Moderne Theoretische Physik III SS Master equation for a two-level atom ( = 40 Punkte, mündlich)

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1 Karlsruher Institut für Technologie Moderne Theoretische Physik III SS 205 Prof. Dr. A. Mirlin Blatt 2, 00 Punkte + 50 Bonuspunkte Dr. U. Karahasanovic, Dr. I. Protopopov Besprechung, Master equation for a two-level atom = 40 Punkte, mündlich) a) We need to solve ṗ = γ 2 p 2 γ 2 p ) ṗ 2 = γ 2 p γ 2 p 2 2) withtheinitial conditionsp i t = 0) = p i0.addingthetwo equationswegetṗ +ṗ 2 = 0. This is of course a generic property of any master equation: any master equation should conserve the normalisation of probability distribution. We can now substitute p 2 = p into the first equation to get ṗ = γ 2 γ 2 +γ 2 )p 3) To construct a general solution of a non-uniform equation we need to find any solution of this equation and add the general solution of the corresponding uniform equation. Thus, γ 2 p = +Ce γ 2+γ 2 )t 4) γ 2 +γ 2 b) At We now use the initial condition p t = 0) = p 0 to get γ 2 p = p 0 γ 2 γ 2 +γ 2 + γ 2 +γ 2 ) e γ 2+γ 2 )t At large times our system relaxes to the steady-state solution p t = ) = γ 2 γ 2 +γ 2, p 2 t = ) = p ) = 5) γ 2 γ 2 +γ 2 6) Of course the steady state solution can be found without the solution of the full master equation. We need just to find p and p 2 nullifying the right-hand sides of Eqs. ) and 2) [taking into account that p +p 2 = ]. From the explicit solution for p it is easy to see that p changes monotonically from p 0 to p ). Thus at arbitrary moment of time 0 < p <. we have γ 2 = γ 2 = Γ. 7) p ) = p 2 ) = /2. 8) This state can be described as a state with infinite temperature. Our system keeps absorbing energy from a classical source until it reaches infinite temperature. Note that it is not possible to reach a state with inverse population negative temperature state) in this way. We will see in exercise 2 how one can achieve inverse population using three-level system.

2 c) We see immediately that γ 2 γ 2 = e ω/k BT = e E 2 E )/k B T 9) and the detailed-balance condition is satisfied. The steady-state distribution is in this case just the thermal distribution p i ) = e E i/k B T e E /k B T +e E 2/k B T. 0) 2. Three-level atom inverse population and lasers = 60 Punkte, mündlich) a) We start from the general master equation ṗ i = k γ ki p k γ ik p i. ) We take into account the possible transitions described in the exercise and find the master equation ṗ = γ 2 p 2 +Γp 3 Γp 2) ṗ 2 = γ 32 p 3 γ 2 p 2 3) ṗ 3 = Γp Γp 3 γ 32 p 3. 4) We need to find the steady-state solution of the master equation. We look for p, p 2 andp 3 nullifyingtheright-handsideandsatisfyingthenormalisationp +p 2 +p 3 =. We find γ 2 γ 32 +Γ) p ) = γ 2 γ 32 +2Γ)+γ 32 Γ γ 32 Γ p 2 ) = γ 2 γ 32 +2Γ)+γ 32 Γ γ 2 Γ p 3 ) = γ 2 γ 32 +2Γ)+γ 32 Γ 5) 6) 7) In particular, for the difference p 2 ) p ) we have p 2 ) p ) = γ 32Γ γ 2 γ 32 +Γ) γ 32 Γ+γ 2 γ 32 +2Γ) 8) We see that p 2 ) p ) > 0 for γ 2 < γ 32 Γ/γ 32 + Γ). The system reaches population inversion in this regime. Our results become especially transparent in the limit of very strong driving and negligible relaxation E 2 E. We have b) We now have the master equation p 3 ) = p ) = 0 p 2 ) = 9) ṗ = gnp 2 +Γp 3 gnp Γp 20) ṗ 2 = gnp +γ 32 p 3 gnp 2 2) ṗ 3 = Γp Γp 3 γ 32 p 3. 22)

3 and the equation for the number of photons ṅ = gnnp 2 p ) κn. 23) All the right-hand sides are zero in the steady-state. Using Eqs. 20), 2), 22) and normalisation condition on probabilities we find gn )γ 32 +Γ) p ) = γ 32Γ +gn )2γ 32 +3Γ) p 2 ) = γ 32Γ+gn )γ 32 +Γ) γ 32Γ +gn )2γ 32 +3Γ) gn )Γ p 3 ) = γ 32Γ +gn )2γ 32 +3Γ) 24) 25) 26) In the limit Γ we get p ) = p 3 ) = gn ) 3gn )+γ 32 27) p 2 ) = gn )+γ 32 3gn )+γ 32 28) As in the exercise 2a in this regime the system is characterised by population inversion with γ 32 p 2 ) p ) = 30) 3gn+γ 32 c) We now use Eq. 23) to find the steady state of the full system. The condition for the steady state reads [ ] gn n ) 3g γ 32 n )+ κ = 0 3) The for κ > gn the only possible solution of this equation is The corresponding occupations are cf. exercise 2a) 29) n ) = 0 32) p 3 ) = p ) = 0 p 2 ) = 33) For κ < gn the solution n ) = 0 can be shown to be unstable. There is, however, another solution n ) = gn κ)γ 32 34) 3gκ In this state our resonator emits coherent light at frequency ω and can serve as a laser. The laser emission rate is κn ) photons per second. The probability distribution of atoms over levels in this regime reads p ) = p 3 ) = 3 κ 3gN p 2 ) = 3 + 2κ 3gM 35) 36) 37)

