Deep lessons from simple Systems.A single oscillator in presence of external forces

The equation for a macroscopic harmonic oscillator, i.e. a mass tied to a spring, in a

dissipative environment is

m d2udt2

= −Ku + Ffrict.t + Ft

taking ω02 = K/m we write. Ft represent the controlled external forces (gravity, electric fields,

etc.), while Ffrict.t represents the stochastic action of the microscopic particles in the

environment (air, water or oil molecules). In equilibrium, the mean value of the position is

⟨ut⟩ = 0, i.e. even in presence of fluctuations their net effect cancels out. But out of equilibrium

it is an experimental fact that friction Ffrict.t will tend to damp the oscillator amplitude in

opposition to the velocity. The simplest expression is:

Ffrict.t ≃ −α. dudt

A microscopic justification of this law implies the consideration of the action of microscopic

particles following a Brownian motion and colliding with the macroscopic oscillator. If the

oscillator is driven by an external force:

d2udt2

+ η dudt

+ ω02u = 1

m Ft

with η = α/m. A general way to solve this equation is to write the force in its Fourier

components:

Ft = ∫ Fωexp−iωt dω2π

Analogously, we write the Fourier component of the displacement: uωexp−iωt form which

we get:

−ω2 − iωη + ω02 uω exp−iωt = Fωexp−iωt

from which

uω =−1/m

ω2 + iωη − ω02

Fω = χωFω

with:

χω =−1/m

ω2 + iωη − ω02

going back to time dependence we get

ut = ∫−∞

∞χt − t ′Ft ′dt ′

where χt is different from cero only for positive times and can be thought as the response to a

δt function:

m d2

dt2+ mη d

dt+ mω0

2 χt = δt

or equivalently

−mω2 − iωηm + mω02 χω = 1

The response function χω has poles when the argument ω takes the values

ω± = ± ω02 − η2/4 − iη = ±ω0 − iη.

The poles occur in the lower half of the complex z =ω plane

χω =mω0

−1

ω − ω0 + iη/2−

mω0

−1

ω + ω0 + iη/2

From χω the time response can be evaluated applying the Fourier transformation

χt = ∫−∞

∞χωexp−iωt dω

2π.

This integral that can be performed by residues.

For t > 0 we can write Fourier transform as a contour integral with the contour closed in the

lower half plane so the exponential in the integral exp−izt will be convergent for Im z < 0 and

t > 0and includes the poles. Therefore the residues have a finite value. For t < 0 in order to

obtain convergence the integral must be close in the upper half plane, Im z > 0, which does not

contain poles and we get zero. Therefore, we get

χt =e−ηt/2 sin ω0

2 − η2/4 t

m ω02 − η2/4

θt

From the analysis of the argument of the square root we see that there are two different

dynamical phases:

The oscillating or under-damped regime which appears provides that ωo > η/2 and the

argument of the square root is positive. Of course if η = 0 no absorption of energy would be

possible. Notice that the damping mechanism has the effect of decreasing the oscillation

frequency into

ω02 = ω0

2 − η2/4

With the small tilde we indicate that the frequency has been corrected by the presence of the

environment.

At the critical value of ω0 = η/2 we are confronted with the limit

sin ω02 − η2/4 t

ω02 − η2/4 ω0

2−η2/4 →0t

hence it gives a response decaying with

χtcritical

tm e−ηt/2θt

which for long time is still dominated by the exponential.

In the other dynamical regime, the over-damped phase, for which ωo < η/2 would render

only the exponential instead of the oscillation meaningless. However the expression above can be

used nevertheless because we know that the analytic continuation of a sine function for an

imaginary argument is just an hyperbolic-sine. Hence

χtoverdamped

→e−ηt/2 sinh t η2/4 − ω0

2

m η2/4 − ω02

θt

We analyze the asymptotic behavior expanding the sinh in terms of exponentials. The decreasing

exponential tends to It is interesting to notice that a general initial condition would have

components decaying with two different rates (imaginary frequencies)

τf−1 = 1

2η 1 + 1 − 4ω0

2/η2ω0 ≪η/2

η

τs−1 = 1

2η 1 − 1 − 4ω0

2/η2ω0 ≪η/2

ω02/η = K/α

τf−1 keeps decreasing with η while τs

−1 reach a minimal value K/α. A general initial condition

with a component involving the last rate is the one that survives the longer times.

One of the lessons we have learned from the analysis of these cases is that analytic

continuation is a key word for understanding the response. Once we have obtained a solution in a

region of parameters we have the behavior in the other regions just by analytic continuation .

This response function can be split into its real and imaginary part as:

Reχ = χ ′ω = 1m

ω2 − ωo2

ω2 − ωo2 2

+ η2ω2

which is associated with the energy dispersion. While Reχ generates a displacement that is in

phase with the force, in the evaluation of the work applied in a period enters the integral of the

force times the derivative of the displacement, which cancels out after integration in a period

To = 2π/ωo . In contrast

Imχ = χ ′′ω = 1m

ηω

ω2 − ωo2 2

+ η2ω2

is associated with the energy absorption. This is because it leads to a displacement that is 90o

out of phase with the force.

