The model of a level crossing with a Coulomb band: exact probabilities of nonadiabatic transitions

0.57 MB·

by  J Lin

The model of a level crossing with a Coulomb band:
exact probabilities of nonadiabatic transitions
4 J Lin1,2 and N A Sinitsyn2
1
1Department of Mathematics, Princeton University, Princeton, NJ 08544, USA
0
2 2TheoreticalDivision, LosAlamosNationalLaboratory, LosAlamos, NM87545, USA

r E-mail: [email protected], [email protected]
a
M
Abstract. Wederiveanexactsolutionofanexplicitlytime-dependentmultichannel

5
modelofquantummechanicalnonadiabatictransitions. Ourmodelcorrespondstothe
1
case of a single linear diabatic energy level interacting with a band of an arbitrary

] N states, for which the diabatic energies decay with time according to the Coulomb
h
law. We show that the time-dependent Schr¨odingier equation for this system can
p
- be solved in terms of Meijer functions whose asymptotics at a large time can be
t
n compactly written in terms of elementary functions that depend on the roots of an
a
Nth order characteristic polynomial. Our model can be considered a generalization
u
q of the Demkov-Osherov model. In comparison to the latter, our model allows one to
[ explore the role of curvature of the band levels and diabatic avoided crossings.

2
v
2
8
6
0
.
1
0
4
1
:
v
i
X
r
a
The model of a level crossing with a Coulomb band 2
1. Introduction
Models that describe multi-channel nonadiabatic processes in explicitly time-dependent
quantum systems are important for the theory of control and characterization of
numerous mesoscopic and atomic systems [1]. Several exact results for explicitly driven
two state systems, such as the Landau-Zener-Stuekkelberg-Majorana formula [2], have
become widely used in practice. However, the physics of nonadiabatic transitions in
multichannel explicitly driven systems is a much more complex topic. Exactly solvable
models can provide the needed intuition about this subject. They also can be used as a
starting point for various approximation schemes and testing numerical algorithms and
analytical methods.
A relatively large class of exactly solvable Landau-Zener-like models with
multichannel interactions has been identified and explored in [3, 4, 5, 6, 7, 8, 9, 10].
These models correspond to quantum mechanical evolution of systems with matrix
Hamiltonian operators of the type
ˆ
C
ˆ ˆ ˆ
H(t) = A+Bt+ , (1)
t
ˆ ˆ ˆ
where A, B and C are constant N ×N matrices. One has to find the scattering N ×N
matrix Sˆ, in which the element S is the amplitude of the diabatic state n(cid:48) at t → +∞,
nn(cid:48)
given that at t → −∞ (or at t = 0) the system was at the state n. The related matrix
Pˆ, with P = |S |2, is called the matrix of transition probabilities.
nn(cid:48) nn(cid:48)
In this article, we present a new exactly solvable system of the type (1) and
determine its matrix of transition probabilities. Schro¨dinger’s equation for the
amplitudes, called b and a , i = 1...N, of this model is given by
0 i
N
db (cid:88)
0
i = βτb + g a (2)
0 j j
dτ
j=1
da k
j j
i = a +g b , 1 ≤ j ≤ N, (3)
j j 0
dτ τ
with constant parameters β, k and g . We will call b (t) the amplitude of the “0th
i i 0
level,” a (t) the amplitude of the “ith level,” etc.; parameters g are called the coupling
i i
constants, and β is called the slope of the 0th diabatic energy level.
