Dynamical Mean Field Theory of the Boson-Fermion Model within the Non Crossing Approximation J. M. Robin MPI-PKS, Dresden, Novembre 1997

I.

Introduction

We considere the Boson Fermion Model within the Dynamical Mean Field Theory. We take the Georges and Gotliar formulation. The corresponding impurity Hamiltonian is given by H =

X

+

X

ε0 c†σ cσ + E0 b† b + g [ c†↑ c†↓ b + b† c↓ c↑ ]

σ

εk c†k,σ ck,σ +

X

vk [ c†k,σ cσ + c†σ ck,σ ]

(1)

k,σ

k,σ

The first part describes the effective site and the second part describes the coupling with the medium. We define the spectral function for the Weiss field by ∆(ω) = 2π

X

vk2 δ(ω − εk )

(2)

k

Then we set H = H0 +HI where HI is the hybridization term between the electrons of the effective site and the auxiliary fermions which describe the Weiss’s field. The Hamiltonian H0 can be diagonalized exactly. Let H0 =

8 X

|ni En hn| +

n=1

X

εk c†k,σ ck,σ

(3)

k,σ

The atomic states which diagobalize H0 are given by |1i = |0i, E1 = 0

(4)

|2i = | ↑i , E2 = ε0

(5)

|3i = | ↓i , E3 = ε0

(6)

|4i = u| ↑↓i − v|•i, E4 = ε0 + E0 /2 − γ

(7)

|5i = v| ↑↓i + u|•i, E5 = ε0 + E0 /2 + γ

(8)

|6i = | ↑ •i, E6 = ε0 + E0

(9)

|7i = | ↓ •i, E7 = ε0 + E0

(10) 1

|8i = | ↑↓ •i, E8 = 2ε0 + E0

(11)

with γ =

h

(ε0 − E0 /2)2 + g 2

1 u = 2

"

2

ε0 − E0 /2 1 − γ

i1/2

(12)

#

1 v = 2

"

2

,

ε0 − E0 /2 1 + γ

#

,

uv =

g 2γ

(13)

The corresponding partition function is Z = T rloc T rmed e−βH

(14)

with T rloc A =

X

hn| A |ni

(15)

n

We introduce the Hubbard operators Xmn Xmn = |mihn|

(16)

We have X

cσ = b =

X

σ Fmn Xmn ,

Bmn Xmn ,

c†σ = b† =

X X

σ Fnm Xmn

Bnm Xmn

(17) (18)

The non-zero terms are ↑ ↑ ↑ ↑ ↑ ↑ F12 = F78 = 1, F34 = F56 = u, F35 = −F46 =v

(19)

↓ ↓ ↓ ↓ ↓ ↓ = −v = F47 = −u, F25 = −F57 = 1, F24 = −F68 F13

(20)

B26 = B37 = 1, B15 = B48 = u, B58 = −B14 = v

(21)

↑ , when spin is not specified. The part which describes the We will write Fmn = Fmn interaction between atomics states and the Weiss’s field becomes

HI =

X

σ σ vk [ Fmn c†k,σ Xmn + Fnm Xmn ck,σ ]

2

(22)

II.

Perturbation Theory

The partition function can be written Z =

Z c

1 dz −βz e Tr 2πi z−H 



(23)

The contour c in the complex plane takes the real axis where are the poles of H. Then we write Z =

Z c

dz −βz X X e hmed| hm|R(z)|mi |medi 2πi med m

(24)

We have introduced R(z). If we write Emed the energy of the state|medi, then we can do z → z + Emed . The contour c is not changed. We obtain Z =

Z c

dz −βz X X −βEmed hmed| hm|R(z + Emed )|mi |medi e e 2πi m med

(25)

We introduce the reduced operator Rm (z) Rm (z) =

1

X

Zmed

med

e−βEmed hmed| hm|R(z + Emed )|mi |medi

(26)

The partition function is then Z

Z = Zmed

c

dz −βz X e Rm (z) 2πi m

(27)

We introduce the spectral functions Am (ω) such that Rm (z) =

Z +∞ −∞

dω Am (ω) 2π z − ω

(28)

We can write Z = Zmed Zloc

(29)

with Zloc =

Z 8 X m=1

dω −βω e Am (ω) 2π

(30)

The problem is now to compute the resolvents Rm (z) by pertubation theory. We use the Dyson identity R(z) =

1 1 1 1 1 1 + HI + HI HI + . . .(31) z − H0 z − H0 z − H0 z − H0 z − H0 z − H0

We define the self energies Σm (z) Rm (z) =

1 z − Em − Σm (z)

(32)

3

III.