Problem. Consider an harmonic force. By defining the average absorbed power ⟨P⟩in a period T = 2π/ω as the work realized by the force in the unit of time:

⟨P⟩ = 1T∫

0

T

Ftdut

dtdt

= 1T∫

0

T

χ ′ω + iχ ′′ωFωe−iωt 2dt

= 12Fω 2ωχ ′′ω

Problem. Analogously to the previous problem show that the energy stored in the system is

E = 12

mω2 + ω02 1

2χ ′ω2 + 1

2χ ′′ω2

= 12m

ω2 + ω02

ω2 − ωo2 2

+ η2ω2

The over-damped regimeThe solution in this limit is equivalent to the complete neglect of the inertial term and it is a

solution of the equation of motion

αu =

spring force

−Ku +

other forces

F .

In the extreme over-damped case discussed above, the susceptibility becomes:

χω ≅ 1m

1ω0

2 − iωη= χ0 1

1 − iωτs

Similarly, the imaginary part χ ′′ω = Imχω becomes

χ ′′ωω ≅ χ0 τs

1 + ω2τs2

We will encounter these expressions again in the context of the diffusion equation.

Epistemological note:-The extremely overdamped regime describes an Aristotelian (i.e. mass independent) motion,

where the notion of inertia is hidden by the role of friction. We see that in between the

Aristotelian, extremely overdamped regime, and the Newtonian regimen of well defined

oscillations there is an analytic discontinuity. Where does it come from? From the

phenomenological use of a friction coefficient. Friction is not fundamental but involves a

particular limit in our vision of Nature. The existence of this discontinuity gives a mathematical

framework to the concept of "incommensurability" of paradigms introduced by Thomas Kuhn to

describe the conflict between the philosophical views of both giants of science.

Mathematical Constraints.

From the solution of the simple oscillator we learnt some general properties of the response

function.

● The response function is analytic (i.e. without poles) in the upper half plane.

● The response function goes to zero faster than 1/ω. This is equivalent to require that

∫0

∞χtdt < ∞ or equivalently that a finite force produces a finite response.

● χ∗ω = χ−ω● The real and imaginary part can be obtained form each other from the Cauchy dispersion

relation:

χ ′ω = − 1π P ∫−∞

∞ χ ′′ω′ω − ω′

dω′

and

χ ′′ω = 1π P ∫−∞

∞ χ ′ω′ω − ω′

dω′

In connection with response functions measured in different spectroscopic techniques, we

remember that they also can be cast in a form that satisfy the Kramers-Kroning relations:

χ ′ω − χ ′∞ = − 2π ∫

0

∞ ω′χ ′′ω′ω2 − ω′2

dω′

and

χ ′′ω = − 2ωπ ∫

0

∞ χ ′ω′ω2 − ω′2

dω′

where χ ′′∞ = 0.

The complete solution.Finally, we remember that the previous paragraphs refer to the response to the force. To

obtain the complete solution of the differential equation one has to sum up the solution of

particular solution (represented above in terms of the response function) plus a solution of the

homogeneous equation which is used to satisfy the initial conditions. Since the equation is

linear and second order in time, we see that a solution of can be written as superposition of two

linearly independent solutions. It is immediate to choose one of them of velocity type:

ut=01

= 0 and mvt=01

= mvo

which yields a solution of the form:

ut =vo

ωosinωot,vt = vo cosωot

Alternatively, we can also have an initial condition of the displacement type

ut=02

= u0 and mvt=02

= 0,

with solution

ut = u0 cosωot; vt = −ωo u0 sinωot.

In this case, there is no much trouble finding the solution of the homogeneous equation.

However, it seems a bit ackwards that after the job done finding the asymptotic solution of the

forced oscillator, we have to start all over to find the solution of the homogenous equation.

However, a hint appears realizing that a solution for the initial condition of velocity type can be

obtained as the response to an impulsive force, that is, a kick or delta function.

u1t = ∫ χt − t ′Fv0δt′dt ′ = χt

after kick

mvt=0+ −

before kick

mvt=0−

= χtmvo

= vo e−ηt/2 sinω0 tω0

θt

In analogous form we can build a force that imposes a displacive initial condition.

Problem. Use the response formalism to show that the solution of the displasive case can be

obtained from a force with the form of a kinck plus an opposed kick:

Fu0t = limτ→0

mu01τ δt + 1

2τ − δt − 1

2τ + ηδt

= mu0∂δt ′∂t ′

+ ηδt ′

the second term warrants that even in presence of friction the initial velocity is zero.The

corresponding Fourier transform results

Fu0ω = −iω + iηmu0

The solution of the dysplasive case represents a double kick on the mass: the one forward

that gives an impulse producing a displacement, followed by another kick backward that leaves

the particle again at rest at position u0. The second term appears in cases where there is a finite

dissipation to allow for the condition of null initial velocity.

This initial condition with the instantaneous double kick is very useful as it is an action local

in space and it can be directly extended to problems involving any number of masses.

Complementary readings

See the description of oscillator in an external force and fluctuating stochastic bath: P. C.

Martin, in “Many-Body Physics”, Ed. C. de Witt and R. Balian eds., Les Houches Summer

School 1967, Gordon and Breach, New York 1968. This is partially reproduced in Linda Reichl

Statistical Mechanics

The Kramers-Kroning relation is discussed in many texts. A good one is Charles Kittel

“Statistical Physics ” Chapter 44, inventory FaMAF 1474.

A Two Body Problem: Coupled Oscillators.Let us consider the case of two oscillators coupled through a spring K.

d2u1

dt2= −ω1

o2u1 − Km1

u1 − u2

= −ω1o2 + K

m1u1 +

Km1

u2

= −ω1o2 + ωx12

2 u1 + M1,2u2

d2u2

dt2= −ω2

2 + ωx212 u2 + M2,1u1

By introducing the trivial notation

− ωx122 = −K/m1 = M1,2

− ωx212 = −K/m2 = M2,1

We get first notice that the equation would be trivial if in the first equation we replace u2 by its

time average value ⟨u2 ⟩ = 0 and in the second u1 → ⟨u1 ⟩ = 0.

