More Related Content Similar to Lie-Trotter-Suzuki分解、特にフラクタル分解について (20) More from Maho Nakata (20) Lie-Trotter-Suzuki分解、特にフラクタル分解について2. exp x(A + B)
exp x(A + B) =
(
exp
x
n
A exp
x
n
B
)
n
+ O
(
x2
n )
exp itH = exp i(ΔtH)N
t = NΔt
3. ̂H = − t
∑
i,σ
( ̂a†
i,σ
̂ai+1,σ + ̂a†
i+1,σ
̂ai,σ) + U
∑
i
̂a†
i↑
̂ai↑ ̂a†
i↓
̂ai↓
U = 0 ̂Ht = − t
∑
i,σ
( ̂a†
i,σ
̂ai+1,σ + ̂a†
i+1,σ
̂ai,σ)
t = 0 ̂HU = + U
∑
i
̂a†
i↑
̂ai↑ ̂a†
i↓
̂ai↓
4. ̂H = − t
∑
i,σ
( ̂a†
i,σ
̂ai+1,σ + ̂a†
i+1,σ
̂ai,σ) + U
∑
i
̂a†
i↑
̂ai↑ ̂a†
i↓
̂ai↓
|Ψ(τ = 0)⟩ τ
|Ψ(τ = τ)⟩ = exp i ̂Hτ|Ψ(τ = 0)⟩ = exp i ̂(Ht + ̂HU)τ|Ψ(τ = 0)⟩
t = 0 U = 0
5. exp i( ̂Ht + ̂HU)τ
exp(A + B) = lim
n→∞ (
exp
A
n
exp
B
n )
n
exp i( ̂Ht + ̂HU)τ = lim
n→∞ (
exp(i ̂Ht
τ
n
)exp i ̂HU(
τ
n
)
)
n
9. exp τ(H1 + H2) =
(
exp
τ
n
H1 exp
τ
n
H2)
n
+ O
(
τ2
n )
n τ2
10. m > 0 exp[x(A + B)] = Sm(x) + O (xm+1
)
Sm(x) = et1A
et2B
et3A
et4B
⋯etM A
exp[x(A + B)] = [Sm(x/n)]
n
+ O (xm+1
/nm
)
11. exp[x(A + B)] ex(A+B)
= exA
exB
+ O(x2
)
ex(A+B)
= I + x(A + B) + O (x2
)
exA
exB
= (I + xA + O(x2
))(I + xB + O(x2
)) = I + x(A + B) + O(x2
)
ex(A+B)
= e(x/2)A
exB
e(x/2)A
+ O(x3
)
1 + x(A + B) +
x2
2
(A + B)2
(
1 +
A
2
x +
(
A
2 )
2
1
2
x2
) (
1 + Bx +
B2
2!
x2
) (
1 +
A
2
x +
(
A
2 )
2
1
2
x2
)
x, x2
13. ex(A+B)
= exA
exB
+ O(x2
)
exA
exB
= ex(A+B)+O(x2
)
ex(A+B)
= e(x/2)A
exB
e(x/2)A
+ O(x3
)
e(x/2)A
exB
e(x/2)A
= ex(A+B)+O(x3
)
ex(A+B)
= ep1xA
ep2xB
ep3xA
ep4xB
⋯epMxB
+ O(xm+1
)
ep1xA
ep2xB
ep3xA
ep4xB
⋯epMxB
= ex(A+B)+O(xm+1
)
14. exA
exB
= ex(A+B)+O(x2
)
exA
exB
= (1 + xA + O(x2
))(1 + xB + O(x2
))
= 1 + x(A + B) + O(x2
)
ex(A+B)+O(x2
)
= 1 + x(A + B) + O(x2
) +
1
2
(x(A + B) + O(x2
))2
= 1 + x(A + B) + O(x2
)
16. ex(A+B)
= exA
exB
+ O(x2
)
exA
exB
= ex(A+B)+O(x2
)
ex(A+B)
= e(x/2)A
exB
e(x/2)A
+ O(x3
)
e(x/2)A
exB
e(x/2)A
= ex(A+B)+O(x3
)
ex(A+B)
= ep1xA
ep2xB
ep3xA
ep4xB
⋯epMxB
+ O(xm+1
)
ep1xA
ep2xB
ep3xA
ep4xB
⋯epMxB
= ex(A+B)+O(xm+1
)
17. S2(x) ≡ e(x/2)A
exB
e(x/2)A
= ex(A+B)+O(x3
)
S2(x)S2(−x) = S2(−x)S2(x)
S2(x)S2(−x) = I = e(x/2)A
exB
e(x/2)A
e−(x/2)A
e−xB
e(−x/2)A
S2(x) = ex(A+B)+x2
R2+x3
R3+x4
R4+⋯
S2(x)S2(−x) = I eA
eB
= eB
eA
eA
eB
= eA+B
S2(x)S2(−x) = I = ex(A+B)+x2
R2+x3
R3+x4
R4+⋯
e−x(A+B)−x2
R2−x3
R3−x4
R4+⋯
= e0
= e2(x2
R2+x4
R4+x6
R6+⋯
19. exp(x + y)A = exp xA + exp xA