This model corresponds to the case of a single linear diabatic energy level crossing
a band of states whose diabatic energies (diagonal elements of the Hamiltonian matrix)
decay as ∼ k /t with time (figure 1). Structurally, the model (2, 3) is very similar
i
to the celebrated Demkov-Osherov model [6]. The latter corresponds to the case of a
single level crossing a band of parallel states. In fact, one can show that the Demkov-
Osherov solution is recovered from our model in a specific limit: k (cid:29) |k −k | ∼ |g |2/β,
i i j i
i,j = 1...N. When this condition is not satisfied, the band has a nonzero curvature
near the avoided level crossings. Hence, our solution makes a unique insight into the
physics of nonadiabatic transitions in the case when a band has a non-zero curvature,
The model of a level crossing with a Coulomb band 3
Figure 1. Time-dependence of the eigenspectrum (adiabatic energies) of the
Hamiltonian with elements H = βt, H = k /t, H = H = g , and zero
00 nn n n0 0n n
otherwise (n=1,...,N; N =8). For any n, all k >0.
n
as shown in figure 1. Another interesting situation corresponds to the case when all k
i
are negative. Our model then corresponds to the system with nonadiabatic transitions
between a level that passes in the vicinity of a band but does not go through avoided
level crossings (figure 2).
We will discuss that not all transition probabilities among diabatic states are well
defined in this system because states of the band become asymptotically degenerate at
t → ∞, so that transitions among them never saturate. Therefore, our main focus will
be on the probability of transition to the 0th level. In particular, we will present a
simple formula (Eq. 27) for the probability to remain in this level after all interactions
if this level was initially populated.
2. Solution of the model
2.1. Transforming Equation (2) into the Meijer equation
We may assume without loss of generality that the k are ordered with k < ··· < k .
j 1 N
The change of variables t = τ2/2, a = τb gives:
j j
N
db (cid:88)
0
i = βb + g b (4)
0 j j
dt
j=1
db
j
2it = (k −i)b +g b , 1 ≤ j ≤ N. (5)
j j j 0
dt
It follows from (5) that
(cid:20) (cid:18) (cid:19)(cid:21)
d 1 k g
j j
t + +i b = b . (6)
j 0
dt 2 2 2i
Then applying (6) to (4), we have
(cid:89)N (cid:20) d (cid:18)1 k (cid:19)(cid:21) d (cid:89)N (cid:20) d (cid:18)1 k (cid:19)(cid:21) (cid:88)N g2 (cid:89)N (cid:20) d (cid:18)1 k (cid:19)(cid:21)
j j j m
i t + +i b = β t + +i b + t + +i b .
0 0 0
dt 2 2 dt dt 2 2 2i dt 2 2
j=1 j=1 j=1 m=1
m(cid:54)=j
(7)
The model of a level crossing with a Coulomb band 4
Figure 2. Time-dependence of the eigenspectrum (adiabatic energies) of the
Hamiltonian with elements H = βt, H = k /t, H = H = g , and zero
00 nn n n0 0n n
otherwise (n=1,...,N; N =8). For any n, all k <0.
n
Define the following polynomial in x:
(cid:89)N (cid:20) (cid:18)1 k (cid:19)(cid:21) (cid:88)N g2 (cid:89)N (cid:20) (cid:18)1 k (cid:19)(cid:21)
j j m
f(x) := β x− −i + x− −i . (8)
2 2 2i 2 2
j=1 j=1 m=1
m(cid:54)=j
Letting x = 1/2+iy, we define the following function of y:
(−i)N (cid:89)N (cid:20) k (cid:21) (cid:88)N g2 (cid:89)N (cid:20) k (cid:21)
j j m
g(y) := f(x) = y + − y + . (9)
β 2 2β 2
j=1 j=1 m=1
m(cid:54)=j
Because the sign of the ordered set {g(−∞),g(−k /2),...,g(−k /2),g(∞)} alternates
N 1
