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§2.1 Single-Particle Green's Functions at Zero Temperature

Hamiltonian with interaction can not always be solved exactly, the first few orders of perturbation theory cannot provide an adequate description of an interacting many-particle system. So we must find other methods to obtain all orders in perturbation theory. Luckily, we can get a formal solution through Green’s functions.

1. Definition, symmetry and physical significance of the single-particle Green’s functions

1.1. Definition

Single-particle Green's function

The single-particle Green’s function is defined as:

(1)Gαβ(x,t;x,t):=iψH0|T^[ψ^Hα(x,t)ψ^Hβ(x,t)]|ψH0,

where |ψH0 is the Heisenberg ground state of the interacting system satisfying

(2)H^|ψH0=Eg0|ψH0,ψH0|ψH0=1.

And ψ^Hα(x,t) is a Heisenberg operator with the time dependence

(3)ψ^Hα(x,t)=eiH^(tt0)/ψ^α(x)eiH^(tt0)/.

The T^ product here represents a generalization of the time-ordering operator,

(4)T^[ψ^Hα(x,t)ψ^Hβ(x,t)]={ψ^Hα(x,t)ψ^Hβ(x,t),t>t,±ψ^Hβ(x,t)ψ^Hα(x,t),t>t,

where the upper (lower) sign refers to bosons (fermions).

More generally, the T^ product of several operators orders them from right to left in ascending time order and adds a factor (1)P, where P is the number of interchanges of fermion operators from the original given order.

1.2. Symmetry

  1. If the Hamiltonian does not depend on time which means that H^(t)=H^, then the Green’s function depends only on the time difference,

    (5)Gαβ(x,t;x,t)=Gαβ(x,x;tt).
  2. If the system is translational invariance (i.e. [H^,P^]=0), then the Green’s function depends only on the position difference,

    (6)Gαβ(x,t;x,t)=Gαβ(xx;t,t).
    Proof of 1. and 2.

    Here we just prove the case of fermions. Expand the definition of the Green’s function,

    iGαβ(x,t;x,t)=ψH0|T^[ψ^Hα(x,t)ψ^Hβ(x,t)]|ψH0=ψH0|T^[eiH^t/ψ^α(x)eiH^t/eiH^t/ψ^β(x)eiH^t/]|ψH0=θ(tt)ψH0|eiH^t/ψ^α(x)eiH^t/eiH^t/ψ^β(x)eiH^t/|ψH0θ(tt)ψH0|eiH^t/ψ^β(x)eiH^t/eiH^t/ψ^α(x)eiH^t/|ψH0=θ(tt)eiωgtψH0|ψ^α(x)eiH^t/eiH^t/ψ^β(x)|ψH0eiωgtθ(tt)eiωgtψH0|ψ^β(x)eiH^t/eiH^t/ψ^α(x)|ψH0eiωgt=θ(tt)ψH0|ψ^α(x)eiH^(tt)/ψ^β(x)|ψH0eiωg(tt)θ(tt)ψH0|ψ^β(xeiH^(tt)/ψ^α(x))|ψH0eiωg(tt).

    It is obviously to see that the Green’s function only depend on the time difference tt. Since the total momentum operator commute with H^, they have common eigenstates

    H^|Eg,P=Eg|Eg,P,P^|Eg,P=P|Eg,P.

    For ground state, we have

    H^|ψH0=Eg|ψH0,P^|ψH0=0.

    And we find

    [ψ^α(x),P^]=[ψ^α(x),βd3xψ^β(x)(i)ψ^β(x)]=βd3x[ψ^α(x),ψ^β(x)(ix)ψ^β(x)]=βd3x{[ψ^α(x),ψ^β(x)]+(ix)ψ^β(x)ψ^β(x)[ψ^α(x),ixψ^β(x)]+}=βd3x[δ(xx)δαβ(ix)ψ^β(x)ψ^β(x)(ix)×0]=iψ^α(x).

    This relation is true for both bosons and fermions. Its solution is

    ψ^α(x)=eiP^x/ψ^α(0)eiP^x/,ψ^α(x)=eiP^x/ψ^α(0)eiP^x/.