Non Crossing Approximation

Zero order gives 0 Rm (z) =

1 z − Em

(33)

and Am (ω) = 2π δ(ω − Em ) Zloc =

8 X

(34)

e−βEm

(35)

m=1

Second order gives 1 1 Rm (z) = + z − Em (z − Em )2



hm| HI

1 z + Emed − H0

HI |mi



(36) med

We apply HI onto the state |mi. We write nm the number of particles in this state, that is n1 = 0, n2 = n3 = 1, n4 = n5 = 2, n6 = n7 = 3, n8 = 4

(37)

The part HI makes a transition towards the state |ni. We have to consider two cases. If nn < nm , it means that we apply the part Fnm c†k Xnm so we obtain σ 2 [1 − nF (εk )] |Fmn |

1 z − En − εk

(38)

The other case corresponds to nn > nm and Fmn Xnm ck . We obtain σ 2 nF (εk ) |Fmn |

1 z − En + εk

(39)

The self energy for the state |mi is Σm (z) =

X

σ 2 |Fnm |

Z

dω ∆(ω) [1 − nF (ω)] Rn (z − ω) 2π

Z

dω ∆(ω) nF (ω) Rn (z + ω) 2π

n,σ

+

X

σ 2 |Fmn |

n,σ

(40)

σ We can introduce ηmn = sign(nn − nm ). We notice that if Fmn 6= 0 then we also have σ Fnm = 0. We obtain

Σm (z) =

X  n,σ

σ 2 |Fnm |

+

σ 2 |Fmn |

 Z

dω ∆(ω) nF (ηmn ω) Rn (z + ηnm ω) 2π

(41)

The 8 self energies for our model are given by Σ1 (z) = Σ2 (z) =

Z

dω ∆(ω)nF (ω) [R2 (z + ω) + R3 (z + ω)] 2π

Z

dω ∆(ω) [nF (−ω)R1 (z − ω) 2π 4

(42)

h

ii

+ nF (ω) u2 R4 (z + ω) + v 2 R5 (z + ω)

(43)

Σ3 (z) = Σ2 (z) Σ4 (z) =

Z

(44)

h dω ∆(ω) u2 nF (−ω) [R2 (z − ω) + R3 (z − ω)] 2π i

(45)

i

(46)

+ v 2 nF (ω) [R6 (z + ω) + R7 (z + ω)] Σ5 (z) =

Z

h dω ∆(ω) v 2 nF (−ω) [R2 (z − ω) + R3 (z − ω)] 2π

+ u2 nF (ω) [R6 (z + ω) + R7 (z + ω)] Σ6 (z) =

Z

dω ∆(ω) [nF (ω)R8 (z + ω) 2π h

ii

+ nF (−ω) v 2 R4 (z − ω) + u2 R5 (z − ω) Σ7 (z) = Σ6 (z) Σ8 (z) =

IV.

Z

(47) (48)

dω ∆(ω)nF (−ω) [R6 (z − ω) + R7 (z − ω)] 2π

(49)

The Green functions

We have to compute the local Green’s function for the fermions and the (hard core) bosons,     

F (τ ) = hT [c↑ (τ ) c†↑ ]i (50) †

B(τ ) = hT [b(τ ) b ]i

We start with the atomic limit. We have F (z) =

i 1 X h −βEm |Fmn |2 e + e−βEn Zloc m,n z + Em − En

(51)

B(z) =

i 1 X h −βEm |Bmn |2 e − e−βEn Zloc m,n z + Em − En

(52)

and

We introduce the free spectral functions Am (ω) = 2πδ(ω − Em ) and the free resolvents Rm (z) = [z − Em ]−1 . We obtain Z 1 X dω −βω 2 F (z) = |Fmn | e [Am (ω)Rm (z + ω) − An (ω)Rm (−z + ω)] (53) Zloc m,n 2π

and the same thing for B(z), but with a minus sign. We obtain the same result for the interacting case, but only inside the NCA. The approximation needed is just 

1 1 hm| |mihn| |ni z − H + Ec z − H + Ec



5

= Rm (z)Rn (z) med

(54)

The spectral functions are given by Z 1 X d˜ ω −β ω˜ 2 AF (ω) = e [1 + e−βω ] Am (˜ ω )An (ω + ω ˜) |Fmn | Zloc m,n 2π

(55)

