝑡 − 𝑘2 ((𝜆1
(𝜆 −𝑘1 )(k1 −λ2 )
𝑛1 (0)
1 −𝜆2 )(𝑘1 −𝑘2 )
=
(λ −k1 )(𝜆1 −𝑘2 )
𝑛2 (0)) 𝑒 −𝜆2𝑡
1 −𝜆2 )(𝑘1 −𝑘2 )
− (𝜆2
(S29)
Division of the Eq. (S29) by 𝑒 −𝜆2 𝑡 gives:
(𝜆 −𝑘1 )(𝜆2 −𝑘2 )
𝑛1 (0)
1 −𝜆2 )(𝑘1 −𝑘2 )
− (𝜆2
(𝜆 −𝑘1 )(𝜆2 −𝑘2 )
𝑛1 (0)
1 −𝜆2 )(𝑘1 −𝑘2 )
− (𝜆2
−𝜆2 ((𝜆1
−𝑘1 ((𝜆1
(𝑘 −𝜆2 )(𝜆1 −𝑘2 )
𝑛2 (0)
1 −𝜆2 )(𝑘1 −𝑘2 )
) − 𝜆2 ((𝜆1
(𝜆 −𝑘1 )(k1 −λ2 )
(λ −k )(𝜆 −𝑘 )
𝑛1 (0) − (𝜆2 1)(𝑘1 2 ) 𝑛2 (0))
1 −𝜆2 )(𝑘1 −𝑘2 )
1 −𝜆2
1 −𝑘2
(𝑘 −𝜆2 )(𝜆1 −𝑘2 )
𝑛2 (0)
1 −𝜆2 )(𝑘1 −𝑘2 )
) − 𝑘2 ((𝜆1
(𝜆 −𝑘1 )(k1 −λ2 )
𝑛1 (0)
1 −𝜆2 )(𝑘1 −𝑘2 )
=
(λ −k1 )(𝜆1 −𝑘2 )
𝑛2 (0))
1 −𝜆2 )(𝑘1 −𝑘2 )
(S30)
− (𝜆2
Eq. (S30) becomes:
(𝜆 −𝑘1 )(𝜆2 −𝑘2 )
1 −𝜆2 )(𝑘1 −𝑘2 )
(𝑘2 −𝜆2 )(𝜆1 −𝑘2 )
(−𝜆2 (𝜆1
𝑘1 (𝜆
1 −𝜆2 )(𝑘1 −𝑘2 )
(𝜆 −𝑘1 )(k1 −λ2 )
1 −𝜆2 )(𝑘1 −𝑘2 )
(λ2 −k1 )(𝜆1 −𝑘2 )
− 𝜆2 (𝜆1
+ 𝑘2 (𝜆
(𝜆 −𝑘1 )(𝜆2 −𝑘2 )
1 −𝜆2 )(𝑘1 −𝑘2 )
+ 𝑘1 (𝜆1
(𝜆 −𝑘1 )(k1 −λ2 )
) 𝑛1 (0)
1 −𝜆2 )(𝑘1 −𝑘2 )
+ 𝑘2 (𝜆1
(𝑘 −𝜆2 )(𝜆1 −𝑘2 )
1 −𝜆2 )(𝑘1 −𝑘2 )
= (−𝜆2 (𝜆2
(λ −k1 )(𝜆1 −𝑘2 )
1 −𝜆2 )(𝑘1 −𝑘2 )
− 𝜆2 (𝜆2
(S31)
) 𝑛2 (0)
1 −𝜆2 )(𝑘1 −𝑘2 )
Eq. (S31) gives:
(
(𝜆1 −𝑘1 )(𝑘1 −𝜆2 )(𝜆2 −𝑘2 )
(𝜆1 −𝜆2 )(𝑘1 −𝑘2 )
−
(𝜆1 −𝑘1 )(k1 −λ2 )(𝜆2 −𝑘2 )
(𝑘2 −𝜆2 )(𝜆1 −𝑘2 )(𝑘1 −𝜆2 )
(𝜆1 −𝜆2 )(𝑘1 −𝑘2 )
(𝜆1 −𝜆2 )(𝑘1 −𝑘2 )
) 𝑛1 (0) = (
−
(𝑘2 −𝜆2 )(𝜆1 −𝑘2 )(𝑘1 −𝜆2 )
(𝜆1 −𝜆2 )(𝑘1 −𝑘2 )
) 𝑛2 (0)
(S32)
Part 3
In Part 3 we find the expressions for Г12 and Г21. We replace c3 in the first row and c1 in the
second row of the system of Eqs. (S5) by (S9) and (S8):
S6
+
