Chapter 3
Linear Response Theory
3.1
Response and Resonance
Consider a damped harmonic oscillator subjected to a time-dependent forcing:
ẍ + 2γ ẋ + ω02 x = f (t) ,
(3.1)
where γ is the damping rate (γ > 0) and ω0 is the natural frequency in the absence of
damping1 . We adopt the following convention for the Fourier transform of a function H(t):
Z∞
dω
Ĥ(ω) e−iωt
2π
(3.2)
dt H(t) e+iωt .
(3.3)
H(t) =
−∞
Z∞
Ĥ(ω) =
−∞
Note that if H(t) is a real function, then Ĥ(−ω) = Ĥ ∗ (ω). In Fourier space, then, eqn.
(3.1) becomes
(ω02 − 2iγω − ω 2 ) x̂(ω) = fˆ(ω) ,
(3.4)
with the solution
x̂(ω) =
ω02
fˆ(ω)
≡ χ̂(ω) fˆ(ω)
− 2iγω − ω 2
(3.5)
where χ̂(ω) is the susceptibility function:
χ̂(ω) =
with
1
ω02
1
−1
=
,
2
− 2iγω − ω
(ω − ω+ )(ω − ω− )
(3.6)
q
ω± = −iγ ± ω02 − γ 2 .
(3.7)
Note that f (t) has dimensions of acceleration.
1
2
CHAPTER 3. LINEAR RESPONSE THEORY
The complete solution to (3.1) is then
Z∞
x(t) =
−∞
fˆ(ω) e−iωt
dω
+ xh (t)
2π ω02 − 2iγω − ω 2
(3.8)
where xh (t) is the homogeneous solution,
xh (t) = A+ e−iω+ t + A− e−iω− t .
(3.9)
Since Im(ω± ) < 0, xh (t) is a transient which decays in time. The coefficients A± may
be chosen to satisfy initial conditions on x(0) and ẋ(0), but the system ‘loses its memory’
of these initial conditions after a finite time, and in steady state all that is left is the
inhomogeneous piece, which is completely determined by the forcing.
In the time domain, we can write
Z∞
x(t) = dt0 χ(t − t0 ) f (t0 )
−∞
Z∞
χ(s) ≡
(3.10)
dω
χ̂(ω) e−iωs ,
2π
(3.11)
−∞
which brings us to a very important and sensible result:
Claim: The response is causal , i.e. χ(t − t0 ) = 0 when t < t0 , provided that χ̂(ω) is analytic
in the upper half plane of the variable ω.
Proof: Consider eqn. (3.11). Of χ̂(ω) is analytic in the upper half plane, then closing in
the UHP we obtain χ(s < 0) = 0.
For our example (3.6), we close in the LHP for s > 0 and obtain
X
1
−iωs
χ(s > 0) = (−2πi)
Res
χ̂(ω) e
2π
ω∈LHP
=
i.e.
χ(s) =
ie−iω+ s
ω+ − ω−
+
ie−iω− s
ω− − ω+
,
p
 −γs
e
2 − γ 2 Θ(s)

√
sin
ω
 ω2 −γ 2
0

 0


−γs

 √e 2
γ −ω02
p
sinh
γ 2 − ω02 Θ(s)
(3.12)
if ω02 > γ 2
(3.13)
if ω02 < γ 2 ,
where Θ(s) is the step function: Θ(s ≥ 0) = 1, Θ(s < 0) = 0. Causality simply means that
events occuring after the time t cannot influence the state of the system at t. Note that, in
general, χ(t) describes the time-dependent response to a δ-function impulse at t = 0.
3.2. KRAMERS-KRONIG RELATIONS
3.1.1
3
Energy Dissipation
How much work is done by the force f (t)? Since the power applied is P (t) = f (t) ẋ(t), we
have
Z∞
P (t) =
dω
(−iω) χ̂(ω) fˆ(ω) e−iωt
2π
dν ˆ∗
f (ν) e+iνt
2π
(3.14)
−∞
−∞
Z∞
∆E =
Z∞
Z∞
dt P (t) =
−∞
2
dω
(−iω) χ̂(ω) fˆ(ω) .
2π
(3.15)
−∞
Separating χ̂(ω) into real and imaginary parts,
χ̂(ω) = χ̂0 (ω) + iχ̂00 (ω) ,
(3.16)
we find for our example
ω02 − ω 2
= +χ̂0 (−ω)
(ω02 − ω 2 )2 + 4γ 2 ω 2
2γω
= −χ̂00 (−ω).
χ̂00 (ω) = 2
(ω0 − ω 2 )2 + 4γ 2 ω 2
χ̂0 (ω) =
(3.17)
(3.18)
The energy dissipated may now be written
Z∞
∆E =
2
dω
ω χ̂00 (ω) fˆ(ω) .
2π
(3.19)
−∞
The even function χ̂0 (ω) is called the reactive part of the susceptibility; the odd function
χ̂00 (ω) is the dissipative part. When experimentalists measure a lineshape, they usually are
referring to features in ω χ̂00 (ω), which describes the absorption rate as a function of driving
frequency.
3.2
Kramers-Kronig Relations
Let χ(z) be a complex function of the complex variable z which is analytic in the upper
half plane. Then the following integral must vanish,
I
dz χ(z)
=0,
(3.20)
2πi z − ζ
C
whenever Im(ζ) ≤ 0, where C is the contour depicted in fig. 3.1.
Now let ω ∈ R be real, and define the complex function χ(ω) of the real variable ω by
χ(ω) ≡ lim χ(ω + i) .
→0+
(3.21)
4
CHAPTER 3. LINEAR RESPONSE THEORY
Figure 3.1: The complex integration contour C.
Assuming χ(z) vanishes sufficiently rapidly that Jordan’s lemma may be invoked (i.e. that
the integral of χ(z) along the arc of C vanishes), we have
Z∞
0=
−∞
Z∞
=
dν
χ(ν)
2πi ν − ω + i
P
dν 0
00
χ (ν) + iχ (ν)
− iπδ(ν − ω)
2πi
ν−ω
(3.22)
−∞
where P stands for ‘principal part’. Taking the real and imaginary parts of this equation
reveals the Kramers-Kronig relations:
χ0 (ω) = P
Z∞
dν χ00 (ν)
π ν−ω
(3.23)
−∞
Z∞
00
χ (ω) = −P
dν χ0 (ν)
.
π ν−ω
(3.24)
−∞
The Kramers-Kronig relations are valid for any function χ(z) which is analytic in the upper
half plane.
If χ(z) is analytic everywhere off the Im(z) = 0 axis, we may write
Z∞
χ(z) =
dν χ00 (ν)
.
π ν−z
(3.25)
−∞
This immediately yields the result
lim χ(ω + i) − χ(ω − i) = 2i χ00 (ω) .
→0+
(3.26)
3.3. QUANTUM MECHANICAL RESPONSE FUNCTIONS
5
As an example, consider the function
χ00 (ω) =
ω2
ω
.
+ γ2
(3.27)
Then, choosing γ > 0,


i/(z + iγ)
Z∞
dω 1
ω
χ(z) =
· 2
=
π ω − z ω + γ2 

−∞
−i/(z − iγ)
if Im(z) > 0
(3.28)
if Im(z) < 0 .
Note that χ(z) is separately analytic in the UHP and the LHP, but that there is a branch
cut along the Re(z) axis, where χ(ω ± i) = ±i/(ω ± iγ).
EXERCISE: Show that eqn. (3.26) is satisfied for χ(ω) = ω/(ω 2 + γ 2 ).
If we analytically continue χ(z) from the UHP into the LHP, we find a pole and no branch
cut:
i
.
(3.29)
χ̃(z) =
z + iγ
The pole lies in the LHP at z = −iγ.
3.3
Quantum Mechanical Response Functions
Now consider a general quantum mechanical system with a Hamiltonian H0 subjected to a
time-dependent perturbation, H1 (t), where
X
H1 (t) = −
Qi φi (t) .
(3.30)
i
Here, the {Qi } are a set of Hermitian operators, and the {φi (t)} are fields or potentials.
Some examples:


−M · B(t)
magnetic moment – magnetic field





R
H1 (t) =
d3r %(r) φ(r, t)
density – scalar potential






 1R 3
− c d r j(r) · A(r, t) electromagnetic current – vector potential
We now ask, what is hQi (t)i? We assume that the lowest order response is linear, i.e.
Z∞
hQi (t)i = dt0 χij (t − t0 ) φj (t0 ) + O(φk φl ) .
(3.31)
−∞
Note that we assume that the O(φ0 ) term vanishes, which can be assured with a judicious
choice of the {Qi }2 . We also assume that the responses are all causal, i.e. χij (t − t0 ) = 0 for
2
If not, define δQi ≡ Qi − hQi i0 and consider hδQi (t)i.
6
CHAPTER 3. LINEAR RESPONSE THEORY
t < t0 . To compute χij (t − t0 ), we will use first order perturbation theory to obtain hQi (t)i
and then functionally differentiate with respect to φj (t0 ):
δ
Q
(t)
i
χij (t − t0 ) =
.
(3.32)
δφj (t0 )
The first step is to establish the result,
Ψ(t) = T exp
i
−
~
Zt
dt0 H0 + H1 (t0 ) Ψ(t0 ) ,
(3.33)
t0
where T is the time ordering operator, which places earlier times to the right. This is easily
derived starting with the Schrödinger equation,
i~
d Ψ(t) = H(t) Ψ(t) ,
dt
where H(t) = H0 + H1 (t). Integrating this equation from t to t + dt gives
Ψ(t + dt) = 1 − i H(t) dt Ψ(t)
~
i
i
Ψ(t0 + N dt) = 1 − H(t0 + (N − 1)dt) · · · 1 − H(t0 ) Ψ(t0 ) ,
~
~
(3.34)
(3.35)
(3.36)
hence
Ψ(t2 ) = U (t2 , t1 ) Ψ(t1 )
U (t2 , t1 ) = T exp
i
−
~
(3.37)
Zt2
dt H(t) .
(3.38)
t1
U (t2 , t1 ) is a unitary operator (i.e. U † = U −1 ), known as the time evolution operator
between times t1 and t2 .
EXERCISE: Show that, for t1 < t2 < t3 that U (t3 , t1 ) = U (t3 , t2 ) U (t2 , t1 ).
If t1 < t < t2 , then differentiating U (t2 , t1 ) with respect to φi (t) yields
i
δU (t2 , t1 )
= U (t2 , t) Qj U (t, t1 ) ,
δφj (t)
~
since ∂H(t)/∂φj (t) = −Qj . We may therefore write (assuming t0 < t, t0 )
δ Ψ(t) i −iH0 (t−t0 )/~
−iH0 (t0 −t0 )/~ Ψ(t0 ) Θ(t − t0 )
e
Q
e
=
j
0
δφj (t ) {φi =0} ~
i
= e−iH0 t/~ Qj (t0 ) e+iH0 t0 /~ Ψ(t0 ) Θ(t − t0 ) ,
~
(3.39)
(3.40)
3.3. QUANTUM MECHANICAL RESPONSE FUNCTIONS
7
where
Qj (t) ≡ eiH0 t/~ Qj e−iH0 t/~
(3.41)
is the operator Qj in the time-dependent interaction representation. Finally, we have
δ
Ψ(t) Qi Ψ(t)
0
δφj (t )
δ Ψ(t)
δ Ψ(t) =
Qi Ψ(t) + Ψ(t) Qi
δφj (t0 )
δφj (t0 )
n i
= −
Ψ(t0 ) e−iH0 t0 /~ Qj (t0 ) e+iH0 t/~ Qi Ψ(t)
~
o
i
+
Ψ(t) Qi e−iH0 t/~ Qj (t0 ) e+iH0 t0 /~ Ψ(t0 )
Θ(t − t0 )
~
i Qi (t), Qj (t0 ) Θ(t − t0 ) ,
(3.42)
=
~
were averages are with respect to the wavefunction Ψ ≡ exp(−iH0 t0 /~) Ψ(t0 ) , with
t0 → −∞, or, at finite temperature, with respect to a Boltzmann-weighted distribution of
such states. To reiterate,
χij (t − t0 ) =
χij (t − t0 ) =
i Qi (t), Qj (t0 ) Θ(t − t0 )
~
(3.43)
This is sometimes known as the retarded response function.
3.3.1
Spectral Representation
We now derive an expression for the response functions in terms of the spectral properties
of the
H0 . We stress that H0 may describe a fully interacting system. Write
Hamiltonian
H0 n = ~ωn n , in which case
i
χ̂ij (ω) =
~
Z∞
dt eiωt Qi (t), Qj (0)
0
Z∞
i
1 X −β~ωm n
dt eiωt
e
m Qi n n Qj m e+i(ωm −ωn )t
=
~
Z m,n
0
o
− m Qj n n Qi m e+i(ωn −ωm )t ,
(3.44)
where β = 1/kB T and Z is the partition function. Regularizing the integrals at t → ∞ with
exp(−t) with = 0+ , we use
Z∞
dt ei(ω−Ω+i)t =
0
i
ω − Ω + i
(3.45)
8
CHAPTER 3. LINEAR RESPONSE THEORY
to obtain the spectral representation of the (retarded) response function3 ,
1 X −β~ωm
χ̂ij (ω + i) =
e
~Z m,n
(
)
m Qj n n Qi m
m Qi n n Qj m
−
ω − ωm + ωn + i
ω + ωm − ωn + i
(3.46)
We will refer to this as χ̂ij (ω); formally χ̂ij (ω) has poles or a branch cut (for continuous
spectra) along the Re(ω) axis. Diagrammatic perturbation theory does not give us χ̂ij (ω),
but rather the time-ordered response function,
i
T Qi (t) Qj (t0 )
~
i
i
=
Qi (t) Qj (t0 ) Θ(t − t0 ) +
Qj (t0 ) Qi (t) Θ(t0 − t) .
~
~
χTij (t − t0 ) ≡
(3.47)
The spectral representation of χ̂Tij (ω) is
1 X −β~ωm
χ̂Tij (ω + i) =
e
~Z m,n
(
)
m Qj n n Qi m
m Qi n n Qj m
−
ω − ωm + ωn − i
ω + ωm − ωn + i
(3.48)
The difference between χ̂ij (ω) and χ̂Tij (ω) is thus only in the sign of the infinitesimal ±i
term in one of the denominators.
Let us now define the real and imaginary parts of the product of expectations values encountered above:
m Qi n n Qj m ≡ Amn (ij) + iBmn (ij) .
(3.49)
That is4 ,
1
1 m Qi n n Qj m +
m Qj n n Qi m
2
2
1 1 m Qi n n Qj m −
m Qj n n Qi m .
Bmn (ij) =
2i
2i
Amn (ij) =
(3.50)
(3.51)
Note that Amn (ij) is separately symmetric under interchange of either m and n, or of i and
j, whereas Bmn (ij) is separately antisymmetric under these operations:
Amn (ij) = +Anm (ij) = Anm (ji) = +Amn (ji)
(3.52)
Bmn (ij) = −Bnm (ij) = Bnm (ji) = −Bmn (ji) .
(3.53)
We define the spectral densities
(
)
%A
1 X −β~ωm Amn (ij)
ij (ω)
≡
e
δ(ω − ωn + ωm ) ,
~Z m,n
Bmn (ij)
%B
ij (ω)
3
4
The spectral representation is sometimes known as the Lehmann representation.
We assume all the Qi are Hermitian, i.e. Qi = Q†i .
(3.54)
3.3. QUANTUM MECHANICAL RESPONSE FUNCTIONS
9
which satisfy
A
%A
ij (ω) = +%ji (ω)
%B
ij (ω)
=
−%B
ji (ω)
,
−β~ω A
%A
%ij (ω)
ij (−ω) = +e
,
%B
ij (−ω)
= −e
−β~ω
%B
ij (ω)
(3.55)
.
(3.56)
In terms of these spectral densities,
Z∞
= P dν
χ̂0ij (ω)
−∞
Z∞
χ̂00ij (ω) = P dν
2ν
0
%A (ν) − π(1 − e−β~ω ) %B
ij (ω) = +χ̂ij (−ω)
ν 2 − ω 2 ij
(3.57)
2ω
00
%B (ν) + π(1 − e−β~ω ) %A
ij (ω) = −χ̂ij (−ω).
ν 2 − ω 2 ij
(3.58)
−∞
For the time ordered response functions, we find
Z∞
χ̂0ijT (ω) = P dν
−∞
Z∞
χ̂00ijT (ω) = P dν
2ν
%A (ν) − π(1 + e−β~ω ) %B
ij (ω)
− ω 2 ij
(3.59)
2ω
%B (ν) + π(1 + e−β~ω ) %A
ij (ω) .
ν 2 − ω 2 ij
(3.60)
ν2
−∞
Hence, knowledge of either the retarded or the time-ordered response functions is sufficient
to determine the full behavior of the other:
χ̂0ij (ω) + χ̂0ji (ω) = χ̂0ijT (ω) + χ̂0jiT (ω)
0
χ̂ij (ω) − χ̂0ji (ω) = χ̂0ijT (ω) − χ̂0jiT (ω) × tanh( 12 β~ω)
00
χ̂ij (ω) + χ̂00ji (ω) = χ̂00ijT (ω) + χ̂00jiT (ω) × tanh( 21 β~ω)
00
χ̂ij (ω) − χ̂00ji (ω) = χ̂00ijT (ω) − χ̂00jiT (ω) .
3.3.2
(3.61)
(3.62)
(3.63)
(3.64)
Energy Dissipation
The work done on the system must be positive! he rate at which work is done by the
external fields is the power dissipated,
d Ψ(t) H(t) Ψ(t)
dt
∂H (t) E
D
X
1
= Ψ(t)
Qi (t) φ̇i (t) ,
Ψ(t) = −
∂t
P =
i
(3.65)
10
CHAPTER 3. LINEAR RESPONSE THEORY
where we have invoked the Feynman-Hellman theorem. The total energy dissipated is thus
a functional of the external fields {φi (t)}:
Z∞
Z∞ Z∞
W = dt P (t) = − dt dt0 χij (t − t0 ) φ̇i (t) φj (t0 )
−∞
−∞ −∞
Z∞
=
dω
(−iω) φ̂∗i (ω) χ̂ij (ω) φ̂j (ω) .
2π
(3.66)
−∞
Since the {Qi } are Hermitian observables, the {φi (t)} must be real fields, in which case
φ̂∗i (ω) = φ̂j (−ω), whence
Z∞
dω
(−iω) χ̂ij (ω) − χ̂ji (−ω) φ̂∗i (ω) φ̂j (ω)
W =
4π
−∞
Z∞
=
dω
Mij (ω) φ̂∗i (ω) φ̂j (ω)
2π
(3.67)
−∞
where
Mij (ω) ≡ 12 (−iω) χ̂ij (ω) − χ̂ji (−ω)
B
(ω)
+
i%
(ω)
.
= πω 1 − e−β~ω %A
ij
ij
(3.68)
Note that as a matrix M (ω) = M † (ω), so that M (ω) has real eigenvalues.
3.3.3
Correlation Functions
We define the correlation function
Sij (t) ≡ Qi (t) Qj (t0 ) ,
which has the spectral representation
h
i
B
Ŝij (ω) = 2π~ %A
ij (ω) + i%ij (ω)
2π X −β~ωm =
e
m Qi n n Qj n δ(ω − ωn + ωm ) .
Z m,n
(3.69)
(3.70)
Note that
∗
Ŝij (−ω) = e−β~ω Ŝij
(ω)
,
∗
Ŝji (ω) = Ŝij
(ω) .
(3.71)
and that
χ̂ij (ω) − χ̂ji (−ω) =
i
1 − e−β~ω Ŝij (ω)
~
(3.72)
This result is known as the fluctuation-dissipation theorem, as it relates the equilibrium
fluctuations Sij (ω) to the dissipative quantity χ̂ij (ω) − χ̂ji (−ω).
3.4. EXAMPLE: S =
1
2
OBJECT IN A MAGNETIC FIELD
11
Time Reversal Symmetry
If the operators Qi have a definite symmetry under time reversal, say
T Qi T −1 = ηi Qi ,
(3.73)
then the correlation function satisfies
Ŝij (ω) = ηi ηj Ŝji (ω) .
3.3.4
(3.74)
Continuous Systems
The indices i and j could contain spatial information as well. Typically we will separate
out spatial degrees of freedom, and write
Sij (r − r 0 , t − t0 ) = Qi (r, t) Qj (r 0 , t0 ) ,
(3.75)
where we have assumed space and time translation invariance. The Fourier transform is
defined as
Z
Ŝ(k, ω) =
Z∞
d r dt e−ik·r S(r, t)
3
(3.76)
−∞
Z∞
1
=
dt e+iωt Q̂(k, t) Q̂(−k, 0) .
V
(3.77)
−∞
3.4
Example: S =
Consider a S =
1
2
1
2
Object in a Magnetic Field
object in an external field, described by the Hamiltonian
H0 = γB0 S z
(3.78)
with B0 > 0. (Without loss
of
generality, we1can take the DC external field B0 to lie along
ẑ.) The eigenstates are ± , with ω± = ± 2 γB0 . We apply a perturbation,
H1 (t) = γS · B1 (t) .
At T = 0, the susceptibility tensor is
− Sβ n n Sα −
− Sα n n Sβ −
γ2 X
χαβ (ω) =
−
~ n
ω − ω− + ωn + i
ω + ω− − ωn + i
β
α
α
− S +
+ S −
− S +
+ Sβ −
γ2
=
−
,
~
ω + γB0 + i
ω − γB0 + i
(3.79)
(3.80)
12
CHAPTER 3. LINEAR RESPONSE THEORY
where we have dropped the hat on χ̂αβ (ω) for notational convenience. The only nonzero
matrix elements are
~γ 2
ω + γB0 + i
−~γ 2
,
χ−+ (ω) =
ω − γB0 + i
χ+− (ω) =
(3.81)
(3.82)
or, equivalently,
1
1
χxx (ω) =
= +χyy (ω)
−
ω + γB0 + i ω − γB0 + i
1
1
2
i
= −χyx (ω) .
+
χxy (ω) = 4 ~γ
ω + γB0 + i ω − γB0 + i
2
1
4 ~γ
3.4.1
(3.83)
(3.84)
Bloch Equations
The torque exerted on a magnetic moment µ by a magnetic field H is N = µ × H, which
is equal to the rate of change of the total angular momentum: J˙ = N . Since µ = γJ ,
where γ is the gyromagnetic factor,
we have µ̇ = γµ × H. For noninteracting spins, the
P
total magnetic moment, M = i µi then satisfies
dM
= γM × H .
dt
(3.85)
Now suppose that H = H0 ẑ+H⊥ (t), where ẑ·H⊥ = 0. In equilibrium, we have M = M0 ẑ,
with M0 = χ0 H0 , where χ0 is the static susceptibility. Phenomenologically, we assume
that the relaxation to this equilibrium state is described by a longitudinal and transverse
relaxation time, respectively known as T1 and T2 :
Ṁx = γMy Hz − γMz Hy −
Mx
T2
(3.86)
Ṁy = γMz Hx − γMx Hz −
My
T2
(3.87)
Ṁz = γMx Hy − γMy Hx −
M z − M0
.
T1
(3.88)
These are known as the Bloch equations. Mathematically, they are a set of coupled linear,
first order, time-dependent, inhomogeneous equations. These may be recast in the form
Ṁ α + Rαβ M β = ψ α ,
−1
−1
with Rαβ (t) = Tαβ
− γ αβδ H δ (t), ψ α = Tαβ
M0β , and