As before, we search for solutions of the equation with the form u1t = e−iωtu1,ω and

u2t = e−iωtu2,ω which would be valid in presence of some dissipation imposed by an imaginary

component in the frequency (that we do not write explicitly).

ω2I−Mo

ω2 − ω1o2 − ωx12

2 0

0 ω2 − ω2o2 − ωx21

2

u1,ω

u2,ω

=0

0

with eigen- frequencies ω12 = ω1

o2 + ωx122 and ω2

2 = ω2o2 + ωx21

2 . with a "steady" neighbor mass

renormalizes the ‘free’ frequency by inserting a new spring to connect the particle 1 to a single

neighbor particle (2) which is taken fix

ω 12 = ω1

o2 + ωx122 = M1,1

and

ω 22 = ω2

o2 + ωx212 = M2,2

This is equivalent to having the mass 2 is pinned to at its mean value at the equilibrium position.

the presence of the spring modifies the frequency of mass 1. The analogous situation is valid for

the shift of frequency 2. We include these “mean-field” renormalization in the free frequency and

forget for the moment their presence.

Any election of u1,ω and u2,ω would give a solution to the equation. The respective

normalized eigenvectors

s1 = exp−iω 1t1

0and s2 = exp−iω 2t

0

1

where the vector 1,0 is the residue of the divergence on the first term.

Trivially, the eigen- frequencies can be obtained from the secular determinant or equivalently

analyzing the resolvent

Doω =

1ω2 − ω1

20

0 1ω2 − ω2

2

which has poles at the natural frequencies ±ω1 and ±ω2 already including part of the interaction

in the Hartree approximation. Sometimes, ω12 and ω2

2 are called Einstein phonons, as they

correspond to isolated vibrational modes, because Einstein modeled the specific heat of s of a

solid with N atoms with 3N independent vibrational models.

Now, we allow the masses to move and as they fluctuate around their equilibrium positions

an interaction appears. The secular equation that must be solved can be cast in the form:

ω2 − ω 12 −M1,2

−M2,1 ω2 − ω 22

u1,ω

u2,ω

= ω2I −Mu1,ω

u2,ω

= 0 #

the matrixM is called the dynamical matrix, it plays a role similar to a Hamiltonian but it is not

necessarily symmetric. While the coupling term is not symmetric the relevant quantity is their

product

ωx4 = M1,2M2,1,

We would like to get the canonical transformation that permits to obtain the normal modes and

the eigen-frequencies.

Decimation Procedure.We will apply a procedure that here it is trivial, but contains the seeds of the important

Renomalization Group procedure.

Imaging that we suspects that we suspects that oscillator 1 is no much modified by the

presence of oscillator 2. Therefore we want to elliminate the degrees of freedom of the later.

0 = −M2,1u1,ω + ω2 − ω 22u2,ω

→ u2,ω =M2,1

ω2 − ω 22

u1,ω

→ 0 = ω2 − ω 12 −

non-linear “shift” due to u2,ω

M1,2M2,1

ω2 − ω 22

u1,ω

0 = ω2 − ω 12 −

Π11ω2

M1,2M2,1

ω2 − ω 22

The last equation must be fullfilled independently of the amplitude u1,ω and is satisfied when

ω is one eigen-frequencies ω12 and ω2

2 of the system. In this case these are the roots of the

quadratic equation:

ω4 − ω2ω 12 + ω 2

2 + ω 12 ⋅ ω 2

2 − ωx4 = 0

are

ω22=

ω 12 + ω 2

2

2+

ω 22 − ω 1

2

2

2

+ ωx4

= ω2o2 + ωx

2 +ωx

4

ω 22 − ω 1

2+ Oωx

6

Similarly,

ω12=

ω 12 + ω 2

2

2− ω 2

2 − ω 12

2

2

+ ωx4

We see that for weak coupling we indeed can expand the square root in power series obtained a

perturbative expansion for the eigenfrequencies in terms of the coupling ωx4 and we get:

ω12=

ω 12 + ω 2

2

2+

ω 22 − ω 1

2

21 + 1

2

4ωx4

ω 22 − ω 1

22+…

= ω1o2 + ωx

2 − ωx4

ω2o2 − ω1

o2+…

ω22= ω2

o2 + ωx2 +

ωx4

ω2o2 − ω1

o2+…

The condition for the convergence of this series in the square bracket being

|Π11ω 12|2 =

2ωx4

ω 22 − ω 1

22≡ 2M1,2

ω22 − ω1

2< 1

This is a form of the Rayleigh − Schrödinger perturbative expansion (the standard form used in

Quantum Mechanics). The perturbation expansion has a restricted convergence. For example a

case with equal masses could not have been dealt with if the exact solution of the secular

equation were not at hand. We observe that the interaction moves both frequencies upwards

while there is a net decrease in the separation of the frequencies.

When the coupling is very strong our physical intuition would like to see the two masses

forming a solid “bar” which oscillates at a frequency which somehow averages the old ones.

Besides, this rigid “body” must have its own internal vibrational mode directly proportional to

ωx.

Check notation to see the definition of the corrected frequencies.