(𝑘 −𝜆 )
{
−𝜆1 𝑐1 = −(Г12 + 𝑘1 )𝑐1 + Г21 (𝜆1−𝑘1 ) 𝑐1
1
−𝜆1 𝑐3 =
(𝜆 −𝑘 )
Г12 (𝑘1 −𝜆2 ) 𝑐3
1
1
2
− (Г21 + 𝑘2 )𝑐3
(S33)
We define λ1 as the left side of the first and second rows of the system of Eqs. (S33)
(𝑘 −𝜆 )
{
−𝜆1 = −(Г12 + 𝑘1 ) + Г21 (𝜆1−𝑘1 )
1
−𝜆1 =
(𝜆 −𝑘 )
Г12 (𝑘1 −𝜆2 ) −
1
1
2
(Г21 + 𝑘2 )
(S34)
The first and the second rows of the system of Eqs. (S34) are equivalent to each other.
Subsequently, we have:
(𝑘 −𝜆 )
(𝜆 −𝑘 )
−(Г12 + 𝑘1 ) + Г21 (𝜆1−𝑘1 ) = Г12 (𝑘1 −𝜆2 ) − (Г21 + 𝑘2 )
1
2
1
1
(S35)
Eq. (S35) gives:
(𝑘 −𝑘2 )
1 −𝜆1 )
Г12 (𝑘1
(𝑘 −𝑘2 )
−
1 −𝑘2 )
= Г21 (𝜆1
(𝑘1 − 𝑘2 )
(S36)
Dividing the Eq. (S36) by (k1-k2), we find:
Г12 = (𝑘1 − 𝜆1 ) (Г21 (𝜆
1
1 −𝑘2 )
− 1)
(S37)
We substitute the n1 = 𝑐2 𝑒 −𝜆2𝑡 and n2 = 𝑐4 𝑒 −𝜆2𝑡 from the general solution (S2) into the system of
Eqs. (S1):
−𝜆 𝑐 = −(Г12 + 𝑘1 )𝑐2 + Г21 𝑐4
{ 2 2
−𝜆2 𝑐4 = Г12 𝑐2 − (Г21 + 𝑘2 )𝑐4
(S38)
By the analogy, we replace c4 in the first row and c2 in the second row of the system of Eqs.
(S38) by (S10) and (S11):
{
(𝑘 −𝜆2 )
𝑐2
2 −𝑘2 )
−𝜆2 𝑐2 = −(Г12 + 𝑘1 )𝑐2 + Г21 (𝜆1
(𝜆 −𝑘 )
−𝜆2 𝑐4 = Г12 (𝑘2 −𝜆2 ) 𝑐4 − (Г21 + 𝑘2 )𝑐4
1
(S39)
2
We define λ2 as the left side of the first and second rows of the system of Eqs. (S39):
(𝑘 −𝜆 )
{
−𝜆2 = −(Г12 + 𝑘1 ) + Г21 (𝜆1−𝑘2 )
2
−𝜆2 =
(𝜆 −𝑘 )
Г12 (𝑘2 −𝜆2 ) −
1
2
2
(Г21 + 𝑘2 )
(S40)
The first and the second rows of the system of Eqs. (S40) are equivalent to each other.
