9-1

Lecture 9- Linear response

Much of what we have done so far has implicitlyassumed low frequency ,

that is,

that we look at the

system on long time scales and coarse grain the dynamics .

µ ✓ macroscopic Irreversible eqs .

new

# linear response*1111141mgFigure adapted from Brenig . statistical theory . t heat

When we weakly perturb a system from equilibrium ,

we can construct a different general approach ,

which

is called linear response .

The reason for the name is

That if the perturbation is small enough ,

the responseof

the system is to a good approximation linear in

the perturbation .

we are in fact veryaccustomed to this :

Examples of linear response

0hm 's law : ] = OE

current a applied electric field

proportionality coefficient o= conductivity

Magnetization : M = XB

9.2

Diffusion ( Fick 's law ) In= - DIN

Heat current Tg. = - HIT

etc .This is in fact how we learn about the properties of

System .

we probe them and measure the responsein order

to learn about the behavior of a system .

More generally,

the response depends on the frequency of

the applied perturbation,

are we expect,

for example,

] ( w ) = olw ) Ecw )

for the frequency dependent response .

Fourier transforming back

to real time :

t

Jlt ) = ) olt - t'

) E ( t'

)dt'

- is

Now,

our task is this : How do we generally

Calculate the response functions ,such as

the

conductivity s de ?

we now derive general formulas for these factors,

often called Kubo formulas .

First we do the

classical case and then the quantum case,

The ideas

are the same,

the details are naturally slightly different .

9-3classical linear

response

Consider the Hamiltonian

HII ,F ) = Holier ) + Ulf

,q)=Ho(I. F) - f It )A( F. I )

Here Ho is the Hamiltonian that describes the unperturbed

system ,

V is the perturbation that we take here to

be proportional to a given macroscopic observable and

f ( E ) is the perturbing field .

( Think of an electric full

where U= -

g-EIHX )

.

Generally the perturbation could

be a sum of such observables,

but it's sufficient to

consider here only one A.

Now,

the system is described bya probability density

f( I. F),

which time evolves according to the Lionville

equationof g= { His } I

- Lg

[Liouvilliauwhere

{ µ, , }=§l9±gio÷i - ftp..FI

. )

is the Poisson bracket.

Since the derivatives are linear,

L is a linear operator ,and therefore

0+g= - hg = { H.tv ,s}={ Ho ,g}+{ vis }

= . hog - hrs

9- y

where ho corresponds to time evolution with Ho and

Lv to time evolution with V.

Assume now that flt )=0 for tato,

i.e.,

that

the perturbation is turned on at to.

Further,

we assume that the system was in equilibrium

at to, glto ) = Beg .

Note, geq

is the equilibrium

distribution function for the unperturbed system

described by Ho.

We now write

9 = seqtorg t OCF )

and do a perturbation expansion to linear order :

0+9 = 0+9%+0+89 = - lhothv )( Segtss ) + ...

now Oegeg.

= - Losey ( actually both terms are zero )

⇒ qsg = - hosg - hrseq

since - h✓8g is higher order in the perturbation .

This has the solution

+- it - t

'

)h°h✓9eqSglt ) =- fdt '

e

to

check this !

-

9-5

We have

-hrseq = { v. seq } = { -

fa,

tziptto }

= . ft { loEio÷* . gifts,÷ ,

= pf e¥t° { A. Ho } = pfseqni

Consider now the response of some observable B.

st BH ) =

(B)It ) - < B >

eq= fdpdifBII.FI ( gltl -

Seat )

t

-H - E) ho

= - JDIDFBSet '

e h✓9eqto

= pfdfdf B§.dt' e-

' t it 'lh°fit

'

) geq it

ho is a linear differential operator . Integrating by parts ( in the djdpypart

t

=

pfdotdpfdt'

fit'

) | et" 't

"h°B ] geqit

to

t

=p fdt' <

Blt- t

'

)

nilo) ) f It '

)eat

to

= pfotdt'

< Bltlitlt '

) >eq

felt'

)

Note : For the density matrix g°÷ - hg orgltte-

htgco ,

but for observables Ft '

- hg and therefore Altl = eht Alo )

=p The linear response is related to an equilibrium correlation

function .

9-6Quantum linear response - Preliminaries

we startby reminding ourselves of quantum statistical

mechanics .

The partition function is

z= § e- Pen= En

< NIIPE In > = Ice- PE )

where LI is the Hamiltonian operator and In ) its eigenvectors

Tt In > = Enln ?

