Optical Feedback Cooling in Optomechanical Systems

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Λαλαγη Ελευθερόπουλος
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1 Optical Feedback Cooling in Optomechanical Systems A brief introduction to input-output formalism C. W. Gardiner and M. J. Collett, Input and output in damped quantum systems: Quantum Stochastic differential equations and the master equation, PRA 31, Quantum damped harmonic oscillator Optomechanical systems, classical description of optical cooling. A quantum treatment of optical cooling. I. Wilson-Rae et al. Theory of Ground State Cooling of a Mechanical Oscillator Using Dynamical Backaction PRL 99, (007) and F. Marquardt et al., Quantum Theory of Cavity-Assisted Sideband Cooling of Mechanical Motion PRL 99, (007). 1

2 H = H sys + H B + H i nt H b = H int = i dω ωb (ω)b(ω) dω κ(ω) b (ω)c c b(ω) This represents a continuum of independant bath harmonic oscillators, and hence: b(ω), b (ω ) = δ (ω ω ) Working in the Heisenberg picture: Â = i Â, H + Â t (0 for normal operators) ḃ(ω) = iωb(ω) + κ(ω)c, Â = i Â, Hsys + dω { b (ω) Â, c Â, c b(ω) } The formal solution for b(ω) is: b(ω) = e iω(t t0) b 0 (ω) + κ(ω) t t 0 c(t )dt, where b 0 (ω) b(ω)(t = 0). Furthermore: Â, { Hsys + dω e +iω(t t0) b 0 (ω) Â, c Â, c } e iω(t t0) b 0 (ω) Â = i t + dω κ(ω) t 0 dt { e +iω(t t ) c Â, c Â, c e iω(t t ) c } Input and output in damped quantum systems: Quantum Stochastic differential equations and the master equation, PRA 31,

3 We now make the first Markov approximation : κ(ω) = γ/π Note the following relations: 1 dω e iω(t t ) = δ(t t ) π t t 0 dt c(t )δ(t t ) = 1 c(t) We can then rewrite the equations of motion for an arbitrary operator, the Quantum Langevin Equation : Â = i Â, Â, ( Hsys c γ c + ) γ γb in (t) +( c + ) Â, γb in (t) c, where we have defined b in (t): b in (t) = 1 dω e iω(t t0) b 0 (ω) π Note also that: b in (t), b in (t ) = δ(t t ). In principle, to calculate the behavior of a real system, we need to know the state of the bath as represented by ρ in. However, in practice we almost always work with weakly damped systems, and treat the input noise as white, i.e.: Tr Tr ρ in b in (t)b in(t ) ρ in b in (t)b in (t ) b in (t)b in(t ) = Nδ(t t ) b in (t)b in (t ) = ( N + 1)δ(t t ) 3

4 Harmonic oscillator: ȧ = iω a a γ a a γ a b in (t) ȧ = +iω a a γ a a γ a b in (t) If the system is linear, we can fourier transform it: iωã = iω a ã γ a ã γ a b in (t) +iωã = +iω a ã γ a ã γ a b in (t), noting that: ã(ω) = 1 π ã (Ω) = 1 π e +iωt a(t)dt e iωt a (t)dt ȧ(t) = 1 iωe iωt ã(ω)dω π b in (ω) b in (ω ) = Nδ(ω ω ) b in (ω) b in ) (ω = ( N + 1 ) δ(ω ω ) So... ã = γa b in i(ω ω a ) γ a = γ a b in χ a (Ω) ã = where χ 1 a γa b in i(ω ω a ) γ a = γ a b in χ a (Ω), = i(ω ω a ) γ a. 4

5 We define the spectral density of the number operator as: S aa (Ω) = dt e +iωt a (t)a(0) = 1 dt dω dω e i(ω ω )t ã (ω )ãω ) π = dω ã (Ω)ã(ω ) = ã (Ω)ã(Ω), where we have assumed: ã (ω)ã(ω ) δ(ω ω ). And hence... b S aa (Ω) = γ a χ(ω) in (Ω) b in (Ω) = Nγ a (Ω ω a ) + ( γ a ) Where we have assumed a white noise input. This is a Lorentzian centered at w a with FWHM γ a this is the expected result for a damped harmonic oscillator. Note also: 1 S aa dω = N, π as you might expect. 5

6 Classical motion: mẍ = kx mγ c ẋ + F P (x, t) th(t) + c Cavity field: So... P 0 (x) ( γa c 1 ) + (x x0 ) Ṗ (t) = γ a P (t) P 0 (x) γ a P (t) P 0 (x 0 ) x dp 0 dx iω P = γ a P + γ ck r c x ( ) cγa k r x P = γ a iω ( ) ckr (1 + i Ω γ = a ) x ( ) Ω 1 + γ a x=0 } {{ } ck r / 6

