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Numerical study on ESR of V15
IIS, U. Tokyo, Manabu MachidaRIKEN, Toshiaki Iitaka
Dept. of Phys., Seiji Miyashita
June 27- July 1, 2005
Trieste, Italy
Nanoscale molecular magnet V15
(http://lab-neel.grenoble.cnrs.fr/)€
K6 V15IVAs6O42 H2O( )[ ] • 8H2O
Vanadiums provide fifteen 1/2 spins.
[A. Mueller and J. Doering (1988)]
Hamiltonian and Intensity
€
H = − Jij
v S i ⋅
v S j
i, j
∑ +v D ij ⋅
v S i ×
v S j( )
i, j
∑ − HS Siz
i
∑
€
I T( ) = I ω,T( ) dω0
∞
∫ =ωHR
2
2′ ′ χ ω,T( ) dω
0
∞
∫€
′ ′ χ ω,T( ) = 1− e−βω( ) Re M x M x t( ) e−iωt dt
0
∞
∫
[H. De Raedt, et al., PRB 70 (2004) 064401]
[M. Machida, et al., JPSJ (2005) suppl.]
€
J = −800K, J1 = −225K, J2 = −350K
€
D1,2x = D1,2
y = D1,2z = 40K
The parameter set
Difficulty
– Its computation time is of(e.g. S. Miyashita et al. (1999))
€
M x M x t( ) =Tr e−βH M x M x t( )
Tr e−βH
– Direct diagonalization requires memory of
€
O N 2( )
€
O N 3( )
difficult!
Two numerical methods
• The double Chebyshev expansion method (DCEM) - speed and memory of O(N) - all states and all temperatures
• The subspace iteration method (SIM) - ESR at low temperatures.
Background of DCEM
The DCEM =a slight modification of the Boltzmann-weighted time-dependent method (BWTDM).
Making use of the random vector technique andthe Chebyshev polynomial expansion
[T. Iitaka and T. Ebisuzaki, PRL (2003)]
DCEM (1)
€
M x M x t( ) =Φ e−βH / 2
( )M x M x t( ) e−βH / 2 Φ( )[ ]av
Φ e−βH / 2( ) e−βH / 2 Φ( )[ ]
av
Random phase vector
€
Φ ˆ X Φ[ ]av
= n ˆ X nn
∑ + e i θ m −θ n( )−δmn[ ]av
n ˆ X mm,n
∑
= Tr ˆ X + Δ ˆ X ≅ Tr ˆ X €
Φ = n e iθ n
n=1
N
∑
DCEM (2)Chebyshev expansions of the thermal and time-evolution operators.
€
e−βH / 2 = I0 − β2( )T0 H( ) + 2 Ik − β
2( )Tk H( )k=1
kmax
∑
€
e− i Ht = J0 t( )T0 H( ) + 2 −i( )kJk t( )Tk H( )
k=1
kmax
∑
€
J
€
HS>> small
Temperature dependence of intensity
[Y.Ajiro et al. (2003)]
Our calculation Experiment
€
Itot β( ) = I ω,β( )dω0
∞
∫
€
′ ′ χ ω,T( ) =π
e−βEm
m
∑ m,n
∑ e−βEm − e−βEn( ) ψ m M x ψ n
2
×δ ω − En − Em( )( )
€
I1 T( ) =πHR
2 HS
8tanh
βHS
2
⎛
⎝ ⎜
⎞
⎠ ⎟
ESR at low temperatures by SIM
We consider the lowest eight levels.
€
R T( ) = I T( ) /I1 T( )
€
I T( ) =ωHR
2
2′ ′ χ ω,T( ) dω
0
∞
∫
Intensity ratio
€
E8 = − 34 J + 3
2 HS
€
E4 = 34 J + 1
2 HS + 32 D
€
E7 = − 34 J + 1
2 HS
€
E6 = − 34 J − 1
2 HS
€
E5 = − 34 J − 3
2 HS
€
E3 = 34 J + 1
2 HS − 32 D
€
E2 = 34 J − 1
2 HS + 32 D
€
E1 = 34 J − 1
2 HS − 32 D
€
HSc0 =
3
2J
Energy levels with weak DM
€
E2
€
E5
€
E1
€
O D( )( )
€
HS⟨⟨HSc 0
€
HSc0⟨⟨HS
€
Rtri T( ) T →0 ⏐ → ⏐ ⏐
€
3
€
1+3D
HS
Intensity ratio of triangle model
At zero temperature