T2 0 0
Tαβ =  0 T2 0  .
0 0 T1
(3.89)
(3.90)
3.4. EXAMPLE: S =
1
2
OBJECT IN A MAGNETIC FIELD
13
The formal solution is written
Zt
M (t) = dt0 U (t − t0 ) ψ(t0 ) + U (t) ψ(0) ,
(3.91)
0
where the evolution matrix,


 Zt

0
0
U (t) = T exp − dt R(t ) ,


(3.92)
0
is given in terms of the time-ordered exponential (earlier times to the right).
We can make analytical progress if we write write M = M0 ẑ + m and suppose |H⊥ | H0
and |m| M0 , in which case we have
mx
ṁx = γ H0 my − γ Hy M0 −
(3.93)
T2
my
(3.94)
ṁy = γ Hx M0 − γ H0 mx −
T2
mz
,
(3.95)
ṁz = −
T1
which are equivalent to the following:
m̈x + 2T2−1 ṁx + γ 2 H02 + T2−2 mx = γ M0 γ H0 Hx − T2−1 Hy − Ḣy
m̈y + 2T2−1 ṁy + γ 2 H02 + T2−2 my = γ M0 γ H0 Hy + T2−1 Hx + Ḣx
and mz (t) = mz (0) exp(−t/T1 ). Solving the first two by Fourier transform,
γ 2 H02 + T2−2 − ω 2 − 2iT2−2 ω m̂x (ω) = γ M0 γ H0 Hx (ω) + (iω − T2−1 )Hy (ω)
γ 2 H02 + T2−2 − ω 2 − 2iT2−2 ω m̂y (ω) = γ M0 γ H0 Hy (ω) − (iω − T2−1 )Hx (ω) ,
(3.96)
(3.97)
(3.98)
(3.99)
from which we read off
χxx (ω) =
γ 2 H0 M0
= χyy (ω)
γ 2 H02 + T2−2 − ω 2 − 2iT2−1 ω
(3.100)
χxy (ω) =
(iω − T2−1 ) γ M0
= −χyx (ω) .
γ 2 H02 + T2−2 − ω 2 − 2iT2−1 ω
(3.101)
Note that Onsager reciprocity is satisfied:
χxy (ω, H0 ) = χtyx (ω, H0 ) = χyx (ω, −H0 ) = −χyx (ω, H0 ) .
(3.102)
The lineshape is given by
χ0xx (ω) =
(γ 2 H02 + T2−2 − ω 2 ) γ 2 H0 M0
2
γ 2 H02 + T2−2 − ω 2 + 4 T2−2 ω 2
(3.103)
χ00xx (ω) =
2 γ H0 M0 T2−1 ω
,
2
γ 2 H02 + T2−2 − ω 2 + 4 T2−2 ω 2
(3.104)
so a measure of the linewidth is a measure of T2−1 .
14
3.5
CHAPTER 3. LINEAR RESPONSE THEORY
Electromagnetic Response
Consider an interacting system consisting of electrons of charge −e in the presence of a timevarying electromagnetic field. The electromagnetic field is given in terms of the 4-potential
Aµ = (A0 , A):
E = −∇A0 −
1 ∂A
c ∂t
B =∇×A .
(3.105)
(3.106)
The Hamiltonian for an N -particle system is
X
N 2
X
1 e
v(xi − xj )
H(Aµ ) =
pi + A(xi , t) − eA0 (xi , t) + U (xi ) +
2m
c
i<j
i=1
Z
Z
2
1
e
= H(0) −
d3x jµp (x) Aµ (x, t) +
d3x n(x) A2 (x, t) ,
(3.107)
c
2mc2
where we have defined
n(x) ≡
N
X
δ(x − xi )
(3.108)
i=1
N
o
e Xn
pi δ(x − xi ) + δ(x − xi ) pi
j (x) ≡ −
2m
(3.109)
j0p (x) ≡ c e n(x) .
(3.110)
p
i=1
Throughout this discussion we invoke covariant/contravariant notation, using the metric


