Linear Response Theory

LINEAR RESPONSE THEORY We have statistically described the time-dependent behavior of an internal variable in an equilibrium system through correlation functions, and now we would like to relate that to experimental observables. That is, how does the system respond, if you drive it from equilibrium? Equilibrium System weak coupling A External Agent internal variable > The system is moved away from equilibrium by external agent. > The system absorbs energy from external agent. What are the time-dependent properties of the system? H = H0 − f (t ) A Internal variable Time-dependence of external agent Hamiltonian of equilibrium state A(t) We average over an ensemble, each member of which is subject to same perturbation. A t0 external force applied at t0 t … ≡ average over equilibrium ensemble … ≡ average over nonequilibrium ensemble moving system from A A ≠ A ( t ) due to interaction Let’s develop A(t ) as an expansion in power of f (t ). ( A ( t ) = terms f ( A(t ) = 0) 1 ∫ + A ) + (terms f ( ) ) +… +∞ −∞ dt0 R ( ) ( t ,t0 ) f ( t0 ) +… 1 R ( ) ( t,t0 ) : Linear Response Function 1 The force is applied at t0 , and we observe the system at t . The linear response function is the quantity that contains the microscopic information that describes how the system responds to the applied force. We will look to find a quantum description of R(1). Rationalization for an expansion of A(t ) in powers of f (t ): Let’s break time up into infinitesimal intervals: ∆ t1 t2 ∆ ∆ t3 t4 ti = i ∆ f ( ti ) = f i A(ti ) = Ai = Ai (…, fi −2 , fi−1 , fi ) Now, Taylor series expand about all fi = 0 A ( ti ) = Ai (… 0 , 0 , 0 ) + A Value with no f applied Linear term:  ∂A ∑j  ∂f i  j  fj   f j =0 lim ∆→0  ∂ Ai  fj  j ≤i  j  f =0 j ∑  ∂f j  1 ∂Ai   fj   j  = ∫ dt j R ( ti ,t j ) f ( t j ) ti −∞ … Sum over change due to force at all times of application ∑ j∆  j∆ ∂f = + Causality: The system cannot respond before the force has been applied. R ( ) ( t,t0 ) = 0 for t < t0 1 The time-dependent change in A is t δ A ( t ) = A ( t ) − A = ∫ dt0 R ( ) ( t,t0 ) f ( t0 ) 1 −∞ Stationarity: The time-dependence of the system only depends on the time interval between application of force and observation. R ( ) ( t ,t0 ) = R ( ) ( t − t0 ) 1 So, 1 t δ A(t ) = ∫ dt0 R(1) (t − t 0 ) f (t 0 ) −∞ The response of the system is a convolution of the material response with the time-development of the applied force. Usually, we define the time interval τ = t − t0 ∞ δ A ( t ) = ∫ dτ R ( ) ( τ ) f ( t − τ ) 1 0 Impulse response. For a delta function perturbation: f (t ) = λδ (t − t 0 ) δ A(t ) = λR(1) (t − t 0 ) Frequency-Domain Representation ∞ δ A ( t ) = ∫ d τ R( ) ( τ ) f ( t − τ ) 1 0 Fourier Transform both sides: δ A ( ω) ∫ +∞ −∞ ∞ 1 dt  ∫ d τ R ( ) ( τ ) f ( t − τ )  eiωt  0  insert δ A ( ω) ∫ +∞ −∞ ∞ dt ∫ d τ e iω( t −τ ) −∞ (e −iωτ e +iωτ ) R( ) ( τ ) f ( t − τ ) 1 setting t′ = t − τ +∞ ∞ −∞ 0 dt′ = dt = ∫ dt ′ eiωt′ f ( t ′ ) ∫ d τ R ( ) ( τ ) eiωt f ( ω) 1 χ ( ω) susceptibility Fourier-Laplace transform F.T. δ A ( ω) = χ ( ω) f ( ω) spectral response A convolution of the force and response in time leads to the product of the force and response in frequency. This is a manifestation of the convolution theorem: ∞ ∞ −∞ −∞ A ( t ) ⊗ B ( t ) ≡ ∫ dτ A ( t − τ ) B ( τ ) = ∫ dτ A ( τ ) B ( t − τ ) = F where A ( ω) = F  A ( t )  and F −1  A ( ω ) B ( ω )  [ ] is a Fourier transform. Note that R(1) (τ ) is real, since