    Then we can rewrite the Green’s function as

    iGαβ(x,t;x,t)=θ(tt)ψH0|eiP^x/ψ^α(0)eiP^x/eiH^(tt)/eiP^x/ψ^β(0)eiP^x/|ψH0eiωg(tt)θ(tt)ψH0|eiP^x/ψ^β(0)eiP^x/eiH^(tt)/eiP^x/ψ^α(0)eiP^x/|ψH0eiωg(tt)=θ(tt)ψH0|ψ^α(0)eiP^(xx)/eiH^(tt)/ψ^β(0)|ψH0eiP(xx)/eiωg(tt)θ(tt)ψH0|ψ^β(0)eiP^(xx)/eiH^(tt)/ψ^α(0)|ψH0eiP(xx)/eiωg(tt),

    where we have used [H^,P^]=0. So it also only depend on the position difference xx. Thus Gαβ(x,t;x,t)=Gαβ(xx,tt). The same is true for the bosons.

  3. If the Hamiltonian is independent of particle’s spin, then

    (7)Gαβ(x,t;x,t)=G(x,t;x,t)δαβ.

    And if α=β, then Gαα(x,t;x,t)=G(x,t;x,t). Assume the spin quantum number is s, then S^2=s(s+1) and α1=2s+1, so

    (8)G(x,t;x,t)=12s+1αGαα(x,t;x,t).
  4. If all three conditions above are met, then

    (9)Gαβ(x,t;x,t)=G(xx,tt)δαβ.

1.3. Physical significance

Without loss of generality, we can assume t>t and let t0=0, then the Green’s function becomes

Gαβ(x,t;x,t)ψH0|ψ^α(x,t)ψ^β(x,t)|ψH0=ψH0|eiH^t/ψα^(x)eiH^t/eiH^t/ψ^β(x)eiH^t/|ψH0=ψH0|eiH^t/Φ2| ψ^α(x)eiH^(tt)/ψ^β(x)eiH^t/|ψH0|Φ1.

Starting from the ground state |ψH0, eiH^t/ represents the evolution to time t. If we apply the field operator ψ^β(x) to this state eiH^t/|ψH0, it means that a particle with spin index β is created at position x. Finally, let this state evolve from time t to t and let the particle created at time t be annihilated by ψ^α(x) at x with a spin index α. So we reach the state |Φ1=ψ^α(x)eiH^(tt)/ψ^β(x)eiH^t/|ψH0.

Similarly for the state Φ2|. Starting from the ground state ψH0|, after evolving to time t, we get the state Φ2|=ψH0|eiH^t/.

So Green’s function is actually proportional to this probability amplitude Φ2|Φ1. That’s why Green’s function is also called as propagator.

2. Relation to observables

2.1. Ground state expectation value of single-particle operators

The expectation value of any single-particle operator in the ground state of the system can be expressed via Green’s function.

The ground state expectation value of a single-particle operator
(10)ψH0|F^(1)|ψH0=±iαβd3xlimxxtt+f^αβ(x,i,s^)Gαβ(x,t;x,t),

where the upper (lower) sign refers to bosons (fermions).

Proof

Recall the previous section, we have obtained the expression of single-particle operator in second quantization:

F^(1)=αβd3xψ^α(x)α|f^(x,p^,s^)|βψ^β(x).

Then the ground state of a single operator F^(1) can be expressed as

ψH0|F^(1)|ψH0=αβd3xψH0|ψ^α(x)f^αβ(x,i,s^)ψ^β(x)|ψH0=αβd3xlimxxf^αβ(x,i,s^)×ψH0|eiH^t/eiEgt/ψ^Hα(x,t)eiH^t/eiH^t/ψ^Hβ(x,t)eiH^t/eiEgt/|ψH0=αβd3xlimxxf^αβ(x,i,s^)ψH0|ψ^Hα(x,t)ψ^Hβ(x,t)|ψH0=αβd3xlimxxtt+f^αβ(x,i,s^)ψH0|ψ^Hα(x,t)ψ^Hβ(x,t)|ψH0

Note that A^(t)B^(t)=T^[A^(t)B^(t)] if t>t. And

T^[A^(t)B^(t)]=T^[±B^(t)A^(t)]=±T^[B^(t)A^(t)],

where the upper (lower) sign refers to bosons (fermions). Then

ψH0|F^(1)|ψH0=αβd3xlimxxtt+f^αβ(x,i,s^)ψH0|T^[ψ^Hα(x,t)ψ^Hβ(x,t)]|ψH0=±αβd3xlimxxtt+f^αβ(x,i,s^)ψH0|T^[ψ^Hβ(x,t)ψ^Hα(x,t)]|ψH0=±αβd3xlimxxtt+f^αβ(x,i,s^)Gβα(x,t;x,t)=±αβd3xlimxxtt+f^αβ(x,i,s^)Gαβ(x,t;x,t).