−1 0 0 0
 0 1 0 0

(3.111)
gµν = g µν = 
 0 0 1 0 ,
0 0 0 1
so that
j µ = (j 0 , j 1 , j 2 , j 3 ) ≡ (j 0 , j)
(3.112)
jµ = gµν j ν = (−j 0 , j 1 , j 2 , j 3 )
(3.113)
jµ Aµ = j µ gµν Aν = −j 0 A0 + j · A ≡ j · A
(3.114)
The quantity jµp (x) is known as the paramagnetic current density. The physical current
density jµ (x) also contains a diamagnetic contribution:
δH
= jµp (x) + jµd (x)
δAµ (x)
e2
e
j d (x) = −
n(x)A(x) = − 2 j0p (x) A(x)
mc
mc
j0d (x) = 0 .
jµ (x) = −c
(3.115)
(3.116)
(3.117)
3.5. ELECTROMAGNETIC RESPONSE
The electromagnetic response tensor Kµν is defined via
Z
Z
c
3 0
jµ (x, t) = −
d x dt Kµν (xt; x0 t0 ) Aν (x0 , t0 ) ,
4π
valid to first order in the external 4-potential Aµ . From
Z
Z
p
i
3 0
d x dt0 jµp (x, t), jνp (x0 , t0 ) Θ(t − t0 ) Aν (x0 , t0 )
jµ (x, t) =
~c
d
e jµ (x, t) = − 2 j0p (x, t) Aµ (x, t) (1 − δµ0 ) ,
mc
15
(3.118)
(3.119)
(3.120)
we conclude
Kµν (xt; x0 t0 ) =
E
4π D p
p 0 0
j
(x,
t),
j
(x
,
t
)
Θ(t − t0 )
µ
ν
i~c2
+
(3.121)
4πe p
j0 (x, t) δ(x − x0 ) δ(t − t0 ) δµν (1 − δµ0 ) .
2
mc
p
The first term is sometimes
known as the paramagnetic response kernel, Kµν
(x; x0 ) =
p
p
(4πi/i~c2 ) jµ (x), jν (x0 ) Θ(t−t0 ) is not directly calculable by perturbation
theory. Rather,
p,T
(x; x0 ) = (4π/i~c2 ) T jµp (x) jνp (x0 ) ,
one obtains the time-ordered response function Kµν
where xµ ≡ (ct, x).
Second Quantized Notation
In the presence of an electromagnetic field described by the 4-potential Aµ = (cφ, A), the
Hamiltonian of an interacting electron system takes the form
XZ
1 ~
e 2
H=
d3x ψσ† (x)
∇ + A − eA0 (x) + U (x) ψσ (x)
2m i
c
σ
Z
Z
1X
(3.122)
+
d3x d3x0 ψσ† (x) ψσ† 0 (x0 ) v(x − x0 ) ψσ0 (x0 ) ψσ (x) ,
2 0
σ,σ
where v(x − x0 ) is a two-body interaction, e.g. e2 /|x − x0 |, and U (x) is the external scalar
potential. Expanding in powers of Aµ ,
Z
Z
1
e2 X
µ
3
p
µ
H(A ) = H(0) −
d3x ψσ† (x) ψσ (x) A2 (x) ,
(3.123)
d x jµ (x) A (x) +
c
2mc2 σ
where the paramagnetic current density jµp (x) is defined by
X
j0p (x) = c e
ψσ† (x) ψσ (x)
(3.124)
σ
ie~ X
†
†
j (x) =
ψσ (x) ∇ ψσ (x) − ∇ψσ (x) ψσ (x) .
2m σ
p
(3.125)
16
3.5.1
CHAPTER 3. LINEAR RESPONSE THEORY
Gauge Invariance and Charge Conservation
In Fourier space, with q µ = (ω/c, q), we have, for homogeneous systems,
c
jµ (q) = − Kµν (q) Aν (q) .
4π
Note our convention on Fourier transforms:
Z 4
dk
Ĥ(k) e+ik·x
H(x) =
(2π)4
Z
Ĥ(k) = d4x H(x) e−ik·x ,
(3.126)
(3.127)
(3.128)
where k · x ≡ kµ xµ = k · x − ωt. Under a gauge transformation, Aµ → Aµ + ∂ µ Λ, i.e.
Aµ (q) → Aµ (q) + iΛ(q) q µ ,
(3.129)
where Λ is an arbitrary scalar function. Since the physical current must be unchanged
by a gauge transformation, we conclude that Kµν (q) q ν = 0. We also have the continuity
equation, ∂ µ jµ = 0, the Fourier space version of which says q µ jµ (q) = 0, which in turn
requires q µ Kµν (q) = 0. Therefore,
X
q µ Kµν (q) =
µ
X
Kµν (q) q ν = 0
(3.130)
ν
In fact, the above conditions are identical owing to the reciprocity relations,
Re Kµν (q) = +Re Kνµ (−q)
(3.131)
Im Kµν (q) = −Im Kνµ (−q) ,
(3.132)
which follow from the spectral representation of Kµν (q). Thus,
gauge invariance ⇐⇒ charge conservation
3.5.2
(3.133)
A Sum Rule
If we work in a gauge where A0 = 0, then E = −c−1 Ȧ, hence E(q) = iq 0 A(q), and
c
ji (q) = − Kij (q) Aj (q)
4π
c
c j
= − Kij (q)
E (q)
4π
iω
≡ σij (q) E j (q) .
(3.134)
Thus, the conductivity tensor is given by
σij (q, ω) =
ic2
Kij (q, ω) .
4πω
(3.135)
3.5. ELECTROMAGNETIC RESPONSE
17
If, in the ω → 0 limit, the conductivity is to remain finite, then we must have
Z
Z∞ D
E
ie2 n
δij ,
d x dt jip (x, t), jjp (0, 0) e+iωt =
m
3
(3.136)
0
where n is the electron number density. This relation is spontaneously violated in a superconductor, where σ(ω) ∝ ω −1 as ω → 0.
3.5.3
Longitudinal and Transverse Response
In an isotropic system, the spatial components of Kµν may be resolved into longitudinal
and transverse components, since the only preferred spatial vector is q itself. Thus, we may
write
Kij (q, ω) = Kk (q, ω) q̂i q̂j + K⊥ (q, ω) δij − q̂i q̂j ,
(3.137)
where q̂i ≡ qi /|q|. We now invoke current conservation, which says q µ Kµν (q) = 0. When
ν = j is a spatial index,
ω
q 0 K0j + q i Kij = K0j + Kk qj ,
(3.138)
c
which yields
c
K0j (q, ω) = − q j Kk (q, ω) = Kj0 (q, ω)
(3.139)
ω
In other words, the three components of K0j (q) are in fact completely determined by Kk (q)
and q itself. When ν = 0,
0 = q 0 K00 + q i Ki0 =
ω
c
K00 − q 2 Kk ,
c
ω
(3.140)
which says
K00 (q, ω) =
c2 2
q Kk (q, ω)
ω2
(3.141)
Thus, of the 10 freedoms of the symmetric 4 × 4 tensor Kµν (q), there are only two independent ones – the functions Kk (q) and K⊥ (q).
3.5.4
Neutral Systems
In neutral systems, we define the number density and number current density as
n(x) =
N
X
δ(x − xi )
(3.142)
i=1
N
j(x) =
o
1 Xn
pi δ(x − xi ) + δ(x − xi ) pi .
2m
i=1
(3.143)
18
CHAPTER 3. LINEAR RESPONSE THEORY
The charge and current susceptibilities are then given by
i n(x, t), n(0, 0)
~
i χij (x, t) =
ji (x, t), jj (0, 0) .
~
χ(x, t) =
(3.144)
(3.145)
We define the longitudinal and transverse susceptibilities for homogeneous systems according to
χij (q, ω) = χk (q, ω) q̂i q̂j + χ⊥ (q, ω) (δij − q̂i q̂j ) .
(3.146)
From the continuity equation,
∇·j+
follows the relation
χk (q, ω) =
∂n
=0
∂t
n
ω2
+ 2 χ(q, ω) .
m q
(3.147)
(3.148)
EXERCISE: Derive eqn. (3.148).
The relation between Kµν (q) and the neutral susceptibilities defined above is then
K00 (x, t) = −4πe2 χ(x, t)
o
4πe2 n n
Kij (x, t) = 2
δ(x) δ(t) − χij (x, t) ,
c
m
(3.149)
(3.150)
and therefore
o
4πe2 n n
−
χ
(q,
ω)
k
c2
m
n
o
2
4πe n
− χ⊥ (q, ω) .
K⊥ (q, ω) = 2
c
m
Kk (q, ω) =
3.5.5
(3.151)
(3.152)
The Meissner Effect and Superfluid Density
Suppose we apply an electromagnetic field E. We adopt a gauge in which A0 = 0, E =
−c−1 Ȧ, and B = ∇ × A. To satisfy Maxwell’s equations, we have q · A(q, ω) = 0, i.e.
A(q, ω) is purely transverse. But then
c
j(q, ω) = − K⊥ (q, ω) A(q, ω) .
(3.153)
4π
This leads directly to the Meissner effect whenever limq→0 K⊥ (q, 0) is finite. To see this,
we write
∇ × B = ∇ (∇ · A) − ∇ 2 A
4π
1 ∂E
=
j+
c
c ∂t
4π c 1 ∂ 2A
=
−
K⊥ (−i∇, i ∂t ) A − 2 2 ,
c
4π
c ∂t
(3.154)
3.5. ELECTROMAGNETIC RESPONSE
19
which yields
∇2 −
1 ∂2 A = K⊥ (−i∇, i ∂t )A .
c2 ∂t2
(3.155)