ω1o = 10; ω2

o = 15

It is interesting to notice the for small perturbation give the effect decrease of difference in

the eigen-frequencies shown in the decrease of their difference:

ω1o = 10; ω2

o = 15

before both frequencies collapse one of them escapes toward infinite and the difference begins to

increase. This is not seen if one plots Δω2 vs. ωx.2 which is a linear plot.

Response of a Two-Body System.The Resolvent or propagator.

As in the case of the isolated mass we are interested in the response of the system to different

external forces.

We notice that in presence of interaction M1,2

Dω =

1

ω2 − ω 12 − M1,2

1ω2 − ω 2

2M2,1

−M1,2

ω2 − ω 12ω2 − ω 2

2 − M1,2M2,1

−M2,1

ω2 − ω 12ω2 − ω 2

2 − M1,2M2,1

1

ω2 − ω 22 − M2,1

1ω2 − ω 1

2M1,2

Let us analyze firstly the response of the mass 1 to a force applied over itself, i.e. the element

D1,1ω.

D1,1ω2 = 1

D1,1oω2

−1− M1,2D2,2

oω2M2,1

=D1,1

oω2

1 − M1,2D2,2oω2M2,1D1,1

oω2

i.e. the exact solution of the resolvent can be written in terms of the unperturbed terms of the

resolvent.

Wigner-Brillouin perturbation for the propagator.The denominator in this equation, in turn, can be expressed as a series in powers of the

interaction parameter between oscillators.

D1,1ω2 = D1,1oω2 + D1,1

oω2 M1,2D2,2oω2M2,1D1,1

oω2

+ D1,1oω2 M1,2D2,2

oω2M2,1D1,1oω2

2+ ⋯.

Dyson Equation and Feynman Diagrams.The series can be expressed in a somewhat more compact notation by defining

Π1,1+ ω22 = M1,2D2,2

oω2M2,1

We can write the equation as:

D1,1ω2 = 1

D1,1oω2

−1− Π1,1

+ ω2

= D1,1oω2 + D1,1

oω2Π1,1+ ω2D1,1

oω2

+ D1,1oω2Π1,1

+ ω2D1,1oω2Π1,1

+ ω2D1,1oω2 +…

= D1,1oω2 + D1,1

oω2Π1,1+ ω2 D1,1

oω2 + D1,1oω2Π1,1

+ ω2D1,1oω2 +…

identifying the bracket with the expression of the second line, we get

D1,1ω2 = D1,1oω2 + D1,1

oω2Π1,1+ ω2D1,1ω2 #

which is called the Dyson equation. Clearly Π1,1+ ω2 represents a “dynamical” shift of frequency

ω1 produced by the presence of the mass 2 at the right. The word dynamical here means that the

shift depends on the frequency one is attempting to test with the external force. This

self-consistent equation has also a diagrammatic representation as Feynman diagrams.

A completely analogous procedure permits to define

Π2,2− ω2 = M2,1D1,1

oω2M1,2,

the dynamical mass renormalization due to the presence of mass 1 at the left. It interesting to

notice that the sum of the series is defined even beyond the radius of convergence, i.e.

M1,2D2,2oω2M2,1D1,1

oω2 < 1

even when the perturbation is not longer a small parameter. This phenomena is sometimes called

the “disappearance of disagreeable divergences”. This is somewhat unexpected if one sticks to

standard Rayleigh-Schrödinger perturbation theory and it is one of the strengths of the Green’s

functions treatment. One can interpret this by thinking that we are doing is a perturbation theory

on the response function. In quantum theory this is equivalent to the Wigner-Brillouin

perturbation theory.

There is also an interesting way to write the non-diagonal term found above.

D1,2 =M1,2

ω2 − ω 12ω2 − ω 2

2 − M1,2M2,1

= 1ω2 − ω 1

2M1,2

1

ω2 − ω 22 − M1,2

1ω2 − ω 1

2M2,1

= D1,11

M1,2D2,22

Here we have introduced the notation D1,11

for the response of mass 1 (the first) in absence of

M1,2. The interaction term D2,22

is the response of the second mass in the presence of 2 (two)

coupled masses.

Meaning of the residues.

Let us assume that we were able to find the normalized normal modes 1 and 2 which can

be expresed in terms of the unperturbed ones |1⟩ and |2⟩ using the projections u 1,1 = 1|1⟩,

u 1,2 = 1|2⟩, etc. i.e. we know a transformation matrix that transforms the non-diagonal matrix

into the normal form.

M = UMU−1=ω2 − ω1

20

0 ω2 − ω22

U−1=

u 1,1 u 2,1

u 1,2 u 2,2

The same transformation, applied to the resolvent would give

Dω2 =

1

ω2 − ω12

0

0 1

ω2 − ω22

in turn, applying back the transformation we get

Dω2 = U−1

1

ω2 − ω12

0

0 1

ω2 − ω22

U

for example

D1,1ω2 =u1,

1u

1,1

ω2 − ω12+

u1,2u

2,1

ω2 − ω22

this means that each pole appears weighted by the components of the normal coordinates in the

old variables. This suggests the definition of a spectral density

ρi,jω2 = − 1π lim

δ→+0ImDi,jω2 + iδ

Hence, we have achieved an algebraic procedure to obtain detailed information on the normal

modes and eigen-frequencies. If we want to be consistent with the definition of η as a rate of

decay used in the first section we write

δ = ω0η

This election of this for imaginary part, as compared with ωη that appears in the realistic model

introduced in the first section, has the advantage that the real component of the frequency

remains unchanged by its insertion. i.e. all the frequencies are just shifted toward the imaginary

component of the complex plane z = ω2 in the same amount δ2. On the other hand, this

simplification has the disadvantage that one could miss the dynamical phase transition allowed

by the imaginary part ωη. However, in a many-body problem this transition would be hard to

describe and for most practical cases the perturbative effect of dissipation introduced by δ2

would suffice.

