1Quantum field theory and Green’s function

† Condensed matter physics studies systems with large numbers of identical particles (e.g. electrons, phonons, photons) at finite temperature.

† Quantum field theory arises naturally if we consider a quantum system composed by a large number of identical particles at finite or zero temperature.

1.1. Quantum statistical physics

1.1.1. Classical statistical physics

Classical statistical physics tells us that in the grand canonical ensemble, the expectation value of a thermal dynamic quantity, which will becalled X here, can be written as:

(1.1)X = n Xn exp-bEn - m Nnn exp-bEn - m Nn

where b = 1 T (here the Boltzmann constant kB is set to unity) and m is the chemical potential. Here, we sum over all possible states of thesystem (n). For the nth state, the total energy of the whole system is En; the particle number is Nn and the value of the quantity X is Xn.

1.1.2. Quantum statistical physics

For a quantum system (at finite temperature T), we can rewrite this formula in terms of quantum operators

(1.2)X =n n X

`exp-bH` - m N

` nn n exp-bH` - m N

` n=

Tr X`

exp-bH` - m N`

Tr exp-bH` - m N`

where n sums over a complete orthonormal basis of the Hilbert space n. H` is the Hamiltonian. N`

is the particle number operator and X`

isthe quantum operator of the quantity that we want to compute. The sum n n … n is also known as a trace Tr. It is easy to verify that theabove formula for X is independent of the choice of basis.

Here, we will not prove this operator formula, because it is not part of the main focus of this lecture. Instead, we will only demonstrate that bychoosing a proper basis, this (quantum) operator formula recovers the classical formula shown above, and therefore, the operator formula is anatural quantum generalization of classical statistical physics.

If H` , N` = 0, we can choose the common eigenstates of H

` and N

` as our basis. In this basis, we have

(1.3)H`

n = En nand

(1.4)N`

n = Nn nwhere En is the eigen-energy and Nn is the eigen-value of the particle number operator.

Using this basis, it is easy to check that

(1.5)exp-bH` - m N` n = exp-bEn - m Nn n

and therefore,

(1.6)X =n n X

`exp-bH` - m N

` nn n exp-bH` - m N

` n=

n n X`

n exp-bEn - m Nnn n n exp-bEn - m Nn

=n Xn exp-bEn - m Nnn exp-bEn - m Nn

Here, we define Xn = n X`

n .

Note: the quantum field theory used in particle physics is the TÆ0 limit of this finite temperature theory.

Q: What happens if H` , N` π 0?

A: If H` , N` π 0, it implies that in this system, the particle number is not conserved. As will be discussed in

later chapters, particle conservation law is related with the U(1) phase symmetry for charge neutral particles (or the U(1) gauge symmetry for charged particles). The absences of particle conservation implies the breaking of phase/gauge symmetry, which means that the system is a superfluid/superconductor.

1.2. Second quantization† The terminology of “second quantization” is due to historical reasons. We are NOT quantizing something for a second time. We are just

using a new basis to handle indistinguishable particles.

Q: Why do we want to use second quantization?

A: It is the most convenient way to handle a large number of indistinguishable particles.

1.2.1. wavefunctions for distinguishable particles

It is easy to write down quantum wavefunctions for systems composed by distinguishable particles. For example, if we have two distinguishableparticles, particle one in state y1 and particle two in y2, the wavefunction is

(1.7)Yr1, r2 = y1r1 y2r2If we have n distinguishable particles, the wavefunction can be written as

(1.8)Yr1, r2, r3, ..., rn = y1r1 y2r2 …ynrnHere, the ith particle is in the quantum state yi for i = 1, 2, …n.

1.2.2. wave functions for indistinguishable particles

For indistinguishable particles, the wavefunction is very complicated, if the particle number is large. This is because we need to ensure that thewavefunction is symmetric (anti-symmetric) for bosons (fermions). If we have two indistinguishable particles, one particle in the state y1 andthe other in the state y2, the wavefunction is

(1.9)Yr1, r2 = y1r1 y2r2 ≤ y2r1 y1r2Here, the + sign is for bosons and the - sign is for fermions. It is easy to check that the wavefunction is symmetric or anti-symmetric

(1.10)Yr1, r2 = ≤Yr2, r1If we have 3 indistinguishable particles, the wavefunction contains 3!=6 terms

(1.11)Yr1, r2, r3 =y1r1 y2r2 y3r3 ≤ y1r1 y3r2 y2r3 ≤ y2r1 y1r2 y3r3 + y3r1 y1r2 y2r3 + y2r1 y3r2 y1r3 ≤ y3r1 y2r2 y1r3

If we have n indistinguishable particles,

(1.12)Yr1, r2, r3, ..., rn =P ≤1P yi1r1 yi2r2 …yinrnwhere P represents all permutations. For a system with n particles, there are n ! different permutations and thus the r.h.s. has n ! terms. For asystem with a large number of indistinguishable particles, it is an extremely complicated to write down its wavefunction in this way. Forexample, for a system with just ten particles, n=10, there are 10!º2.6 million terms, which is impossible to write down. In condensed matter

2 Phys540.nb

physics, a typical system has huge number of particles (~1023) and thus we need a better way to write down our quantum theory.

Note: in particle physics, although one typically studies systems with a very small number of particles (e.g. two particles collides with each other in a collider), however, it is still necessary to consider large numbers of particles because we have “virtue particles”. For example, the E&M interactions between two electrons is realized by exchanging virtual photons. If one takes into account these virtual particles, the particle number is not very small, and thus we shall use second quantization.

1.2.3. Fock space

The reason why it is hard to write down wavefunctions for indistinguishable particles is because when we write do the wavefunciton, we needto specify which particle is in which quantum state. For example, yir j means the particle number j is in the quantum state yi. This procedure

is natural for distinguishable particles, where we do know that particle i is in a certain quantum state. However, for indistinguishable particles,we don’t actually know which particle is in this state, because we cannot distinguish the particles. In other words, the traditional way to writedown a wavefunction is designed for distinguishable particles. For indistinguishable particles, we need to first treat them as distinguishableparticles, and then repeat the wavefunciton using all different permutations to make the wavefunction symmetric/anti-symmetric. This proce-dure is not natural and is very complicated.

For indistinguishable particles, it is more natural to use the occupation number basis, which is known as the Fock space. In the Fock space, amany-body quantum state is written in terms of occupation numbers:

(1.13)Y = n1, n2, n3 …nNwhere ni is the number of particles in state yi (ni is known as the occupation number). Here we don’t specify which particle is in the stateyi. Instead, we just count the number of particles in this state. In this approach, the particle are indistinguishable automatically and thus all

the complicates mentioned above are avoided.

1.2.4. Creation and annihilation operators:

In the Fock space, all physical operators can be written in terms of creation and annihilation operators. The creation operator increases theoccupation number by 1 and the annihilation operator reduces the occupation number by 1.

For bosons, we have

(1.14)bi† n1, n2, n3 …nN = ni + 1 n1, n2, , …ni + 1, …nN

(1.15)bi n1, n2, n3 …nN = ni n1, n2, , …ni - 1, …nN and it is easy to check that these operators obey the following commutation relations:

(1.16)bi, b j = bi†, b j

† = 0

(1.17)bi, b j† = dij

For fermions, due to the Pauli exclusive principle, each quantum state can at most have one particle (so ni = 0 or 1).

(1.18)ci† …, 1, … = 0

(1.19)ci† …, 0, … = -1j=1

i-1ni …, 1, …(1.20)ci …, 1, … = 0

(1.21)ci† …, 0, … = -1j=1

i-1ni …, 0, …It is easy to check that fermion creation/annihilation operators obey anti-commutation relations:

(1.22)ci, c j = ci†, c j

† = 0

(1.23)ci, c j† = dij

Phys540.nb 3

Note: the factor -1j=1i-1ni is dropped in many textbooks, because +|y and -|y represent the same

quantum state. However, this factor is curial to ensure the anti-commutation relation ci, cj† = dij. We will

study this factor later when we examine 1D quantum systems.

1.2.5. Particle number operator

For bosons, using the definition above, it is easy to check that

(1.24)bi† bi …, ni, … = ni …, ni, …

So the eigenvalues of the operator bi† bi is the number of particles in quantum the state i, which is known as the occupation number. In other

words, the particle number quantum operator for bosons in the quantum state yi is(1.25)ni = bi

† bi

The total particle number operator is

(1.26)N`=

ini =

ibi

† bi

For fermions, it is easy to notice that

(1.27)ci† ci …, 0, … = 0

(1.28)ci† ci …, 1, … = 1

Therefore, the particle number operator for fermions in the quantum state yi is(1.29)ni = ci

† ci

As a result, the total particle number operator is

(1.30)N`=

ini =

ici

† ci

In the next a couple of sections, we will only consider fermions as our example (c and c†), but the same conclusions are applicable for bosons (b

and b†)

1.2.6. Quantum states

In the Fock space, all quantum states can be written in terms of creation and annihilation operators. First, we need to define the vacuum(ground states in high energy physics) 0 by assuming that there is one and only one state in the Fock space that is annihilated by anyannihilation operators. This state is our vacuum

(1.31)ci 0 = 0 for any ci

It is easy to check that this quantum states has zero particle

(1.32)N`

0 =ici

† ci 0 = 0

Then, for an arbitrary states with one particle (a one-particle state), we can write it as

(1.33)y =iai ci

† 0where ais are some complex numbers.