In the static limit, ∇ 2 A = K⊥ (i∇, 0)A, and we define
1
≡ lim K (q, 0) .
λ2L q→0 ⊥
(3.156)
λL is the London penetration depth, which is related to the superfluid density ns by
ns ≡
mc2
4πe2 λ2L
(3.157)
= n − m lim χ⊥ (q, 0) .
q→0
(3.158)
Ideal Bose Gas
We start from
i ji (q, t), jj (−q, 0)
~V
~ X
(2ki + qi ) ψk† ψk+q .
ji (q) =
2m
χij (q, t) =
(3.159)
(3.160)
k
For the free Bose gas, with dispersion ωk = ~k2 /2m,
i(ω −ω
)t
ji (q, t) = (2ki + qi ) e k k+q ψk† ψk+q
~2 X
i(ω −ω
)t
ji (q, t), jj (−q, 0) =
(2ki + qi )(2kj0 − qj ) e k k+q
4m2 0
k,k
× ψk† ψk+q , ψk† 0 ψk0 −q
(3.161)
(3.162)
Using
[AB, CD] = A [B, C] D + AC [B, D] + C [A, D] B + [A, C] DB ,
(3.163)
we obtain
~2 X
i(ω −ω
)t ji (q, t), jj (−q, 0) =
(2ki +qi )(2kj +qj ) e k k+q n0 (ωk )−n0 (ωk+q ) , (3.164)
2
4m
k
where n0 (ω) is the equilibrium Bose distribution5 ,
n0 (ω) =
5
Recall that µ = 0 in the condensed phase.
1
eβ~ω e−βµ
−1
.
(3.165)
20
CHAPTER 3. LINEAR RESPONSE THEORY
Thus,
n0 (ωk+q ) − n0 (ωk )
~ X
χij (q, ω) =
(2ki + qi )(2kj + qj )
4m2 V
ω + ωk − ωk+q + i
k
1
1
~n0
−
qi qj
=
4m2 ω + ωq + i ω − ωq + i
Z 3
0
0
~
d k n (ωk+q/2 ) − n (ωk−q/2 )
+ 2
ki kj ,
m
(2π)3 ω + ω
−
ω
+
i
k−q/2
k+q/2
(3.166)
(3.167)
where n0 = N0 /V is the condensate number density. Taking the ω = 0, q → 0 limit yields
χij (q → 0, 0) =
n0
n0
q̂i q̂j + δij ,
m
m
(3.168)
where n0 is the density of uncondensed bosons. From this we read off
χk (q → 0, 0) =
n
m
,
χ⊥ (q → 0, 0) =
n0
,
m
(3.169)
where n = n0 + n0 is the total boson number density. The superfluid density, according to
(3.158), is ns = n0 (T ).
In fact, the ideal Bose gas is not a superfluid. Its excitation spectrum is too ‘soft’ - any
superflow is unstable toward decay into single particle excitations.
3.6
Density-Density Correlations
In many systems, external probes couple to the number density n(r) =
we may write the perturbing Hamiltonian as
Z
H1 (t) = − d3r n(r) U (r, t) .
PN
i=1 δ(r − ri ),
and
(3.170)
The response δn ≡ n − hni0 is given by
Z
hδn(r, t)i =
Z
3 0
d r dt0 χ(r − r 0 , t − t0 ) U (r 0 , t0 )
hδ n̂(q, ω)i = χ(q, ω) Û (q, ω) ,
(3.171)
(3.172)
3.6. DENSITY-DENSITY CORRELATIONS
21
where
χ(q, ω) =
1 X −β~ωm
e
~VZ m,n
1
=
~
S(q, ω) =
2
( m n̂q n ω − ωm + ωn + i
Z∞
dν S(q, ν)
−
2 )
m n̂q n ω + ωm − ωn + i
1
1
−
ω + ν + i ω − ν + i
(3.173)
−∞
2
2π X −β~ωm e
m
n̂
q n δ(ω − ωn + ωm ) .
VZ m,n
(3.174)
Note that
n̂q =
N
X
e−iq·ri ,
(3.175)
i=1
and that n̂†q = n̂−q . S(q, ω) is known as the dynamic structure factor . In a scattering experiment, where an incident probe (e.g. a neutron) interacts with the system via a potential
U (r − R), where R is the probe particle position, Fermi’s Golden Rule says that the rate at
which the incident particle deposits momentum ~q and energy ~ω into the system is given
by
I(q, ω) =
=
2
2π X −β~ωm e
m; p H1 n; p − ~q δ(ω − ωn + ωm )
~Z m,n
2
1 Û (q) S(q, ω) .
~
(3.176)
2
The quantity Û (q) is called the form factor. In neutron scattering, the “on-shell” condition requires that the incident energy ε and momentum p are related via the ballistic
dispersion ε = p2 /2mn . Similarly, the final energy and momentum are related, hence
ε − ~ω =
p2
2mn
− ~ω =
(p − ~q)2
2mn
=⇒
~ω =
~q · p
mn
−
~2 q 2
2mn
.
(3.177)
Hence, for fixed momentum transfer ~q, ω can be varied by changing the incident momentum
p.
Another case of interest is the response of a system
to a foreign object moving with trajectory
R(t) = V t. In this case, U (r, t) = U r − R(t) , and
Z
Û (q, ω) =
Z
d r dt e−iq·r eiωt U (r − V t)
3
= 2π δ(ω − q · V ) Û (q)
(3.178)
hδn(q, ω)i = 2π δ(ω − q · V ) χ(q, ω) .
(3.179)
so that
22
3.6.1
CHAPTER 3. LINEAR RESPONSE THEORY
Sum Rules
From eqn. (3.174) we find
Z∞
2
dω
1 X −β~ωm ω S(q, ω) =
e
m n̂q n (ωn − ωm )
2π
VZ m,n
−∞
=
1 X −β~ωm e
m n̂q n n [H, n̂†q ] m
~VZ m,n
1 1 n̂q [H, n̂†q ] =
n̂q , [H, n̂†q ] ,
(3.180)
~V
2~V
where the last equality is guaranteed by q → −q symmetry. Now if the potential is velocity
independent, i.e. if
N
~2 X 2
H=−
(3.181)
∇i + V (r1 , . . . , rN ) ,
2m
i=1
P
iq·ri we obtain
then with n̂†q = N
i=1 e
=
N
[H, n̂†q ] = −
=
n̂q , [H, n̂†q ] =
~2 X 2 iq·ri ∇i , e
2m
i=1
N X
~2
q·
2im
~2
2im
q·
∇i eiq·ri + eiq·ri ∇i
i=1
N X
N h
X
e−iq·rj , ∇i eiq·ri + eiq·ri ∇i
i
i=1 j=1
2 2
= N ~ q /m .
We have derived the f -sum rule:
Z∞
(3.182)
dω
N ~q 2
ω S(q, ω) =
.
2π
2mV
(3.183)
(3.184)
−∞
Note that this integral, which is the first moment of the structure factor, is independent of
the potential !
n times
z
}|
{
Z∞
D
h
iE
dω n
1
ω S(q, ω) =
n̂q H, H, · · · [H, n̂†q ] · · ·
.
(3.185)
2π
~V
−∞
Moments with n > 1 in general do depend on the potential. The n = 0 moment gives
Z∞
dω n
1 S(q) ≡
ω S(q, ω) =
n̂q n̂†q
2π
~V
−∞
Z
1
=
d3r hn(r) n(0)i e−iq·r ,
(3.186)
~
which is the Fourier transform of the density-density correlation function.
3.7. DYNAMIC STRUCTURE FACTOR FOR THE ELECTRON GAS
23
Compressibility Sum Rule
The isothermal compressibility is given by
κT = −
1 ∂V 1 ∂n = 2
.
V ∂n T
n ∂µ T
(3.187)
Since a constant potential U (r, t) is equivalent to a chemical potential shift, we have
hδni = χ(0, 0) δµ
=⇒
1
κT =
lim
~n2 q→0
Z∞
dω S(q, ω)
.
π
ω
(3.188)
−∞
This is known as the compressibility sum rule.
3.7
Dynamic Structure Factor for the Electron Gas
The dynamic structure factor S(q, ω) tells us about the spectrum of density fluctuations.
P
The density operator n̂†q = i eiq·ri increases the wavevector by q. At T = 0, in order for
† n n̂q G to be nonzero (where G is the ground state, i.e. the filled Fermi sphere), the
state n must correspond to a particle-hole excitation. For a given q, the maximum excitation
frequency is obtained by taking an electron just inside the Fermi sphere, with wavevector
k = kF q̂ and transferring it to a state outside the Fermi sphere with wavevector k + q. For
|q| < 2kF , the minimum excitation frequency is zero – one can always form particle-hole
excitations with states adjacent to the Fermi sphere. For |q| > 2kF , the minimum excitation
frequency is obtained by taking an electron just inside the Fermi sphere with wavevector
k = −kF q̂ to an unfilled state outside the Fermi sphere with wavevector k + q. These cases
are depicted graphically in fig. 3.2.
We therefore have
ωmax (q) =
ωmin (q) =
~q 2 ~kF q
+
2m
m