Z 1 X d˜ ω −β ω˜ 2 AB (ω) = e [1 − e−βω ] Am (˜ ω )An (ω + ω ˜) |Bmn | Zloc m,n 2π

(56)

and

For fermions we obtain ω −β ω˜ 1 Z d˜ e [1 + e−βω ] [A1 (˜ ω )A2 (ω + ω ˜ ) + A7 (˜ ω )A8 (ω + ω ˜) AF (ω) = Zloc 2π + u2 [A3 (˜ ω )A4 (ω + ω ˜ ) + A5 (˜ ω )A6 (ω + ω ˜ )] i

+ v 2 [A3 (˜ ω )A5 (ω + ω ˜ ) + A4 (˜ ω )A6 (ω + ω ˜ )]

(57)

We notice that for hard core bosons, it is better to compute the associate fermionic Green’s ˜ ) function (put a sign plus instead of a minus sign). We define the Green’s function B(τ the same way as for F (τ ). The two Green’s functions are related by 1 A˜B (ω) = AB (ω) cotanh( βω) 2

(58)

The sum rule is Z

dω ˜ AB (ω) = 1 2π

(59)

instead of Z

dω AB (ω) = 1 − 2hb† bi 2π

(60)

we have also hb† bi =

Z

dω nF (ω) A˜B (ω) 2π

(61)

Z

dω nB (ω) AB (ω) 2π

(62)

instead of hb† bi =

This is much better.

6

V.

Numerical Solution

We solve the equations for the symmetric case, for a Bethe’s lattice of width unity. The DMFT gives ∆(ω) = t2 AF (ω)

(63)

with t = 0.25. A possible way to solve these equations is the following. First, there is no problems to compute the propagators Rm (z) self consistently and to obtain the final solution. We can do it for z = ω + iη. The problem is to compute the spectral functions. We consider AF (ω). We write it like Z dx −βx 1 X 2 e [Am (x)An (ω + x) + An (x)Am (x − ω)] |Fm,n | AF (ω) = Zloc m,n 2π

(64)

It appears the quantity am (ω) = Am (ω) eβ(Floc −ω)

(65)

so that AF (ω) =

2

X

|Fm,n |

Z

m,n

dx [am (x)An (ω + x) + an (x)Am (x − ω)] 2π

(66)

Usually the numerical accuracy is to bad to compute am (ω). The idea is to compute these quantities directly, that is without using the factors exp β(Floc − ω). We go back to the equation for the self energy and we introduce the notation Σm (ω) = Fm (ω) − iΓm (ω)/2 for the retarded part. We obtain Γm (ω) =

X

[

σ 2 |Fmn |

+

σ 2 |Fnm |

]

Z

n,σ

dx ∆(x) nF (ηmn x) An (ω + ηmn x) 2π

(67)

The real part can be obtained via the Hilbert’s transformation Fm (ω) = P P

Z

dx Γm (x) 2π ω − x

(68)

Numerically the stability is very good, but we can also compute the real part and the imaginary part by adding a small iη. The spectral function can be written Am (ω) =

Γm (ω) [ω − Em − Fm (ω)]2 + [Γm (ω)/2]2

(69)

We shall use this formulation. We introduce γm (ω) = Γm (ω) eβ(Floc −ω)

(70)

we obtain a first relation γm (ω) =

X n,σ

[

σ 2 |Fmn |

+

σ 2 |Fnm |

]

Z

dx ∆(x) nF (−ηmn x) an (ω + ηmn x) 2π

7

(71)

where we use e±βω nF (±ω) = nF (∓ω). A second relation is given by am (ω) =

γm (ω) [ω − Em − Fm (ω)]2 + [Γm (ω)/2]2

(72)

We see that for a given Σm (ω), we have two relations for am (ω), via γm (ω). In fact, these P two relations do not depend of the value of Floc . If we start with Zloc = exp(−βEm ), then we can hope that that we get the right value at the end. It is better to use the sum rule Z

1 =

dω X am (ω) 2π m

(73)

A possible way is then (1) We start with the free spectral functions. Am (ω) = 2πδ(ω − Em )

,

am (ω) = Am (ω) eβ(Floc −Em )

(74)

with "

Floc

X 1 e−βEm = − Ln β m

#

(75)

If we compute AF (ω), we must find the atomic limit. (2) We do the first loop. First Γm (ω) =

X

. . . An (ω + ηmn x)

(76)

. . . an (ω + ηmn x)

(77)

n,σ

γm (ω) =

X n,σ

then Am (ω) = Γm (ω)/ . . .

