Interacting Field Theory: Time-ordered Green's Function

Basic Introduce

Perturbation Theory

[Definition] : In quantum field theory, the perturbation theory is a power series respect to the coupling parameter (parameter of interaction term) of the theory. Namely, for a quantity A(λ) with λ is the coupling parameter, when λ=0 it is the free field theory. Then the perturbation theory give us:

A(λ)A(0)(1+c1λ+c2λ2+c3λ3+)

Where A(0) is the quantity in Free Field Theory. And coefficient cn can be obtained by perturbation calculation(like Feynman Diagram)

One should note that this expansion usually is NOT convergent. But it is treated as an asymptotic expansion respect to λ , namely:

limλ01λn(A(λ)A(0)i=0nciλi)=0

For first some orders it will improve the calculation, though it is divergent actually.

Scale, Cutoff and Lagrangian

A field system has (uncountable) infinite degree of freedom , which leads to a large class of divergence in Quantum Field Theory. To overcome this, one needs a cutoff in the theory. The cutoff demand that there are only finite energy-momentum modes in any integral, equivalently it leads to a discrete real space.

The cutoff might come from the fundamental property of the nature: the nature is separated into different parts according to the scale . And for each part we have a effective theory to describe it. One can think the Quantum Field Theory just work for a finite range of energy, with the upper bound Λ , which is the cutoff. The cutoff make all quantities convergent, or the Interacting Field Theory is regularized.

With the cutoff, dimensional-analysis, and symmetry analysis, one can determinate the form of Lagrangian for the Interacting Field Theory. Especially at the energy scale which is far from the cutoff.

lambda-phi^4 Theory

For free Klein-Gordon Field, the Lagrangian is constructed by the quantities with dimensional M4 , and field has the dimensional of M2 .. (( HX3M2(ϕ)2M ,, with k , has dimensional X1=M , we have ϕM ) If we need the system has the symmetry of ϕϕ , Then we can write the Lagrangian as:

Linteracting=12μϕμϕm22ϕ2λϕ4λΛ2ϕ6λΛ4ϕ8

Where λ,λ, are dimensional-less. At the energy scale far from Λ , one can find only ϕ4 term works. That is why it called as λϕ4 theory, or Scalar Field self-interaction theory:

L=12μϕμϕm22ϕ2λ4!ϕ4

Those omitted operators with high dimensional usually are called as irrelevant operators. Or nonrenormalizable operators . A theory with no nonrenormalizable operators are called renormalizable.

Yukawa Theory

For Dirac spinor Field, similarly Lagrangian has the dimensional M4 , Then M4M(ψ)2 , we know Dirac Field has the dimensional M3/2 , ψ,ψ¯ has the same dimensional. So the lowest order of interaction between scalar Field and Dirac Field should have the form of ψ¯ϕψ , then we can write down the Lagrangian:

L=12μϕμϕmscalar22ϕ2λ4!ϕ4+ψ¯(iγμμmDiracgϕ)ψ

Where g is the coupling parameter of scalar-Dirac field interaction.

This theory is also known as Yukawa Theory.

Quantum ElectroDynamics

The interaction between Dirac Spinor Field and Maxwell Field is described by Quantum Electrodynamics (QED) , it has the form:

L=ψ¯(iγμμeγμAμm)ψ14FμνFμν

Here we do not use the form of Maxwell Field’s Lagrangian of Gupta-Bleuler Quantization.

It can also be obtained by dimensional-analysis, if one notes Aμ has the dimensional of M1 like Scalar field.

The interaction term of QED, with jμ=ψ¯γμψ is the conserved current of U(1) symmetry of Dirac Field, one can write it as:

Lint=eAμjμ

One can check QED Lagrangian is invariant under gauge transform:

ψ(x)eiθ(x)ψ(x) ; Aμ(x)Aμ(x)1eμθ(x)

If we demand the Gauge invariance , there will not be any self-interaction term of Maxwell Field like A4,AμAμνAν, .

Proof Under Gauge transform: ΔL=ψ¯γμμθψ+ψ¯γμμθψ+ΔLMaxwell=ΔLMaxwell While for LMaxwell=14FμνFμν , we have: Fμν=μAννAμ=Fμν1eμνθ+1eνμθ=Fμν That is to say ΔLMaxwell=0 . q.e.d. For self-interaction of Maxwell Field, one can check they will not have gauge invariance.

