Suceptibility

Given a free energy functional, F=F0+Ff compare to Thermodynamics: F~=Ffx with a generalized force and displacement f,x. So, Ff=dxf(x,t)s(x,t).

Examples:

f S
H M
E P

For \(t'

Fourier transforming this tω,xk. s~(k,ω)=χ~(k,ω)f~(k,ω).

χ(k,ω)=dxdtexp(iωt)exp(ikx)χ(x,t) note the signs only are due to us wanting the plane wave to look like expi(ωtkx). Then we get, χ~(k,ω)=χ(k,ω)+iχ(k,ω). So, χ~=χ~(k,ω).

A system with inversion symmetry, x=x, χ~(k,ω)=χ~(k,ω). χ is even in ω and χ is odd in ω.

Then, χ~(k,ω)=dxdtexp(iωt)exp(ikx)χ(x,t).

C.f. the AC suceptibility P~(k,ω)=χ~(k,ω)E~(k,ω) and M~(k,ω)=χ~(k,ω)H~(k,ω).

Obtaining the work, W=FdS=Fstdt=Fvdt. Power: P=limT1T0Tf(t)stdt=limt1T{0T(s(t)ftdt)s(t)f(t)|0T}, with our generalized force and displacement. The boundary terms go to zero, f(t)={fωexp(iωt)}=12(fωexp(iωt)+fωexp(iωt))=fωcosωt+fωsinωt. f(t)=cosω0t,f(ω)=π(δ(ωω0)+δ(ω+ω0))f(t)=12πf(ω)exp(iωt)dω. Then, s(ω)=χ~(ω)f(ω). s(t)=12πs(ω)exp(iωt)dω=12(exp(iω0t)χ(ω0)+exp(iω0t)χ(ω0))=12(exp(iω0t)(χ(ω0)+iχ(ω0))+exp(iω0t)(χ(ω0)iχ(ω0)))=χ(ω0)cos(ω0t)+χ(ω0)sin(ω0t). Then the power becomes, P=limT1T0Tdts(t)12(iω)(fωexp(iωt)fωexp(iωt))dt. s(t)=tdtχ(tt)f(t). Since there is no response from the future, χ(tt)=0 for \(t (1)p(ω)=dτiωτ4limT1T0Tdt(fω2exp(iω(2tτ))fω2exp(iω(2tτ))+|fω|2(exp(iωτ)exp(iωτ)))(2)=dτiωτ4limT1T(1)0Tdt(|fω|2(exp(iωτ)exp(iωτ)))(3)=ω2|fω|2dτχ(τ)sin(ωτ)(4)=ω2|fω|2{F(χ(τ))}(5)=ω2|fω|2{χ~(τ)}(6)=ω2|fω|2~χ(τ)>0.

Then, ωχ(ω)>0 which means each one must be greater than zero. Know that χ(ω) is odd.

Note: χ(k,ω)=χ(ω) shorthand was used.

Example: Calculate the current through a wire. Suppose we have V(t)=Vωcos(ωt). p=V2R=Vω2cos2ωtR=Vω22R. Suppose we have a wire of length L and cross-section A, conductivity σ=1ρ. R=ρLA=LσA The force (electric field) due to the voltage is then E(t)=Eωcosωt=VωLcosωt. The power density, power per unit volume, p=pV=Vω22R(LA)=Vω22ρL2=Vω2σ2L2=12σEω2=12ω|fω|2χ(ω). Thus, the suceptibity is chi(ω)=σω and |fω|2=Eω2. α(ω), the polarizability is, α=σω. p=limω0p(ω)σ=limω0ωα(ω) gives the DC conductivity on the left and the AC polarizability on the right.

Suppose we have an order parameter smacro(x,t)=sdxG(xx,t)(smacro(x,t)s). Transforming this, s~macro(k,ω)=s~dkG~(kk,ω)(s~macro(k,ω)s~)

Compare if we have a Schrodinger equation and a propogator: Hφ=itφφ(x,t)=dtdxU(xx,tt)φ(xx).

Static Susceptibility

Removing the time dependence.

χ0(r). The response to this continuous force (a lean rather than a kick) is then s(x)=dxχ0(xx)f(x).

We will get: χ~(k)=skeqfk|f=0.

Ff(t)=dxf(x,t)s(x,t). For a static system, Ff=xf(x)s(x). Then, s~keq=Tr(s~kexp(β(F0Ff)))Tr(exp(β(F0Ff)))=1βVlnZfk. χ~0(k)=skeqfk|f=0=βC^(k,0). The hat here means just a fourier transform for space.