For two particle states, they can be written as

(1.34)y =aij c j† ci

† 0For n particle states, they can be written as

(1.35)y =ai1 …in cin† …ci2

† ci1† 0

1.2.7. quantum operators

Any quantum operator can be written in terms of creation and annihilation operators c s and c†s:

4 Phys540.nb

(1.36)X = fk1,k2 …km,q1,q2 …qm'ck1 ck2 ... ckm cq1

† cq2† …cqm'

†

If every term in this operator has m annihilation operators and m’ creation operators, this operator is known as a (m+m’) fermion operator.

1.2.8. Physical observables and correlation functions

Since all quantum operators can be written in terms of creation and annihilation operators, the expectation value of any physical observable canbe written in terms of the expectation values of creation and annihilation operators:

(1.37)X = X = fk1,k2 …km,q1,q2 …qm'ck1 ck2 ... ckm cq1

† cq2† …cqm'

†Therefore, in second quantization, anything we need to compute reduces to objects like this: ck1 ck2 ... ckn cq1

† cq2† …cqm

†. This type of objects

are known as correlation functions. If there are N creation and annihilation operators, it is known as a N-point correlation function. Here,N = m + m '.

Q: What are the simplest nontrivial correlation functions? Here, “simplest” means that we want the number of creation and annihilation operatorto be as small as possible. “nontrivial” means the correlation function need to be nonzero.

A: If the particle number is conserved (H` , N` = 0, which is true for most of the cases we study), they are the two-point correlation functions.

This is because, if H` , N` = 0, we can prove that ck1 ck2 ... ckm cq1

† cq2† …cqm'

† = 0 if m∫m’. In other words, if we want to have a nontrivial

correlation function, it must have the same number of creation and annihilation operators. Therefore, we only need to consider N-pointcorrelation functions when N is even. The smallest positive even integer is 2, so the simplest nontrivial correlation functions are two-point

correlation functions ck cq†. The next one is four-point correlation ck1 ck2 cq1

† cq2†.

Proof: As shown above, if H` , N` = 0, we know that particle number is a conserved quantity and thus we can choose the common eigenstates of

H`

and N`

as our basis. In this basis, we have

(1.38)ck1 ck2 ... ckm cq1† cq2

† …cqm'† =

n n ck1 ck2 ... ckm cq1† cq2

† …cqm'† n exp-bEn - m Nn

n exp-bEn - m NnBecause n is the eigenstate of the total particle number operator N

`, we know that the quantum state n has Nn particles.

Define

(1.39)y = ck1 ck2 ... ckm cq1† cq2

† …cqm'† n

because a creation operator increase the particle number by 1, while an annihilation operator reduce it by 1, it is easy to note that y hasNn + m ' - m particles.

If m ' ∫ m, Nn + m ' - m ∫ Nn, which means that the quantum states n and y has different number of particles, i.e. they are both eigen states

of N`

, but they have different eigenvalues. In quantum mechanics, we learned that two eigenstates of the same operator are orthogonal, if they

have different eigenvalues, so we know immediately that n y = 0. And therefore, n ck1 ck2 ... ckm cq1† cq2

† …cqm'† n = 0. As a result, the

correlation function ck1 ck2 ... ckm cq1† cq2

† …cqm'† = 0.

Note: If H` , N` π 0, we will have nonzero correlation functions with m πm '. For example, in superfluids,

b π 0, and in superconductors ck1ck2

π 0. These cases will be considered later.

1.2.9. Summary† The terminology of “second quantization” is due to historical reasons. We are NOT quantizing something for a second time. We are just

using a new basis to handle indistinguishable particles.

† In both high energy and condensed matter physics, quantum field theory utilize the “second quantization” construction. The reason is because quantum field theory deals with more than one indistinguishable palaces, and the second quantization formulas are the most natural way to describe this type of physics.

† In second quantization (same is true for quantum field theory), any physical quantities are reduced to computing various correlation functions.

† If particle number is conserved, only correlation functions with same number of creation and annihilation operators are nontrivial.

† The order matters: ck cq† ∫ cq

† ck. Therefore, depending on the way to order the creation/annihilation operators, we can define different

correlation functions. This topics will be address in the next section.

Phys540.nb 5

1.3. Hamiltonian in second quantization

Let’s consider a Hamiltonian with three terms: the kinetic energy HK , the potential energy HP and interactions between different particles HI .

(1.40)H = HK + HP + HI

1.3.1. Potential energy

For particles in a potential UrØ, the total potential energy (summed over all particles) is

(1.41)P.E. = „ rØ

UrØ rr

Ø

where rrØ is the particle density at r

Ø. In second quantization, particle density is

(1.42)rrØ = y†r

Ø yrØ

Here, y†rØ creates an electron at r

Ø and y(r

Ø) annihilate an electron at r

Ø. Therefore, the potential energy part of the Hamiltonian is

(1.43)HP = „ rØ

UrØ y†r

Ø yrØ

1.3.2. Kinetic energy

Momentum space

Total kinetic energy for a system with many particles is

(1.44)K.E. =„ kØ

2 pdekØ nk

Ø

Here, ekØ is the dispersion relation of a single particle and nk

Ø is the number of particles with momentum k. In the denominator, d is the

spatial dimension of the system. In second quantization, the quantum operator for nkØ is

(1.45)nkØ = y†k

Ø ykØ

Here, y†kØ creates an electron at k

Ø and y(k

Ø) annihilate an electron at k

Ø. Therefore,

(1.46)HK =„ kØ

2 pdekØ y†k

Ø ykØ

If we consider nonrelastivistic particles ekØ = k2 2 m,

(1.47)HK =„ kØ

2 pd

k2

2 my†k

Ø ykØ

Q: What shall we do for relativistic particles?

A: We cannot just set e(k)=c|k| or ek = c2 k2 + m02 c4 . This is because |x| or x are not analytic functions (they are singular at

k = 0). For fermions, it turns out that we need two species of fermions to have a relativistic dispersion. This topic will be discussedlatter, when we study the Dirac theory.

Real space

The momentum space creation/annihilation operators and real space creation/annihilation operators are connected by the Fourier transforma-tion. For the annihilation operator,

6 Phys540.nb

(1.48)ykØ = „ r

؉- k

Ø

ÿ rØ

yrØ

The inverse transformation is

(1.49)yrØ =

„ kØ

2 pd‰Â k

Ø

ÿ rØ

ykØ

The transformation for the creation operators can be obtained by taking the Hermitian conjugate

(1.50)y†kØ = „ r

؉ k

Ø

ÿ rØ

y†rØ

(1.51)y†rØ =

„ kØ

2 pd‰-Â k

Ø

ÿ rØ

y†kØ

Therefore, it is easy to check that

(1.52)HK =„ kØ

2 pd

k2

2 my†k

Ø ykØ = „ r

Ø“y†r

Ø“yrØ

2 m

(1.53)

Hk = „ rØ“y†r

Ø“yrØ

2 m= „ r

Ø 1

2 m“

„ kØ

2 pd‰-Â k

Ø

ÿ rØ

y†k“ „ k 'Ø

2 pd‰Â k'

Ø

ÿ rØ

yk 'Ø

= „ rØ „ k

Ø

2 pd

„ k 'Ø

2 pd

kØÿ k 'Ø

2 my†k yk ' exp r

Øk 'Ø- kØ ÿ rØ =

„ kØ

2 pd

„ k 'Ø

2 pd

kØÿ k 'Ø

2 my†k yk ' „ r

Øexp r

Øk 'Ø- kØ ÿ rØ

= „ kØ

2 pd

„ k 'Ø

2 pd

kØÿ k 'Ø

2 my†k yk ' 2 pd dk

Ø- k '

Ø =„ kØ

2 pd

kØÿ kØ

2 my†k

ØykØ =

„ kØ

2 pdekØ nk

Ø

1.3.3. Lattice systems

In a lattice, if we ignore the interactions between electrons, the Hamiltonian contains two terms.