If f^αβ=α|f^|β is independent of particle’s spin, i.e. f^αβ=f^δαβ, then

(11)ψH0|F^(1)|ψH0=±iαd3xlimxxtt+f^(x,i)Gαα(x,t;x,t).
Examples
  • For total kinetic energy operator:

    ψH0|T^|ψH0=±iαβd3xlimxxtt+δαβ(222m)Gαβ(x,t;x,t)(12)=±iαd3xlimxxtt+(222m)Gαα(x,t;x,t).
  • For total momentum operator:

    (13)ψH0|P^|ψH0=±iαd3xlimxxtt+(i)Gαα(x,t;x,t).
  • For local density operator:

    ψH0|n^(y)|ψH0=αψH0|ψ^α(y)ψ^α(y)|ψH0=αψH0|ψ^Hα(y,t)ψ^Hα(y,y)|ψH0=αlimyytt+ψH0|ψ^Hα(y,y)ψ^Hα(y,t)|ψH0(14)=±iαlimyytt+Gαα(y,t;y,t).

    And total particle number operator:

    ψH0|N^|ψH0=ψH0|d3y n^(y)|ψH0=d3yψH0|n^(y)|ψH0(15)=±iαd3ylimyytt+Gαα(y,t;y,t).
  • For spin density operator:

    (16)ψH0|s^(y)|ψH0=±iαβα|s^|βlimyytt+Gβα(y,t;y,t).

    And total spin operator:

    (17)ψH0|S^|ψH0=±iαβα|s^|βd3ylimyytt+Gβα(y,t;y,t).
  • For electrical current density operator:

    ψH0|J^(x)|ψH0=ie2mα{ψH0|ψ^α(x)ψ^α(x)|ψH0ψH0|(ψ^α(x))ψ^α(x)|ψH0}=ie2mα{ψH0|ψ^Hα(x,t)ψ^Hα(x,t)|ψH0ψH0|(ψ^Hα(x,t))ψ^Hα(x,t)|ψH0}=ie2mα{limxxtt+xψH0|ψ^Hα(x,t)ψHα(x,t)^|ψH0limxxtt+xψH0|ψ^Hα(x,t)ψ^Hα(x,t)|ψH0}=ie2m±iαlimxxtt+(xx)Gαα(x,t;x,t)(18)=±e2mαlimxxtt+(xx)Gαα(x,t;x,t).

2.2. Ground state energy

We have obtain the total kinetic energy of ground state in (12), so what about the potential energy V^ and thereby determine the total ground state energy? The Hamiltonian H^=H^0+V^ of system in second quantization is

H^=αd3xψ^α(x)(222m)ψ^α(x)+12ααββd3xd3xψ^α(x)ψ^β(x)Vαα,ββ(x,x)ψ^β(x)ψ^α(x).

We assume that V^ is a two-particle operator which is independent of particles’ spin. The ground state energy can be expressed as

(19)Eg=ψH0|H^|ψH0=ψH0|H0^|ψH0+ψH0|V^|ψH0.

where

(20)ψH0|H0^|ψH0=ψH0|T^k|ψH0=±iαd3xlimxxtt+(222m)Gαα(x,t;x,t),(21)ψH0|V^|ψH0=±i2d3xlimxxtt+α[it+222m]Gαα(x,t;x,t).
The ground state energy

Then we can express the total ground state energy solely in terms of the single-particle Green’s function,

(22)Eg=±i2αd3xlimxxtt+(it222m)Gαα(x,t;x,t).
Proof

Consider the Heisenberg equation of Heisenberg field operator ψ^Hα(x,t):

itψ^Hα(x,t)=[ψ^Hα(x,t),H^]=eiH^t/[ψ^α(x),H^]eiH^t/,

where

[ψ^α(x),H^]=βd3z[ψ^α(x),ψ^β(z)T^kψ^β(z)]+12ββγγd3zd3z×[ψ^α(x),ψ^β(z)ψ^γ(z)Vββ,γγ(z,z)ψ^γ(z)ψ^β(z)].