(78)

am (ω) = γm (ω)/ . . .

(79)

We norm X =

Z

dω X am (ω) 2π m

(80)

am (ω) → am (ω)/X

(81)

We compute the spectral function AF (ω) =

X

...

(82)

m,n

(3) We repeat step (2) until we reach the fixed point. We can improve this algorithm in many way.

8

VI.

References

Applications of the NCA to DMFT problems can be found in P. Lombardo, M. Avignon, J. Schmalian and K. H. Bennemann, Dynamical mean-field theory for perovskites, Phys. Rev. B 54, 5317 (1996). and T. Schork and S. Blawid, The periodic Anderson model with correlated conduction electrons, cond-mat. The numerical solution of the equations is given by N. E. Bickers, D. L. Cox and J. W. Wilkins, Self-consistent large-N expansion for normalstate properties of dilute magnetic alloys, Phys. Rev. B 36, 2036 (1987). M. H. Hettler, J. Krohn and S. Hershfield, Non-equilibrium Dynamics of the Anderson Impurity Model, cond-mat. The auxiliary propagators are introduced and used by E. M¨ uller-Hartmann, Self-consistent Pertubation Theory of the Anderson Model: Ground State Properties, Z. Phys. B 57, 281 (1984). The complete derivation of the Green’s function within the NCA is given in N. E. Bickers, Review of techniques in the large-N expansion for dilute magnetic alloys, Rev. of Modern Phys. 59, 845 (1987). Finally, a very good introduction can be found in A. C. Hewson, The Kondo problem to Heavy Fermions, Cambridge University Press, Cambridge, 1993. and T. Pruschke and N. Grewe, The Anderson model with finite coulomb repulsion, Z. Phys. B 74, 439 (1989).

9

February 4, 1998

Dynamical Mean Field Theory of the Boson-Fermion Model within the Non Crossing Approximation. II J. M. Robin MPI-PKS, Dresden

We look at the possibility of non diagonal local propagators following the idea of Alfonso. First we write down the calculation of the Green function. We see the differents approximations needed to obtain the so called NCA result. It appears that non diagonal propagators enters in the evaluation. Then we rederive the perturbation equations for these propagators. In the case of the Boson-Fermion Model, it seems that it does really matter, while in the Anderson-Hubbard Model, it does not matter. We should check if there is such terms in the models used by Avignon and co and Schork and Blawid.

I.

Approximations for the Green’s function

We consider the Green function with imaginary time GAB (τ ) = hT [A(τ )B]i

(1)

A(τ ) = eτ H A e−τ H

(2)

with

Then we go to Matsubara frequencies such that GAB (iωM ) =

Z β 0

dτ eiτ ωM GAB (τ )

(3)

With this trick, we can forget the part [−β; 0]. We use the identity Z

1 dz −βz e Tr 2πi z−H 



n

= T r e−βH

o

(4)

Where the contour encloses the real axis in counterclockwise fashion. We obtain   Z 1Z dz dz 0 z(τ −β) −τ z0 1 1 iτ ωM GAB (iωM ) = dτ e e e Tr A B Z 2πi 2πi z − H z0 − H

(5)

We integrate over τ , and we introduce ξ = eiβωM such that ξ = −1 for fermions and ξ = 1 for bosons. 0   1 Z dz dz 0 ξe−βz − e−βz 1 1 Tr A B GAB (iωM ) = Z 2πi 2πi iωM + z − z 0 z − H z0 − H

1

(6)

We integrate one part over z and the other part over z 0 .   1 1 Z dz 0 −βz0 1 GAB (iωM ) = ξe Tr 0 A B Z 2πi z − iωM − H z 0 − H   1 Z dz −βz 1 1 + e Tr A B Z 2πi z − H z + iωM − H

(7)

If we set z 0 = z + iωM , then the get the same form, but one is with a contour surrounding the real axis with poles z = En and the other shifted with poles z = En − iωM . So we write   1 1 1 Z dz −βz e Tr A B (8) GAB (iωM ) = Z 2πi z − H z + iωM − H This is exactly the derivation of Bickers (Rev. Mod. Phys). Next we consider the case where B = A† and A = A† =

X

Fmn Xmn

X

(9)

Fmn Xnm

(10)

