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Molecular Nanomagnets$

Dante Gatteschi, Roberta Sessoli, and Jacques Villain

Print publication date: 2006

Print ISBN-13: 9780198567530

Published to Oxford Scholarship Online: September 2007

DOI: 10.1093/acprof:oso/9780198567530.001.0001

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(p.343) APPENDIX H PROOF OF THE LANDAU–ZENER–ST üCKELBERG FORMULA

(p.343) APPENDIX H PROOF OF THE LANDAU–ZENER–ST üCKELBERG FORMULA

Source:
Molecular Nanomagnets
Publisher:
Oxford University Press

The solution of system (8.12) which satisfies the initial condition X(0) = 1, Y (0) = 0 can be written as

(H.1)
{ X ( t ) = 1 T 0 t dt exp [ iU ( t ) ] Y ( t ) ( a ) Y ( t ) = T 0 t dt exp [ iU ( t ) ] X ( t ) ( b )
Eliminating Y (t) in (H.1) one obtains
(H.2)
X ( t ) = 1 ω T 2 0 t dt exp [ iU ( t ) ] 0 t dt exp [ iF ( t ) ] U ( t ) .
The lower integration bound can be replaced by -∞ because there are no transitions between t = -∞ and t = 0.

Equation (H.2) can be solved by iteration. One obtains a series which will only be written for t = ∞, namely

(H.3)
X ( ) = 1 + Σ n = 1 ( ω T 2 ) n I n
where
(H.4)
I n = dt 2 n t 2 n dt 2 n 1 t 2 n dt 1 expf ( { t } )
with
(H.5)
f ( { t } ) = i j = 1 n ξ ( t 2 j 2 t 2 j 1 2 )
and ξ = v/(2ℏ).

The following change of variables, introduced by Kayanuma (1984), will now be made:

(H.6)
{ x 1 = t 1 (a) x p = t 1 + j = 1 p 1 ( t 2 j + 1 t 2 j ) ( 2 p n ) (b) y p = t 2 p t 2 p 1 ( 1 p n ) (c)
(p.344) The inverse transformation is
(H.7)
{ t 2 p = x p + y 1 + y 2 + + y p (a) t 2 p + 1 = x p + 1 + y 1 + y 2 + + y p (b)
It follows that
t 2 j 2 t 2 j 1 2 = y j ( t 2 j + t 2 j 1 ) = 2 y j ( x j + y 1 + y 2 + + y j 1 + y j / 2 )
and f({t}) = g({x}, {y}), with
(H.8)
g = ( { x } , { y } ) = i ξ [ 2 Σ j = 1 n x j y j + ( Σ j = 1 n y j ) 2 ] .

Since t pt p−1, the quantities xj and yj satisfy the relations

(H.9)
x 1 x 2 x j x j 1 x n
and
(H.10)
y 1 , y 2 , , y j , , y n 0
so that
(H.11)
I n = dx 1 x 1 dx 2 x n 1 dx n 0 dy 1 0 dy 2 × 0 dy n exp g ( { x } , { y } ) .

The derivation of (H.11) makes use of the fact that the Jacobian of the transformation is equal to 1, as can be easily checked. This relation can also be written as

(H.12)
I n = ( D ) Dx Dy exp g ( { x } , { y } )
where the integration domain (D) is defined by (H.9) and (H.10).

It will now be shown that the variables xj can be replaced in (H.9) by any perturbation (thus changing the domain (D)) without modifying the integral (H.12). Indeed, the integral is not modified by replacing the variables xj and yj since this is just changing the name of the variables. But the function (H.8) is symmetric, and does not change if the inverse permutation is done. This proves the proposition. For instance, in the case n = 2, the definition (H.12) is I 2 = dx 1 x 1 dx 2 0 dy 1 0 dy 2 exp g ( x 1 , x 2 , y 1 , y 2 , ) . Changing the name of the variables yields I 2 = dx 2 x 2 dx 1 0 dy 1 0 dy 2 exp g ( x 2 , x 1 , y 2 , y 1 , ) and since g is a symmetric function, this is equal to the expected expression

I 2 = dx 2 x 2 dx 1 0 dy 1 0 dy 2 exp g ( x 1 , x 2 , y 1 , y 2 , ) .

(p.345) It follows that (H.11) can be written as

(H.13)
I n = 1 n ! ( j = 1 n 0 dy j ) exp [ i ξ ( Σ j = 1 n y j ) 2 ] [ j = 1 n dx j exp ( 2 i ξ x j y j | y j | ) ]
where ε = 0. Indeed that integral can be written as a sum of n!integrals of the form (H.12), with n! different definitions of the integration domain (D) corresponding to all permutations of the xj, and all those integrals are equal.

In (H.13), the parameter ε = 0 has been introduced to facilitate the calculation. Integrating over the xj, one obtains

(H.14)
I n = 1 n ! ( j = 1 n 0 dy j ) exp [ i ξ ( Σ j = 1 n y j ) 2 ] j = 1 n 2 2 + 4 ξ 2 y j 2
In the limit ε → 0, the fraction becomes a delta function and the exponential may be replaced by 1. It follows that
(H.15)
I n = 1 n ! j = 1 n ( 0 dy j 2 2 + 4 ξ 2 y j 2 ) = 1 n ! ( 0 dy 2 2 + 4 ξ 2 y 2 ) n = 1 n ! ( ( 1 / ξ ) 0 dy 1 1 + y 2 ) n = 1 n ! ( π 2 ξ ) n .
Using this result in (H.3) and replacing ξ by v/(2ħ), one obtains
(H.16)
X ( ) = 1 + Σ n = 1 1 n ! ( ∏ħω T 2 υ ) n = exp ( πħω T 2 v )
The probability 1−δ P of still being in the left-hand well at t = ∞ is the square of this quantity. Formula (8.17) follows.