Antiresonance in the forced pairLet us consider two weakly coupled harmonic oscillators, where only one of them is driven

by a periodic force [see Fig. 2(a)]. In such a system, one expects two resonances located close to

unperturbed eigenfrequencies ω1 and ω2. One of the resonances of the forced oscillator

demonstrates . Amplitude on the second oscillator shows two symmetric line-shapes, described

by a Lorentzian functions, and is known as Breit-Wigner resonance near its eigenfrequency ω1,

and ω2. However, when looking at the amplitude of the first oscillator, one sees a first resonance

of a Breit-Wigner form while the second resonance exhibits unusual sharp suppression of the

amplitude near the eigenfrequency of the second oscillator ω2 [see Fig. 2(b,c)]. This dip

represents the destructive interference between between the forces of the second oscillator and

the external force.

Problem. Complete the analysis of this case.

Epistemological note: Using an harmonic oscillator as a model for the atom Hendrik Antoon

Lorentz suggested that the oscillations of these charged particles were the source of light. In

spectroscopy, the theory of polarization of resonance radiation was developed simultaneously by

Breit, Joos, Enrique Gaviola and Peter Pringsheim. Lorentzian shapes

The asymmetric shapes with the dip was first recognized and interpreted in the ionization

cross section by Ugo Fano, then it was known a Fano shape. In vibrational spectroscopy they

where found in Raman spectrum by Fernando Cerderia and collab. In the context of electrical

conduction this phenomenon has been dubbed antiresonance [D’Amato, Pastawski, Weisz,

PRB89]. All these phenomena, which seemed unrelated at first sigth, are now know to be a form

of interference.

The solution of the three body problemHere we solve the three body problem to obtain the basic procedure that already hints on the

solution of the more general situation of N-body problem and even the case N → ∞D1,1ω2 = D1,1

oω2 + D1,1oω2Π1,1

+3ω2D1,1oω2

= 1

D1,1oω2

−1+ M2,1

1

D2,2oω2

−1− M2,3D3,3

oω2M3,2

M1,2

i.e. it is defined in terms of a continued fraction. The supra index m indicates that the quantity

is calculated in a finite m-body system. The frequency corrections are:

Π2,2− ω2 = M2,1D1,1

oω2M1,2

Π2,2+ ω2 = M2,3D3,3

oω2M3,2.

Notice that the perturbation series for this term becomes:

D2,2ω2 = D2,2oω2

× 1 + Π2,2+ ω2D2,2

oω2 + Π2,2+ ω2D2,2

oω2Π2,2+ ω2D2,2

oω2 + ⋯

× 1 + Π2,2− ω2D2,2

oω2 + Π2,2− ω2D2,2

oω2Π2,2− ω2D2,2

oω2 + ⋯

i.e. that the infinite (possibly divergent series) is written as the product of two infinite (possibly

divergent series) and still it can be summed up to an exact, finite value.

Finally, the corrected local Greens function is

D2,2ω2 = D2,2oω2 + D2,2

oω2Π2,2+ ω2 + Π2,2

− ω2D2,2ω2

= 1

D2,2oω2

−1− Π2,2

+ ω2 − Π2,2− ω2

Problem: Show that we also can calculate this correlation function:

D1,3 = D1,11

M1,2D2,22

M2,3D3,33

This formula will be generalized for the more general N body problem.

The solution of the N body problemHere, we obtain the analytical results.

We introduce the

The solution of the X-talConsider an homogeneous infinite chain or a finite set of N particles in a ring configuration

this last will be called Born-von Karmann boundary condition. We start with the mean field

frequency

ω n2 = ω0

2 + 2ωx2 for all n

Here the 2 in represents the 2 neighbors (Notice: this would not hold for a mass at the edge of

chain which has only one neighbor).

Mn,n−1un−1 + ω2 − ω n2un + Mn,n+1un+1 = 0

with ω n2 ≡ ω n

2 and Mn,n+1 ≡ M

Standard Bloch Theorem.

We discuss firstly the solutions of the Bloch theorem and the appearance of the continuum

spectrum. Let us identify the normal modes with the wave number q.

un = uoe iqna

which proposes solutions that are travelling waves having an equivalent weight for every site in

the chain. Every element in the column vector ω2I −Muω = 0 has the same form aside from a

trivial factor

ω2 − ω 02 + 2ωx

2 cosqa = 0

hence replacing ω 02

ωk2 = ω0

2 + 2ωx21 − cosqa

= ω02 + 2ωx

2 sin2qa/2

The typical case considered in Solid State textbooks is when there is no natural frequency but

that from those springs are connecting masses, hence ω02 = 0 and

ωq = ±2ωx2 sinqa/2

= ±2 Km sinqa/2

Notice that:

1- There are positive frequencies which correspond to positive and negative wave vectors

respectively.

2-Also notice that this dispersion relation presents translational symmetry in the q space, i.e.

ωq = ωq+Qnwith = ±2nπ/a

Qn is the magnitude of the reciprocal lattice vector corresponding to the chain.

The plane wave solution represents states extended over all the masses (sites). Combining

them one can build wave packets moving with a wave velocity dωq /dq. Notice that when ωq

reaches its maximum the group velocity becomes zero.

