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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
̂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↓
̂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
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
̂H =
∑
ij
vija†
i
aj +
∑
ijkl
wijkla†
i
a†
j
alak
exp i ̂Hτ|ΨHF⟩ = c0 exp iE0τ|Ψ0⟩ + c1 exp iE1τ|Ψ1⟩ + ⋯
̂UN⋯ ̂U2
̂U1
̂U0 |ΨHF⟩ =
∏
i
Ui |ΨHF⟩
̂H =
∑
ij
vija†
i
aj +
∑
ijkl
wijkla†
i
a†
j
alak
̂H =
∑
ij
vija†
i
aj +
∑
ijkl
wijkla†
i
a†
j
alak
̂H =
∑
ij
̂Hij +
∑
ijkl
Hijkl
exp i ̂Hτ = exp i(
∑
ij
̂Hijτ +
∑
ijkl
̂Hijklτ) = lim
n→∞ ∏
ij
exp(i ̂Hij
τ
n
)
∏
ijkl
exp(i ̂Hijkl
τ
n
)
n
exp τ(H1 + H2) =
(
exp
τ
n
H1 exp
τ
n
H2)
n
+ O
(
τ2
n )
n τ2
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
)
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
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
)
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
)
e(x/2)A
exB
e(x/2)A
= ex(A+B)+O(x3
)
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
)
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+⋯
eA
eB
= eB
eA
eA+B
= eA
eB
eA
eB
= exp
[
A + B +
1
2
[A, B]
]
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
)
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
O(x4
)
O(x4
)
O(x4
)
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
)
[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⋯
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⋯
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
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
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
)
Lie-Trotter-Suzuki分解、特にフラクタル分解について

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Lie-Trotter-Suzuki分解、特にフラクタル分解について

  • 1.
  • 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
  • 6. ̂H = ∑ ij vija† i aj + ∑ ijkl wijkla† i a† j alak exp i ̂Hτ|ΨHF⟩ = c0 exp iE0τ|Ψ0⟩ + c1 exp iE1τ|Ψ1⟩ + ⋯
  • 7. ̂UN⋯ ̂U2 ̂U1 ̂U0 |ΨHF⟩ = ∏ i Ui |ΨHF⟩ ̂H = ∑ ij vija† i aj + ∑ ijkl wijkla† i a† j alak
  • 8. ̂H = ∑ ij vija† i aj + ∑ ijkl wijkla† i a† j alak ̂H = ∑ ij ̂Hij + ∑ ijkl Hijkl exp i ̂Hτ = exp i( ∑ ij ̂Hijτ + ∑ ijkl ̂Hijklτ) = lim n→∞ ∏ ij exp(i ̂Hij τ n ) ∏ ijkl exp(i ̂Hijkl 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
  • 12.
  • 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+⋯
  • 18. eA eB = eB eA eA+B = eA eB eA eB = exp [ A + B + 1 2 [A, B] ]
  • 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 )