Equilibrium expectation values of osservaslcs A

< A) = tz § < nialu ) e- PE '

= tz § < nine #In )

= Tr ( Sega ) = Tr ( Ageq )

with

geq = tz e- P "

the density matrix

Note that

trseq = zttr ( e- PHI = I § < ulip 'tIn >

= I § e- Pen= 1

In general , anynon . equilibrium situation is described by a

density matrix g,

which does not have to be diagonal .

Note : Density matrix Is said to bepure ,

if it can be written

as g= 1+7<+1 for Some state 1+7.

In this case

< a > = Trlga ) = { 41+7<41 Ain ) = < tlalffln> < NDH)

=11

= KIAH ?

9-7

A density matrix that is not pure is called mixed .

In order for trg= 1,

we need < +1+7=1.

Therefore ,a pure density matrix further satisfies :

g?

= 1+741+79+1 = It > 5+1=9

NI : Suppose our Hilbert space is a product space

It = Ha @ HB and It > EH.

It ? is called a

product state if it can be written as

H7 = Ita ) Q 14,37 with Ha ? Etta and His ? EHB .

The reduce density matrix of A is obtained by

tracing out the degrees of freedom of B,

i. a .

9A= Tr

, g

If { Inn.

) 0 Inn ) } is abasis in H

,

and therefore

{ Ina ) } a basis for Ha and { In ,3 ) } a basis for H,

,

then a general state It ? EH is ( we assume cnn.lu 'a7

=8n,n '

It ? =[ than

,14A > 0 In , } )

and sinkfor

B)n AMB

corresponding to a densitymatrix

*

9 =[ E tnanat ( ha ) @ Ina ) ) ( < MAI @ < MBI )

NAMB Ma ,Mp

MAMB

The reduced density matrix for A is :

9A = Trog = § 41319 Ina )

9-8or

% = n§ma,n,

than than,

Ina > < mal

The entanglement entropy between A and B is defined

as the vonNeumann of 9A ,

i.e.

,

s =- Traffog 9A

EI " The entanglement entropybetween A and B should

be the same as the onebetween B and A

.

That is

s= - Trasabg 9n.

=- Tr

,9,36g 5,3

show that this is true

EI : If Ha = { IT ),

It 7 } = HB is the Hilbert space

of a single spin- tz

,

then lta 0 An =H is the

Hilbert space of two spins . Calculate the entanglement

entropy of the following States

Hs7=r÷( 11^7×0117 - 1+7<41 )

1++7 = 11^7×0117

It. ) = tz ( 1T ) @ a) + It ) o It ) + It ? a lilt 117×0114

Do youunderstand the results ? Can you

unite Ito >

as a product state ?

After this excursion into reduced density matrices of9-9

bipartite systems ,

lets keep goingwith our story .

The time evolution of 91£ ) is governed by the quarter

version of the Lionville equation,the von

Neumann eg .

¥ =- Lg = ¥ [ His ] ( put t=1 )

This follows from the Schroedinger equation ,if we write

9= In Sun In > < ml

So

idtg = §mqm¢iotln ? )< ml + In > i0t< my

Now So

iotln ) = H In >

idtg = I Snn ( Hlu > < ml - In > < ml H )

= Hg - g H = [ His ]

In the schiddiugu picture all the time dependence is in the

States

idc In ) = Hln ) ⇒ Inch ) = Uk ) In )

- i Ht

where if H independent of time Uk ) =e

Then

glt ) = In 9am Into ) < mltll = UH )g5' et )

In the Heisenberg picture the States are time independent ,

but the operators have a time dependence

Alt ) = W CH Ault )

9-10

while the States and g are time independent .

These pictures are equivalent ,since they give the same

expectation value of all observables

< AXH = Trlgltla ) = Tr ( ultlgutlt ) A )

= Tr ( gutltlaultl ) = Tr 19A 't ) )T

cyclic invariance of trace

Tr IABCI = Tr ( BCA )=Tr ( CAB )

The third option is the interaction Picture . Here we

assumewe have a Hamiltonian

H = Hotv

To be explicit , sayltslt ) ) is the state in the

Schroedinger picture . Then

idtltsltl ) = HIYSIH )

Define the state in the interaction picture as

It ,=lt ) ) = ei #°tltslt ) ) = Uolt ) ltslt ) )

Then

idtlttltl ) = ( H - Ho ) eiltotltslt ) >

= V It .

,_It ) )

and therefore

iotg = [ ✓, 9 ,=lH } where V may

dependI

explicitly on t.

9- 11Now

, for everything to be consistent,

we also need

4 tslt ) 1 As 1 tslt ) ) = { ttltl 1 A±H ) ITIK ) ) =

= < tslt ) leiiltot A ,=lt ) e'

'

Hot1µs ( t ) )

or

A ,=lt ) = eittot As e

- ittot

and therefore

id+A±lt ) = [ A±lH ,Ho ]