One can also write the so-called Scalar QED Lagrangian to describe the interaction between charged Klein-Gordon Field and Maxwell Field:

L=12(μeAμ)ϕ(μeAμ)ϕm22|ϕ|214FμνFμν

Perturbation Expansion

Interaction Picture

If interaction term contain no time-differential of Field, then canonical momentum is the same as Free field case , that is to say, in Schrodinger Picture, field-operators and canonical momentum operators of interacting Field system have the same form of those which in non-interacting system.

In quantum mechanics, we know the relationship between interaction picture and Schrodinger picture:

With H^=H^0+H^int , we have:

|ψI(t)=eiH^0t|ψS(t)O^I(t)=eiH^0tO^SeiH^0t

So that αI(t)|O^I(t)|βI(t)=αS(t)|O^S|βS(t) . One can find that the operator in Interaction Picture is the same as the operator in Heisenberg Picture for Free system.

Then we have the equation of motion in Interaction Picture:

iddt|ψI(t)=H^int,I(t)|ψI(t)

Or the time-evolution operator, which is defined by |ψI(t)=U^I(t,t0)|ψI(t0) .

itU^I(t,t0)=H^int,I(t)U^I(t,t0)

One can write the Dyson Series of time-evolution operator:

U^I(t,t0)=1+(i)t0tdt1H^int,I(t)+(i)2t0tdt1t0t1dt2H^int,I(t1)H^int,I(t2)+

Or with the time-order product, one can write it as:

U^I(t,t0)=Texp(it0tdτH^int,I(τ))1+(i)t0tdτH^int,I(τ)+(i)22!t0tdτ1t0tdτ2T{H^int,I(τ1)H^int,I(τ2)}    +(i)33!t0tdτ1t0tdτ2t0tdτ3 T{H^int,I(τ1)H^int,I(τ2)H^int,I(τ3)}+

Where we treat H^int,I(τ) as bosonic, that is to say:

T{H^int,I(t1)H^int,I(tn)}=σSnΘ(tσ(1)tσ(2))Θ(tσ(2)tσ(3))Θ(tσ(n1)tσ(n))H^int,I(tσ(1))H^int,I(tσ(n))

Proof One can rewrite the differential equation: itU^I(t,t0)=H^int,I(t)U^I(t,t0) as integral equation, with U^I(t0,t0)=1 :: U^I(t,t0)=1+(i)t0tdt1 H^int,I(t1)U^I(t1,t0) Then use this equation recurrently: U^I(t,t0)=1+(i)t0tdt1 H^int,I(t1)+(i)2t0tdt1t0t1dt2H^int,I(t1)H^int,I(t2)+ The third term at right-hand-side above, can be rewritten by time-ordered product, noting that the integral is calculated at the range of t1>t2 : t0tdt1t0t1dt2H^int,I(t1)H^int,I(t2)=12!(t0tdt1t0tdt2 Θ(t1t2)H^int,I(t1)H^int,I(t2)+Θ(t2t1)H^int,I(t2)H^int,I(t1))=12!t0tdt1t0tdt2T(H^int,I(t1)H^int,I(t2)) Where the factor (2!)1 comes from the time-ordered product of n operators has n! terms. Then for high order terms, the relation is the same. Then we have: U^I(t,t0)=Texp(it0tdτH^int,I(τ)) q.e.d.

With the time-evolution operator, we can write the operators in Heisenberg Picture for interacting system:

O^H(t)=U^I(t,0)O^I(t)U^I(t,0)

Proof The time-evolution operator for interaction picture we defined above is: U^I(t,t0)=U^I(t,t0)=eiH^0(tt0)U^S(t,t0)eiH^0t0 For: eiH^0t|ψS(t)=|ψI(t)=U^I(t,t0)|ψI(t0)=U^I(t,t0)eiH^0t0|ψS(t0) That is: U^S(t,t0)=eiH^0tU^I(t,t0)eiH^0t0 Then: O^H(t)=U^S(t,0)O^SU^S(t,0)=U^I(t,0)eiH^0tO^SeiH^0tU^I(t,0)=U^I(t,0)O^I(t)U^I(t,0)

Interacting Vacuum

[Theorem(Gell-Mann-Low)] : In the interacting field system, the vacuum is generally not the vacuum state of direct product of free fields. If denote the vacuum without interaction as |0 , and vacuum with interaction is |Ω . The interacting Hamiltonian is H^=H^0+gH^int , with g is the coupling parameter .