We are left with 1 Z dz −βz e T r {R(z)|mihn|R(z + iωM )|n0 ihm0 |} Fmn Fm0 n0 G(iωM ) = Z 2πi We write the trace as a sum over |αi|medi. X 1 X Z dz −βz G(iωM ) = Fmn Fm0 n0 e Z med 2πi X

hmed|hm0 |R(z)|mihn|R(z + iωM )|n0 i|medi

(11)

(12)

We want to introduce the reduced resolvents Rm (z), so we shift by Emed . X 1 X Z dz −βz −βEmed e e G(iωM ) = Fmn Fm0 n0 Z med 2πi hmed|hm0 |R(z + Emed )|mihn|R(z + Emed + iωM )|n0 i|medi Let us define some non diagonal resolvents 1 X −βEmed Rmm0 (z) = e hmed|hm|R(z + Emed )|m0 i|medi Zmed med

(13)

(14)

The approximation we need to make, is a decoupling, which might correspond to negelect vertex functions, so we have X 1 Z dz −βz e Rm0 m (z) Rnn0 (z + iωM ) (15) G(iωM ) = Fmn Fm0 n0 Zloc 2πi If we further make the approximation Rmm0 (z) = δm,m0 Rm (z), then we have the usual NCA formula for the Green function. If we introduce the spectral functions Amm0 (ω) such that Z dω Amm0 (ω) Rmm0 (z) = (16) 2π z − ω then we obtain X 1 Z dω −βω G(z) = Fmn Fm0 n0 e [Am0 m (ω)Rnn0 (ω + z) Zloc 2π + ξAnn0 (ω)Rm0 m (ω − z)] 2

(17)

II.

New Perturbation theory

We rewrite the old equations and add one index, or two. The partition function is Zloc =

XZ m

dz −βz e Rmm (z) 2πi

(18)

dω −βω e Amm (ω) 2π

(19)

or Zloc =

XZ m

Then we try to compute Rmm0 (z) with the help of the Dyson equation, 1 1 1 1 1 1 1 = + HI + HI HI z−H z − H0 z − H0 z − H0 z − H0 z − H0 z − H

(20)

The zero order gives δm,m0 z − Em

0 Rmm 0 (z) =

(21)

The second order gives Rmm00 =

0 Rmm 00

+

0 Rmm



hm|HI

1 z + Emed − H0

0

HI |m i



Rm0 m00

(22)

med

If we take HI = Fn0 m0 c†k Xn0 m0 for the right and HI = Fnm ck Xmn for the left, we get 0 0 Rmm00 = Rmm 00 + Rmm Rm0 m00

X

0 Fnm Fn0 m0 vk2 nF (−εk ) Rnn 0 (z − εk )

(23)

If we add the other part we get, in our previous notations X

Σmm0 (z) =

+

σ Fnm Fnσ0 m0

X

Z

σ Fmn Fmσ 0 n0

dω ∆(ω) nF (−ω) Rnn0 (z − ω) 2π Z

dω ∆(ω) nF (ω) Rnn0 (z + ω) 2π

(24)

0 Rm (z) Σmm0 (z) Rm0 m00 (z)

(25)

The Dyson equation is now 0 Rmm00 (z) = δm,m00 Rm (z) +

X

or in matrix form R−1 (z) = [R0 (z)]−1 − Σ(z)

III.

(26)

Non diagonal propagators

Let us put by hand the symmetry Rmm0 (z) ∼ δm,m0 and see if we do obtain Σmm0 (z) ∼ σ σ σ σ δm,m0 . We are looking for non zero coupling products like Fnm Fnm 0 or Fmn Fm0 n . So we ↑ ↑ ↓ ↓ ↓ ↓ ↑ ↑ find four such pairs of coupling, F34 , F35 ; F24 , F25 ; F57 , F47 ; F56 , F46 which give Σ45 and Σ54 . If we look at the values, at the lowest order, these contributions do not cancel. They are equal. So if we start with diagonal propagators at the first iteration, we obtain some non diagonal ones at the next iteration. 3

IV.

The case of Anderson-Hubbard Model

In this case the local states are given by |1i = |0i

E1 = 0

(27)

|2i = | ↑i

E2 = ε0

(28)

|3i = | ↓i

E3 = ε0

(29)

|4i = | ↑↓i

E4 = 2ε0 + U

(30)

In this case we obtain ↑ ↑ F12 = F34 = 1

(31)

↓ ↓ = 1 = −F24 F13

(32)

So if we start with diagonal propagators, no off diagonal self energies arise.

