Causality and the Kramers Kronig relations

Transcription

1 Causality and the Kraers Kronig relations Causality describes the teporal relationship between cause and effect. A bell rings after you strike it, not before you strike it. This eans that the function that describes the response of a bell to being struck ust be zero until the tie that the bell is struck. Consider a particle of ass oving in a viscous fluid. The differential equation that describes this syste is, dv dt + bv = F(t). Here is the daping constant, is the velocity, and is a driving force. A special case for the driving force is a δ function force which strikes the syste at t =. The solution to the differential equation for a δ function drive force is called the ipulse response function g(t). The sybol g is used because the ipulse response function is soeties called the Green's function. b v F(t) dg dt + bg = δ(t). The solution to this equation is, g(t) = exp( t/τ), Where τ is the decay tie τ = /b g t/τ The utility of the ipulse response function is that any driving force can be thought of as being built up of any δ function forces. /7

2 F(t) = δ(t )F( )d By superposition, the response to a driving force functions. A special driving force is a haronic driving force, occur at the sae frequency as the driving force, haronic force into the equation above. is a su of the ipulse response Since the integral is over t, a factor of e iωt can be put inside the integral.. The response will. To show this, insert a Make a change of variables: t t, d t = dt, and reverse the liits of integration. The only tie dependence of is the factor of because the variable gets integrated out. Thus a haronic driving force F(ω)e iωt produces a haronic response v(ω)e iωt where, The generalized susceptibility χ is the ratio of response to driving force. t t t The generalized susceptibility is the Fourier transfor of the ipulse response function. For the case of a particle oving in a viscous fluid, Another way to calculate the generalized susceptibility is to assue that the driving force and 2/7 F(t) v(t) = g(t )F( )d t t t F(t) = F(ω)e iωt v(t) = v(ω)e iωt v(t) = g(t )F(ω) d t e iωt t v(t) = e iωt g(t t )F(ω) e iω(t ) d = t v(t) = e iωt F(ω) g( ) d t e iωt t v(t) e iωt t v(ω) = F(ω) g(t) e iωt dt v(ω) F(ω) χ(ω) = = g(t) e iωt dt τ χ(ω) =. ( iωτ) +ω 2 τ 2 t t

3 the response both have a haronic tie dependence,. Substituting this for into the differential equation yields, This can be solved for the generalized susceptibility. v(t) = v(ω)e iωt iωv(ω) + bv(ω) = F(ω). v(ω) F(ω) χ(ω) = = τ iωτ =. iω+b +ω 2 τ 2 F(t) = F(ω)e iωt.5. Re I τ χ(ω) ωτ There is a subtle issue with inus signs here. It is equally valid to assue that the haronic dependencies of the drive and the response have the for v(t) = v(ω)e iωt, F(t) = F(ω)e iωt. Notice the inus sign that has appeared in the exponent. With this choice, the iaginary part of the susceptibility changes sign: χ(ω) = = iω+b Either descriptions of the haronic dependence e iωt or e iωt are equally valid and there is no consistent choice ade in the literature. Here we continue with assuing a haronic dependence of e iωt. Be aware that the sign of the iaginary part of the susceptibility ight be different fro forulas found in other sources. The causal nature of the ipulse response function (it has to be zero for ) has consequences for the for of the susceptibility. Any function can be written in ters of an even coponent and an odd coponent. E(t) O(t) Since the ipulse response function ust be zero for, the even and the odd 3/7 τ g(t) = E(t) + O(t). t < +iωτ +ω 2 τ 2 t <

4 coponents ust add to zero for. t < g E O g t/τ Note that if we know the either the even coponent or the odd coponent we construct the other. E(t) = sgn(t)o(t) = O(t) = sgn(t)e(t) = (g( t) + g(t)) Repeating what was stated above, the susceptibility is the Fourier transfor of the ipulse response function, The integral of an odd function over an even interval ( ) is zero so the real part of the susceptibility is the Fourier transfor of the even coponent, and the iaginary part is the Fourier transfor of the odd coponent, Moreover is an even function while is an odd function. 4/7 2 2 ( g( t) + g(t)) χ(ω) = g(t) e iωt dt = (E(t) + O(t))(cos( ωt) + i sin( ωt))dt., Re[χ] = = E(t) cos(ωt)dt, I[χ] = = O(t) sin(ωt)dt. (ω) = ( ω) (ω) = ( ω)