Counting states.

Consider a finite ring with N masses. The solution above holds. However one notices that

uN = uoe iqNa = uo

qNa = 2πs for s = 0,±1,±2,⋯

i.e. the and only N of them give different solutions. Hence

qs = s 2πNa

i.e., the q states are equally distributed with a distance Δq = 2πNa

resulting in a uniform density.

In consequence, the spectral density is

Jq = Na2π

From this we calculate the density of frequency per mass per unit volume

Jqdq = Jωdω

hence:

Jω = 1π

1dωdq

= 12π

1ωx cos qa

2

= 2π

1

4ωx2 − ω2

here, the frequency can take either positive or negative values.

We see that there is a singularity associated with the vanishing group velocity at q = πa .

Sometimes it is more simple to use ω2 as a variable to analyze the spectrum.

Jqdq = ρω2dω2

ρω2 = 1π

1dω2

dq

= 1πωx

21

1 − ω2 − ω02

2ωx2

2

Here, it is clear that around the initial Hartree frequency ω02 the collective interaction develops a

band of width 4ωx2

Then, we analyze the dynamical consequences of the existence of these collective states. We

would like to know how a localized excitation would spread out. The

m d2

dt2un = Kun+1 − 2un + un−1

is very similar to the equation for the cylindrical Bessel functions Jn

ddtJn = − 1

2Jn+1 − Jn−1

which can be applied twice to get.

d2

dt2Jn = 1

4Jn+1 − 2Jn + Jn−1

The only difference with the equation for the amplitudes above is the time scale

ω02 = ω0

2 + 2ωx2 = 4K/m

unt = Jnω0t

Hence, the solution at every site is a Bessel functions. By plotting the square amplitude one plots

the potential energy Unt stored in each site. In particular at site 0, one sees that:

U0t = 12

mω02|u0t|

2

t→∞∼ 1/t

One gets the result seen in the work by Danieli.

Dyson Equation.

Finally, let us discuss the extended states from the point of view of the Dyson equation.

Π± =|M|2

ω2 − ω02 − Π±

which might have a complex solutions.

ω2−ω02

2− ω2−ω0

2

2

2

− |M|2 ifω2−ω0

2

2> |M|

ω2−ω02

2− i |M|2 − ω2−ω0

2

2

2

ifω2−ω0

2

2

2

< |M|2

ω2−ω02

2+

ω2−ω02

2

2

− |M|2 ifω2−ω0

2

2> |M|

We selected the sign in the square root requiring that

ImΠ ≤ 0,

this means that the amplitude should not diverge with time. Also we require that the integral in

the semicircle of the complex plane must be complex

lim|ω |→∞

Π = 0

From this we evaluate the Density of States

J ω2 = 1π

1

|M|2 − ω2−ω02

2

2

from which we calculate the actual Spectral Density

ρω2dω2 = ρω22ωdω = Jωdω

hence

Jω = 12π

2ω

|M|2 − ω2−ω02

2

2

Problem: Generalize the above results for a dimerized chain (equal masses and different

spring constants) and plot the density of states.

Problem: Idem as above results for a dimeric chain (different masses and equal spring

constants)

Impurity States: Localized modesHere we discuss localized states and the situation of disordered equation. Follow the section

in Taylor.

First consider a simple alternative where the a different mass at site 0th

ω2I −M =

⋱ M

M ω2 − ω02 M

αM ω2 − αω02 αM

M ω2 − ω02 M

M ⋱

The the self-frequency can be obtained by analizing the poles of the local Green’s function

D0,0ω2 = 1ω2 − αω0

2 − α2Πω2

If the “impurity” mass is lighter than the others, α = m/m0 > 1 . Then, one expects that this

site tends to vibrate at a higher frequency than the others. This gives rise to a localized state at

high frequencies.

ωloc.2 =

4ωx.2 α

2α2 − 1for α = m

mo> 1

Now, consider a situation where the “impurity” at the origin has a heavier mass. The

frequency can not be lower that 0. This lower local frequency, then does not couples so good

with the continuous states. We identify an isolated pole at some point to the frequency plane and

this gives rise to a broad resonance.

Problem. Find the Density of states for both situations with a mass 1/4 and 4 times the

others. Plot the states and the density of states. Interpret.

Further readings:

Green functions and impurities in harmonic chains, D. Prato and C. A. Condat, Am. J. Phys.

51, 140 (1983)

The single oscillator with an “environment”.This situation discused above can be considered as a particular case of the heavy impurity

with natural frequency ω02. Consider that it is coupled to an infinite chain of much lighter

particles i.e. instead of the Brownian bath. In absence of mutual interactions these particles have

natural frequencies (ωi2 ≡ 0 with i = ±1,±2,… .When they interact a full continuous spectrum

starting from ω = 0 and ending at 4ωx.

The exact dynamics is obtained from

D0,0ω2 = 1ω2 − αω0

2 − α2Πω2.

Alternatively, one can perform the broad band (fast fluctuations approximation) by

considering at an early stage the limit K → ∞ and m → 0 so ωx2 is very big. A first rough

approximation, sometimes called first pole approximation would be to evaluate the

self-frequencies at the original pole, i.e.

D0,0ω2 ≃ 1ω2 − ω0

2 + α2ωx2 + α2ReΠω2|ω2=ω0

2+αωx2 + iα2 ImΠω2|ω2=ω0

2+αωx2

.