(1.54)H = HK + HP = „ rØ“y†r

Ø“yrØ

2 m+ Ur

Ø y†rØ yr

Ø

where the first term is the kinetic energy and the second term is the lattice potential, where UrØ is a periodic potential. In principle, we could

use this Hamiltonian. However, it is not the most convenient way to handle a lattice systems. A more convenient way utilizes Bloch waves and

the band structure. In a lattice system, the kinetic energy of a single particle is described by the band structure enkØ. Here n is the band index

and kØ

is a momentum point in the (reduced) Brillouin zone. Therefore, total energy of the system (if we ignore interactions is)

(1.55)E =n

BZ

„ kØ

2 pdenk

Ø nnkØ

Here, we sum over all bands n and the integral is over the first Brillouin zone BZ. nnk

Ø is the occupation number for the quantum state in

band n at momentum kØ

. In second quantization, the quantum operator for nnkØ is

Phys540.nb 7

(1.56)nnkØ = yn

†kØ ynk

Ø

where y†nk creates a particles in band n with momentum k

Ø. In other words, it is the creation operator for a Bloch wave. ynk is the annihila-

tion operator for a Bloch wave. Therefore, the Hamiltonian for non-interacting electrons is

(1.57)Hk =n

BZ

„ kØ

2 pdenk yn

†k ynk

For simplicity, we will not consider lattices in this chapter. Instead, we will only consider particles moving in free space with dispersionk2 2 m. But please keep in mind that for a lattice system, we just need to substitute e = k2 2 m into the corresponding dispersion for Blochwaves.

Both the kinetic energy part of the Hamiltonian and the potential energy part contain one creation and one annihilation operator. Sothey are both known as the quadratic terms, or two-fermion terms. If the Hamiltonian only contains these two terms, we call thesystem a non-interacting system, because there is no interaction between particles in this Hamiltonian.

1.3.4.Interactions

Let’s consider Coulomb interaction between electrons. The total Coulomb energy is

(1.58)EInt =1

2 „ r

Ø„ r 'Ø

V rØ- rØ

' rrØ rr

Ø'

where V r = e2 r and rr is the particle density. In second quantization, we know that particle density is

(1.59)rrØ = y†r

Ø yrØ

Therefore, the interaction part of the Hamiltonian is

(1.60)HI =1

2 „ r

Ø„ r 'Ø

V rØ- rØ

' y†rØ yr

Ø y†rØ

' yrØ

'

The interaction term contains four creation/annihilation operators, and thus this term is called a quartic term or a four-Fermi term.

Typically, we reorder the operators using the anti-commutation relation between creation/annihilation operators.

(1.61)

HI =1

2 „ r

Ø„ r 'Ø

V rØ- rØ

' y†rØ yr

Ø y†r 'Ø yr '

Ø

=1

2 „ r

Ø„ r 'Ø

V rØ- rØ

' y†rØ y†r '

Ø yr 'Øyr

Ø + 1

2 „ r „ r ' V r

Ø- rØ

' dr - r ' y†rØ yr

Ø'

=1

2 „ r

Ø„ r 'Ø

V rØ- rØ

' y†rØ y†r '

Ø yr 'Øyr

Ø + 1

2 „ r V 0 y†r

Ø yrØ

=1

2 „ r

Ø„ r 'Ø

V rØ- rØ

' y†rØ y†r '

Ø yr 'Øyr

Ø + V 0 N

2

where N is the total number of particles in the system. The last term may looks problematic (it is singular), because for the Coulomb interaction,V 0 = 1 0 = ¶, but this term will not cause any trouble. We can simply drop it, because this term just shifts the total energy by a constant. In asolid state material, this term is canceled by the potential energy from the nucleons.

Very typically, when we write down the Hamiltonian, we put the operators in such an order that creation operators are put on the left andannihilation operators are on the right. This order is called normal order and the procedure to put operators in such an order is known as"normal ordering".

Typically, one use two colons to represent normal ordering. If we put the product of some creation and annihilation operators operator betweentwo colons, it means that we reorder these operators into normal order. For bosons, we just reorder the operators. For fermions, we need anextra factor +1 or -1, depending on whether we have even or odd number of permutations to reorder the operators. For example,

(1.62): y†r yr ' := y†r yr '(1.63): yr ' y†r := -y†r yr '

8 Phys540.nb

(1.64): yr ' yr ''y†r := y†r yr ' yr ''For the interaction terms, instead of using the most straightforward formula,

(1.65)HI =1

2 „ r

Ø„ r 'Ø

V rØ- rØ

' y†rØ yr

Ø y†r 'Ø yr '

Ø

we typically uses the normal ordered interaction term

(1.66)HI =1

2 „ r

Ø„ r 'Ø

V rØ- rØ

' : y†rØ yr

Ø y†r 'Ø yr '

Ø :=1

2 „ r

Ø„ r 'Ø

V rØ- rØ

' y†rØ y†r '

Øyr 'Ø yr

Ø

These two definitions differ by a constant, which doesn’t play any fundamental role. In other words, using normal order shifts the total energyby a constant. In fact, the reason we use normal ordering is because by doing so, the energy of the vacuum state is set to zero. In normal order,annihilation operator is put to the right side. By definition, the vacuum state is destroyed by any annihilation operator

(1.67)y 0 = 0

Therefore,

(1.68): H : 0 = … y 0 = 0 = 0 0So the vacuum state is an eigenstate of : H : with zero eigen energy.

1.3.5. Summary

In this chapter, we consider the simplest case

(1.69)H = „ rØ“y†r

Ø“yrØ

2 m+

1

2 „ r

Ø„ r 'Ø

V rØ- rØ

' y†rØ y†r '

Ø yr 'Ø yr

Ø

The quadratic term: kinetic energy for non-relativistic particles e = k2 2 m. The quartic term: normal ordered four-Fermi interactions.

However, it is important to keep in mind that the conclusions and methods can be generalized easily to other more complicated cases.

1.4. Equation of motion for correlation functions

In the Heisenberg picture, the equation of motion of an operator X is

(1.70)Â∑X t∑ t

= X t, Ht

Here, we set the Planck constant Ñ to unity for simplicity.

1.4.1. Equation of motion for annihilation operators

For the Hamiltonian

(1.71)H = „ rØ“y†r

Ø“yrØ

2 m+

1

2 „ r

Ø„ r 'Ø

V rØ- rØ

' y†rØ y†r '

Ø yr 'Ø yr

Ø

the equation of motion for the annihilation operator y is

(1.72)

Â∑yr0

Ø, t

∑ t= yr0

Ø, H = „ r“dr

Ø- r0

Ø“yr2 m

+

1

2 „ r

Ø„ r 'ØdrØ- r0

Ø V rØ- r '

Ø y†r 'Ø yr '

Ø yrØ + 1

2 „ r

Ø„ r 'Ø

V rØ- r '

Ø dr 'Ø- r0

Ø y†rØ yr '

Ø yrØ

= -“2yr0

Ø, t

2 m+ „ r V r0

Ø- rØ y†r

Ø, t yr

Ø, t yr0

Ø, t

Phys540.nb 9

Similarly, for the conjugate operator y†, we have

(1.73)-Â∑y†r0

Ø, t

∑ t= -

“2y†r0Ø

, t2 m

+ y†r0Ø „ r V r0

Ø- rØ y†r

Ø yrØ

For noninteracting particles V = 0, this equation is very similar to the Schrodinger equation. But please keep in mind that y and y† here areoperators, instead of wavefunctions.

1.4.2. Equation of motion for correlation functions

Define two point correlation functions

(1.74)G>r1,Ø

t1; r2Ø

t2 =1

Âyr1

Ø, t1 y†r2

Ø, t2

Here, the factor 1 Â is introduced for historical reason. For systems with translational symmetry in space and time, G>r1,Ø

t1; r2Ø

t2 only depends

the time and position difference between 1 and 2

(1.75)G>r1,Ø

t1; r2Ø

t2 = G>r1Ø- r2

Ø, t1 - t2 = G>r

Ø, t

Here, we define rØ= r1

Ø- r2

Ø and t = t1 - t2

Q: What is the equations of motion for G>r, t?

(1.76)

∑t G>rØ

.t = ∑t1 G>rØ

.t1 - t2 = ∑t1

1

Âyr1

Ø, t1 y†r2

Ø, t2 = -Â ∑t1 yr1

Ø, t1 y†r2

Ø, t2

= --“2yr1

Ø, t1

2 m+ V r0

Ø- rØ y†r

Ø, t1 yr

Ø, t1 yr1

Ø, t1 y†r2, t2

=1

2 m“r1

2 yr1Ø

, t1 y†r2, t2 - V y† y† y y = Â

2 m“2 G>r, t - V < y† y† y y >

If V = 0 (free systems), we have a closed partial differential equation for two fermion correlation functions.