4

February 17, 1998

Dynamical Mean Field Theory of the Boson-Fermion Model within the Non Crossing Approximation. III J. M. Robin MPI-PKS, Dresden

We derive the new equations when non diagonal terms are taken into account. Expressions of the local self energies and local propagators are given. We show that there is only minor modifications in all expressions. However, the expression of the spectral function for fermions (which gives the density of states) is obtained in a form which does not appear positive definite at first sight. May be some more work is needed to put these equations in a safer form. Anyway, since there is only few more work to solve the present equations, we should first see what it gives.

I.

New equations

We start with the expression of the local self energy Σmm0 (z). The general expression when we take into account the non diagonal propagators, is X

Σmm0 (z) =

+

σ Fnm Fnσ0 m0

X

Z

σ Fmn Fmσ 0 n0

dω ∆(ω) nF (−ω) Rnn0 (z − ω) 2π Z

dω ∆(ω) nF (ω) Rnn0 (z + ω) 2π

(1)

In order to solve the equations, we start with the bare propagators Rmm0 (z) = δm,m0 [z − Em ]−1 . So the idea is to put this symmetry in the expression of the self energy. If we do this, we obtain only two non diagonal self energies, Σ45 (z) = Σ54 (z). This implies that we obtain two non diagonal propagators, R45 (z) and R54 (z). So we play the game again and include this symmetry. We obtain that these new propagators only gives contributions to four diagonal self energies, Σ66 , Σ77 , Σ22 and Σ33 . So the game stops here. We give this expression of all the non zero self energies. Σ11 (z) = Σ22 (z) =

Z

dω ∆(ω)nF (ω) [R22 (z + ω) + R33 (z + ω)] 2π

Z

dω ∆(ω) [nF (−ω)R11 (z − ω) 2π

h

(2)

ii

+ nF (ω) u2 R44 (z + ω) + v 2 R55 (z + ω) + uvR45 (z + ω) + uvR54 (z + ω) 1

(3)

Σ33 (z) = Σ22 (z) Σ44 (z) =

Z

(4)

h dω ∆(ω) u2 nF (−ω) [R22 (z − ω) + R33 (z − ω)] 2π i

+ v 2 nF (ω) [R66 (z + ω) + R77 (z + ω)] Σ55 (z) =

Z

(5)

h dω ∆(ω) v 2 nF (−ω) [R22 (z − ω) + R33 (z − ω)] 2π i

+ u2 nF (ω) [R66 (z + ω) + R77 (z + ω)] Σ66 (z) =

Z

(6)

dω ∆(ω) [nF (ω)R88 (z + ω) 2π

h

ii

+ nF (−ω) u2 R55 (z − ω) + v 2 R44 (z − ω) − uvR45 (z − ω) − uvR54 (z − ω) Σ77 (z) = Σ66 (z) Σ88 (z) =

(7) (8)

Z

dω ∆(ω)nF (−ω) [R66 (z − ω) + R77 (z − ω)] 2π

Z

dω ∆(ω) [nF (ω) [−uvR66 (z + ω) − uvR77 (z + ω)] 2π

(9)

and Σ45 (z) =

+ nF (−ω) [uvR22 (z − ω) + uvR33 (z − ω)]] Σ54 (z) = Σ45 (z)

(10) (11)

From these expressions, we obtain all non zero local propagators R11 (z) =

1 z − E1 − Σ11 (z)

(12)

R22 (z) =

1 z − E2 − Σ22 (z)

(13)

R33 (z) =

1 z − E3 − Σ33 (z)

(14)

R44 (z) =

z − E5 − Σ55 (z) [z − E4 − Σ44 (z)][z − E5 − Σ55 (z)] − Σ45 (z)Σ54 (z)

(15)

R55 (z) =

z − E4 − Σ44 (z) [z − E5 − Σ55 (z)][z − E4 − Σ44 (z)] − Σ54 (z)Σ45 (z)

(16)

R66 (z) =

1 z − E6 − Σ66 (z)

(17)

R77 (z) =

1 z − E7 − Σ77 (z)

(18) 2

R88 (z) =

1 z − E8 − Σ88 (z)

(19)

R44 (z) =

Σ45 (z) [z − E4 − Σ44 (z)][z − E5 − Σ55 (z)] − Σ45 (z)Σ54 (z)

(20)

R55 (z) =

Σ54 [z − E5 − Σ55 (z)][z − E4 − Σ44 (z)] − Σ54 (z)Σ45 (z)

(21)

and

We notice that R22 (z) = R33 (z)

(22)

R66 (z) = R77 (z)

(23)

R45 (z) = R54 (z)

(24)

Up to this point, we can solve the problem, for a given Weiss self energy, ∆(ω). So, this is the first thing to do. Does the system still converge with these new terms?