exactly N times, g has exactly N real roots. Let these roots be l ,...,l , and let
1 N
ξ = 1/2+il be the corresponding roots of f(x). We note that
j j
(cid:88)N (cid:88)N (cid:18)g2 k (cid:19)
j j
l = − . (10)
j
2β 2
j=1 j=1
With these definitions in place, (7) becomes
N (cid:20) (cid:18) (cid:19)(cid:21) (cid:18) (cid:19)
(cid:89) d 1 k d d
j
i t + +i b = f t +1 . (11)
0
dt 2 2 dt dt
j=1
Because the operator
d
t +1 (12)
dt
commutes with itself and multiplication by a constant, (11) becomes
(cid:32) (cid:33)
N (cid:20) (cid:18) (cid:19)(cid:21) N (cid:20) (cid:21)
(cid:89) d 1 k d (cid:89) d
j
i t + +i −β t +1−ξ b = 0. (13)
j 0
dt 2 2 dt dt
j=1 j=1
The model of a level crossing with a Coulomb band 5
We use the identity
d2 d (cid:18) d (cid:19)
t = t −1 (14)
dt2 dt dt
and multiply (13) by it to get the desired form of the differential equation for b :
0
(cid:32) (cid:33)
N (cid:20) (cid:18) (cid:19)(cid:21) N (cid:20) (cid:21)
d (cid:89) d 1 k (cid:89) d
j
−t t − −i −βit t +1−ξ b = 0. (15)
j 0
dt dt 2 2 dt
j=1 j=1
Equation (15) is an (N +1)th order linear differential equation, which can be found in
Section 5.8 of [11], equation 1. It has a fundamental set of N +1 Meijer’s G solutions.
2.2. Solving for b
0
For notational convenience, define
1 k
j
h = −i , (16)
j
2 2
and the following three sets:
ξ = ξ ,...,ξ (17)
N 1 N
h = h ,...,h (18)
N 1 N
h r = h ,...,h ,h ,...,h . (19)
N 1 r−1 r+1 N
Equation (15) has the following fundamental set of N +1 Meijer’s G solutions:
G1,N (cid:0) ξN (cid:12)(cid:12)βit(cid:1),(cid:110)G1,N (cid:16) ξN (cid:12)(cid:12)βit(cid:17)(cid:111)r=N . (20)
N,N+1 0,hN N,N+1 hr,0,hNr(cid:12) r=1
Each of these solutions remains bounded as t → 0, and furthermore, all but the first go
to zero as t → 0 (Appendix A.3i).
2.3. Finding the vector of solutions
Each of the solutions in (20) for b specifies a unique vector of solutions to the original
0
system of equations, and the general solution to the system (4, 5) is a linear combination
The model of a level crossing with a Coulomb band 6
of these vectors. Manipulating the system (4, 5) gives the following N equations:
N
(cid:88) db
0
g b = i −βb
j j 0
dt
j=1
N (cid:20) (cid:21) N
(cid:88) d db (cid:88)
g (k −i)b = 2it i 0 −βb − g2b
j j j dt dt 0 j 0
j=1 j=1
(cid:34) (cid:35)
N (cid:20) (cid:21) N N
(cid:88) d d db (cid:88) (cid:88)
g (k −i)2b = 2it 2it i 0 −βb − g2b − g2(k −i)b
j j j dt dt dt 0 j 0 j j 0
j=1 j=1 j=1
.
.
.
(cid:34) (cid:34) (cid:35) (cid:35)
N (cid:20) (cid:21) N N
(cid:88) d d db (cid:88) (cid:88)
g (k −i)N−1b = 2it ··· 2it i 0 −βb − g2b ··· − g2(k −i)N−2b .
j j j dt dt dt 0 j 0 j j 0
j=1 j=1 j=1
(21)
Multipying through by (2t)1/2 gives equations for the a :
j
N (cid:18) (cid:19)
(cid:88) db
g a = (2t)1/2 i 0 −βb
j j 0
dt
j=1
(cid:32) (cid:33)
N (cid:20) (cid:21) N
(cid:88) d db (cid:88)
g (k −i)a = (2t)1/2 2it i 0 −βb − g2b
j j j dt dt 0 j 0
j=1 j=1
(cid:32) (cid:34) (cid:35) (cid:33)
N (cid:20) (cid:21) N N
(cid:88) d d db (cid:88) (cid:88)
g (k −i)2a = (2t)1/2 2it 2it i 0 −βb − g2b − g2(k −i)b
j j j dt dt dt 0 j 0 j j 0
j=1 j=1 j=1
.
.
.
(cid:32) (cid:34) (cid:34) (cid:35) (cid:35) (cid:33)
N (cid:20) (cid:21) N N
(cid:88) d d db (cid:88) (cid:88)
g (k −i)N−1a = (2t)1/2 2it ··· 2it i 0 −βb − g2b ··· − g2(k −i)N−2b .
j j j dt dt dt 0 j 0 j j 0
j=1 j=1 j=1
(22)
The left-hand side of (22) is an N ×N Vandermonde matrix
 