For definiteness, consider the fermion case which is more complicated, so

[ψ^α(x),H^]=T^k(x)ψ^α(x)12ββγd3zψ^β(z)Vββ,αγ(z,x)ψ^γ(x)ψ^β(z)+12βγγd3zψ^γ(z)Vαβ,γγ(x,z)ψ^γ(z)ψ^β(x)=T^k(x)ψ^α(x)+βγγd3zψ^γ(z)Vαβ,γγ(x,z)ψ^γ(z)ψ^β(x).

So the Heisenberg equation becomes

(itT^k(x))ψ^Hα(x,t)=βγγd3zψ^Hγ(z,t)Vαβ,γγψ^Hγ(z,t)ψ^Hβ(x,t).

Multiply both sides left by ψ^Hα(x,t) and then take the ground state expectation value

(itT^k(x))ψH0|ψ^Hα(x,t)ψ^Hα(x,t)|ψH0=βγγd3zψH0|ψ^Hα(x,t)ψ^Hγ(z,t)Vαβ,γγ(x,z)ψ^Hγ(z,t)ψ^Hβ(x,t)|ψH0.

In the limit xx,tt+, integrating x and summing α, then we get the expression for ψH0|V^|ψH0 in fermion case

ψH0|V^|ψH0=(it+222m)d3xαψH0|ψ^Hα(x,t)ψ^Hα(x,t)|ψH0=i2d3xlimxxtt+α[it+222m]Gαα(x,t;x,t).

For boson case, the negative sign becomes positive sign. Add the total kinetic energy, and we finally get the total ground state energy,

Eg=±i2αd3xlimxxtt+(it222m)Gαα(x,t;x,t).

The ground state energy (with Hellman- Feynman theorem)

We can write the ground state energy in another form

(23)Eg=Eg0+01dλλψH0(λ)|λV^|ψH0(λ)=Eg0+01dλλ[±i2d3xlimxxtt+α[it+222m]Gαα(x,t;x,t).]

where the Hamiltonian is written with a variable coupling constant λ as H^(λ)=T^k+λV^. Then H^(1)=H^ and H^(0)=H^0=T^k, T^k|ϕ0=Eg0|ϕ0 where |ϕ0 is the ground state without interaction V^.

Proof

The time-independent Schrödinger equation for an arbitrary value λ is

H^(λ)|ψH0(λ)=Eg(λ)|ψH0(λ).

Take the derivative of λ on both sides, then

ddλ[H^(λ)|ψH0(λ)]=[ddλH^(λ)]|ψH0(λ)+H^(λ)[ddλ|ψH0(λ)],ddλ[Eg(λ)|ψH0(λ)]=[ddλEg(λ)]|ψH0(λ)+Eg(λ)[ddλ|ψH0(λ)].

Multiply both sides left by ψH0(λ)|,

ψH0(λ)|[ddλH^(λ)]|ψH0(λ)+ψH0(λ)|H^(λ)[ddλ|ψH0(λ)]=ψH0(λ)|[ddλEg(λ)]|ψH0(λ)+Eg(λ)ψH0(λ)|[ddλ|ψH0(λ)].

Thus we have proof the Hellman-Feynman theorem,

ψH0(λ)|[ddλH^(λ)]|ψH0(λ)=ddλEg(λ)

Substitute H^=T^k+λV^, then

ψH0(λ)|V^|ψH0(λ)=ddλEg(λ)1λψH0(λ)|λV^|ψH0(λ)=ddλEg(λ).

Integrate λ from zero to one on both sides and note that Eg(0)=Eg0 and Eg(1)=Eg

Eg=Eg0+01dλλψH0(λ)|λV^|ψH0(λ).

If the Hamiltonian does not depend on time, and the system is homogeneous, we can write the ground state energy in a simpler form. First, take the Fourier transform of the single-particle Green’s function

(24)Gαβ(xx,tt)=1Vkdω2πGαβ(k,ω)eik(xx)eiω(tt).

And the corresponding inverse Fourier transform is

(25)Gαβ(k,ω)=d3(xx)d(tt)Gαβ(xx,tt)eik(xx)eiω(tt).

Then the ground state energy is

Eg=±i2αd3xlimxxtt+(it222m)Gαα(xx,tt)=±i2αd3xlimxxtt+(it222m){1Vkdω2πGαα(k,ω)eik(xx)eiω(tt)}=±i2Vαd3xlimxxtt+kdω2π(ω+2k22m)Gαα(k,ω)eik(xx)eiω(tt)(26)=±i2limη0+αkdω2π(ω+2k22m)Gαα(k,ω)eiωη.