Define the time-dependent Hamiltonian H^ϵ=H^0+eϵ|t|gH^int , which effectively interpolates between H^ and H^0 in the limit ϵ0+,t . Let U^ϵ,I(t,t0) be the time-evolution operator of this Hamiltonian in the Interaction picture, and H^0|Ψ0=E0|Ψ0 .. The Gell-Man & Low Theorem asserts that if the limit as ϵ0+ of:

|Ψϵ(±)=U^ϵ,I(0,±)|Ψ0Ψ0|U^ϵ,I(0,±)|Ψ0

exists, then |Ψϵ(±) are eigenstates of H^ .

Note that the theorem does not guarantee that if |Ψ0=|0 is the vacuum of H^0 the evolved state will be the ground state. But in perturbation theory, we believe it is indeed the ground state.

Proof In Schrodinger Picture, we have the equation for time-evolution operator: it1U^ϵ,S(t1,t2)=H^ϵ(t1)U^ϵ,S(t1,t2)it2U^ϵ,S(t1,t2)=U^ϵ,S(t1,t2)H^ϵ(t2) Or the integral equation, with g=eϵθ :: U^ϵ,S(t1,t2)=1+(i)t2t1dt(H^0+eϵ(θ|t|)H^int)U^ϵ,S(t,t2) For the case 0t1t2 , with changing the variables τ=t+θ :: U^ϵ,S(t1,t2)=1+(i)t2+θt1+θdτ(H^0+eϵτH^int)U^ϵ,S(τθ,t2) For the propagator of H^(±)(t)=H^0+e±ϵtH^int is U^(±)(t,t) ,, which satisfies the equation: U^(±)(t1+θ,t2+θ)=1+(i)t2+θt1+θdτ(H^0+e±ϵτH^int)U^(±)(τ,t2+θ) With the same initial condition: U^ϵ,S(t1=t2,t2)=1=U^(+)(t1+θ=t2+θ,t2+θ) Then we have: U^ϵ,S(t1,t2)=U^(+)(t1+θ,t2+θ) . That is to say: θU^ϵ,S(t1,t2)=t1U^ϵ,S(t1,t2)+t2U^ϵ,S(t1,t2) Because θ enters in the operator U^(+)(t1+θ,t2+θ) only in its temporal variables. If t1t20 , one has U^ϵ,S(t1,t2)=U^()(t1θ,t2θ) , which induces: θU^ϵ,S(t1,t2)=t1U^ϵ,S(t1,t2)t2U^ϵ,S(t1,t2) An identity for t10t2 can be obtained by writing U^ϵ,S(t1,t2)=U^ϵ,S(t1,0)U^ϵ,S(0,t2) . Then with θ=ϵgg , one can write it as: iϵggU^ϵ,S(t1,t2)={H^ϵ(t1)U^ϵ,S(t1,t2)U^ϵ,S(t1,t2)H^ϵ(t2)0t1t2H^ϵ(t1)U^ϵ,S(t1,t2)+U^ϵ,S(t1,t2)H^ϵ(t2)t1t20 Or in the Interaction Picture: iϵggU^ϵ,I(t1,t2)={H^ϵ,I(t1)U^ϵ,I(t1,t2)U^ϵ,I(t1,t2)H^ϵ,I(t2)0t1t2H^ϵ,I(t1)U^ϵ,I(t1,t2)+U^ϵ,I(t1,t2)H^ϵ,I(t2)t1t20 Where H^ϵ,I(t)=eiH^0tH^ϵ(t)eiH^0t . Apply it with t2=,t1=0 or t1=+,t2=0 on an eigenstate |Ψ0 of H^0 , note that H^ϵ,I(±)=H^0,H^ϵ,I(0)=H^ϵ(0)=H^ , one has: iϵggU^ϵ,I(0,)|Ψ0=H^U^ϵ,I(0,)|Ψ0E0U^ϵ,I(0,)|Ψ0iϵgΨ0|gU^ϵ,I(+,0)=E0Ψ0|U^ϵ,I(+,0)Ψ0|U^ϵ,I(+,0)H^ Or with U^ϵ,I(+,0)=U^ϵ,I(0,+) , we can write them as: (H^E0±iϵgg)U^ϵ,I(0,±)|Ψ0=0 With the definition of |Ψϵ(±) ,, one can write down: (H^E0)|Ψϵ(±)=iϵggU^ϵ,I(0,±)|Ψ0Ψ0|U^ϵ,I(0,±)|Ψ0=iϵgg|Ψϵ(±)    (±iϵgU^ϵ,I(0,±)|Ψ0(Ψ0|U^ϵ,I(0,±)|Ψ0)2Ψ0|gU^ϵ,I(0,±)|Ψ0)=iϵgg|Ψϵ(±)U^ϵ,I(0,±)|Ψ0(Ψ0|U^ϵ,I(0,±)|Ψ0)2Ψ0|(E0H^)U^ϵ,I(0,±)|Ψ0=iϵgg|Ψϵ(±)|Ψϵ(±)Ψ0|(E0H^)|Ψϵ(±) Or, with E(±)=E0Ψ0|(E0H^)|Ψϵ(±) , we have: (H^E(±))|Ψϵ(±)=iϵgg|Ψϵ(±) Do the limit with ϵ0+ , with the assumption, gg|Ψϵ(±) will be finite, but with the factor ϵ it will vanish. That is to say, at the limit: H^|Ψ(±)=E(±)|Ψ(±) It is the eigenstate of interacting system. q.e.d.