1 1 ··· 1
 k −i k −i ··· k −i 
1 2 N
 
V =  (k −i)2 (k −i)2 ··· (k −i)2 , (23)
N  1 2 N 
 ... ... ... ... 
(k −i)N−1 (k −i)N−1 ··· (k −i)N−1
1 2 N
(whose inverse is well known; for example, a general expression for the inverse can be
found in [12]), multiplied with the column vector [g a ,...,g a ]T.
1 1 N N
The model of a level crossing with a Coulomb band 7
3. Transition probabilites
3.1. Initially populating b
0
Consider the first solution in (20):
b = G1,N (cid:0) ξN (cid:12)(cid:12)βit(cid:1)
0 N,N+1 0,hN
 
(cid:89)N Γ(1−ξ ) (cid:18) 1 k 1 k (cid:19). (24)
j 1 N
=  (cid:16) (cid:17)NFN 1−ξ1,...,1−ξN; +i ,..., +i ;−βit
Γ 1 +ikj 2 2 2 2
j=1 2 2
Then b (t → 0) < ∞, and since the kth derivative of a hypergeometric function is a
0
multiple of another hypergeometric function (Appendix A.3ii), we have
(cid:12)
dk (cid:12)
b (cid:12) < ∞. (25)
dtk 0(cid:12)
t→0
Then it follows from (22) that with this choice of b , a (0) = ··· = a (0) = 0. We now
0 1 N
set
 (cid:16) (cid:17)
N Γ 1 +ikj
b0 = (cid:89) Γ(21−ξ2) GN1,,NN+1(cid:0)0ξ,hNN(cid:12)(cid:12)βit(cid:1), (26)
j
j=1
in order that b may be initially populated with probability 1, and using (24) and the
0
asymptotics from (A.8), we find (Appendix A.3iii):
N (cid:18) (cid:19)
(cid:89) exp(−2πl )+1
P = |b (t → ∞)|2 = j . (27)
00 0
exp(πk )+1
j
j=1
With a (0) = ··· = a (0) = 0 and b chosen as in (26), we may simply solve (5)
1 N 0
for each j, using the antiderivative in (A.9), to find
a = −iQg√jt−ikj/2 (cid:90) tx(1−hj)−1G1,N (cid:0) ξN (cid:12)(cid:12)βix(cid:1) dx (28)
j 2 N,N+1 0,hN
0
−iQg t−ikj/2 (cid:16) (cid:12) (cid:17)
= √j t1−hjG1,N ξN (cid:12)βit , (29)
2 N,N+1 0,hNj,hj−1(cid:12)
where
 (cid:16) (cid:17)
N Γ 1 +ikj
(cid:89) 2 2
Q =  . (30)
Γ(1−ξ )
j
j=1
The resultant asymptotic is (Appendix A.3iv):
P = |a |2
0j j
∼ |Q|2gj2 (cid:12)(cid:12)(cid:12)(cid:88)N e−πls/2(βt)ils Γ(1−ξs) (cid:32)(cid:81)Nr=1,r(cid:54)=sΓ(ξs −ξr)(cid:33)(cid:12)(cid:12)(cid:12)2. (31)
2β (cid:12)(cid:12)s=1 Γ(1+ξs −hj) (cid:81)Nr=1,r(cid:54)=jΓ(ξs −hr) (cid:12)(cid:12)
The model of a level crossing with a Coulomb band 8
Equation (31) shows that, unlike P , the probability P does not converge at t → ∞
00 0j
in the general case. Physically, this behavior follows from the fact that diabatic energies
of levels with j (cid:54)= 0 become asymptotically degenerate at t → ∞ with a characteristic
level splitting behaving as ∼ 1/t with time. On the other hand, couplings between
those levels, in the leading order of the perturbation expansion, also decay as ∼ 1/t, so
that both diagonal and off-diagonal terms in the Hamiltonian projected on the N state
subspace of levels with j (cid:54)= 0 remain of the same order in magnitude. Hence transitions
between such diabatic states never saturate. One can explore the problem of scattering
in the adiabatic basis but we will not pursue it here.
3.2. Initially populating a to find P (arbitrary N)
q q0
As mentioned earlier (see Appendix A.3i) all but the first solution in (20) go to 0 when
t → 0; furthermore, by (A.11), it follows that
(cid:12)
dk (cid:12)
tk−1/2 b (cid:12) < ∞, (32)
dtk 0(cid:12)
t→0
so in the limit t → 0, (22) reduces to
(cid:40) (cid:34) (cid:34) (cid:35)(cid:35) (cid:41)N
(cid:88)N (cid:18) d (cid:19)m−1 db
g (k −i)m−1a (0) = (2t)1/2i(2i)m−1 t 0 . (33)
j j j
dt dt
j=1 t→0 m=1
We have that
(cid:40) (cid:41)
(cid:18) d (cid:19)m (cid:88)m m dj
t = tj , (34)
dt j dtj
j=1
where
(cid:40) (cid:41)
m
= S(m,j) (35)
j
denotes the Stirling numbers of the second kind, which satisfy the recurrence
(cid:40) (cid:41) (cid:40) (cid:41) (cid:40) (cid:41)
m+1 m m
= j + , (36)
j j j −1
and are equal to the number of ways to partition m labelled elements into j nonempty
unlabelled sets. Then (33) becomes
 (cid:34) (cid:34) (cid:40) (cid:41) (cid:35)(cid:35) N
(cid:88)N m(cid:88)−1 m−1 dj+1 
g (k −i)m−1a (0) = (2t)1/2i(2i)m−1 tj b .
j j j j dtj+1 0
 
