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SMALLEST REFRIGERATOR WITHOUT MOVING PARTS Lajos Disi KFKI - - PowerPoint PPT Presentation
SMALLEST REFRIGERATOR WITHOUT MOVING PARTS Lajos Disi KFKI - - PowerPoint PPT Presentation
SMALLEST REFRIGERATOR WITHOUT MOVING PARTS Lajos Disi KFKI Research Institute for Particle and Nuclear Physics H-1525 Budapest 114, POB 49, Hungary CONTENTS Linden, Popescu, Skrzypczyk: How small thermal machines can be? LSP: The smallest
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SMALLEST REFRIGERATOR: 3-LEVEL-SYSTEM
Hot and cold reservoirs: Th > Tc. Refrigerator will yield T0 < Tc. Transition |0 → |1 is heated by Th, |0 → |2 is cooled by Tc. Let ǫc > ǫh! −−ǫc−− exp(−ǫc/kBTc) −−ǫc−− exp[−(ǫc/kBTc) + (ǫh/kBTh)] −−ǫh−− exp(−ǫh/kBTh) −−ǫh−− 1 −−0−− − 1 Make exp[−(ǫc/kBTc) + (ǫh/kBTh)] = exp[−(ǫc − ǫh)/kBT0] ⇒ Effective temperature of the TLS |1h , |1c:
T0 = 1 − ǫh
ǫc
1 − ǫh
ǫc Tc Th
Tc (< Tc).
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2nd SMALLEST REFRIGERATOR: 2xTLS
Hot and cold reservoirs: Th > Tc. Refrigerator will yield T0 < Tc. Transition |0 → |1 is heated by Th, |0 → |2 is cooled by Tc. Let ǫc > ǫh! |1c −−ǫc−− exp(−ǫc/kBTc) |1h −−ǫh−− exp(−ǫh/kBTh) |0c −−0c−− 1 |0h −−0h−− 1 Make exp[−(ǫc/kBTh) + (ǫh/kBTc)] = exp[−(ǫc − ǫh)/kBT0] ⇒ Effective temperature of the TLS |1h , |1c:
T0 = 1 − ǫh
ǫc
1 − ǫh
ǫc Tc Th
Tc (< Tc).
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TLS THERMALIZATION DYNAMICS
TLS: ˆ a = |0 1| , ˆ a† = |1 0| , ˆ H = ǫˆ a†ˆ a; Heat bath: β = 1/kBT. Thermalization master equation: d ˆ ρ dt = − i ǫ[ˆ a†ˆ a, ˆ ρ]+Γ
- ˆ
aˆ ρˆ a† − 1 2{ˆ a†ˆ a, ˆ ρ}
- +e−βǫΓ
- ˆ
a†ˆ ρˆ a − 1 2{ˆ aˆ a†, ˆ ρ}
- .
2nd term: spontaneous decay |1 → |0 at rate Γ. 3rd term: thermal excitation |0 → |1 at rate Γ×Boltzmann factor. Competition ⇒ Gibbs stationary state at (inverse) temperature β: ρ − → |0 0| + exp(−βǫ) |1 1| (t ≫ eβǫ/Γ). MLS: Any TL subspace may likewise be thermalized. ˆ a = |n m| , ǫ = ǫn − ǫm > 0, Γ = Γnm Each TL subspace may have different temperatures Tnm. If some are equilibrated by reservoirs, the rest obtains calculable ‘effective temperatures’.
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2nd SMALLEST Q-REFRIGERATOR DYNAMICS
No external resources of energy just heat flow Th → Tc. Refrigerator: 2xTLS, in contact with Th, Tc where Th > Tc and ǫc > ǫh. Develops a temperature T0 < Tc for the TLS subspace |1h , |1c. Can cool a ‘thermometer’ to temperature T0 < Tc. Thermometer: third TLS ˆ a3, ˆ a†
3,
ǫ3 = ǫ0 = ǫc − ǫh. Coupled to ˆ a0 = |1h 1c| , ˆ a†
0 = |1c 1h| of the refrigerated subspace.
Master eq. in interaction picture: d ˆ ρ dt = Γc
- ˆ
ac ˆ ρˆ a†
c − 1
2{ˆ a†
cˆ
ac, ˆ ρ}
- + e−βcǫcΓc
- ˆ
a†
c ˆ
ρˆ ac − 1 2{ˆ acˆ a†
c, ˆ
ρ}
- +
+Γh
- ˆ
ah ˆ ρˆ a†
h − 1
2{ˆ a†
hˆ
ah, ˆ ρ}
- + e−βhǫhΓh
- ˆ
a†
h ˆ
ρˆ ah − 1 2{ˆ ahˆ a†
h, ˆ
ρ}
- +
−i g
- ˆ
a†
3ˆ
a0 + ˆ a†
0ˆ
a3, ˆ ρ
- If coupling g ≪ Γc, Γh then ˆ
ρ3 → |03 03| + exp(−ǫ3/kBT0) |13 13|.
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2nd SMALLEST HEAT ENGINE: 2xTLS
We want a negative T0 (population inversion). Change the role of Th and Tc in refrigerator: ⇒ T0 may be negative! Reorganized refrigerator becomes heat engine. Transition |0h → |1h is heated by Th, |0c → |1c is cooled by Tc. Let ǫh > ǫc now (opposite than for refrigerator)! |1h −−ǫh−− exp(−ǫh/kBTh) |1c −−ǫc−− exp(−ǫc/kBTc) |0h −−0h−− 1 |0c −−0c−− 1 Make exp[−(ǫh/kBTh) + (ǫc/kBTc)] = exp[−(ǫh − ǫc)/kBT0] ⇒ Negative effective temperature of the TLS |1h , |1c:
T0 = 1 − ǫc
ǫh
1 − ǫc
ǫh Th Tc
Th < 0 if Th Tc > ǫh ǫc > 1.
Negative T0 means population inversion between |1c and |1h. It can ’lift a weight’ at constant speed!
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2nd SMALLEST HEAT ENGINE DYNAMICS
Resource: heat flow Th → Tc. Engine: 2xTLS, in contact with Th, Tc where Th/Tc > ǫh/ǫc > 1. Develops population inversion T0 < 0 for the TLS subspace |1c , |1h. Can ‘lift a weight’ at stationary power. Weight: harmonic ocillator ˆ a3, ˆ a†
3,
ǫ3 = ǫ0 = ǫh − ǫc. Coupled to ˆ a0 = |1c 1h| , ˆ a†
0 = |1h 1c| of the population inverted TLS.
Master eq. in interaction picture (formally same as refrigerator’s, just [ˆ a3, ˆ a†
3] = 1):
d ˆ ρ dt = Γc
- ˆ
ac ˆ ρˆ a†
c − 1
2{ˆ a†
cˆ
ac, ˆ ρ}
- + e−βcǫcΓc
- ˆ
a†
c ˆ
ρˆ ac − 1 2{ˆ acˆ a†
c, ˆ
ρ}
- +
+Γh
- ˆ
ah ˆ ρˆ a†
h − 1
2{ˆ a†
hˆ
ah, ˆ ρ}
- + e−βhǫhΓh
- ˆ
a†
h ˆ
ρˆ ah − 1 2{ˆ ahˆ a†
h, ˆ
ρ}
- +
−i g
- ˆ
a†
3ˆ
a0 + ˆ a†
0ˆ
a3, ˆ ρ
- If coupling g ≪ Γc, Γh then, for Th/Tc > ǫh/ǫc, the oscillator energy
ǫ3ˆ a†
3ˆ