(1.77)Â ∑t G0>r.t + “2

2 mG0

>r, t = 0

The sub-index 0 here implies that we are considering a non-interacting system without interaction. We can solve this partial differentialequation (with proper initial conditions and boundary conditions), and obtain the correlation function G0

>

For interacting systems (V ∫ 0), the story is not as simple.

(1.78)Â ∑t G>r.t + “2

2 mG>r, t = -Â V < y† y† y y >

So we have an inhomogeneous partial differential equation.

(1.79)Â ∑t G>r.t + “2

2 mG>r, t = f r, t

The terminology inhomogeneous equation means that the r.h.s. of the equation is nonzero. Inhomogeneous equation may look complicated,because for a different f r.t, it seems that we will need to solve a different equation. However, this is not the case. We just need to solve onequestion and then for any f r., t, we can get the solution directly.

Q: How do we solve an inhomogeneous partial differential equation?

A: Let’s look at the E&M textbook. The Green’s function method.

1.4.3. E&M: electric potential f(r) for charge distribution r(r)

Gauss’s law

(1.80)“ ÿEØr

Ø = rrØ

10 Phys540.nb

We know that

(1.81)EØr

Ø = -“frØ

so

(1.82)“2frØ = rr

ØThis is an inhomogeneous equation. How do we solve it? We first solve a different equation:

(1.83)“r12 Gr1

Ø, r2Ø = dr1

Ø- r2

Ø

Gr1Ø

, r2Ø is the electric potential at point r1

Ø induced by a point charge located at position r2

Ø. We know the solution of this equation, which is just

the Coulomb’s law

(1.84)Gr1Ø

, r2Ø = 1

4 p

1

r1Ø- r2

Ø

After we find Gr1Ø

, r2Ø, the solution for the inhomogeneous equation “2fr

Ø = rrØ can be obtained easily as

(1.85)frØ = „ r0

ØGr

Ø, r0Ø rr0

Ø = „ r0Ø 1

4 p

1

rØ- r0

Ørr0

Ø

Mathematicians call Gr1Ø

, r2Ø the Green’s function. And this methods of solving inhomogeneous PDEs are known as the Green’s function

approach. In general, one first substitute the inhomogeneous part with a delta function. Then, one solve this new PDE, whose solution is theGreen’s function. Once the Green’s function is obtained, one can write down the solution of the inhomogeneous PDE very easily using anintegral. For the correlation function G>, the equation of motion is

(1.86)Â ∑t G>r.t + “2

2 mG>r, t = -Â V < y† y† y y >

What we will need to do here is to substitute the r.h.s. by a delta function

(1.87)Â ∑t+“2

2 mGr, t = dr dt

and Gr, t is our Green’s function.

1.4.4.time ordering: a trick to get delta functions

Define time-ordered correlation functions (the Green’s functions)

(1.88)Gr1, t1; r2 t2 =1

ÂTyr1, t1 y†r2, t2

Here, Ty†r1, t1 yr2, t2 is known as the time - ordered product.

(1.89)T yr1, t1y†r2, t2 = yr1, t1 y†r2, t2 if t1 > t2

≤y†r2, t2 yr1, t1 if t1 < t2

For bosons, we use the + sign and for fermions we use the - sign. Another way to write down the same product is

(1.90)T yr1, t1y†r2, t2 = yr1, t1 y†r2, t2 ht1 - t2 ≤ y†r2, t2 yr1, t1 ht2 - t1where hx is the step function hx = 1 for x > 0 and hx = 0 for x < 0. Now, let’s consider the EOM for Gr, t.

(1 91)

∑t Gr.t = ∑t1

1

ÂT yr1, t1 y†r2, t2 = -Â ∑t1 yr1, t1 y†r2, t2 ht1 - t2 ≤ y†r2, t2 yr1, t1 ht2 - t1

= -Â ∑t1 yr1, t1 y†r2, t2 ht1 - t2 ≤ y†r2, t2 -Â ∑t1 yr1, t1 ht2 - t1 -Â yr1, t1 y†r2, t2 ¡ y†r2, t2 yr1, t1 dt1 - t2

Phys540.nb 11

=1

2 m∑r1

2 Tyr1, t1 y†r2, t2 - V T y† y† y y - Â dr1 - r2 dt1 - t2

=Â

2 m“2 Gr, t - V T y† y† y y - Â dr1 - r2 dt1 - t2

If V = 0 (non-interacting systems)

(1.92)Â ∑t G0r.t + 1

2 m“2 G0r, t = dr dt

Bottom line, by time-ordering, we automatically get a delta function in the equation of motion, which makes G r, t a Green’s function.

1.4.5. Green’s function for free particles: G0r , tIn this section, we consider non-interacting particles at zero temperature. What happens at finite tempearture and in the presence of interactionswill be discussed in latter part of the lecture.

For the EOM, we have

(1.93)Â ∑t+“2

2 mG0r, t = dr dt

it is easy to solve in the momentum space.

(1.94) „ r „ t exp- k r +  w t  ∑t+“2

2 mG0r, t = „ r „ t exp-Â k r + Â w t dr dt

(1.95) „ r „ t -Â ∑t+“2

2 mexp-Â k r + Â w t G0r, t = 1

(1.96)w -k2

2 m „ r „ t exp-Â k r + Â w t G0r, t = 1

Define

(1.97)G0k, w = „ r „ t exp-Â k r + Â w t G0r, t

It is easy to check that

(1.98)G0r, t = „ k

2 pd

„w

2 pexp k r -  w t G0k, w

which is the inverse transformation.

Using G0k, w, we get

(1.99)w -k2

2 mG0k, w = 1

and therefore

(1.100)G0k, w = 1

w -k2

2 m

More generally, for non-interacting particles with more complicated dispersion.

(1.101)G0k, w = 1

w - ekFor example, in a lattice system, if we ignore interactions, the Green’s function for particles in band n is

12 Phys540.nb

(1.102)G0nk, w = 1

w - enkHere, we assume that both the creation and annihilation operators are for Bloch waves in band n. If the two operators are for different bands, thecorrelation function is zero.

1.4.6. Note: there are many other reason to use T.

(1) Path integral leads to T naturally.

(2) The evolution operator U t = Texp „ t Ht t(3) With T, bosons and fermions are unified together. Same theory with two different boundary conditions.

Due to time-limitation, we will not address (1) and (2) in details in this course. But for theorists, it is very important to understand the pathintegral formula. The third point will be addressed in the next section.

1.5. Boundary condition and connections between different Green’s functions

1.5.1. Other correlation functions and the boundary condition

As mentioned in Sec. 1.2.9, because two quantum operators are in general non-commutative AB ∫ BA, the order of operators matters a lotwhen we define the correlation functions. Depending on the order of operators, we can define many different correlation functions. Forexample, the time-ordered correlation function defined above. In addition, we can define the following two correlation functions (and othercorrelation functions).

(1.103)G>1, 2 = 1

Ây1 y†2

(1.104)G<1, 2 = ≤ 1

Ây†2 y1

Here, for simplicity, we use “1” to represent r1, t1 and “2” to represent r2, t2. Please notice that we cannot use the simple commutation/anti-commutation relation

(1.105)y1y†2 ¡ y†2 y1 ∫ d1 - 2This is because the commutation/anti-commutation relations we are familiar with are “equal time” commutation/anti-commutation relations.

(1.106)yr1, ty†r2, t ¡ y†r2, t yr1, t ∫ dr1 - r2Please notice that all the operators are at the same time (remember that we are using the Heisenberg picture, so that the operators are time-dependent).

What we considered here are operators at different time points

(1.107)y1y†2 ≤ y†2 y1 = yr1, t1 y†r2, t2 ≤ y†r2, t2 yr1, t1 ∫ d1 - 2Therefore, in general, it is not d(1-2). And thus

(1.108)G>1, 2 - G<1, 2 ∫ 1

Âd1 - 2

In other words, we cannot simply relate these two Green’s function using equal time commutation/anti-commutation relations.

For statistical average, we have Boltzmann factor ‰bH . For time-evolution, we have the evolution operator ‰ÂH t. It seems that inverse tempera-ture b is just the imaginary time. Let’s try this idea by allowing time to be complex.