j=1 j=1
t→0 m=1
(37)
If we put
(cid:16) (cid:12) (cid:17)
b = tikr/2G1,N ξN (cid:12)βit , (38)
0 N,N+1 hr,0,hNr(cid:12)
The model of a level crossing with a Coulomb band 9
we have the following result, using (A.11) and (A.13):
dj+1 (cid:89)N (cid:18)Γ(1+h −ξ )(cid:19)
limtj+1/2 b = (Γ(h −j))−1 r s (39)
t→0 dtj+1 0 r Γ(1+hr −hs)
s=1
N (cid:18) (cid:19)
(cid:89) Γ(1+h −ξ )
= (h −1) (Γ(h ))−1 r s , (40)
r j r
Γ(1+h −h )
r s
s=1
where (h −1) denotes the falling factorial (h −1)···(h −j +1)(h −j) and where
r j r r r
(h −1) = h −1. Then (37) becomes
r 0 r
(cid:40) (cid:41)N
N
(cid:88)
g (k −i)m−1a (0) = γ , (41)
j j j m,r
j=1
m=1
where, for notational convenience, we let
(cid:34) (cid:40) (cid:41) (cid:35)
m−1 N (cid:18) (cid:19)
(cid:88) m−1 (cid:89) Γ(1+h −ξ )
γ = (2)m−1/2(i)m(Γ(h ))−1 (h −1) r s . (42)
m,r r r j
j Γ(1+h −h )
r s
j=1 s=1
The Stirling numbers of the second kind satisfy the relation
(cid:40) (cid:41)
m
(cid:88) m
(x) = xm−1, (43)
j
j
j=1
which simplifies (42) to
N (cid:18) (cid:19)
(cid:89) Γ(1+h −ξ )
γ = (2)m−1/2(i)m(Γ(h ))−1(h −1)m−1 r s . (44)
m,r r r
Γ(1+h −h )
r s
s=1
Using notation from (23), with b as in (38),
0
   
g a (0) γ
1 1 1,r
. .
 ..  = V−1 .. . (45)
  N  
g a (0) γ
N N N,r
Setting ζ (a vector) equal to the right-hand side of (45), we let c be the qth column of
r q
 
 . . −1 g 0 ··· 0
. . 1
. .
0 g ··· 0 
  2
ζ.1 ··· ζ.N  ... ... ... ... , (46)
. .
. .
0 0 ··· g
N
and denote the components of c by
q
 
c
1,q
.
cq =  .. . (47)
c
N,q
The model of a level crossing with a Coulomb band 10
If we put
(cid:88)N (cid:16) (cid:12) (cid:17)
b = c tikr/2G1,N ξN (cid:12)βit , (48)
0 r,q N,N+1 hr,0,hNr(cid:12)
r=1
this corresponds to the vector of states where b (0) = a (0) = ··· = a (0) = a (0) =
0 1 q−1 q+1
··· = a (0) = 0, and a (0) = 1.
N q
Again applying the asymptotics from (A.8) provides an asymptotic for P
q0
(Appendix A.3v):
(cid:32) (cid:33)(cid:12) (cid:12)2
(cid:88)N g2 (cid:12)(cid:88)N (cid:12)
P = |b |2 ∼ exp −π j (cid:12) c tikr/2eπkr/2(cid:12) . (49)
q0 0 2β (cid:12) r,q (cid:12)
(cid:12) (cid:12)
j=1 r=1
Amplitudes for the states a are given in Appendix B.
j
4. Discussion
The solution of our model is expressed through the roots of the polynomial g in (9),
which generally cannot be obtained explicitly. To provide better intuition about the
transition probabilities in our model, we will explore special situations that allow us to
obtain explicit expressions for the transition probabilities.
4.1. Degenerate band
Consider the case of all the k identical, i.e. k = k for all i = 1...N. This case
i i
corresponds to
(cid:32) (cid:33)
(cid:18) k(cid:19)N−1 k (cid:88)N g2
g(y) = y + y + − i , (50)
2 2 2β
i=1
with simple roots l = −k/2 for i = 1,...,N − 1 and l = −k/2 + (cid:80)N g2/2β.
i N i=1 i
Substituting this into (27) we find
 (cid:18) (cid:20) N (cid:21)(cid:19) 
(cid:80) g2
exp π k − i +1
β
P =  i=1 . (51)
00
 exp(πk)+1 
Interestingly, thissolutionshowsthat, nomatterhowlargeN isandhowstrongcoupling
constants are, if the band levels are degenerate the survival probability is always non-
(cid:0) (cid:1)
vanishing, i.e. P > 1/ eπk +1 . Such a behavior can occur only when the curvature
00
of the band is non-vanishing. It was first noticed in the Demkov-Osherov model with a
piecewise linearly changing slope of the 0th level [13]. It is confirmed now for a model
with fully continuous time-dependence of diabatic energy levels. We also note that,
in the case of a degenerate band, a linear transformation can reduce the degenerate
states to one coupled to the 0th state and N − 1 uncoupled, leading to a two-state
curve-crossing problem [14].