Subsequently, we have:
S7
(𝑘 −𝜆 )
(𝜆 −𝑘 )
−(Г12 + 𝑘1 ) + Г21 (𝜆1−𝑘2 ) = Г12 (𝑘2 −𝜆2 ) − (Г21 + 𝑘2 )
2
2
1
(S41)
2
Eq. (S41) gives:
(𝑘 −𝑘 )
(𝑘 −𝑘 )
Г12 ((𝑘1 −𝜆2)) = Г21 (𝜆1−𝑘2) − (𝑘1 − 𝑘2 )
1
2
2
(S42)
2
Dividing the Eq. (S42) by (k1-k2), we get:
Г12 = (𝑘1 − 𝜆2 ) (Г21 (𝜆
1
2 −𝑘2 )
− 1)
(S43)
The Eqs. (S37) and (S43) are equivalent. Subsequently, we have:
(𝑘1 − 𝜆1 ) (Г21 (𝜆
1
− 1)
−𝑘
1
2)
= (𝑘1 − 𝜆2 ) (Г21 (𝜆
1
−
−𝑘
2
2)
1)
(S44)
Eq. (S44) gives:
(Г21 (𝜆
1
−
−𝑘
1
2)
(𝑘 −𝜆 )
1) = (𝑘1 −𝜆2 ) (Г21 (𝜆
1
1)
(S45)
(𝑘1 −𝜆1 )−(𝑘1 −𝜆2 )
(𝑘1 −𝜆1 )
(S46)
1
(𝑘1 −𝜆1 )(𝜆2 −𝑘2 )−(𝑘1−𝜆2 )(𝜆1 −𝑘2 )
)
(𝑘1 −𝜆1 )(𝜆2 −𝑘2 )(𝜆1 −𝑘2 )
Г21 (
=
(𝑘1 𝜆2 −𝑘1 𝑘2 −𝜆1 𝜆2 +𝜆1 𝑘2 )−(𝑘1 𝜆1 −𝑘1 𝑘2 −𝜆1 𝜆2 +𝜆2 𝑘2 )
)
(𝑘1 −𝜆1 )(𝜆2 −𝑘2 )(𝜆1 −𝑘2 )
Г21 (
Г21 ((𝑘
(𝑘1 −𝑘2 )(𝜆2 −𝜆1 )
1
−
−𝑘
2
2)
(𝜆 −𝜆 )
= (𝑘2 −𝜆1 )
1
(S47)
1
(𝜆 −𝜆 )
1 −𝜆1 )(𝜆2 −𝑘2 )(𝜆1 −𝑘2 )
) = (𝑘2 −𝜆1 )
1
(S48)
1
We define Γ21 as the left side of Eq. (S48):
Г21 =
(𝜆2 −𝑘2 )(𝜆1 −𝑘2 )
(S49)
(𝑘1 −𝑘2 )
Substituting Eq. (S49) into Eq. (S37), we find Γ12:
(𝜆 −𝑘 )(𝜆 −𝑘 )
1
1
− 1) = (𝑘1 − 𝜆1 ) ( 2 (𝑘2 −𝑘1) 2 (𝜆 −𝑘 ) −
)
−𝑘
1
2
1
2
1
2
(𝜆2 −𝑘2 )−(𝑘1−𝑘2 )
(𝜆2 −𝑘1 )(𝑘1 −𝜆1 )
)
𝜆1 (
)=
(𝑘1 −𝑘2 )
(𝑘1 −𝑘2 )
Г12 = (𝑘1 − 𝜆1 ) (Г21 (𝜆
(𝑘1 −
(𝜆 −𝑘 )
1) = (𝑘1 − 𝜆1 ) ((𝑘2 −𝑘2 ) − 1) =
1
2
(S50)
Part 4
In part 4 we find expressions for λ1 and λ2. We define λ2 as the left side of Eq. (S49):
(𝑘 −𝑘 )
𝜆2 = Г21 (𝜆1−𝑘2) + 𝑘2
1
(S51)
2
Replacing λ2 in Eq. (S50) by Eq. (S51) gives:
(𝜆2 −𝑘1 )(𝑘1 −𝜆1 )
Г12 = (
(𝑘1 −𝑘2 )
)=(
(𝑘 −𝑘2 )
+𝑘2 −𝑘1 )(𝑘1 −𝜆1 )
1 −𝑘2 )
(Г21 (𝜆1
(𝑘1 −𝑘2 )
S8
) = (Г21 (𝜆
1
1 −𝑘2 )
− 1) (𝑘1 − 𝜆1 )
(S52)
Eq. (S52) gives:
Г12 (𝜆1 − 𝑘2 ) = Г21 (𝑘1 − 𝜆1 ) − (𝑘1 − 𝜆1 )(𝜆1 − 𝑘2 )
(S53)
