Lefschetz-thimble inspired analysis of the Dykhne-Davis-Pechukas method and an application for the Schwinger Mechanism
iTHEMS, RIKEN
- Aug. 21, 2020
Takuya Shimazaki
based on TS, K. Fukushima, Ann. Phys. 415 168111 (2020) The University of Tokyo
Lefschetz-thimble inspired analysis of the Dykhne-Davis-Pechukas - - PowerPoint PPT Presentation
Lefschetz-thimble inspired analysis of the Dykhne-Davis-Pechukas method and an application for the Schwinger Mechanism iTHEMS, RIKEN Aug. 21, 2020 Takuya Shimazaki based on TS, K. Fukushima, Ann. Phys. 415 168111 (2020) The University of Tokyo
based on TS, K. Fukushima, Ann. Phys. 415 168111 (2020) The University of Tokyo
V(z) = − Ezz .
We consider an electric field along the direction.
Γ ≃ exp (− πm2 eE ) .
The decay width is given by Energy
Energies are tilted by
Dirac sea Particle states
For A(t) = − Et , H is called the Landau-Zener model.
⊥ = m2 + k2 x + k2 y ,
Energy The transition probability is analytically calculable with an initial condition, P(k) = |ψ+(∞)|2 = exp (− πm2
⊥
eE ) This is the Landau-Zener formula. ψ+(−∞) = 0 .
δE ≡ E+ − E− ≠ 0
2 + m2 ⊥ .
tc
2 + m2 ⊥ = − πm2 ⊥
2 + m2 ⊥ and tc = − kz
P(k) = exp (− πm2
⊥
eE ) = PDDP
tc
∞ −∞
−(t) ·
2 .
i=±
Re t
∂S ∂z
z=zi
= 0
thimbles
zs,i .
dz dτ = ± ∂S ∂z with z(τ = 0) = zs,i
d dτ Im S[z(τ)] = 0 , d dτ Re S[z(τ)] ≥ 0 ( ≤ 0)
Good properties of the flow
= Their dual thimbles intersect the original integration contour.
1 2
1 2 Re z Im z
1 2
1 2 Re z Im z
arg ℏ > 0 arg ℏ < 0
We complexify and focus on the blue line.
∞ −∞
∞ −∞
t=tc
+
+ (∞) = ∑ i
tc = iτ but tpole = iT .
Re t
Re t
× × ×
DDP
1 2
1 2 Re t Im t
× × ×
1
1 Re t Im t
Lines{ Solid: thimbles Dashed: dual thimbles Dotted: DDP contour Dots{ Cross: poles Filled: saddle points Open: closing points
The red thimble
Full Our method DDP 1.2 1.4 1.6 1.8
log10[P(τ, T = 3 − τ, Λ = 10)]
When A(t) = − Et , it reduces the Landau-Zener model.
There exist a closing point and a pole tc = 1 ω tanh−1 (− γkz m + i γm⊥ m ) tpole = i π 2ω .
⊥
We define by
⊥
γ ≡ mω eE
DDP Full Our method 1 2 3 4 5 1.0 1.5 2.0 2.5
m = 3 , eE = 3 γ ≡ mω eE ( = ω) For simplicity, we set k = 0 . Then, tc = i 1 ω tan−1 γ , tpole = i π 2ω .
When γ → ∞ , tc gets closer to tpole .
1 2 3 2 3 S0
γ = π/2
m ≫ ω , m2 ≫ eE are assumed . Poles of tanh(ωt) play pivotal roles .
m = 3 , eE = 3 , E = 10ε
γ = mω eE = ω
+ (∞) = ∑ i
+