In this case the frequency never gets into the over-damped regime.However a much better

approximation is to solve the self-consistent equation for the real part and plug this solution in

the evaluation of the imaginary part. By keeping the complete expresion and expanding the

self-energies at the lowes order in ω2 we get

D0,0ω2 ≃ 1ω2 − ω0

2

Δω2∼α ω2−ωc2

+αωx2 + α2ReΠω2 +i

Γω2∼ωη1−α

α2 ImΠω2|ω2=ωr2

=1/1 − α

ω2 − ωr2 + iωη

The exact pole solution, with a low order expantion of the self-energy could be called the Self

Consistent Fermi Golden Rule, consists in the exact evaluation of the pole when both real part

and the imaginary part are allowed to take complex argument. Still the dynamics becomes

exponentially attenuated as it occurred in the introductory example.

If one would decide to keep the complete expresion for the self-energy, a cross-over from

from the exponential to a power-law should be observed.

Problem. Work out the above model and analyze it numerically or analytically to find that

the spectral density ρω2 is very well approximated by a Lorentzian and that the time

dependence of D0,0ω2 and ωD0,0ω2 [see bellow] is well approximated by an exponential.

Epistemological note (VERY IMPORTANT): We started this chapter analyzing an

oscillator and proposing a phenomenological model for the dissipative action of it environment.

Now, we found the exact solution of a precisely defined many-body system, where a portion of it

can be identified as the system, and the rest was considered the environment. From the exact

solution we arrives, under precise conditions, to the initial phenomenological model.

A Classical Field Theory: Continuum limitConsider again a finite ordered chain arranged as a ring. Let’s us go back and be a little

more formal.

H = ∑n=1

N

12m

pn2 + ωx

2un − un+12

here pn ≡ mun , hence the equations of movement are obtained thought the Hamilton’s canonical

equations

mun = ∂H∂pn

= pn

pn = − ∂H∂un

= ωx2un+1 − 2un + un−1

In the long wavelength limit (qa ≪ 1) the phase and group velocity become cp = cg, the

sound velocity. If we look at the system from far away, as we look a guitar string ignoring the

atoms, we only notice the overall behavior of the string without any reference to constituent

atoms.

x = na and un → uxa1/2

and

ωx → ca

hence

un+1 = a1/2ux + a ≃ a1/2 ux + a∂ux∂x

+ 12

a2 ∂2ux∂x2

For the case ωo2 = 0 the equation of motion becomes:

∂2u∂t2

− c2 ∂2u∂x2

= 0,

which is the standard equation for sound waves.

While for ωo2 > 0

∂2u∂t2

− c2 ∂2u∂x2

+ ωo2u2x, t = 0.

which describes the optical branches of acoustic waves.

Of course these equations can be obtained from a Hamiltonian formalism. In order to

transform the Hamiltonian general term into a Hamiltonian density we need:

∑n

→ 1a ∫ dx

n → x = na

un → φx

The Hamiltonian becomes

H = ∫0

L

dx 12

p2x, t + v2 ∂φ∂x

2

+ ωo2φ2 x, t

= ∫0

L

dx Hpx, t,φx, t

in order to find the equations we must introduce the functional derivatives by means of the

Dirac’s δ function

δφxδφx ′

= δx − x ′

δδφx ′

∂φx∂x

= ∂∂x

δx − x ′

hence the Hamilton’s equations

un = ∂H∂pn

pn = − ∂H∂un

we used above are generalized to

px = − δHδφx

= −v2 ∫0

L

dx ′ ∂φx ′∂x ′

∂∂x

δx − x ′ = v2 ∂2

∂x2φx

φx = − δHδpx

= px

Epistemological note: Here, as far as I know, there is a gap in the literature. One would

need to describe a classical field which, having a form of energy dissipation that degrades the

propagating waves (easily describe through non-hermitian components) and the gradual

dominance of the diffusion modes which are conservative as in a concentration. i.e. the question

is how to arrive to a diffusion equation from a purely reversible dynamics. We will leave open

this link between the microscopic reversibly world and the macrospic dissipative one and we will

jump directly to macroscopic equation.

More Green’s Functions: Fick’s Law and DiffusionEquation.

Before analyzing the Green’s function for the quantum system let us consider the Diffusion

Equation

∂∂t

n + D∇2n = 0

It says that if the density n is in a maximum (i.e. ∂2/∂x2n < 0 then the density in this point

decreases. We recall that if we have an initial localized spot at Cx, 0 = δx at time t = 0, then

the solution for later times is

Cx, t = Px, t; 0, 0

= 1

4πDtexp− x2

4Dtθt

which provides the evolution of a mass that is conserved

∫ nx, tdx = 1

that spreads with time over an area determined by the mean square dispersion

⟨x2 ⟩ = ∫ x2nx, tdx = 2Dt.

We also notice that if we want to describe an initial condition with an arbitrary initial density

n0x1, t1 at a given time t1 the solution for later times is

Cx, t = ∫ Px, t; x1, t1n0x1, t1dx1

This means that the density at site x2 and time t2is the superposition of all the densities originated

at different sites x1.

The Green’s function and Dynamical response:

Now, consider a case where we keep dropping ink for a lapse of time at an injection rate

described for a function ninicx1, t1 the net concentration at later times is:

nx, t = ∫−∞

t

dt1 ∫ dx1Px, t; x1, t1ninj.x1, t1dx1dt1

Clearly, the Propagator Px, t; x1, t1 contains all the information we want to know about the

diffusion equation. Notice that

∂∂t

+ D∇2 Px, t; x1, t1 = δx − x1δt2 − t1.