exp[x(A + B)] = exp[sx(A + B)]exp[(1 − 2s)x(A + B)]exp[sx(A + B)]
S3(x) ≡ S2(sx)S2((1 − 2s)x)S2(sx) = e
s
2 xA
esxB
e
1 − s
2 xA
e(1−2s)xB
e
1 − s
2 xA
esxB
e
s
2 xA
S3(x) ≡ e
s
2 xA
esxB
e
1 − s
2 xA
e(1−2s)xB
e
1 − s
2 xA
esxB
e
s
2 xA
= ex(A+B)
+ O(x4
)
20. S3(x) ≡ S2(sx)S2((1 − 2s)x)S2(sx) = e
s
2 xA
esxB
e
1 − s
2 xA
e(1−2s)xB
e
1 − s
2 xA
esxB
e
s
2 xA
S2(x) ≡ e
x
2 A
exB
e
x
2 A
= ex(A+B)+x3
R3+x5
R5+⋯
= ex(A+B)+x3
R3+O(x5
)
R2, R4
O(x4
)
S3(x) = S2(sx)S2((1 − 2s)x)S2(sx)
= esx(A+B)+s3
x3
R3+O(x5
)
e(1−2s)x(A+B)+(1−2s)3
x3
R3+O(x5
)
esx(A+B)+s3
x3
R3+O(x5
)
= ex(A+B)+[2s3
+ (1 − 2s)3
]R3+O(x5
)
R3
2s3
+ (1 − 2s)3
= 0
22. s =
1
2 −
3
2
= 1.35120719195965…
S3(x) = exp
(
1
2
sxA
)
exp(sxB)exp
[
1
2
(1 − s)xA
]
exp[(1 − 2s)xB]exp
[
1
2
(1 − s)xA
]
exp(sxB)exp
(
1
2
sxA
)
S3(x) = ex(A+B)
+ O(x5
)
23. [0,t]
S4(x) ≡ S2(s2x)2
S2((1 − 4s2)x)S2(s2x)2
, s2 =
1
4 −
3
4
= 0.414490771794375⋯
S6(x) ≡ S4(s4x)2
S4((1 − 4s4)x)S4(s4x)2
s4 =
1
4 −
5
4
= 0.373065827733272⋯
S8(x) ≡ S6(s6x)2
S6((1 − 4s6)x)S6(s6x)2
, s6 =
1
4 −
7
4
= 0.359584649349992⋯
24. S4(x)
S4(x) ≡ S2(s2x)2
S2((1 − 4s2)x)S2(s2x)2
,
= e
s2
2 xA
es2xB
es2xA
es2xB
e
1 − 3s2
2 xA
e(1−4s2)xB
e
1 − 3s2
2 xA
es2xB
es2xA
es2xB
e
s2
2 xA
s2 =
1
4 −
3
4
= 0.414490771794375⋯
25. S4(x) ≡ S2(s2x)2
S2((1 − 4s2)x)S2(s2x)2
S2(sx) = esx(A+B)+s3
x3
R3+O(x5
)
S4(x) = S2(s2x)2
S2((1 − 4s2)x)S2(s2x)2
= e2sx(A+B)+2s3
x3
R3+O(x5
)
S2((1 − 4s2)x)S2(s2x)2
= e2sx(A+B)+2s3
x3
R3+O(x5
)
e(1−4s)x(A+B)+(1−4s)3
x3
R3+O(x5
)
S2(s2x)2
= e2sx(A+B)+2s3
x3
R3+O(x5
)
e(1−4s)x(A+B)+(1−4s)3
x3
R3+O(x5
)
e2sx(A+B)+2s3
x3
R3+O(x5
)
= e(A+B)(2sx+(1−4s)x+2sx)+4s3
x3
R3+(1−4s)3
x3
R3+O(x5
)
= e(A+B)x+4s3
x3
R3+(1−4s)3
x3
R3+O(x5
)
= e(A+B)x+(4s3
+(1−4s)3
)x3
R3+O(x5
)
4s3
+ (1 − 4s)3
= 0
26. S6(x) ≡ S4(sx)2
S4((1 − 4s)x)S4(sx)2
S4(sx) = esx(A+B)+s5
x5
R5+O(x7
)
S6(x) = S4(sx)2
S4((1 − 4s)x)S4(sx)2
= e2sx(A+B)+2s5
x5
R5+O(x7
)
e(1−4s)x(A+B)+(1−4s)5
x5
R5+O(x7
)
e2sx(A+B)+2s5
x5
R5+O(x7
)
= e(A+B)(2sx+(1−4s)x+2sx)+4s5
x5
R5+(1−4s)5
x5
R5+O(x7
)
= e(A+B)x+4s5
x5
R5+(1−4s)5
x5
R5+O(x7
)
= e(A+B)x+(4s5
+(1−4s)5
)x5
R5+O(x7
)
4s5
+ (1 − 4s)5
= 0
27. exp[x(A + B)] = exp[sx(A + B)]exp[(1 − 2s)x(A + B)]exp[sx(A + B)]
exA
exB
= ex(A+B)+O(x2
)
e(x/2)A
exB
e(x/2)A
= ex(A+B)+O(x3
)