0


 ~q2 −
2m
~kF q
m
(3.189)
if q ≤ 2kF
(3.190)
if q > 2kF .
This is depicted in fig. 3.3. Outside of the region bounded by ωmin (q) and ωmax (q), there
are no single pair excitations. It is of course easy to create multiple pair excitations with
arbitrary energy and momentum, as depicted in fig. 3.4. However, these multipair states
do not couple to the ground state G through a single application of the density operator
n̂†q , hence they have zero oscillator strength: n n̂†q G = 0 for any multipair state n .
24
CHAPTER 3. LINEAR RESPONSE THEORY
Figure 3.2: Minimum and maximum frequency particle-hole excitations in the free electron
gas at T = 0. (a) To construct a maximum frequency excitation for a given q, create a hole
just inside the Fermi sphere at k = kF q̂ and an electron at k0 = k + q. (b) For |q| < 2kF the
minumum excitation frequency is zero. (c) For |q| > 2kF , the minimum excitation frequency
is obtained by placing a hole at k = −kF q̂ and an electron at k0 = k + q.
3.7.1
Explicit T = 0 Calculation
We start with
S(r, t) = hn(r, t) n(0, 0)
Z 3 Z 3 0
dk
d k ik·r X −ik·ri (t) ik0 ·rj e
.
e
e
=
(2π)3 (2π)3
(3.191)
(3.192)
i,j
The time evolution of the operator ri (t) is given by ri (t) = ri + pi t/m, where pi = −i~∇i .
Using the result
1
eA+B = eA eB e− 2 [A,B] ,
(3.193)
which is valid when [A, [A, B]] = [B, [A, B]] = 0, we have
e−ik·ri (t) = ei~k
2 t/2m
e−ik·r e−ik·pi t/m ,
(3.194)
hence
Z
S(r, t) =
Z 3 0
d3k
d k i~k2 t/2m ik·r X −ik·ri ik·pi t/m ik0 ·rj e
e
e
e
e
.
(2π)3 (2π)3
i,j
(3.195)
3.7. DYNAMIC STRUCTURE FACTOR FOR THE ELECTRON GAS
25
Figure 3.3: Minimum and maximum excitation frequency ω in units of εF /~ versus wavevector q in units of kF . Outside the hatched areas, there are no single pair excitations.
We now break the sum up into diagonal (i = j) and off-diagonal (i 6= j) terms.
For the diagonal terms, with i = j, we have
−ik·r ik·p t/m ik0 ·r 0
i
i
(3.196)
= e−i~k·k t/m ei(k−k)·ri eik·pi t/m
e i
e
e
3
X
(2π)
0
= e−i~k·k t/m
δ(k − k0 )
Θ(kF − q) e−i~k·qt/m ,
NV
q
since the ground state G is a Slater determinant formed of single particle wavefunctions
√
ψk (r) = exp(iq · r)/ V with q < kF .
For i 6= j, we must include exchange effects. We then have
−ik·r ik·p t/m ik0 ·r i
j
e
e i
e
=
=
XX
1
Θ(kF − q) Θ(kF − q 0 )
N (N − 1) q
q0
Z 3 Z 3
d ri d rj −i~k·qt/m −ik·ri ik0 rj
×
e
e
e
V
V
i(q−q 0 −k)·ri i(q 0 −q+k0 )·rj
−e
e
XX
(2π)6
Θ(kF − q) Θ(kF − q 0 )
N (N − 1)V 2
q
q0
n
o
× e−i~k·qt/m δ(k) δ(k0 ) − δ(k − k0 ) δ(k + q 0 − q) . (3.197)
26
CHAPTER 3. LINEAR RESPONSE THEORY
Figure 3.4: With multiple pair excitations, every
part of (q, ω) space is accessible. However,
these states to not couple to the ground state G through a single application of the density
operator n̂†q .
Summing over the i = j terms gives
Z
Sdiag (r, t) =
d3k ik·r −i~k2 t/2m
e
e
(2π)3
Z
d3q
Θ(kF − q) e−i~k·qt/m ,
(2π)3
(3.198)
while the off-diagonal terms yield
Z 30
d3q
dq
=
Θ(kF − q) Θ(kF − q 0 )
3
(2π) (2π)3
3
+i~k2 t/2m −i~k·qt/m
0
× (2π) δ(k) − e
e
δ(q − q − k)
Z 3
Z 3
dq
d k ik·r +i~k2 t/2m
2
e
e
Θ(kF − q) Θ(kF − |k − q|) e−i~k·qt/m ,
=n −
3
(2π)
(2π)3
(3.199)
Z
Soff−diag
d3k ik·r
e
(2π)3
Z
and hence
d3q
~k2 ~k · q S(k, ω) = n (2π) δ(k) δ(ω) +
Θ(k
−
q)
2π
δ
ω
−
−
F
(2π)3
2m
m
2
~k
~k · q
− Θ(kF − |k − q|) 2πδ ω +
−
(3.200)
2m
m
Z 3
dq
~k2 ~k · q 4 2
= (2π) n δ(k)δ(ω) +
Θ(k
−
q)
Θ(|k
+
q|
−
k
)
·
2πδ
ω
−
−
.
F
F
(2π)3
2m
m
2
4
Z
3.7. DYNAMIC STRUCTURE FACTOR FOR THE ELECTRON GAS
27
For k, ω 6= 0,, then,
1
S(k, ω) =
2π
ZkF
Z1
p
~k 2 ~kq 2
dq q dx Θ k 2 + q 2 + 2kqx − kF δ ω −
−
x
2m
m
−1
0
m
=
2π~k
kF
Z
0
m
=
4π~k
r
Z1
2mω
k
mω 2
dq q Θ
q +
dx δ x +
− kF
−
~
2q ~kq
−1
2
kF
Z
0
k mω 2 2mω
2
du Θ u +
.
− kF Θ u − −
~
2
~k
(3.201)
The constraints on u are
2mω k mω 2
kF ≥ u ≥ max kF −
, −
.
~
2
~k
2
2
(3.202)
Clearly ω > 0 is required. There are two cases to consider.
The first case is
kF2 −
2mω k mω 2
≥ −
~
2
~k
=⇒
0≤ω≤
~kF k ~k 2
−
,
m
2m
which in turn requires k ≤ 2kF . In this case, we have
2mω m
2
2
k − kF −
S(k, ω) =
4π~k F
~
2
m ω
=
.
2π~2 k
(3.203)
(3.204)
The second case
kF2 −
2mω k mω 2
≤ −
~
2
~k
=⇒
ω≥
~kF k ~k 2
−
.
m
2m
(3.205)
However, we also have that
k mω 2
−
≤ kF2 ,
2
~k
(3.206)
~k
~k
|k − 2kF | ≤ ω ≤
|k + 2kF | .
2m
2m
(3.207)
hence ω is restricted to the range
The integral in (3.201) then gives
k mω 2 m
2
S(k, ω) =
k − −
.
4π~k F
2
~k
(3.208)
28
CHAPTER 3. LINEAR RESPONSE THEORY
Figure 3.5: The dynamic structure factor S(k, ω) for the electron gas at various values of
k/kF .
Putting it all together,