With the Gell-Mann & Low Theorem, one can express the vacuum of interacting system by the vacuum of free system:

|Ω=limϵ0+U^ϵ,I(0,±)|00|U^ϵ,I(0,±)|0

Time-ordered Green’s Function

[Definition] : In Field Theory, we usually need to compute the time-ordered Green's Function, which is defined as:

G(x1,,xn)=Ω|T{ϕ1(x1)ϕn(xn)}|Ω

Where |Ω is the vacuum(ground state of interacting system), and ϕi is the i --th field operator(in Heisenberg Picture), and xi=(xi0,xi) is spacetime point.

With the Gell-Mann & Low Theorem, we can express it as:

G(x1,,xn)=limϵ0+0|T{ϕ1,I(x1)ϕn,I(xn)exp(i+dτ eϵ|τ|H^int,I(τ))}|00|exp(i+dτ eϵ|τ|H^int,I(τ))|0

Where |0 is the vacuum of non-interacting system.

Proof Without loss of generality, we assume x10>x20>>xn0 , other cases can be simply obtained by exchange the field operators and add sign (±) for fermionic operators. Then with the `Gell-Mann & Low theorem` and for the n=2 case: G(x1,x2)=limϵ0+0|U^ϵ,I(+,0)ϕ1(x1)ϕ2(x2)U^ϵ,I(0,)|00|U^ϵ,I(+,)|0=limϵ0+10|U^ϵ,I(+,)|0{0|U^ϵ,I(+,0)U^ϵ,I(0,x10)ϕ1,I(x1)U^ϵ,I(x10,0)    U^ϵ,I(0,x20)ϕ2,I(x2)U^ϵ,I(x20,0)U^ϵ,I(0,)|0}=limϵ0+0|U^ϵ,I(+,x10)ϕ1,I(x1)U^ϵ,I(x10,x20)ϕ2,I(x2)U^ϵ,I(x20,)|00|U^ϵ,I(+,)|0=limϵ0+10|U^ϵ,I(+,)|0{0|T{exp(ix10+dτ eϵ|τ|H^int,I(τ)ϕ1,I(x1)    x20x10dτ eϵ|τ|H^int,I(τ))ϕ2,I(x2)x20dτ eϵ|τ|H^int,I(τ)}|0}=limϵ0+0|T{ϕ1,I(x1)ϕ2,I(x2)dτ eϵ|τ|H^int,I(τ)}|00|U^ϵ,I(+,)|0 Where the normalization factor can be simply obtained by compute Ω|Ω by Gell-Mann & Low theorem. And similarly for cases of more than two operators. Some times people will simply write it as: G(x1,,xn)=0|T{ϕ1,I(x1)ϕn,I(xn)exp(i+dτ H^int,I(τ))}|00|exp(i+dτ H^int,I(τ))|0 And deal with the divergence at the final of all computation.

Wick Theorem

[Definition] : Normal-ordered product is a rule to handle the order-problem of field-operators’ product. In QFT, any field operators has the creation-part and annihilation-part, Normal-ordered product let all annihilation parts be at the right side of creation parts. When one needs to exchange two fermionic operators, there should be sign by the permutation:

N{ϕ1(a)ϕ2(c)ϕ3(c)ϕ4(a)}=(±)σ((1234)(23,14))ϕ2(c)ϕ3(c)ϕ1(a)ϕ4(a)N{A1A2+B1B2+}=N{A1A2}+N{B1B2}+

Where σ means the number of exchanges between fermionic operators.