𝜆1 2 − (𝑘1 + 𝑘2 + Г12 + Г21 )𝜆1 + 𝑘1 𝑘2 + Г21 𝑘1 + Г12 𝑘2 = 0
(S54)
Eq. (S53) becomes:
Now we define λ1 as the left part of Eq. (S49):
(𝑘 −𝑘 )
𝜆1 = Г21 (𝜆1−𝑘2) + 𝑘2
2
(S55)
2
Substitution Eq. (S55) into Eq. (S50) gives:
Г12 = (
(𝑘 −𝑘 )
(𝜆2 −𝑘1 )(𝑘1 −Г21 1 2 −𝑘2 )
(𝜆2 −𝑘2 )
(𝑘1 −𝑘2 )
(𝜆 −𝑘 )
) = ((𝜆2 − 𝑘1 ) − Г21 (𝜆2 −𝑘1 )) =
2
2
(𝜆2 −𝑘1 )(𝜆2 −𝑘2 )−Г21 (𝜆2 −𝑘1 )
(𝜆2 −𝑘2 )
𝜆22 −𝑘1 𝜆2 −𝑘1 𝜆2 +𝑘1 𝑘2 −Г21 (𝜆2 −𝑘1 )
(𝜆2 −𝑘2 )
=
(S56)
Eq. (S56) becomes:
𝜆22 − (𝑘1 + 𝑘2 + Г21 + Г12 )𝜆2 + 𝑘1 𝑘2 + Г21 𝑘1 + Г12 𝑘2 = 0
(S57)
Eqs. (S54) and (S57) are equivalent. Subsequently, solutions for these equations are the same. A
general form of these equations can be written as:
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0
(S58)
A solution of Eq. (S58) can be found using a well-known formula:
𝑥=
−𝑏±√𝐷
(S59)
2𝑎
where 𝐷 = √𝑏 2 − 4𝑎𝑐
Now we find solutions of Eqs. (S54) and (S57) using the Eq. (S59):
𝐷 = (Г12 + Г21 + 𝑘1 + 𝑘2 )2 − 4(Г12 𝑘2 + Г21 𝑘1 +𝑘1 𝑘2 ) = (Г12 + Г21 )2 + (𝑘1 + 𝑘2 )2 + 2(𝑘1 +
𝑘2 )(Г12 + Г21 ) − 4(Г12 𝑘2 + Г21 𝑘1 +𝑘1 𝑘2 ) = (Г12 + Г21 )2 + (𝑘1 + 𝑘2 )2 + 2(𝑘1 Г12 + 𝑘1 Г21 +
𝑘2 Г12 + 𝑘2 Г21 ) − 4(Г12 𝑘2 + Г21 𝑘1 +𝑘1 𝑘2 ) = (Г12 + Г21 )2 + (𝑘1 − 𝑘2 )2 + 2(𝑘1 Г12 − 𝑘1 Г21 −
𝑘2 Г12 + 𝑘2 Г21 ) = (Г12 + Г21 )2 + (𝑘1 − 𝑘2 )2 + 2(𝑘1 − 𝑘2 )(Г12 − Г21 )
(S59)
𝜆1,2 =
(Г12 +Г21 +𝑘1 +𝑘2 )±√(Г12 +Г21 )2 +(𝑘1 −𝑘2 )2 +2(𝑘1 −𝑘2 )(Г12 −Г21 )
2
(S60)
Taking into account that λ1 and λ2 are reciprocals of lifetimes τ1 and τ2, we find:
𝜏1,2 =
1
𝜆1,2
=
2
(Г12 +Г21 +𝑘1 +𝑘2 )±√(Г12 +Г21 )2 +(𝑘1 −𝑘2 )2 +2(𝑘1 −𝑘2 )(Г12 −Г21 )
S9
(S61)
References
(1) Adkins, W. A.; Davidson, M. G. Ordinary differential equations; Springer: New York,
2012.
S10