II.

The fermionic Green function

Our goal is in fact to compute the Green function for the fermion. Using the formula which take account for non diagonal term, we have Z X dx AF (ω) = Fmn Fm0 n0 [am0 m (x)Ann0 (x + ω) 2π + ann0 (x)Am0 m (x − ω)] (25) We have introduced the auxiliary spectral function amm0 (ω) = Amm0 (ω) eβ(Floc −ω)

(26)

We find ten different contributions instead of six. Z dx AF (ω) = [ a11 (x)A22 (x + ω) + a22 (x)A11 (x − ω) 2π + a77 (x)A88 (x + ω) + a88 (x)A77 (x − ω) + u2 [ a33 (x)A44 (x + ω) + a44 (x)A33 (x − ω)] + u2 [ a55 (x)A66 (x + ω) + a66 (x)A55 (x − ω)] + v 2 [ a33 (x)A55 (x + ω) + a55 (x)A33 (x − ω)] + v 2 [ a44 (x)A66 (x + ω) + a66 (x)A44 (x − ω)] + uv[ a33 (x)A45 (x + ω) + a45 (x)A33 (x − ω)] + uv[ a33 (x)A54 (x + ω) + a54 (x)A33 (x − ω)] − uv[ a45 (x)A66 (x + ω) + a66 (x)A45 (x − ω)] − uv[ a54 (x)A66 (x + ω) + a66 (x)A54 (x − ω)]] There should be a way to avoid the uv terms, since the sign can be negative. 3

(27)

III.

Numerical calculation

As before, we write the retarded part of the self energy as Σmn = Hmn − iΓmn /2

(28)

and we introduce γmn (ω) = Γmn (ω) eβ(Floc −ω)

(29)

First we consider the regular propagators, that is all diagonal propagators but 44 and 55. In this case we proceed as before. We compute both Z

dx ∆(x)nF (x) [A22 (x + ω) + A33 (x + ω)] 2π

(30)

Z

dx ∆(x)nF (−x) [a22 (x + ω) + a33 (x + ω)] 2π

(31)

Amm (ω) =

Γmm (ω) [ω − Em − Hmm (ω)]2 + [Γmm (ω)]2 /4

(32)

amm (ω) =

γmm (ω) [ω − Em − Hmm (ω)]2 + [Γmm (ω)]2 /4

(33)

Γ11 (ω) = γ11 (ω) =

Then we compute

Next we considere the other terms. We first notice that we can always compute Γ44 , Γ45 ... and γ44 , γ45 ... What we need is a further relation between, say, A45 and a45 . In order to get it, we define for the common denominator D45 (ω) = [ω − E4 − Σ44 (ω)][ω − E5 − Σ55 (ω)] − Σ45 (ω)Σ54 (ω)

(34)

Then we take the real and imaginary part in the way R D45 (ω) = D45 (ω) +

i I D (ω) 2 45

(35)

We obtain 1 1 R D45 = [ω − E4 − H44 ][ω − E5 − H55 ] − Γ44 Γ55 − H45 H54 + Γ45 Γ54 4 4

(36)

I D45 = [ω − E5 − H55 ] Γ44 + [ω − E4 − H44 ] Γ55 + Γ45 H54 + Γ54 H45

(37)

where Hij means Hij (ω). We obtain the expressions of these propagators as R44 (ω) =

ω − E5 − H55 + iΓ55 /2 R I D45 + iD45 /2

(38)

R45 (ω) =

H45 − iΓ45 /2 R I D45 + iD45 /2

(39)

R54 (ω) =

H54 − iΓ54 /2 R I D45 + iD45 /2

(40) 4

R55 (ω) =

ω − E4 − H44 + iΓ44 /2 R I D45 + iD45 /2

(41)

Then we obtain the spectral functions as A44 (ω) =

R I − Γ55 D45 [ω − E5 − H55 ]D45 R 2 I 2 [D45 ] + [D45 ] /4

(42)

I R H45 ]D45 + Γ45 D45 A45 (ω) = R 2 I 2 [D45 ] + [D45 ] /4

(43)

A54 (ω) =

R I + Γ54 D45 H54 ]D45 I 2 R 2 ] + [D45 ] /4 [D45

(44)

A55 (ω) =

I R [ω − E4 − H44 ]D45 − Γ44 D45 R 2 I 2 [D45 ] + [D45 ] /4

(45)