7 Optomechanical Hamiltonian: ( H = ω a aa + 1 ) ( + ω c cc + 1 ) + H int Radiation Pressure Interaction Optical Cavity Mechanical Resonator L We now assume we are in the adiabatic limit: 0 1 c w c. In this case we can approximate the interaction term with: ) L 0 ω a ω a L 0 + x w a (1 xl0 mω ( w a 1 c + c ) } L {{ 0 } g H ω a 1 g ( c + c ) aa + ω c cc Now we include the bath couplings and a driving term: H = w a a a 1 σ 0 ( ) c + c + w c c c L + Ae ( iωt a + a ) + H B,a + H int,a + H B,c + H }{{ int,c } Optical driving term Optical bath Mechanical bath 7

8 We can calculate the equations of motion: ȧ = iω a a 1 g ( c + c ) iae iωt γ a a γ a a in ċ = iω c c + iω a ga a γ c c γ c c in, where a in b a,in and c in b c,in. We make the substitution a e iwt (ā + d), where ā is a constant and d represents the quantum fluctuations of the system. The equation of motion for d is: d = i (ā + d)+iω a g (ā + d) ā where = ω ω a is the optical detuning. ( c + c ) ia γ (ā + d) e+iwt γ a a }{{ in, } γ a d in If we assume d ā, we can linearize the equation. Furthermore we see the natural choice for A: ( A i γ ) a + ā d = i d + iα(c + c ) γ a d γ a d in α āω a g Additionally: ċ = iω c c + iω a g ( ā ā + ā d + d ā + d d ) γ c c γ c c in ā +ā d+ād = iω cc + i ( α d + αd ) γ c c γ c c in, where I have ignored the ā term, which is just a displacement of the mean position of the mechanical resonator. 8

9 What about a fourier transform? iω d(+ω) = i d(+ω) + iα c(+ω) + c ( Ω)! γ a d(+ω) γ a d in (+Ω) iω c(+ω) = iω c c(+ω) + i α d(+ω) + α d( Ω)! γ c c(+ω) γ c c in (+Ω) So... χ 1 d (Ω) d + + iα( c + + c ) = γ a d + in (Ω) c + i(α c + + α c ) = γ a c + in χ 1 c Let s define a system operator, A: d d + A d c ; Ã = d c + c c χ 1 d where: (Ω) 0 iα iα 0 χ 1 d ( Ω) iα iα iα iα χ 1 c (Ω) 0 iα iα 0 χ 1 c χ 1 a χ 1 c = i(ω ) γ a = i(ω ω c ) γ c ( Ω) ÃT = γa d + in γa d in γc c + in γc c in 9

10 The solution is: ( γd ( ) d = χ a d in + χ a ξ c α d in α d in + i ) γ c α χ c c in χ c c in / ( ) 1 + α ξ a ξ c ( γc ( ) c = χ c c in + α χ c ξ a c in c in + i ) γ d χ c α χ a d in αχ a d in / ( ) 1 + α ξ a ξ c where: ξ a/c (Ω) = χ a/c ( Ω) χ a/c(ω). What is the input? d in d in c in c in = 0 Ground state (a is coherent state!) = n th Thermal state S cc (Ω) = χ c 1 + α ξ a ξ c Modified mechanical response n th γ c 1 α χ c ξ a Main thermal excitation γ a αχ a + Zero point fluc. + ( n th + 1) γ c α χ c ξ a Back action 10

11 From Gardiner and Collett: a out = a in + γ a a So the optical field leaving the cavity is: S aa,out (Ω) = dte a iωt out (t)a out(0) S aa,out (Ω + ω a ) = dte iωt d in (t) + γ (ā a + d (t) ) d in (0) + γ a (ā + d(0)) = d in (Ω) d in (Ω) + dt e iω t γa ( d in (Ω) d(ω ) + d (Ω) d in (ω ) ) + ( γ a ā + d (Ω)ã(ω ) ) S aa,out (Ω + ω a ) = π ā 1 δ(ω) + γ a 1 + α ξ a ξ c γ a α χ a χ a ξ c + n th γ c αχ a χ c }{{ } ( n th + 1) γ αχa c χ c Stokes (Heating) Anti-Stokes (Cooling) + 11

12 = ω c n th = 1000 n = 1 π dω S cc (Ω) Mean Phonon Number ω c / γ a = 0. ω / γ = 0.5 c a ω / γ = 1 c a ω / γ = c a Input Power (α) Anti Stokes/Stokes Ratio Input Power (α) 1

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