For G>

(1.109)G>1, 2 = 1

Ây1 y†2 =

Tr exp-bH` - m N` yr1, t1y†r2, t2

 Tr exp-bH` - m N`

= Tr exp-bH` - m N` expÂH

`t1 yr1 exp-ÂH

`t1 expÂH

`t2 y†r2 exp-ÂH

`t2 Â Tr exp-bH` - m N

`

Phys540.nb 13

If we use eigenenergy states to compute the sum

(1.110)

G>1, 2 = Tr exp-bH` - m N` expÂH

`t1 yr1 exp-ÂH

`t1 expÂH

`t2 y†r2 exp-ÂH

`t2 Â Tr exp-bH` - m N

` =

nexpb m Nn expEn t1 - b n yr1 exp-ÂH

`t1 expÂH

`t2 y†r2 n exp-Â En t2 Â Tr exp-bH` - m N

` =

nexpb m Nn expEn t1 -  t2 - b n yr1 exp-ÂH

`t1 expÂH

`t2 y†r2 n  Tr exp-bH` - m N

` Because En has no upper bound, as En Ø +¶, to keep the factor expEn t1 -  t2 - b converge, we need to haveRe t1 t2 0, In other words, Im t1 - t2 > -b.

In the same time, using 1 =m m m , we have

(1.111)

G>1, 2 = n

expb m Nn expEn t1 -  t2 - b n yr1 m

m m exp-ÂH`

t1 expÂH`

t2 y†r2 n Tr exp-bH` - m N

` =

n,mexpb m Nn expEn t1 -  t2 - b n yr1 m m y†r2 n exp Emt2 - t1 Tr exp-bH` - m N

` To make sure that Em Ø +¶ shows no singularity, Imt1 - t2 < 0.

So, 0 > Im t1 - t2 > -bFor G<, we have 0 < Im t1 - t2 < b

Let’s come back to G>

(1.112)

G>r1, t1 - Â b; r2, t2 =Tr exp-bH` - m N

` expbH` expÂH

`t1 yr1 exp-ÂH

`t1 exp-bH

` expÂH`

t2 y†r2 exp-ÂH`

t2 Â Tr exp-bH` - m N

` =Tr expb m N

` expÂH`

t1 yr1 exp-ÂH`

t1 exp-bH` expÂH

`t2 y†r2 exp-ÂH

`t2 Â Tr exp-bH` - m N

` =Tr expÂH

`t1 yr1 expb m N` - 1 exp-ÂH

`t1 exp-bH

` expÂH`

t2 y†r2 exp-ÂH`

t2 Â Tr exp-bH` - m N

` =e-bmTr expÂH

`t1 yr1 exp-ÂH

`t1 exp-bH` - m N

` expÂH`

t2 y†r2 exp-ÂH`

t2 Â Tr exp-bH` - m N` =

e-bmTr y1 exp-bH` - m N

` y†2Â Tr exp-bH` - m N

` = e-bm

Tr exp-bH` - m N` y†2 y1

 Tr exp-bH` - m N`

= ≤e-bm G<1, 2

Therefore, we find that

(1.113)eb m G>r1, t1 - Â b; r2, t2 = ≤G<r1, t1; r2, t2

1.5.2. BC for the time-ordered Green’s function.

The time-ordered Green’s function can be defined along the imaginary axis also

(1.114)T yr1, t1y†r2, t2 = yr1, t1 y†r2, t2 if Imt1 < Imt2≤y† r2, t2 y r1, t1 if Im t1 > Imt2

So,

(1.115)T G = G> if Imt1 < Im t2G< if Imt1 > Im t2

It is easy to check that for Gr, t, we have

(1.116)≤‰bm Gr, t -  b = Gr, tThis is a boundary condition for imaginary time. At m=0, Bosons have a periodic BC, and fermions have an anti-periodic BC.

1.5.3. the k,w space

Define

14 Phys540.nb

(1.117)

G>k, w = Â „ r „ t ‰-Â k r+Â w t G>r, t = yk,w yk,w†

G<k, w = ≤ „ r „ t ‰- k r+ w t G<r, t = yk,w† yk,w

(1.118)

G<k, w = ≤ „ r „ t ‰- k r+ w t G<r, t =

 eb m „ r „ t ‰- k r+ w t G>r, t -  b =  eb m „ r „ t ' ‰- k r+ w t'+ b G>r, t ' = ‰-bw-m G>k, w

Define the “spectral function”

(1.119)Ak, w = G>k, w ¡ G<k, wBecause G<k, w = ‰-bw-m G>k, w

(1.120)Ak, w = G>k, w ¡ G<k, w = G>k, w ¡ ‰-bw-m G>k, wSo

(1.121)G>k, w = Ak, w1 ¡ ‰-bw-m

= Ak, w ‰ bw-m

‰ bw-m ¡ 1= Ak, w1 ≤ 1

‰ bw-m ¡ 1 = Ak, w1 ≤ f w

(1.122)G<k, w = ‰-bw-m G>k, w = ‰-bw-m Ak, w1 ¡ ‰-bw-m

= Ak, w 1

‰ bw-m ¡ 1= Ak, w f w

Here f w is the boson/fermion distribution function.

Note: Here, the only thing we assume is that the indistinguishable particles commute or anti-commute with each other. In Green’sfunction, these two choices leads to two different boundary conditions, from which the Bose-Einstein and Fermi-Dirac distributionsarise naturally.

1.5.4. G in the k,w space part I: the Matsubara frequencies

Remember that Gr, t = ≤‰bm Gr, t -  b, which is (almost) a PBC (anti-PBC) for t. We know that boundary conditions implies quantization(discrete w).

For example, for a periodic function f t = f t + T, we know that in the frequency space, f w is defined only on a discrete set of frequencypoints, w = 2 p n T . For anti-PBC, f t = - f t + T, w = 2 n + 1 p T .

For Gr, t = ≤‰bm Gr, t -  b, if we go to the k,w space and define

(1.123)Gr, t = „ k „w ‰Â k r-Âw t Gk, w

The condition Gr, t = ≤‰bm Gr, t -  b implies that

(1.124) „ k „w ‰Â k r- w t Gk, w = ≤‰bm ‰-w b „ k „w ‰Â k r- w t Gk, w

So, =1

(1.125)‰bw-m = ≤1

For bosons,

(1.126)bw - m = 2 n p Â

(1.127)w =2 n p Â

b+ m

Following the historical convention, we write w = Â wn

(1.128)wn =2 n p

b- Â m

These discrete frequency points wn are known as the Matsubara frequencies

Phys540.nb 15

For fermions,

(1.129)bw - m = 2 n + 1 p Â

(1.130)w =2 n + 1 p Â

b+ m

(1.131)wn =2 n + 1

bp - Â m

1.5.5. G in the k,w space part II: analytic continuation

After we find Gk, Âwn, where wn takes discrete values, now let’s define a new function Gk, z by simply replacing the discrete number  wn

into a complex number z that various continuously . For the function Gk, z, it is a function well-defined at every point on the complex z plane(there may be some singularity points). At the Matsubara frequencies, this new function Gk, z coincides with Gk, Âwn. This procedure isknown as analytic continuation. This new function is very useful. Here, I will show you that by defining Gk, z, we can find a very easy way toget G> and G< from G. Later, we will use this Gk, z to compute Gk, Âwn.

(1.132)

Gk, Â wn = 0

-Â b

„ t ‰-wn t Gk, t =

0

-Â b

„ t ‰-wn t G>k, t = 0

-Â b

„ t ‰-wn t „w

2 p ‰- w t G>k, w =

„w

2 p

‰-Â w-wn -Â b - 1

w - Â wn

Ak, w1 ¡ ‰-bw-m

=

„w

2 p

1 - ‰-w-Â wn b

w - Â wn

Ak, w1 ¡ ‰-bw-m

= „w

2 p

1 ¡ ‰-w-m b

 wn - w

Ak, w1 ¡ ‰-bw-m

= „w

2 p

Ak, w wn - w

Therefore,

(1.133)Gk, z = „w

2 p

Ak, wz - w

Now, we substitute z by w + Âd

(1.134)Gk, w + Â e = „W

2 p

Ak, Ww + Â e - W

(1.135)

ImGk, w + Â e = Gk, w + Â e - Gk, w - Â e2 Â

=1

2 „W

2 p Ak, Ww + Â e - W

-Ak, W

w - Â e - W =

1

2 „W

2 p

Ak, Ww - W2 + e2

-2 Â e = -„W

2 p

e

w - W2 + e2Ak, W = -1

2 „W dw - W Ak, W = -1

2Ak, w

So,

(1.136)Ak, w = -2 ImGk, w + Â e

1.5.6. Example: free particles

(1.137)Â ∑t G0r.t + 1

2 m“2 G0r, t = dr dt

in the k,w space

(1.138)Â wn -k2

2 mG0k, Â wn = 1

(1.139)G0k, Â wn =1

 wn -k2

2 m

More generic case: dispersion relation e(k)

16 Phys540.nb

(1.140)G0k, Â wn =1

 wn - ekAnalytic continuation:

(1.141)G0k, z = 1

z - ekthe spectral function:

(1.142)Ak, w = -2 ImGk, w + Â e = -2 Im 1

w + Â e - ek =2 e

w - ek2 + e2= 2 p dw - ek

particular number:

(1.143)nkt = yk†t ykt = G<k, t - t =

„w

2 pG<k, w =

„w

2 p2 p dw - ek f w = f ek

(1.144)1 ≤ nkt = ykt yk†t = G>k, t - t =

„w

2 pG>k, w =

„w

2 p2 p dw - ek1 ≤ f w = 1 ≤ f ek

1.5.7. Interacting particles.

Once we found Gk, Âwn, we can get Ak, w using

(1.145)Ak, w = -2 ImGk, w + Â eThen we can get all other correlation functions like G<k, t - t and G>k, t - t using

(1.146)G>k, w = Ak, w1 ¡ ‰-bw-m

= Ak, w ‰ bw-m

‰ bw-m ¡ 1= Ak, w1 ≤ 1

‰ bw-m ¡ 1 = Ak, w1 ≤ f w

(1.147)G<k, w = ‰-bw-m G>k, w = ‰-bw-m Ak, w1 ¡ ‰-bw-m

= Ak, w 1

‰ bw-m ¡ 1= Ak, w f w

Bottom line: if we know one correlation function, we can get other. Therefore, among all the different correlation functions, we justneed to focus on the one that is easiest to compute. The easiest one is the Green’s function (time-ordered correlation functions), becauseit give us delta functions.

1.6. Feynman diagram

1.6.1. multi-particle Green’s functions

Single particle Green’s function:

(1.148)G1, 1 ' = 1

ÂTy1y†1 '

Two-particle Green’s function:

(1.149)G21, 2; 1 ', 2 ' = 1

Â

2

Ty1 y2 y†2 ' y†1 '

three-particle Green’s function:

(1.150)G31, 2, 3; 1 ', 2 ', 3 ' = 1

Â

3

Ty1 y2 y3 y†3 ' y†2 ' y†1 '

n-particle Green’s function:

Phys540.nb 17

(1.151)Gn1, 2, …, n; 1 ', 2 ', …, n ' = 1

Â

n

Ty1 y2 …yn y†n ' …y†2 ' y†1 '

1.6.2. the equations of motion of the Green’s functions

(1.152)

Â∑yr0, t

∑ t= yr0, H = „ r

“dr - r0“yr2 m

+

1

2 „ r „ r ' dr - r0 V r - r ' y†r ' yr ' yr + 1

2 „ r „ r ' V r - r ' dr ' - r0 y†r yr ' yr

= -“2yr0, t

2 m+ „ r V r0 - r y†r yr yr0

(1.153)Â ∑t+1

2 m“2 G1, 2 = dr1 - r2 dt1 - r2 +

1

 „ r V r0 - r Ty†r1, t1 yr1, t1 yr0, t1 y†r2, t2

(1.154)

 ∑t+1

2 m“2 G1, 1 ' = d1 - 1 ' + Â „ r V r1 - r2 1

Â

2

T y†r2, t1 yr2, t1 yr1, t1y†r1', t1' =

d1 - 1 ' + Â „ r V r1 - r2 1

Â

2

Ty†r2, t1 + d yr2, t1 yr1, t1 y†r1', t1' =

d1 - 1 ' ≤ Â „ r V r1 - r2 1

Â

2

Tyr1, t1 yr2, t1 y†r2, t1 + d y†r1', t1' =

d1 - 1 ' ≤ Â „ r V r1 - r2 G21, 2; 1 ', 2+ t1=t2

Here 2+ means that the time argument for 2+ is slightly larger than 2, to keep the operators in the right order t2+ = t2 + d.

(1.155)Â ∑t+1

2 m“2 G1, 1 ' = d1 - 1 ' ≤ Â „ r2 V r1 - r2 G21, 2; 1 ', 2+ t1=t2

This equation tells us that in order to get G1, 2, we need to know G21, 2; 1 ', 2 '. So we need to write down the EOM for G2

(1.156)Â ∑t+1

2 m“2 G21, 2; 1 ', 2 ' = d1 - 1 ' G2, 2 ' ≤ d1 - 2 ' G2, 1 ' ≤ Â „ r2 V r1 - r2 G31, 2, 3; 1 ', 2 ', 3+ t1=t3

This equation tells us that in order to get G2, we need to know G3. Repeat the same procedure, we find that if we want to know Gn, we need toknow Gn+1.

(1.157)

 ∑t+1

2 m“2 Gn1, 2, …, n; 1 ', 2 ', …, n ' = d1 - 1 ' Gn-12, …, n; 2 ', …, n ' ≤ d1 - 2 ' Gn-12, …, n; 1 ', 3 ', …, n ' +

… ≤ Â „ r2 V r1 - r2 Gn+11, 2, …, n + 1; 1 ', 2 ', …, n + 1 t1=tn+1

So we cannot get a close set of equations. In order words, the number of unknowns is always the number of equations+1, and thus thereis no way to solve these equations.

1.6.3. Q: How to solve this equation? A: the perturbation theory

G1, 1 ' depends on interaction strength V . Let’s expand G as a power series of V

(1.158)G1, 1 ' = G01, 1 + OV + OV 2 + OV 3 + …

Notice that Gn-1 is related to V μGn. Therefore, to get G up to the order of OV n, we just need to keep G2 to the order of OV n-1, and G3 toOV n-2 … to OV 0 and set Gn+2 = 0

1.6.4. the zeroth-order approximation (free-particle approximation, or say non-interacting approximation)

If we want to get G to the zeroth order, we need to set G2 = 0

18 Phys540.nb

(1.159)Â ∑t1 +1

2 m“r1

2 G01, 1 ' = d1 - 1 '

One equation, one unknown. It can be solved easily.

(1.160)G0k, Âwn =1

Âwn -k2

2 m

1.6.5. the first-order approximation (the Hartree-Fock approximation)

If we want to get G to the first order, we need to set G3 = 0 and keep G2 to OV 0

(1.161)Â ∑t1 +1

2 m“r1

2 G1, 1 ' = d1 - 1 ' ≤ Â „ r2 V r1 - r2 G21, 2; 1 ', 2+ t1=t2

(1.162)Â ∑t1 +1

2 m“r1

2 G21, 2; 1 ', 2 ' = d1 - 1 ' G2, 2 ' ≤ d1 - 2 ' G2, 1 ' + OV G3 = d1 - 1 ' G02, 2 ' ≤ d1 - 2 ' G02, 1 '

two equations and two unknowns. The solution for the second equation is very simple:

(1.163)G21, 2; 1 ', 2 ' = G01, 1 ' G02, 2 ' ≤ G01, 2 ' G02, 1 'Let’s check it

(1.164)Â ∑t1 +1

2 m“r1

2 G01, 1 ' G02, 2 ' = d1 - 1 ' G02, 2 '

(1.165)Â ∑t1 +1

2 m“r1

2 G01, 2 ' G02, 1 ' = d1, 2 ' G02, 1 '

So we can go back to the first equation to get G.

(1.166)

 ∑t1 +1

2 m“r1

2 G1, 1 ' =

d1 - 1 ' ≤ Â „ r2 V r1 - r2 G01, 1 ' G02, 2+ t1=t2 +Â „ r2 V r1 - r2 G01, 2+ G02, 1 ' t1=t2

(1.167)Â ∑t1 +1

2 m“r1

2 G1, 1 ' = d1 - 1 ' + f 1, 1 '

This equation can be separated into two equations:

(1.168)Â ∑t1 +1

2 m“r1

2 G01, 1 ' = d1 - 1 '

(1.169)Â ∑t1 +1

2 m“r1

2 G11, 1 ' = f 1, 1 '

and G = G0 + G1.

The first equation have been solved before. It is just the zeroth-order equation (the free theory)

(1.170)Â ∑t1 +1

2 m“r1

2 G01, 1 ' = d1 - 1 '

For the second equation, the solution is straightforward using the Green’s function technique

(1.171)G11, 1 ' = „ r3 „ t3 G01, 3 f 3, 1 '

Let’s check this

Phys540.nb 19

(1.172)Â ∑t1 +1

2 m“r1

2 „ r3 „ t3 G01, 3 f 3, 1 ' = „ r3 „ t3 d1 - 3 f 3, 1 ' = f 1, 1 '

So

(1.173)

G11, 1 ' = „ r3 „ t3 G01, 3 f 3, 1 ' = ≤ „ r3 „ t3 „ r2 G01, 3 V r3 - r2 G03, 1 ' G02, 2+ t3=t2

+Â „ r3 „ t3 „ r2 G01, 3 V r3 - r2 G03, 2+ G02, 1 ' t3=t2

For G1, 1 '

(1.174)

G1, 1 ' = G01, 1 ' ≤ Â „ r3 „ t3 „ r2 „ t2 G01, 3 V r3 - r2 dt3 - t2 G03, 1 ' G02, 2+ +

 „ r3 „ t3 „ r2 „ t2 G01, 3 V r3 - r2 dt3 - t2 G03, 2+ G02, 1 '

The first term is the free propagator, the second therm is known as the Hartree term and the last term is the Fock term. Here, we use the deltafunction dt3 - t2 to enforce the constrain that t3 = t2.