[Theorem] : Normal-ordered product has the property:

N{ϕσ(1)ϕσ(2)ϕσ(n)}=(1)σN{ϕ1ϕn}

Where σ means the number of exchanges between fermionic operators.

Proof We will use mathematical induction to prove this. For n=2 case: N{ϕ1ϕ2}=(±1)0N{ϕ1ϕ2} And with the property of that [ϕ1(a),ϕ2(a)]ξ=[ϕ1(c),ϕ2(c)]ξ=0 , where ξ=1 for Boson and ξ=1 for Fermion (this demand these two operators are all fermionic.) , where [A,B]ξ=ABξBA . . Then we find that: N{ϕ2ϕ1}=N{ϕ2(c)ϕ1(c)+ϕ2(a)ϕ1(c)+ϕ2(c)ϕ1(a)+ϕ2(a)ϕ1(a)}=ϕ2(c)ϕ1(c)+ξϕ1(c)ϕ2(a)+ϕ2(c)ϕ1(a)+ϕ2(a)ϕ1(a)=ξϕ1(c)ϕ2(c)+ξϕ1(c)ϕ2(a)+ϕ2(c)ϕ1(a)+ξϕ1(a)ϕ2(a)=ξ(ϕ1(c)ϕ2(c)+ϕ1(c)ϕ2(a)+ξϕ2(c)ϕ1(a)+ϕ1(a)ϕ2(a))=ξN{ϕ1ϕ2}=(±1)1N{ϕ1ϕ2} Where the factor 1 appears only when ϕ1,ϕ2 are all fermionic operators. That is to say it works for n=2 case. We assume it works for all nk cases, then for n=k+1 case, we first express it with some normal-ordered product of nk operators: N{ϕ1ϕkϕk+1}=N{ϕ1ϕkϕk+1(c)}+N{ϕ1ϕk}ϕk+1(a) The first term in right-hand-side can be computed term by term. By the definition, one can check that: N{ϕ1ϕkϕk+1(c)}=(±1)kϕk+1(c)N{ϕ1ϕk} in which we need to move ϕk+1(c) to the head of the queue of annihilation parts, which will contribute the factor (±1)l , where l is the number of annihilation parts (if exchange two fermionic operators) in that term. Then we move it to the head of queue of creation parts, every exchange between operators will contribute the factor (±1) for the (anti-)commutator relation of (Fermion)Boson operators. Then the total factor should be (±1)k where k is the number of fermionic operators in ϕ1,,ϕk , and only when ϕk+1 is fermionic it is (1) . Then we have: N{ϕ1ϕkϕk+1}=(±1)kϕk+1(c)N{ϕ1ϕk}+N{ϕ1ϕk}ϕk+1(a) Then obviously that for the permutation among 1k , we have: N{ϕσ(1)ϕσ(k)ϕk+1}=(±1)σN{ϕ1ϕkϕk+1} And for the permutation of k+1 and k , , we can check: N{ϕ1ϕk+1ϕk}=N{ϕ1ϕk1(ϕk+1(c)ϕk(c)+ϕk+1(c)ϕk(a)+ϕk+1(a)ϕk(c)+ϕk+1(a)ϕk(a))}=ξN{ϕ1ϕk1(ϕk(c)ϕk+1(c)+ξϕk+1(c)ϕk(a)+ξϕk+1(a)ϕk(c)+ϕk(a)ϕk+1(a))}=ξN{ϕ1ϕk1ϕk(c)ϕk+1(c)}+ξN{ϕ1ϕk1ϕk(a)ϕk+1(a)}    +N{ϕ1ϕk1ϕk+1(c)ϕk(a)}+N{ϕ1ϕk1ϕk+1(a)ϕk(c)} In which: N{ϕ1ϕk1ϕk+1(c)ϕk(a)}=(ξ)k1ϕk+1(c)N{ϕ1ϕk1}ϕk(a)=ξN{ϕ1ϕk1ϕk(a)ϕk+1(c)}N{ϕ1ϕk1ϕk+1(a)ϕk(c)}=(ξ)kϕk(c)N{ϕ1ϕk1}ϕk+1(a)=ξN{ϕ1ϕk1ϕk(c)ϕk+1(a)} Then we proved: N{ϕ1ϕk+1ϕk}=ξN{ϕ1ϕkϕk+1} Then, for arbitrary element σk+1 of k+1 permutation group, can always be represent as σk(kk+1)σk ,, σk,σk will contribute two factors as (ξ)σk,(ξ)σk , and exchange (kk+1) contributes factor ξ , where ξ=1 only when ϕk,ϕk+1 are fermionic. Then we finally proved: N{ϕσ(1)ϕσ(k+1)}=(1)σN{ϕ1ϕk+1} By the mathematical induction, the theorem has been proved.