We also can define the Fourier transform:

Pq, t = ∫ dx Px, t; 0, 0 e−iqx = e−D2qtθt

Pq,ω = ∫ dt Pq, t; 0, 0 e+iωt = −iω + Dk2 −1

Px, t; 0, 0 = ∫ dω2π

dq

2πPk,ω e+iqx−iωt

Notice the particular election we have done of the normalization factor and the phases in the

exponential in the Fourier transform. Pk,ω will satisfy the equation:

−iω + Dk2 Pk,ω = 1.

i.e. the analytical structure of Pk,ω reflects the physical properties of the equation: the absence

of propagating waves and the dominance of the diffusion modes.

The response function

Hext = ∫ dr

density or

displacement

nr, t

generalized

“force”

δμr

turned off at

positive times

θ−tadiabatic

turn on

eηt

an example of this in the linear harmonic oscillator is:

Hext = ∫ dr nr, tδμrθteηt

we know that we can evaluate the displacement at time t = 0 as the response to the applied force.

At this time the external force cancels out the internal force. There is displacement but there is no

velocity.

δnq, t = 0 = ∫−∞

0

χq, t ′δμqeηt ′dt ′

= χq,ω = 0δμq

The question is: which is the expression for Pq,ω, defined as

δnq,ω = Pq,ωδnq, t = 0

that provides the response for positive times of this perturbation applied to negative times? i.e.

How is this density perturbation related to the dynamical susceptibility?

The density perturbation at time t after the interaction is turned off:

δnq, t = ∫−∞

0

dt ′χq, t − t ′eηt ′δμq

and its Fourier transform is:

δnq,ω = ∫0

∞dt e iωt ∫

−∞

0

dt ′ ∫ dω′

2π2iχ ′′q,ω′e iω′t−t ′eηt ′δμq

= ∫ dω′

πi

χ ′′q,ω′ω′ − ωω′ − iη

δμq

= ∫ dω′

πiχ ′′q,ω′ 1

ω1

ω′ − ω− 1

ωδμq

= 1iω χq,ω − χq,ω = 0δμq

= 1iω χq,ω − χq,ω = 0

δnq, t = 0χq,ω = 0

from which

Pq,ω = 1iω

χq,ωχq, 0

− 1

We can use this equation in diverse situations.

As a first example we consider the linear oscillator, where we knew the susceptibility is

χω =−1/m

ω2 − ω02 + iωη

and we want the propagator.

Pω = 1iω

χωχ0

− 1

=iω + iη

ω2 + iωη − ω02

Which has the Fourier transform:

Pt = u0 cosω0 t +η

2 ω0

sinω0 t e− 1

2ηt × Θt

i.e. it corresponds to the initial condition of displacement type (injection) where vt=0 = 0 and

ut=0 = u0.

By looking at the Green’s function in the frequency representation we came to identify the

Fourier transform of the force required to give such initial condition it is precisely the Fourier

transform of the kick-antikick we developed as an exercise.:

Fu0ω = −iω + iηmu0

Another example for the application of the relation among χ and P is the diffusion equation

discussed before. There, we knew the density-density propagator using the propagator we have

obtained above:

χq,ω/χq, 0 =Dq2

−iω + Dq2.

and the imaginary par is:

χ ′′q,ωω = χq, 0

Dq2

ω2 + Dq22

which is a Lorentzian with integrated intensity χq, 0, height χq, 0/Dq2 and width that goes to

zero in the large wave length limit of q →0.

NOTE: This last quantity is important since it is related to the density-density correlation function Pn,nq,ωat temperature kBT through the fluctuation-dissipation theorem:

χ ′′q,ωω =

Pn,nq,ωkBT

.

These results on the Green’s functions and response function for the diffusive case are

important for two main reasons:

1- They will enable to make useful analogies with the solutions and propagators appearing in

the solution of the Schrödinger equation.

2- They constitute an objective all by themselves since they correspond to the expected

hydrodynamic behavior of excitation expected in macroscopic limit of large wave lengths and

long times. At this moment is not obvious how but it is clear that we should be able to obtain

such behavior from Quantum Mechanics.

Chapter summaryIn this chapter have addressed two concepts.

1-The Green’s function or propagator is related to the intrinsic dynamics of a classical

system, accounting on how does it reacts to initial conditions and injections of excitations and

how does the system reacts to external forces.

2- Dissipation processes firstly smuggled into our description through our knowledge of the

response of a “dissipative” classical system as a phenomenological friction force or equivalently

as an imaginary part in the frequency. Later on, we saw that this effectively emerges from the

proper consideration of the thermodynamic limit of an “environment” composed by infinitely

many degrees of freedom, as occurred in the environment provided by a chain of other linear

oscillators. An alternative model for dissipation is an oscillator subjected to a random

environment or brownian bath whose stochastic collitions can be described as a Langevin force.

This last alternative is subjec to more obscure approximations on the nature of the correlations of

such environment. We are going to search for justification for such behavior once we are

immersed in the quantum world.

Used Formulas (CHECK)limη→+0

1ω − ω′ ± iη

= P 1ω − ω′

∓ iπδω − ω′

where the principal value is defined

P 1ω − ω′

= limη→0

ω − ω′

ω − ω′2+ η2

and the delta function is

δω − ω′ = 1π lim

η→+0

η

ω − ω′2+ η2

= ∫ e iω−ω′ tdt check

and

δt − t ′ = ∫ e−iωt−t ′ dω2π

check