5 The Kraers Kronig relations The Kraers Kronig relations describe how the real and iaginary parts of the susceptibility are related to each other. If either the real part or the iaginary part of the susceptibility is known for positive frequencies ω >, the entire susceptibility can be calculated at all frequencies. Suppose we know for ω >. Then for all frequencies can be constructed because (ω) = ( ω). The even coponent of the ipulse response function can be found by inverse Fourier transforing. The odd coponent of the ipulse response function is related to the even coponent by O(t) = sgn(t)e(t). The iaginary part of the susceptibility can then be constructed since it is the Fourier transfor of the odd coponent. 2 (ω) = E(t) cos(ωt)dt E(t) = (ω) cos(ωt)dω O(t) = sgn(t)e(t) E(t) = sgn(t)o(t) 2 (ω) = O(t) sin(ωt)dt O(t) = (ω) sin(ωt)dω The equations in the box above are known as the Kraers Kroning relations. This is the representation of the Kraers Kronig relations in the tie doain. Many observable quantities obey the Kraers Kroning relations. For instance the electric susceptibility describes the electric polarization of a aterial responds to an applied electric field. This response ust be causal so the real and iaginary parts of the electric susceptibility ust be related by the Kraers Kronig relations. This is also true for the agnetic susceptibility, the electrical conductivity, the theral conductivity, and the dielectric constant. A plane wave oving in the positive x direction has the for e ikx ωt. If the frequency is negative, the wave oves in the negative x direction. Typically in an experient, only the positive frequencies are easured where the waves ove fro a source to the detector. This presents no difficulty since all of the inforation is contained in the positive frequencies. Soeties it is experientally easier to easure the real part or the iaginary part of the susceptibility. The Kraer Kronig relations can then be used to calculate the part that is difficult to easure. If both real and iaginary parts can be easured, it is possible to check for experiental errors using the Kraers Kronig relations. If a susceptibility is calculated theoretically, it is a good idea to check and see if it satisfies the Kraers Kronig relations. It is considered a serious error to present a result that violates causality. It is traditional to write the Kraers Kronig relations in the frequency doain. This unfortunately introduces a singularity in the forula. The singularity in the integral akes the for that is given below less suitable for a nuerical evaluation of the Kraers Kronig relation. Nevertheless, it coonly appears in the literature and is given for copleteness. i Since the Fourier transfor of sgn(t) is, we can use the convolution theore to take the ω Fourier transfor of the equations and, O(t) = sgn(t)e(t) E(t) = sgn(t)o(t) 5/7

6 = i ( i ), ω i = i. ω Here '*' represents convolution. Using the definition of convolution yields the Kraers Kronig relations in the frequency doain. = P d, ( ω ) ω ω ( ω ) ω ω = P d. Here the P before the integral indicates that one should use the Cauchy principle value of the integral. This is necessary because of the singularity that the integral contains. The advantage of this for is that one sees iediately that the real part of the susceptibility can be deterined fro the iaginary part and vice versa without transforing to the tie doain. The Kraers Kronig relations are often put in another for where the integrals only involve positive frequencies. The integral for is split into two parts. In the first ter ake a change of variables ω, use the fact that is an odd function:, and reverse the liits of integration. The integrals can be cobined. Rewriting the factor, ( ω) = ( ) ω ( ω ) ω ω ω = P d P d, (ω) ( ω ) ω +ω ( ω ) ω ω = P d P d, = P ( + ) ( )d ω +ω ω ω ω ( + ) =, ω +ω ω ω 2ω ω 2 ω 2 ( ) 6/7

7 the Kraers Kronig relations can also be written, = 2 P d, 2ω ω ( ω ) ( ω 2 ω ) ω 2 ( ) ( ω 2 ω ) ω 2 χ = P d. Note that the singularity is stronger in this for aking it less suitable for a nuerical evaluation.. Classical linear response theory is described in Response and Stability by A. B. Pippard, Cabridge University Press (985). 2. A discussion of causality and separating the ipulse response function into even and odd parts is found in The Fourier Transfor and Its Applications by R. N. Bracewell, McGraw Hill (978). 7/7