mkF


· πω

π 2 ~2 2v k


F





πkF
ω
F
S(k, ω) = mk
1
−
−
·
2
2

4k
π ~
vF k








0
if 0 < ω ≤ vF k −
k
2kF
2 if vF k −
~k2 2m if ω ≥ vF k +
~k2
2m
≤ ω ≤ vF k +
~k2
2m
~k2
2m
(3.209)
.
Integrating over all frequency gives the static structure factor,
Z∞
1
† S(k) =
n n =
V k k
dω
S(k, ω) .
2π
(3.210)
−∞
The result is
S(k) =

3k

−


 4kF




n







 2
Vn
k3
3
16kF
n
if 0 < k ≤ 2kF
if k ≥ 2kF
if k = 0 ,
where n = kF3 /6π 2 is the density (per spin polarization).
(3.211)
3.8. CHARGED SYSTEMS: SCREENING AND DIELECTRIC RESPONSE
3.8
3.8.1
29
Charged Systems: Screening and Dielectric Response
Definition of the Charge Response Functions
Consider a many-electron system in the presence of a time-varying external charge density
ρext (r, t). The perturbing Hamiltonian is then
Z
Z
n(r) ρext (r, t)
3
H1 = −e d r d3r0
r − r 0 Z 3
d k 4π
= −e
n̂(k) ρ̂ext (−k, t) .
(2π)3 k2
(3.212)
The induced charge is −e δn, where δn is the induced number density:
δ n̂(q, ω) =
4πe
χ(q, ω) ρ̂ext (q, ω) .
q2
(3.213)
We can use this to determine the dielectric function (q, ω):
∇ · D = 4πρext
(3.214)
∇ · E = 4π ρext − e hδni .
(3.215)
In Fourier space,
iq · D(q, ω) = 4π ρ̂ext (q, ω)
(3.216)
iq · E(q, ω) = 4π ρ̂ext (q, ω) − 4πe δ n̂(q, ω) ,
(3.217)
so that from D(q, ω) = (q, ω) E(q, ω) follows
1
iq · E(q, ω)
δ n̂(q, ω)
=
=1−
(q, ω)
iq · D(q, ω)
Z n̂ext (q, ω)
(3.218)
4πe2
χ(q, ω) .
q2
(3.219)
4πe2
χ(q, 0) = 1 .
q→0 q 2
(3.220)
=1−
A system is said to exhibit perfect screening if
(q → 0, ω = 0) = ∞
=⇒
lim
Here, χ(q, ω) is the usual density-density response function,
χ(q, ω) =
1 X
2ωn◦
n n̂q 0 2 ,
2
2
~V n ωn◦ − (ω + i)
(3.221)
where we content ourselves
to work at T = 0, and where ωn◦ ≡ ωn − ω◦ is the excitation
frequency for the state n .
30
CHAPTER 3. LINEAR RESPONSE THEORY
From jcharge = σE and the continuity equation
iq · hĵcharge (q, ω)i = −ieωhn̂(q, ω)i = iσ(q, ω) q · E(q, ω) ,
(3.222)
(3.223)
we find
4π ρ̂ext (q, ω) − 4πe δ n̂(q, ω) σ(q, ω) = −iωe δ n̂(q, ω) ,
or
δ n̂(q, ω)
1 − −1 (q, ω)
4πi
σ(q, ω) =
= (q, ω) − 1 .
=
ω
−1 (qω)
e−1 ρ̂ext (q, ω) − δ n̂(q, ω)
(3.224)
Thus, we arrive at
1
4πe2
= 1 − 2 χ(q, ω)
(q, ω)
q
,
(q, ω) = 1 +
4πi
σ(q, ω)
ω
(3.225)
Taken together, these two equations allow us to relate the conductivity and the charge
response function,
iω
e2 χ(q, ω)
σ(q, ω) = − 2
.
(3.226)
q 1 − 4πe2 2 χ(q, ω)
q
3.8.2
Static Screening: Thomas-Fermi Approximation
Imagine a time-independent, slowly varying electrical potential φ(r). We may define the
‘local chemical potential’ µ
e(r) as
µ≡µ
e(r) − eφ(r) ,
(3.227)
where µ is the bulk chemical potential. The local chemical potential is related to the local
density by local thermodynamics. At T = 0,
2/3
~2 2
~2 2
kF (r) =
3π n + 3π 2 δn(r)
2m
2m
2
~
2 δn(r)
2 2/3
=
(3π n)
1+
+ ... ,
2m
3 n
µ
e(r) ≡
(3.228)
hence, to lowest order,
δn(r) =
3en
φ(r) .
2µ
(3.229)
This makes sense – a positive potential induces an increase in the local electron number
density. In Fourier space,
hδ n̂(q, ω = 0)i =
3en
φ̂(q, ω = 0) .
2µ
(3.230)
3.8. CHARGED SYSTEMS: SCREENING AND DIELECTRIC RESPONSE
31
Poisson’s equation is −∇2 φ = 4πρtot , i.e.
iq · E(q, 0) = q 2 φ̂(q, 0)
= 4π ρ̂ext (q, 0) − 4πe hδ n̂(q, 0)i
= 4π ρ̂ext (q, 0) −
6πne2
µ
(3.231)
φ̂(q, 0) ,
(3.232)
and defining the Thomas-Fermi wavevector qTF by
2
qTF
≡
6πne2
,
µ
(3.233)
we have
φ̂(q, 0) =
4π ρ̂ext (q, 0)
,
2
q 2 + qTF
(3.234)
hence
e hδ n̂(q, 0)i =
2
qTF
· ρ̂ext (q, 0)
2
q 2 + qTF
=⇒
(q, 0) = 1 +
2
qTF
q2
(3.235)
Note that (q → 0, ω = 0) = ∞, so there is perfect screening.
−1
The Thomas-Fermi wavelength is λTF = qTF
, and may be written as
λTF =
π 1/6 √
√
rs aB ' 0.800 rs aB ,
12
(3.236)
where rs is the dimensionless free electron sphere radius, in units of the Bohr radius aB =
~2 /me2 = 0.529Å, defined by 34 π(rs aB )3 n = 1, hence rs ∝ n−1/3 . Small rs corresponds to
high density. Since Thomas-Fermi theory is a statistical theory, it can only be valid if there
1/3
are many particles within a sphere of radius λTF , i.e. 43 πλ3TF n > 1, or rs <
∼ (π/12) ' 0.640.
TF theory is applicable only in the high density limit.
In the presence of a δ-function external charge density ρext (r) = Ze δ(r), we have ρ̂ext (q, 0) =
Ze and
hδ n̂(q, 0)i =
2
ZqTF
2
q 2 + qTF
=⇒
hδn(r)i =
Z e−r/λTF
4πr
(3.237)
Note the decay on the scale of λTF . Note also the perfect screening:
e hδ n̂(q → 0, ω = 0)i = ρ̂ext (q → 0, ω = 0) = Ze .
3.8.3
(3.238)
High Frequency Behavior of (q , ω)
We have
−1 (q, ω) = 1 −
4πe2
χ(q, ω)
q2
(3.239)
32
CHAPTER 3. LINEAR RESPONSE THEORY
and, at T = 0,
1
1 X † 2
1
−
,
χ(q, ω) =
j n̂q 0
~V
ω + ωj0 + i ω − ωj0 + i
j
(3.240)
where the number density operator is
n̂†q =
P
iq·r