1.6.6. second order:

If we want to get G to the order of OV 2, we set G4 = 0, keep G3 to OV 0 and G2 to OV 1

(1.175)Â ∑t1 +1

2 m“r1

2 G1, 1 ' = d1 - 1 ' ≤ Â „ r2 V r1 - r2 G21, 2; 1 ', 2+ t1=t2

(1.176)Â ∑t+1

2 m“2 G21, 2; 1 ', 2 ' = d1 - 1 ' G2, 2 ' + d1 - 2 ' G2, 1 ' ≤ Â „ r2 V r1 - r2 G31, 2, 3; 1 ', 2 ', 3+ t1=t3

(1.177)Â ∑t1 +1

2 m“r1

2 G31, 2, 3; 1 ', 2 ', 3 ' = d1 - 1 ' G22, 3; 2 ', 3 ' ≤ d1 - 2 ' G22, 3; 1 ', 3 ' + d1 - 3 ' G22, 3; 1 ', 2 '

The last equation give us (keep only OV 0 terms)

(1.178)G31, 2, 3; 1 ', 2 ', 3 ' = G01, 1 ' G02, 2 ' G03, 3 ' ≤ G01, 1 ' G02, 3 ' G02, 3 ' ≤ G01, 2 ' G02, 1 ' G03, 3 ' +

G01, 2 ' G02, 3 ' G03, 1 ' + G01, 3 ' G02, 1 ' G03, 2 ' ≤ G01, 3 ' G02, 2 ' G03, 1 'So we can get G2 and then G

† Too complicated!

† Hard to compute when n is large.

There is a simple way to directly get the finally answer we want, thanks the very smart technique designed by Feynman.

1.6.7. Feynman diagrams and Feynman rules† Each integration coordinate is represented by a point;

† A propagator, G01, 1 ', is represented by a solid line:

† A creation operator is represented by a solid line attached to the point with an arrow from the point;

† An annihilation operator is represented by a solid line attached to the point with an arrow to the point;

† The interaction V 1 - 2 is represented by a dashed line connecting two points (r1 and r2). And each ending point has a creation operator

and an annihilation operator:

† Connect the lines, and keep the direction of the arrows

Find all the diagrams obey the rules described above, each of them represent a term in the power series expansion of G. Zeroth order termscontains no V , first order terms contains just one V , nth order term contains n V . Here we consider two point correlation functions G1, 1 ' as

an example. Because G1, 1 ' = 1

 Ty1 y†1 ' has one creation and one annihilation term, we have one ending point and one starting point

for solid lines. At the zeroth order, we don’t have V , and these two points are the only thing we have in our diagrams. As a result the onlydiagram we have is a solid line connecting these two points.

20 Phys540.nb

The first order terms has one V , and it is easy to note that there are only two ways to connect the diagrams

Therefore, up to the first order

(1.179)

G1, 1 ' = G01, 1 ' ≤ Â „ r3 „ t3 „ r2 „ t2 G01, 3 V r3 - r2 dt3 - t2 G03, 1 ' G02, 2+ +

 „ r3 „ t3 „ r2 „ t2 G01, 3 V r3 - r2 dt3 - t2 G03, 2+ G02, 1 ' t3=t2

If we want higher order terms, we just add more Vs to the diagrams. For example, if we have two Vs (second order terms), we have

We just need to find all the plots and then write them into integrals (and the compute the integrals).

Summary:

† Draw all possible diagrams up to order OV n† Give each point a label m, which includes position rm and time tm

† For each fermionic loop, we get an extra factor of (-1).

† For each dashed line (interaction V 1 - 2), we multiply by a prefactor Â.

† And remember that the Coulomb interaction is instantaneous, so that we need to multiply dti - t j for each V ri - r j. For high-energy

physics (in the relativistic regime), interaction has time-dependent built in, V ri - r j, ti - t j, so we don’t need to add the constraint

dti - t j there.

1.6.8. k,w-space

If the system have momentum and energy conservation laws, it is typically much easier to compute Green’s functions in the momentum-energyspace. In the k,w-space, we use the same diagrams and in addition, we also:

† Assign momentum and frequency to each line and keep the momentum and energy conservation law at each ending point or crossing points.

† For each loop, there is a pair of unknown q and  Wn and we need to integrate/sum over them.

† For each fermionic loop, we get an extra factor of (-1), which comes the commutation relation.

† For mth order diagram, we get a factor -1m

For example, using the same diagram above, we can compute the two point correlation functions as

(1.180)

Gk, Âwn = G0k, Âwn ¡ „q

2 p3

1

b

 WnG0k, Âwn G0k, Âwn V 0, 0 G0k,  Wn -

„ q

2 p3

1

b

 WnG0k, Âwn G0k, Âwn G0k + q, Âwn +  Wn V q,  Wn + ...

Phys540.nb 21

1.7. the Dyson’s equation

If we look at the diagrams, there are many repeating structures. These repeating structures can be utilized to simplify the calculation.

1.7.1. The sum of a geometric series

How do we compute the sum of a geometric series:

(1.181)X = a + a q + a q2 + a q3 + …

First, we notice that

(1.182)q X = a q + a q2 + a q3 + a q4 + …

then, we rewrite X as

(1.183)X = a + a q + a q2 + a q3 + … = a + q X

So,

(1.184)X - q X = a

(1.185)X =a

1 - q

We can use the same trick to sum many Feynman diagrams.

1.7.2. example 1: the Hartree approximation

For diagrams, we can use the same trick. For example, the following diagrams can be summed together

…

Here we sum over these diagrams and ignore others. This approximation is known as the Hartree approximation.

(1.186)Gk, Âwn º G0k, Âwn + G0k, Âwn SH k, Âwn G0k, Âwn + G0k, Âwn SH k, Âwn G0k, Âwn SH k, Âwn G0k, Âwn +

G0k, Âwn SH k, Âwn G0k, Âwn SH k, Âwn G0k, Âwn SH k, Âwn G0k, Âwn + …

where

(1.187)SH k, Âwn = = ¡„q

2 p3

1

b

 WnV 0, 0 G0q,  Wn

This term is known as the self-energy correction. It is the same diagram we considered above but with external legs removed. Using the sametrick mentioned above, we find

(1.188)GH k, Âwn = G0k, Âwn1 - SH k, Âwn G0k, Âwn-1 = G0k, Âwn-1 - SH k, Âwn -1=

1

Âwn - ek - SH k, ÂwnThis technique and this formula is known as the Dyson’s equation. Here, we find that the Green’s function for interacting particles is verysimilar to the free Green’s function G0. The only thing interactions does is to change the single particle energy ek into ek + SH k, Âwn. Inother words, the interactions changes the “energy” of the particle by SH k, Âwn. This is the reason why this term is called a self-energycorrection. However, it is important to keep in mind that this term is NOT really a shift in ek, because it is also a function of frequency.

1.7.3. example 2: the Hartree-Fock approximation

We can get more accurate results by adding more diagrams into our calculation. For example,

22 Phys540.nb

2 …

The approximation that sums over these diagrams (and ignore others) are known as the Hartree-Fock approximation.

(1.189)Gk, Âwn º G0k, Âwn + G0k, Âwn SHFk, Âwn G0k, Âwn + G0k, Âwn SHFk, Âwn G0k, Âwn SHFk, Âwn G0k, Âwn +

G0k, Âwn SHFk, Âwn G0k, Âwn SHFk, Âwn G0k, Âwn SHFk, Âwn G0k, Âwn + …

here the Hartree-Fock self-energy correction SHFk, Âwn is

(1.190)

SHFk, Âwn =+ = ¡

„ q

2 p3

1

b

 WnV 0, 0 G0q,  Wn -

„ q

2 p3

1

b

 WnG0k + q, Âwn +  Wn V q,  Wn

Using the same trick, (the Dyson’s equation), we find that

(1.191)GHFk, Âwn = G0 k, Âwn1 - SHFk, Âwn G0k, Âwn-1= G0k, Âwn-1 - SHFk, Âwn -1

=1

 wn - ek - SHFk, ÂwnThe final result is almost the same as the Hartree approximation. We just need to change SH k, Âwn into SHFk, Âwn.

