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Excitons and Cooper PairsTwo Composite Bosons in Many-Body Physics$

Monique Combescot and Shiue-Yuan Shiau

Print publication date: 2015

Print ISBN-13: 9780198753735

Published to Oxford Scholarship Online: March 2016

DOI: 10.1093/acprof:oso/9780198753735.001.0001

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(p.465) Appendix A Some Mathematical Results

(p.465) Appendix A Some Mathematical Results

Source:
Excitons and Cooper Pairs
Author(s):

Monique Combescot

Shiue-Yuan Shiau

Publisher:
Oxford University Press

A.1 Kronecker symbol and delta function

The Kronecker symbol and the delta function are fundamentally the same: they both force the quantity at hand, whether quantized or continuous, to be equal. The associated prefactors however differ when switching from a discrete sum over k quantized in 2π/L to an integral

(A.1)
k  L2π3d3k.

These prefactors can be (2π)D, LD, or ND, where D is the space dimension, L is the sample size, and N is the number of terms in the 1D discrete sum. While these prefactors can be easily guessed from dimensional arguments, they of course follow from hard algebra.

A.1.1 For R and Q quantized in a size L = Na sample

  1. (i) For 1D vectors R = na and Q=p(2π/L), with L = Na, and (n,p,N) integers, the sum

    (A.2)
    S(Q)=ReiQR=n=0N1eiQna

    (p.467) reduces to N for Q = 0 while, for Q0,

    (A.3)
    S(Q0)=1eiQNa1eiQa=1eiQL1eiQa=0,

    since Q=p(2π/L). So,

    (A.4)
    ReiQR=NδQ,0.

    For 3D vectors, we split the R sum through QR=QxRx+QyRy+QzRz and then use Eq. (A.4) for NxNyNz=N3. This gives

    (A.5)
    ReiQR=N3δQ,0,

    with N3 replaced by N2 for 2D vectors. So, for space dimension D, we end with

    (A.6)
    ReiQR=NDδQ,0.

  2. (ii) In the same way, for 1D vectors R = na and Q=p(2π/L), with L = Na,

    (A.7)
    QeiQR=p=0N1eiRp2π/L.

    The above sum reduces to N for R = 0 while, for R0, it reads 1eiRN2π/L/1eiR2π/L. As the numerator cancels for R = na with n integer, the sum in Eq. (A.7) reduces to NδR,0, which extends in D dimensions as

    QeiQR=NDδR,0.

A.1.2 For r continuous and Q quantized in 2π/L

  1. (i) In 1D, the integral 0LdreiQr is equal to L for Q = 0, while for Q0, it reads (eiQL1)/iQ, which cancels for Q quantized in 2π/L. This result extends in D dimensions as

    LDdDreiQr=LDδQ,0.

  2. (ii) In 1D, the discrete sum

    S(r)=QeiQr=p=0N1eirp2π/L

    (p.468) is equal to N for r = 0, while for r0, it reads

    S(r0)=1eirN2π/L1eir2π/L=eiπNr/Leiπr/L sin(πNr/L)sin(πr/L).

    We then note that, for |r|L=Na,

    sin(πNr/L)sin(πr/L)=sin(πr/a)sin(πr/L)sin(πr/a)πr/LLδa(r),

    where δa(r)=(πr)1sin(πr/a) is a possible definition of the delta function when the width a goes to 0. By noting that S(r=0)=N=L/a while δa(0)=1/a, we end with

    QeiQr=Lδa(r).

    This result extends in D dimensions as

    QeiQr=LDδa(r).

A.1.3 For r and Q continuous in an infinite sample

The delta distribution is mathematically defined, in 1D, through

(A.15)
drf(r)δ(r)=f(0)

for any nonsingular function f(r). A possible representation of δ‎(r) is

(A.16)
δ(r)=12πdqeiqr.

A simple way to show that the above integral has the (A.15) property is to add a convergence factor to the exponential. We then find that

(A.17)
dqeiqrη|q|=0dqeq(ir+η)+0dqeq(irη)=1ir+η1irη=2ηr2η2.

For η‎ very small, η/(r2+η2) is very much peaked on r = 0; so, for f(r) nonsingular, we can replace f(r) with f(0) in the product f(r)2η/(r2+η2). We are left with

(A.18)
dr2ηr2+η2=2ηπη=2π,

which leads to Eq. (A.15) for δ‎(r) given by Eq. (A.16).

(p.469) Extension to D dimensions gives

(A.19)
dDq eiqr=(2π)Dδ(r)

and, in the same way,

(A.20)
dDr eiqr=(2π)Dδ(q).

A.2 Fourier transform and series expansion

A.2.1 Fourier transform

Let r be a continuous vector, and |r be the set of states normalized by

(A.21)
r|r=δ(rr).

Their closure relation reads in D dimensions

(A.22)
I=dDr|rr|,

as seen from I2=I.

We now introduce the states |q in the dual space, defined as

(A.23)
r|q=eiqr(2π)D/2.

They are normalized as

(A.24)
q|q=dDrq|rr|q=dDr(2π)Dei(qq)r=δ(qq),

their closure relation reading

(A.25)
I=dDq|qq|.

By writing the function f(r) of the continuous variable r as r|f, we find that

(A.26)
f(r)=r|f=dDqr|qq|f=dDqeiqr(2π)D/2 fq.