For the uniform case, thermodynamically, χ=(ss)2.

Fluctuation-Dissipation Theorem

p(ω)=ω|fω|22χ(ω). Want χC, connect fluctuations.

Cases: f(x,t)χ(x,t) and f(x)χ(x).

Starting from forced equilibrium at t=0 due to f(x). Now we turn off the force. So, f(x,t>0)=0 so what is our χ?

From our response, s(x,tt0)=dxG(xx,tt0)s(x,t0). For t0=0, s(x,t)=dxG(xx,t)s(x,0). Then, s(x,t)=dxtdtχ(xx,tt)f(x,t). We choose, f(x,t)=f(x) for t<0 and 0 otherwise. Then, s(x,t0)=dx0dtχ(xx,tt)f(x). Let τ=tt, dt=dτ, t{,0} so τ{+,t}. Then, s(x,t)=dxf(x)tdτχ(xx,τ) [2b].

For the ideal gas, χ0(r)=δ(r)α. Then, inserting this, s(x,0)=ρ(x,0)=f(x)α. Which gives, ρ(x,t)=dxf(x)αG(xx,t). Cid(x,t)=G(x,t)βα. Then, ρ(x,t)=dxf(x)αβαCid(xx,t)=dxf(x)βCid(xx,t). Comparing this to the last statement in the previous paragraph [2b], βCid(xx,t)tdτχ(xx,τ). Then, βCidt=χ(x,)χ(x,t)=χ(x,t). Which gives us the FD theorem in the time domain, for t>0,

(7)χ(x,t)=βCid(x,t)t

And is zero for t<0.

Note that C(x,t)=C(x,t). Now we can calculate χ(ω). χ(r,ω)=χ~(ω)=±dtχ(t)exp(iωt) which is zero for t<0. So, χ~(ω)=0dtχ(t)exp(iωt)=β0C(t)texp(iωt). Integrating by parts, =β(C(t)exp(iωt)|00dtC(t)(iω)exp(iωt))=χ~+iχ~. The first term is real, thus, χ~=β0dtC(t)(iω)exp(iωt)=βω0dtC(t)cos(ωt). Since C(t)=C(t) and cos(t)=cos(t), thus the integrand is even. Thus, χ~=βω2dtC(t)cosωt=βω2dtC(t)cosωt+iβω2dtC(t)sinωt=βω2dtC(t)exp(iωt)=ωβ2C~(ω).

Then we get the FD theorem in frequency domain for classical systems,

(8)χ~(ω)=ωβ2C~(ω)

Then, p(ω)=βω2|fω|24C~(ω). So in the continuum limit for matter, we get density fluctuations. For solids (crystals), this would be lattice vibrations.

The QM version,

(9)χ~(k,ω)=12(1exp(βω))C~(k,ω).

Starting point is QM version of linear response, Kudo formula, A^(t)=A^0it0tdt[A^(t),V^(t)], 0 with respect to equilibrium and A^=Tr[ρ^A^]Tr[ρ^].

Kramers-Kronig Relations

We have, and abbreviate, χ(r,t)χ(t), real and t>0. Then, χ(r,ω)=χ(ω)+iχ(ω). Then, χ~(ω)=χ~(ω). 2 real valued functions on half line, ω>0.

Let χ(ω)χ(u+iv)=0dtχ(t)exp(iut)exp(vt). The zero coming from the causality condition. From Cauchy’s theorem for a function analytic in a region and a contour in the region, Cf(z)dz=0. Cauchy’s integral formula for a simple pole is cf(z)zzdz=2πif(z).

Then, cχ(ω)ωωdω=0 by making a loop that goes near ω but cuts it out with a small half circle. Taking the off-axis loop to infinity, we get zero. The integral on the axis, limε0=(ωε+ω+ε)χ~(ω)dωωω=PVχ(ω)ωωdω The semicircle integral, χ~(ω)ωωdω=χ~(ω)ωωdω=χ(ω)ln(ωω)|ω+ϵexp(iπ)ω+ϵexp(i0)=iπχ~(ω). Then, the principle value of the integral is iπχ~(ω). Hence, χ~(ω)=1πiχ~(ω)ωωdω. So, χ~(ω)={χ~(ω)}=1πχ(ω)ωωdω.

Since χ~(ω)=χ~(ω), χ~(ω)=χ~(ω). So, χ~(ω)=1π0χ(ω)(1ωω1ωω)=2π0χ(ω)ωω2ω2dω.

Alternatively, χ~(ω)=2ωπ0χ(ω)ω2ω2dω.

Author: Christian Cunningham

Created: 2024-05-30 Thu 21:17

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