 ie i
(1st quantized)
(3.241)
†
k ψk+q

P
ψk
nd
(2
quantized:
{ψk , ψk† 0 }
= δkk0 ) .
Taking the limit ω → ∞, we find
Z∞ 0
2
dω 0
2 X † 2
ωj0 = − 2
j n̂q 0
ω S(q, ω 0 ) .
χ(q, ω → ∞) = −
2
~V ω
~ω
2π
j
(3.242)
−∞
Invoking the f -sum rule, the above integral is n~q 2 /2m, hence
χ(q, ω → ∞) = −
nq 2
,
mω 2
and
−1 (q, ω → ∞) = 1 +
ωp2
,
ω2
(3.243)
(3.244)
where
r
ωp ≡
4πne2
m
(3.245)
is the plasma frequency.
3.8.4
Random Phase Approximation (RPA)
The electron charge appears nowhere in the free electron gas response function χ0 (q, ω).
An interacting electron gas certainly does know about electron charge, since the Coulomb
repulsion between electrons is part of the Hamiltonian. The idea behind the RPA is to
obtain an approximation to the interacting χ(q, ω) from the noninteracting χ0 (q, ω) by
self-consistently adjusting the charge so that the perturbing charge density is not ρext (r),
but rather ρext (r, t) − e hδn(r, t)i. Thus, we write
4πe2 RPA
χ (q, ω) ρ̂ext (q, ω)
q2
n
o
4πe2
= 2 χ0 (q, ω) ρ̂ext (q, ω) − e hδ n̂(q, ω)i ,
q
e hδ n̂(q, ω)i =
(3.246)
(3.247)
3.8. CHARGED SYSTEMS: SCREENING AND DIELECTRIC RESPONSE
33
Figure 3.6: Perturbation expansion for RPA susceptibility bubble. Each bare bubble contributes a factor χ0 (q, ω) and each wavy interaction line v̂(q). The infinite series can be
summed, yielding eqn. 3.249.
which gives
χRPA (q, ω) =
χ0 (q, ω)
1+
4πe2 0
χ (q, ω)
q2
(3.248)
Several comments are in order.
1. If the electron-electron interaction were instead given by a general v̂(q) rather than
the specific Coulomb form v̂(q) = 4πe2 /q 2 , we would obtain
χRPA (q, ω) =
χ0 (q, ω)
.
1 + v̂(q) χ0 (q, ω)
(3.249)
2. Within the RPA, there is perfect screening:
4πe2 RPA
χ (q, ω) = 1 .
q→0 q 2
lim
(3.250)
3. The RPA expression may be expanded in an infinite series,
χRPA = χ0 − χ0 v̂ χ0 + χ0 v̂ χ0 v̂ χ0 − . . . ,
(3.251)
which has a diagrammatic interpretation, depicted in fig. 3.6. The perturbative
expansion in the interaction v̂ may be resummed to yield the RPA result.
4. The RPA dielectric function takes the simple form
RPA (q, ω) = 1 +
4πe2 0
χ (q, ω) .
q2
(3.252)
34
CHAPTER 3. LINEAR RESPONSE THEORY
5. Explicitly,
"
(
2
(ω − ~q 2 /2m)2
1 kF
qTF
ω − vF q − ~q 2 /2m 1−
+
ln Re (q, ω) = 1 + 2
ω + vF q − ~q 2 /2m q
2 4q
(vF q)2
#)
(ω − ~q 2 /2m)2
ω − vF q + ~q 2 /2m ln + 1−
(3.253)
ω + vF q + ~q 2 /2m (vF q)2

q2
πω

· qTF
if 0 ≤ ω ≤ vF q − ~q 2 /2m

2

2vF q






2
2
2
RPA
Im (q, ω) = πkF 1 − (ω−~q /2m) qTF
if vF q − ~q 2 /2m ≤ ω ≤ vF q + ~q 2 /2m
4q
q2

(vF q)2








0
if ω > vF q + ~q 2 /2m
(3.254)
RPA
6. Note that
RPA (q, ω → ∞) = 1 −
ωp2
,
ω2
(3.255)
in agreement with the f -sum rule, and
RPA (q → 0, ω = 0) = 1 +
2
qTF
,
q2
(3.256)
in agreement with Thomas-Fermi theory.
7. At ω = 0 we have
RPA
q2
(q, 0) = 1 + TF
q2
(
)
q + 2kF 1 kF
q2
,
+
1 − 2 ln 2 2q
4kF
2 − 2kF
(3.257)
which is real and which has a singularity at q = 2kF . This means that the long-distance
behavior of hδn(r)i must oscillate. For a local charge perturbation, ρext (r) = Ze δ(r),
we have
Z∞
Z
1
hδn(r)i = 2
dq q sin(qr) 1 −
,
(3.258)
2π r
(q, 0)
0
and within the RPA one finds for long distances
hδn(r)i ∼
Z cos(2kF r)
,
r3
rather than the Yukawa form familiar from Thomas-Fermi theory.
(3.259)
3.8. CHARGED SYSTEMS: SCREENING AND DIELECTRIC RESPONSE
3.8.5
35
Plasmons
The RPA response function diverges when v̂(q) χ0 (q, ω) = −1. For a given value of q, this
occurs for a specific value (or for a discrete set of values) of ω, i.e. it defines a dispersion
relation ω = Ω(q). The poles of χRPA and are identified with elementary excitations of the
electron gas known as plasmons.
To find the plasmon dispersion, we first derive a result for χ0 (q, ω), starting with
i (3.260)
χ0 (q, t) =
h n̂(q, t), n̂(−q, 0) i
~V
h
i
X †
X †
i
=
h
ψk,σ ψk+q,σ ,
ψk0 ,σ0 ψk0 −q,σ0 i ei(ε(k)−ε(k+q))t/~ Θ(t) ,
(3.261)
~V
0 0
kσ
k ,σ
where ε(k) is the noninteracting electron dispersion. For a free electron gas, ε(k) =
~2 k2 /2m. Next, using
AB, CD = A B, C D − A, C BD + CA B, D − C A, D B
(3.262)
we obtain
χ0 (q, t) =
i X
(fk − fk+q ) ei(ε(k)−ε(k+q))t/~ Θ(t) ,
~V
(3.263)
kσ
and therefore
0
Z
χ (q, ω) = 2
fk+q − fk
d3k
(2π)3 ~ω − ε(k + q) + ε(k) + i
.
(3.264)
Here,
fk =
1
e(ε(k)−µ)/kB T
+1
(3.265)
is the Fermi distribution. At T = 0, fk = Θ(kF − k), and for ω vF q we can expand
χ0 (q, ω) in powers of ω −2 , yielding
q2
3 ~kF q 2
kF3
0
χ (q, ω) = − 2 ·
1+
+ ... ,
(3.266)
3π mω 2
5 mω
so the resonance condition becomes
4πe2
0 = 1 + 2 χ0 (q, ω)
q
ωp2
3 vF q 2
=1− 2 · 1+
+ ... .
ω
5 ω
(3.267)
This gives the dispersion
3 vF q 2
ω = ωp 1 +
+ ... .
10 ωp
(3.268)
Recall that the particle-hole continuum frequencies are bounded by ωmin (q) and ωmax (q),
which are given in eqs. 3.190 and 3.189. Eventually the plasmon penetrates the particle-hole
continuum, at which point it becomes heavily damped since it can decay into particle-hole
excitations.