the response of a system is an observable. The susceptibility χ (ω ) is complex. We will relate C AA ( τ ) to R(1) (τ ) and σ abs ( ω) to χ (ω ) . χ ( ω) = χ′ ( ω) + iχ′′ ( ω) ∞ χ′ χ ( ω) = ∫ d τ R ( ) ( τ ) eiωτ 1 χ′′ 0 ∞ ∞ = ∫ d τ R ( ) ( τ ) cos ωτ + ∫ d τ R ( ) ( τ ) sin ωτ 1 0 1 0 χ′: even in frequency ( ) 1 χ′ ( ω) = Re  F R ( ) ( τ )    χ′′: odd in frequency ( χ′′ ( ω) = Im  F R (  ω 1) ( τ))   0 χ ′(ω ) = χ ′(−ω ) χ ′′(ω ) = − χ ′′(−ω ) Notice also ∴ χ (− ω ) = χ (ω ) * χ ′(ω ) = 12 [χ (ω ) + χ (−ω )] χ ′′(ω ) = 1 2i [χ(ω ) − χ (−ω )] χ * (ω ) KRAMERS-KRÖNIG RELATIONS A consequence of causality is that χ ′(ω ) is not independent of χ ′′(ω ) : χ′ ( ω) = 1 +∞ χ′′ ( ω′ ) P dω′ π ∫-∞ ω′ − ω χ′′ ( ω) = 1 +∞ χ′ ( ω′ ) P dω′ π ∫−∞ ω′ − ω These are obtained from ∞ 1 χ′ ( ω) = ∫ R ( ) ( t ) cos ωt dt 0 and R( ) ( t ) = 1 1 +∞ χ′′ ( ω) sin ωt dt π ∫−∞ Substituting: χ′ ( ω) = = +∞ 1 ∞ dt cos ωt ∫ χ′′ ( ω′ ) sin ω′t d ω′ ∫ −∞ π 0 +∞ R 1 lim ∫ dω′χ′′ ( ω′ ) ∫ cos ωt sin ω′t dt 0 π R→∞ −∞ using cos ax sin bx = 12 sin ( a + b ) x + 12 sin ( b − a ) x = +∞  − cos ( ω′ + ω) R +1 cos ( ω′ − ω) R +1 1 − lim P ∫ dω′χ′′ ( ω) 12   −∞ ω′ − ω π R→∞ ω′ + ω   If we choose R → ∞ , the cosine terms vanish since they oscillate rapidly. This is equivalent to averaging over a monochromatic field. If we instead average over a single cycle: R = 2π / ( ω′ − ω) , we obtain χ′ ( ω) = χ′′ ( ω′ ) 1 +∞ P ∫ dω′ π −∞ ω′ − ω The other relation can be derived the same way. Example: Classical Response Model absorption of radiation by dipoles with a forced damped harmonic oscillator: x + γ x + ω02 x = F ( t ) For an E.M. wave: F ( t ) = F0 cos ωt = x (t ) = qE0 cos ωt m qE0 1 1 cos ( ωt + δ ) 2 m  2 2 2 2 2 ( ω0 − ω ) + γ ω  sin δ = 2γ ω ( ω2 − ω2 )2 + γ 2ω2   0  1 2 ∞ An impulsive driving force gives the response function: x ( t ) = ∫ d τ R ( ) ( τ ) f ( t − τ ) if F(t ) = F0δ (t − t0 ), then x (t ) = F0 R(1) (t ): R( ) ( τ ) = 1 χ ( ω) = χ′′ ( ω) = 1  γ  exp  − τ  sin Ωτ mΩ  2  1 0 Ω = ω02 − γ 2 4 1 m ( ω − ω2 − iγω) 2 0 2γω m ( ω02 − ω2 ) + γ 2 ω2    2 for γ << ω0 γ χ= χ ( ω) ≈ 1m ω02 − ω2 + i γω 1 1 2mω0 ω − ω0 + i γ / 2 for ω ≈ ω0 ω2 − ω02 = ( ω + ω0 )( ω − ω0 ) ≈ 2ω0 ( ω − ω0 ) ≈ 2ω ( ω − ω0 ) ω0 ω χ≈ 1 1 1 = 2mω0 ( ω0 − ω) + i γω 2mω0 ω − ω + iγ ( 0 ) 2 Nonlinear Response Functions If the system does not respond in a manner linearly proportional to the applied force, we can include nonlinear terms: the higher expansion orders in A(t ) . Let’s look at second order: δ A (t ) ( 2) = ∫ dt1 ∫ dt2 R ( 2) ( t;t1 ,t2 ) f1 ( t1 ) f 2 ( t2 ) Again we are integrating over the entire history of the application of two forces f1 and f2, including any quadratic dependence on f . apply f1 apply f2 observe t1 t2 t time In this case, we will enforce causality through a time ordering that requires (1) that all forces must be applied before a response is observed and (2) that the application of f2 must follow f1: R( 2) ( t;t1 ,t2 ) ⇒ R( 2) ⋅ Θ ( t − t2 ) ⋅ Θ ( t2 − t1 ) = ∫ dt2 ∫ dt1 R ( 2) ( t;t1 ,t2 ) f1 ( t1 ) f 2 ( t2 ) t ≥ t2 ≥ t1 δ A(t ) ( 2) or t t2 −∞ −∞ Now we will call the system stationary so that we are only concerned with the time intervals between interactions. δ A(t ) ( 2) t = ∫ dt2 −∞ ∫ t2 −∞ dt1 R ( 2) ( t − t2 ,t2 − t1 ) f1 ( t1 ) f 2 ( t2 ) If we define the intervals between adjacent interactions τ1 = t2 − t1 τ 2 = t − t2 τ1 t1 ∞ = ∫ d τ1 0 ∫ ∞ 0 d τ2 R( 2) ( τ1 , τ2 ) f1 ( t − τ1 − τ2 ) f 2 ( t − τ2 ) τ2 t2 t