1.7.4. example 3: include more diagrams

If we include more diagrams, the Green’s function still takes the same structure

(1.192)Gk, Âwn = G0k, Âwn + G0k, Âwn Sk, Âwn G0k, Âwn + G0k, Âwn Sk, Âwn G0k, Âwn Sk, Âwn G0k, Âwn +

G0k, Âwn Sk, Âwn G0k, Âwn Sk, Âwn G0k, Âwn Sk, Âwn G0k, Âwn + …

Here, the self-energy correction contains more diagrams

(1.193)Sk, Âwn = + + + + + + +

Using Dyson’s equation, we find that

(1.194)Gk, Âwn =G0k, Âwn

1 - Sk, Âwn G0k, Âwn=

1

1

G0k,Âwn - Sk, Âwn=

1

Âwn - ek - Sk, Âwn

1.7.5. One-particle irreducible diagrams

If we use the Dyson’s equation, the key is to avoid double counting. For example, the second order diagram has been

included in the Hartree approximation when we include in the self-energy. Therefore, when we compute second order self-energy

corrections, we should not include this diagram again. The rule to avoid double counting is to use 1-particle irreducible diagrams in the self-energy S. One-particle irreducible diagrams means that if we cut one internal link (solid line), the diagram is still connected. For example, thefollowing diagrams are 1-particle irreducible diagrams, and thus they should be include din the self-energy:

This diagram is not 1-particle irreducible, and thus we should not include them in the self-energy correction (it has already been taken care of

by the first order term )

Phys540.nb 23

1.7.6. Summary† Draw all possible 1-particle-irreducible diagrams.

† Remove external legs to get the self energy (note: each fermion loop contributes a factor -1. For nth order diagram, we need an factor -1n in the momentum space formula).

† Use Dyson’s equation to get the full Green’s function. GÂ wn = 1

Âwn-ek-Sk,Âwn† Other particles can be treated using the same approach (photons, phonons, etc.)

1.8. Physical meaning (fermions)

The physical meaning of the spectral function Ak, w is: if we have a particle with momentum k, Ak, w 2 p is the probability for this particleto have energy w.

1.8.1. Ak, w ≥ 0

The proof can be found in the book of Mahan (page 151). We will not show it here.

1.8.2. ‚w

2 pAk, w=1

(1.195)Ak, w = G>k, w + G<k, wIntegrate over all w

(1.196)-¶

¶ „w

2 pAk, w =

-¶

¶ „w

2 pG>k, w +

-¶

¶ „w

2 pG<k, w

Notice that

(1.197)-¶

¶ „w

2 p‰Âw t G>k, w = ÂG>k, t = yk, t0 + t y†k, t0

If we set t = 0, we find that by integrate over all w, we get the equal-time correlation function

(1.198)-¶

¶ „w

2 pG>k, w = yk, t0 y†k, t0

Similarly,

(1.199)-¶

¶ „w

2 pG<k, w = y†k, t0 yk, t0

Therefore,

(1.200)-¶

¶ „w

2 pAk, w =

-¶

¶ „w

2 pG>k, w +

-¶

¶ „w

2 pG<k, w = yk, t0 y†k, t0 + y†k, t0 yk, t0 = 1 = 1

In other words, -¶¶ „w

2 pAk, w measures the commutator/anit-commutator between y and y†, which is unity.

1.8.3. Free particles

For particles without interactions, the Green’s function is

(1.201)G0k, Âwn =1

Âwn - ekThe spectral function Ak, w is obtained by

24 Phys540.nb

(1.202)Ak, w = -2 limeØ0 ImGk, w + Â e = -2 limeØ0 Im 1

w + Â e - ek = limeØ0

2 e

w - ek2 + e2= 2 p dw - ek

For a fixed k, if we plot Ak, w as a function of w, we find a delta peak at w = ek. This delta peak means that for a free particle with momentumk, the energy of this particle can only take one value: w = ek. And it is easy to check that

(1.203)-¶

¶ „w

2 pAk, w = 1

1.8.4. Interacting particles

When interactions are taken into account,

(1.204)Gk, Âwn =1

Âwn - ek - Sk, Âwn=

1

Âwn - ek - ReSk, Âwn - Â ImSk, ÂwnSometimes, the real part of S is labeled as S1, where the imaginary part is S2

(1.205)Sk, z = S1k, z + Â S2k, zTherefore, the spectral function is

(1.206)

Ak, w = -2 limeØ0 ImGk, w + Â e = -2 limeØ0 Im 1

w + Â e - ek - S1k, w + Â e - Â S2k, w + Â e =

limeØ0

2 S2k, w + Â ew - ek - S1k, w + Â e2 + S2k, w + Â e2

=2 S2k, w

w - ek - S1k, w2 + S2k, w2

1.8.5. The real part of self-energy correction

Here, we consider the real part of Sk, w. For simplicity, we consider the limit that ImSk, wØ 0. When S2(k,w)Ø0

(1.207)Ak, w = 2 S2k, ww - ek - S1k, w2 + S2k, w2

= dw - ek - S1k, w

If S1k, w only depends on k and is independent of w

(1.208)Ak, w = dw - ek - S1kHere, the spectrum function is also a delta function, which implies that for a particle with momentum k the energy of this particle isek + S1k. In other words, interactions between electrons change the energy of a particle from ek into ek + S1k. The physical meaninghere is that for free particles, the energy of a particle is just the kinetic energy ek. But for interacting particles, the energy of a particle containsboth kinetic energy ek and contributions from interactions S1k.In reality, S1k, w also depends on w. Here, for a fixed k, we can solve the following equation

(1.209)Ak, w = dw - ek - S1k, w = 1

1 - ∑wS1k, w w=eèk

dw - eèk

where eèk is the solution of the equation w - ek - S1k, w = 0 (we assume that there is only one solution for simplicity). Here, we used thefact that

(1.210)d f x =i

dx - xif ' xi

Notice that at fixed k, 1

1-∑wS1k,w w=eèk is a function of k, which we will call Zk.

(1.211)Ak, w = Zk dw - eèkHere, we learned that the real part of the self-energy can do two things: (1) renormalizing the dispersion relation and (2) changing the coeffi-cient in front of the delta function.

But remember that

Phys540.nb 25

(1.212)-¶

¶ „w

2 pAk, w = 1

while

(1.213)-¶

¶ „w

2 pZk dw - eèk = Zk ∫ 1

This means that if Zk ∫ 1, ImSk, w cannot be zero for all frequency. So we must consider ImSk, w to fully understand interaction effects.

1.8.6. The imaginary part of self-energy correction

If the imaginary part of the self-energy is none zero, the spectral function is no longer a delta function.

(1.214)Ak, w = 2 S2k, ww - ek - S1k, w2 + S2k, w2

If S2 is small, Ak, w still shows a peak when we fix k and plot Ak, w as a function of w. The location of the peak is at eèk, where again eèkis the solution of the equation w - ek - S1k, w = 0. The width of the peak is S2k, w.The fact that Ak, w is not a delta peak implies that for a fixed k, the energy of the particle is not a unique value, but has a distribution. In otherwords, there is an uncertain in energy. This comes from the uncertainty principle for time and energy. The uncertainty principle tells us that tomeasure the energy with infinite accuracy, it must take infinite long time. Here, what we are trying to do is to measure the energy of a particlewith a fixed momentum k. If the momentum of this particle remains k for infinite long time, we can determine its energy accurately. But if themomentum of the particle varies with time, we will not have enough time to measure the energy, and thus the energy has an uncertainty Ñ/t,where t is the time span in which the momentum can remain unchanged. For non-interacting systems, the momentum of the particle neverchange, so that we can determine the energy. Therefore, the spectral function is a delta function. For interacting systems, because a particle withmomentum k will collide with other particles, its momentum can only remain invariant between to collisions. If the average time span betweentwo collisions is t, which is known as the collision time or the life time of an electron, the energy would get an uncertainty Ñ/t. This is thewidth of the peak, i.e. the imaginary part of the self-energy.

In summary, the real part of the self-energy modifies the dispersion relation while the imaginary part tells us the inverse of the life time of thisparticle.

If S2k, w<< eè k, the width of the peak is small, and thus we can think this peaks as “almost” free particles. Peaks in the spectral function ofthis type are called quasi-particles. The are not the particles that we originally consider. They have a different dispersion eèk and they havefinite life time 1 S2.

If S2k, w >> eè k, the concept of a particle becomes ill-defined. There, particles scatter with each other so frequently, i.e. particles arecorrelated so strongly with each other, that we cannot separate a single particle from the environment.

Experimentally/theoretically, to determine whether a system contains quasi-particles or not, we plot Ak, w and search for peaks.

1.8.7. Diagrams

In QED, the interaction V in our diagrams is substitute by the Green’s function of photons. There, we can understand the diagram of interactionas an electron shots a photon, which is then absorbed by another electron.

26 Phys540.nb