4 3. Langevin-Gleichung: LRC-Schwingkreis = 50 Punkte, mündlich ) LRC-Schwingkreis mit Bewegungsgleichung C V + V R + V L I 0 = δ I 38) wobei I = V/R δi der Strom durch den Widerstand ist; δi beschreibt Nyquist- Rauschen, d.h. es gilt δit)δit ) = 2k BT R δt t ). a) Fouriertransformation der Bewegungsgleichung: Vt) dω 2π Vω)eiωt führt auf die folgende Form der Gleichung 38): ω 2 C +iω R + ) Vω) iωi 0 ω) = iωδiω). L Aufgelöst nach Iω) = I 0 ω)+δiω): erhält man die Impedanz Iω) = iω ω 2 C +i ω R + L ) Vω), b) Wegen Zω) = Vω) Iω) = iω ω 2 C + iω R + L = iωc + R + iωl Vω)Vω ) = Zω)Zω ) Iω)Iω ) ist also = Zω)Zω ) I 0 ω)i 0 ω )+I 0 ω) δiω ) + δiω) I }{{}}{{} 0 ω )+ δiω)δiω ) =0 =0 = V 0 ω)v 0 ω )+Zω)Zω ) δiω)δiω ) δvω)δvω ) = Zω)Zω ) δiω)δiω ). δiω)δiω ) = dt dt e iωt e iω t δit)δit ) }{{} 2k B T R δt t ) = 2k BT dt e iω+ω )t R }{{} 2πδω ω ) = 4πk BT R δω ω ),

5 und somit δvω)δvω ) = Zω)Zω ) δiω)δiω ) = Zω)Z ω) }{{} Zω) 2 = 4πk BT R c) Korrelationen des Spannungsrauschens δvt)δvt ) = 2π) 2 πr πr πc 2 R dω dω 4πk B T R δω +ω ) + ωc ) 2 δω +ω ) R 2 ωl dω e iωt e iω t δvω)δvω ) dω e iω ω t ) Zω) 2 δω +ω ) dωe iωt t ) R 2 + ωcr R ) 2 ωl dωe iωt t ) ω 2 ω 2 + ω CR) 2 2 LC ) 2. 39) Das Integral kann man mit Hilfe des Residuensatzes berechnen hierzu benötigt man die Polstellen erster Ordnung im Integranden, also die Nullstellen des Nenners. Sei f eine Funktion mit einfachen Nullstellen z k, so sind die Residuen von /f durch Res ) f,z k = f z k ) gegeben. Der Residuensatz beschreibt das Integral entlang eines geschlossenen Weges γ in der komplexen Ebene, wobei der Weg selbst keine Singularität trifft, durch die Summe der vom Weg eingeschlossenen Residuen: γ dz fz) = 2πi N Resf,z k ). 40) Der Nenner in Gl. 39) ist ein quartisches Polynom in ω, hat also 4 komplexe Nullstellen: ω2 LC +i ω 2 ) ) CR = ω i 2CR + ω i 2CR ) ) ω +i 2CR + ω +i 2CR, wobei = 4C 2 R 2 + LC ist. Da wir nur den Fall )2 > LC betrachten, ist > 0 und somit reell. Das Integral kann als k= dω e iωt t ) ω 2 ω ω )ω ω 2 )ω ω 3 )ω ω 4 ) geschrieben und mit dem Residuensatz berechnet werden. Die exponentiell gedämpften Lösungen erhält man für Reω k > 0 und t t > 0 bzw. Reω k < 0 und t t < 0. Im ersten Fall muss die Kontur in der oberen, im zweiten Fall in der unteren komplexen Halbebene geschlossen werden.

6 Abbildung : Voltage-voltage correlation function δvt)δvt ) as a function of t t δvt)δvt ) t>t = k B T πc 2 R t t 2πi e 2C e t t C t t e + ) 2 e it t ) i 2 ) +e it t ) + i i i RC RC +2 2 i RC [ ) + i e it t ) ) i + e it t ) ] [ cos t t ) ) sin t t ) ) ] Analoges ergibt sich für den zweiten Fall, sodass kombiniert [ cos t t ) ) folgt. δvt)δvt ) t t C e sin t t ) ] ) 2 i RC 2 )