[Definition] : The Wick Contraction between two field operators are defined as:

ϕ1(x1)ϕ2(x2):=0|T{ϕ1(x1)ϕ2(x2)}|0{[ϕ1(a)(x1),ϕ2(c)(x2)]ξx10>x20ξ[ϕ2(a)(x2),ϕ1(c)(x1)]ξx20>x10

Where ξ=±1 , and [A,B]ξ=ABξBA . . For ξ=1 , it is the commutator and for ξ=1 it is anti-commutator. We used the fact that (anti-)commutator between creation/annihilation part of (Fermionic) Bosonic field operators are c-number. So the right-hand-side always equals to its expectation on free-vacuum, if we let ξ=1 when A,B are fermionic and ξ=1 otherwise.

[Theorem(Wick)] : Time-ordered product of operators can be written as normal-ordered product and sum of all possible Wick’s contraction:

T{ϕ1ϕn}=N{ϕ1ϕn}    +N{ϕ1ϕ2ϕ3ϕn}+N{ϕ1ϕ2ϕ3ϕn}++N{ϕ1ϕ2ϕn}    +N{all terms with two Wick's contractions}    ++N{all terms with full Wick's contractions}

Proof One can use the Mathematical induction to prove this. For n=2 case, if x10>x20 T{ϕ1ϕ2}=ϕ1(c)ϕ2(c)+ϕ1(a)ϕ2(c)+ϕ1(c)ϕ2(a)+ϕ1(a)ϕ2(a)=ϕ1(c)ϕ2(c)+ξϕ2(c)ϕ1(a)ξϕ2(c)ϕ1(a)+ϕ1(a)ϕ2(c)+ϕ1(c)ϕ2(a)+ϕ1(a)ϕ2(a)=N{ϕ1ϕ2}+[ϕ1(a),ϕ2(c)]ξ And if x20>x10 : T{ϕ1ϕ2}=ξT{ϕ2ϕ1}=ξN{ϕ2ϕ1}+ξ[ϕ2(a),ϕ1(c)]ξ=N{ϕ1ϕ2}+ξ[ϕ2(a),ϕ1(c)]ξ with the theorem about normal-order product above. Then we find that: T{ϕ1ϕ2}=N{ϕ1ϕ2}+ϕ1ϕ2 That is to say, the equality works for n=2 case. Then we will use the Mathematical induction, assume it works for n=k case, without loss of generality, we let xk+10<xk0<<x10 , T{ϕ1ϕkϕk+1}=T{ϕ1ϕk}ϕk+1=N{all possible contractions}(ϕk+1(c)+ϕk+1(a)) We need to check how ϕk+1 will enter the normal-ordered product in every term: First, the term with no Wick's contraction, we need to compute N{ϕ1ϕk}ϕk+1 , consider the term: ϕ1(c)ϕ2(c)ϕk1(c)ϕk(a)ϕk+1(c)=ξϕ1(c)ϕ2(c)ϕk1(c)ϕk+1(c)ϕk(a)+ϕ1(c)ϕ2(c)ϕk1(c)ϕkϕk+1 According to this, one can infer that: N{ϕ1ϕk}ϕk+1(c)=N{ϕ1ϕkϕk+1(c)}+N{ϕ1ϕk1ϕkϕk+1}+N{ϕ1ϕk1ϕkϕk+1} Then: N{ϕ1ϕk}(ϕk+1(c)+ϕk+1(a))=N{ϕ1ϕkϕk+1}+N{ϕ1ϕk1ϕkϕk+1}+N{ϕ1ϕk1ϕkϕk+1} Same for other terms with soem Wick's contractions. Then we can infer that: N{all possible contractions}(ϕk+1(c)+ϕk+1(a))=N{all possible contractionsϕk+1}+N{all possible contractions withϕk+1}=N{all possible contractions of k+1 operators} Then we have proved the k , to k+1 . q.e.d.

With Wick's Theorem , we can convert the problem of computing the time-ordered Green’s function to the problem of computing some Wick’s contractions’ sum. And this will be solved with the Feynman Diagram and Feynman rule.