In the same way, the Fourier transform fq of f(r) reads in terms of f(r) as

(A.27)
fq=q|f=dDrq|rr|f=dDreiqr(2π)D/2 f(r).

(p.470) This compact approach to Fourier transform leads to an equal splitting of the prefactor (1/2π)D between f(r) and fq. It however fails to address the fundamental restriction on f(r) or fq convergence, as illustrated in Section A.3 for the Coulomb potential: 4πe2/q2 decreases fast enough at large q to have a Fourier transform, but e2/r does not.

A.2.2 Series expansion

We now consider a function of the continuous variable r in a sample volume LD and with periodic boundary conditions, φ(r)=φ(r+L). Because of this periodicity, we can restrict φ(r)=r|φ to r inside the LD volume. The closure relation for the relevant |r states then reads

(A.28)
I=LDdDr|rr|.

A possible complete basis for such periodic functions is made of |Q states with wave function

(A.29)
r|Q=eiQrLD/2.

The states |Q have the L periodicity, r|Q=r+L|Q for eiQL=1, that is, for Q quantized in 2π/L. These states are normalized as

(A.30)
Q|Q=LDdDr Q|rr|Q=LDdDrLD ei(QQ)r=δQ,Q,

their closure relation reading

(A.31)
I=Q|QQ|,

as seen from I2=I.

The periodic function φ(r) expands on the |Q states as

(A.32)
φ(r)=r|φ=Qr|QQ|φ=QeiQrLD/2φQ,

or in the dual space as

(A.33)
φQ=Q|φ=LDdDr Q|rr|φ=LDdDr eiQrLD/2φ(r).

For φ(r) equal to a constant ϕ0, this series expansion has one term only, φQ=δQ,0φ0LD/2, while the average value of φ(r) is related to φQ=0 as

(A.34)
LDdDrLDφ(r)=φQ=0LD/2.

(p.471) A.3 Coulomb scatterings

A.3.1 Bulk systems

We here show that, in a 3D sample volume L3, the Coulomb scattering associated with nonzero momentum transfer Q reads

(A.35)
VQ0(3)=4πe2L3Q2.

Although commonly done, dropping the sample volume by setting L3=1 is unwise. Indeed, keeping the volume factor L3 is necessary to check homogeneity in the obtained results: potential scatterings must scale as an energy, that is, as e2/[L]. Keeping L3 also allows us to identify overextensive terms, which must cancel out exactly, and underextensive terms, which can be neglected in extensive quantities like the system energy.

Equation (A.35) follows from calculating

(A.36)
1(2π)3d3q 4πe2q2eiqr=1(2π)302πq2dq 4πe2q20πsinθdθ eiqrcosθ=e2π0dq11dueiqru=2e2πr0sinqrqdq.

The integral over q being equal to π/2, we end up with

(A.37)
1(2π)3d3q 4πe2q2eiqr=e2r.

At this stage, we can note that discrete sums over Q quantized in 2π/L are commonly calculated by turning to an integral according to

(A.38)
QfQ  L2π3d3qfq.

This would lead to

(A.39)
e2r=L2π3d3q 4πe2L3q2eiqr=QVQeiQr,

with VQ=4πe2/L3Q2. However, this procedure is not fully satisfactory, because 4πe2/L3Q2 is infinite for Q = 0. So, the Q = 0 term has to be excluded from the Q sum. This singularity can be directly seen from

(A.40)
d3r eiQre2r,

which diverges for Q = 0 because of the long-range character of Coulomb potential.

It is possible to give a meaning to Eq. (A.40) by restricting the integral to a finite volume, namely,

(A.41)
VQ(3)=L3d3rL3 eiQre2r,

(p.472) For Q = 0, this gives V0(3)=2πe2/L while, for Q0, Eq. (A.37) leads to

(A.42)
VQ0(3)=L3d3rL3eiQr 1(2π)3d3q 4πe2q2eiqr.

For q continuous, the integral over r gives (2π)3δ(qQ), as shown in Eq. (A.20). So, we end with

(A.43)
VQ0(3)=4πe2L3Q2,

in agreement with Eq. (A.35).

A.3.2 Two–dimensional quantum wells

Addressing problems dealing with semiconductor quantum wells requires considering Coulomb scatterings in 2D. Following the same procedure, we first calculate

(A.44)
1(2π)2d2q 2πe2qeiqr=1(2π)20qdq2πe2q02πdθeiqrcosθ=e22π40dq0π/2dθcos(qrcosθ).

The integral over θ‎ gives (π/2)J0(qr), while the integral of the Bessel function J0(x) from 0 to ∞ gives 1. So, we end up with

(A.45)
1(2π)2d2q 2πe2qeiqr=e2r.

Since the Fourier transform of e2/r is infinite when calculated in an infinite volume, we introduce, as for 3D,

(A.46)
VQ(2)=L2d2rL2 eiQre2r.

For Q = 0, we again find that V0(2)=2πe2/L while, for Q0, Eq. (A.45) leads to

(A.47)
VQ0(2)=L2d2rL2eiQr 1(2π)2d2q2πe2qeiqr.

For q continuous, the integral over r gives (2π)2δ(qQ). So, we end with

(A.48)
VQ0(2)=2πe2L2Q.

Note that the factors L3 or L2 makes the VQ0(D) scatterings scale as e2/[L]; so, these scatterings are energy-like quantities, as required.