And another form of the ground state energy is

Eg=Eg0±i201dλλd3xlimxxtt+α[it+222m]Gαα(x,t;x,t)(27)=Eg0±i201dλλlimη0+αkdω2π(ω2k22m)Gαα(k,ω)eiωη.

3. Free fermions

As an example of the above formalism, we will consider the single-particle Green’s function for a noninteracting homogeneous system of fermions. The Hamiltonian of the system is

(28)H^=T^k=αd3xψ^α(x)(222m)ψ^α(x).

In plane wave representation, we have

ψ^α(x)=1VkC^kαeikx,ψ^α(x)=1VkC^kαeikx,H^=T^k=kαϵk0C^kαC^kα,

where ϵk0=2k2/2m. The ground state is defined as

(29)|Φ0=|k|kFαC^kα|0,H^|Φ0=T^k|Φ0=Eg0|Φ0.

Then

kαC^kαC^kαϵk0|Φ0=kαn^kαϵk0|Φ0=kαϵk0θ(kF|k|)|Φ0=Eg0|Φ0,

where n^kα|Φ0=C^kαC^kα|Φ0=θ(kF|k|)|Φ0. So the ground state energy is

(30)Eg0=kαθ(kF|k|)ϵk0=|k|kFαϵk0.

With box normalization which means that

k()=V(2π)3d3k().

Then the summation of k can be converted to integral of k

Eg0=αV(2π)3d3kθ(kF|k|)ϵk0=αV(2π)34π0kFk2dk2k22m=2V(2π)34π12m5kF55,

and Φ0|N^|Φ0=αV(2π)34π3kF33=N, EF=2kF22m. Thus the ground state energy of Eg0 can be expressed as

(31)Eg0=35NEF.

These are well-known results in previous study (solid state physics). Now we will use single-particle Green’s function to re-obtain above results to verify that our theory is correct. The single-particle Green’s function for fermions is defined as

iGαβ0(x,t;x,t)=Φ0|T^[ψ^Hα(x,t),ψ^Hβ(x,t)]|Φ0=θ(tt)Φ0|ψ^Hα(x,t)ψ^Hβ(x,t)|Φ0θ(tt)Φ0|ψ^Hβ(x,t)ψ^Hα(x,t)|Φ0,

where we use G0 to represent the Green’s function of the noninteracting system. The field operator in Heisenberg picture is

ψ^Hα(x,t)=eiT^kt/ψ^α(x)eiT^kt/=eiT^kt/1VkC^kαeikxeiT^kt/(32)=1VkeikxeiT^kt/C^kαeiT^kt/=1VkeikxC^H,kα(t),

where we have set t0=0. There are two methods to calculate C^H,kα(t)=eiT^kt/C^kαeiT^kt/ which will be left as an exercise.

Exercise: Calculate C^H,kα
  1. Use Baker-Campbell-Hausdorff formula

    eλA^B^eλA^=n=0λnn!C^n,where C^0=B^,C^n=[A^,C^n1].
  2. Solve Heisenberg equation with condition C^H,kα(t=0)=C^kα,

    itC^H,kα=[C^H,kα,H^].

The result is

(33)C^H,kα(t)=C^kαeiϵk0t/=C^kαeiωkt,

where ϵk0=ωk. Then

iGαβ0(x,t;x,t)=θ(tt)1VkkΦ0|C^kαC^kβ|Φ0eikxiωkteikx+iωktθ(tt)1VkkΦ0|C^kβC^kα|Φ0eikxiωkteikx+iωkt,

where

Φ0|C^kαC^kβ|Φ0=δkkδαβΦ0|C^kαC^kα|Φ0=δkkδαβΦ0|(1C^kαC^kα)|Φ0=δkkδαβΦ0|1n^kα|Φ0=δkkδαβθ(|k|kF),Φ0|C^kβC^kα|Φ0=δkkδαβΦ0|C^kαC^kα|Φ0=δkkδαβθ(kF|k|).

Then we find

iGαβ0(x,t;x,t)=θ(tt)δαβVkeik(xx)eiωk(tt)θ(|k|kF)θ(tt)δαβVkeik(xx)eiωk(tt)θ(kF|k|).
Single-particle Green's function for free fermions

Thus we finally obtain the single-particle Green’s function for free fermions

(34)iGαβ0(x,t;x,t)=δαβVk[θ(tt)θ(|k|kF)θ(tt)θ(kF|k|)]eik(xx)eiωk(tt)=δαβd3k(2π)3[θ(tt)θ(|k|kF)θ(tt)θ(kF|k|)]eik(xx)eiωk(tt).

The Fourier transform of Green’s function is

(35)Gαβ0(k,ω)=δαβ[θ(|k|kF)ωωk+iη+θ(kF|k|)ωωkiη]=δαβ1ωωk+i sgn(|k|kF).
Proof

We can use an integral representation for the step function

θ(tt)=dω2πieiω(tt)ω+iη.

If t>t, we choose the lower-half ω plane. If t<t, we choose the upper-half ω plane.

step function contour step function contour Contour for the step function.

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So the Green’s function in (34) can be rewritten as

Gαβ0(x,t;x,t)=d3kdω(2π)4eik(xx)eiω(tt)×δαβ[θ(|k|kF)ωωk+iη+θ(kF|k|)ωωkiη],

which immediately yields

Gαβ0(k,ω)=δαβ[θ(|k|kF)ωωk+iη+θ(kF|k|)ωωkiη].

And the ground state energy can be calculated via (22)

Eg0=i2αd3xlimxxtt+(it222m)Gαα0(x,t;x,t)=i2αd3xlimxxtt+(it222m){iVk[θ(tt)θ(|k|kF)θ(tt)θ(kF|k|)]eik(xx)eiωk(tt)}=i2αd3xlimxxtt+iVk[θ(tt)θ(|k|kF)θ(tt)θ(kF|k|)](ϵk0+2k22m)eik(xx)eiωk(tt)=12αkθ(kF|k|)(2ϵk0)=αkθ(kF|k|)ϵk0,

which is the same as (30). Of course we can use another expression (26), then

Eg=i2limη0+αkdω2π(ω+2k22m)Gαα(k,ω)eiωη=i2limη0+αkdω2π(ω+2k22m)[θ(|k|kF)ωωk+iη+θ(kF|k|)ωωkiη]eiωη=i2αklimη0+dz2π(z+k22m)[θ(|k|kF)zωk+iη+θ(kF|k|)zωkiη]eizη=i2αklimη0+12π2πi(ϵk0+iη+2k22m)θ(kF|k|)ei(ωk+iη)η=αkϵk0θ(kF|k|),

where we choose the upper-half z plane in third line.

upper-half z plnae upper-half z plnae Contour for the third line integral.

TikZ code
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\documentclass[border=10pt]{standalone}
\usepackage{tikz}
\usetikzlibrary{arrows.meta,decorations.markings,calc,bending,shadows}

\begin{document}

\begin{tikzpicture}[>={Kite[inset=0pt,length=0.32cm,bend]},
    decoration={markings,
    mark= at position .1 with {\arrow{>}},
    mark= at position .3 with {\arrow{>}},
    mark= at position .45 with {\arrow{>}},
    mark= at position 0.8 with {\arrow{>}},
    }]
    \def\radius{4}

\filldraw[thick,postaction={decorate}, fill=gray!40] 
    (-\radius,0)--(\radius,0) arc (0:180:\radius)--cycle;
    
\fill (1.5,0.8) node[above right]{$\epsilon_{\mathbf{k}} +i\eta$}  circle (2pt) ;
\fill (1.5,-0.8) node[below right]{$\epsilon_{\mathbf{k}} +i\eta$}  circle (2pt) ;

\draw[-Latex] (-\radius*1.5,0) -- (\radius*1.5,0) node[below]{$\mathrm{Re}(z)$};
\draw[-Latex] (0,-\radius*0.5) -- (0,\radius*1.2)node[right]{$\mathrm{Im}(z)$} ;
\end{tikzpicture}

\end{document}

This result is also the same as (30).

4. References

  1. Fetter, A. L., Walecka, J. D. Quantum Theory of Many-Particle Systems. Courier Corporation (2002).
  2. Mahan, G. D., Many-Particle Physics. Springer Science+Business Media, LLC. (2000).
  3. Altland, A., Simons, B. D., Condensed Matter Field Theory. Cambridge University Press. (2010)
  4. 金彪. 量子多体理论. 2024年春
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