So all we have to do to get the corresponding amn is to define dI45 (ω) = [ω − E5 − H55 ] γ44 + [ω − E4 − H44 ] γ55 + γ54 H45 + γ45 H54

(46)

Then we find directly a44 (ω) =

R [ω − E5 − H55 ]dI45 − γ55 D45 R 2 I 2 [D45 ] + [D45 ] /4

(47)

a45 (ω) =

R H45 ]dI45 + γ45 D45 R 2 I 2 [D45 ] + [D45 ] /4

(48)

a54 (ω) =

R H54 ]dI45 + γ54 D45 R 2 I 2 [D45 ] + [D45 ] /4

(49)

a55 (ω) =

R [ω − E4 − H44 ]dI45 − γ44 D45 R 2 I 2 [D45 ] + [D45 ] /4

(50)

We see in fact that the result is very similar. There is always an imaginary part somewhere with the good stuff inside. At first sight, the algorithm remains the same. We still need to find a better way to write the fermionic spectral function. Concerning the sum rules of the non diagonal spectral functions amm0 (ω), I suppose we must renormalize them the same way as the diagonal ones in the numerical calculation. Does each have a zero sum rule? In the case of many non diagonal terms, it might be good to find a way to obtain directly the expression for the amn . It should not be difficult.

5

March 13, 1998

Few remarks about the conductivity J. M. Robin MPI-PKS, Dresden

We want to compute the optical conductivity given by the Kubo formula, the bubble for the current-current Green function. Z 1 dε Z dω 0 N0 (ε) A(ε, ω 0 ) A(ε, ω + ω 0 ) [nF (ω 0 ) − nF (ω + ω 0 )] (1) σ(ω) = σ0 2π 2π ω and for ω = 0, Z 1 1 dε Z dω 0 N0 (ε) A(ε, ω 0 )2 (2) σ(0) = σ0 2 2π 2π 4kB T cosh ( 12 βω 0 ) Here the spectral function are normalized to 2π and the bare density of states is given by 1 √ 2 N0 (x) = 4t − x2 (3) 2πt2 The NCA computation gives the density of states N (ω) which is normalized to one. First we can construct the associed spectral function A(ω) = 2πN (ω). We compute the real part of the retarded Green function (the local Green function), Z dx A(x) R(ω) = P P (4) 2π ω − x The retarded Green function is then given by i G(ω) = R(ω) − A(ω) (5) 2 We also obtain the Weiss self energy (retarded), ΣW (ω) = t2 G(ω)

(6)

Now we use the canonical form of the Green function in order to extract the interacting self energy, Σ(ω), [G(ω)]−1 = ω − ΣW (ω) − Σ(ω)

(7)

Where we use the notation the the symmetric case. We write this self energy as i Σ(ω) = H(ω) − Γ(ω) (8) 2 It is intersting to plot the width Γ(ω) since it contains all the many body properties of the fermionic system. Now the spectral function are constructed as Γ(ω) A(ε, ω) = (9) [ω − ε − H(ω)]2 + [Γ(ω)/2]2 That is all we need. It is important to obtain the Weiss self energy and the spectral function from the same result, the density of states. 1

March 13, 1998

Few remarks about the specific heat J. M. Robin MPI-PKS, Dresden

We start with the Hamiltonian in momentum space which is given by X X g X † [ck+q,↑ c†−k,↓ bq + b†q c−k,↓ ck+q,↑ ] H = εk c†k,σ ck,σ + E0 b†q bq + √ M k,q q k,σ

(1)

In my notation, we define the Green fonction G(k, τ ) = hT [ck,↑ (τ ) c†k,↑ ]i. The equation of motion gives ∂τ G(k, τ ) = δ(τ ) − εk G(k, τ ) − L(k, τ ) where the new Green function is g X L(k, τ ) = √ hT [c†−k+q,↓ (τ )bq (τ ) c†k,↑ ]i M q

(2)

(3)

It is a fermionic Green function, so we obtain hH1 i = 2

X Z k

dω nF (ω) AL (k, ω) 2π

(4)

where we have introduced the spectral function associated to L. The equation of motion gives AL (k, ω) = (ω − εk ) A(k, ω)

(5)

Now we have to obtain the value hH0 i. We have for the fermions 2

XZ k

dω nF (ω) εk A(k, ω) 2π

(6)

and for the bosons XZ q

dω nB (ω) E0 AB (q, ω) 2π

(7)

or if we use the modified Green function XZ q

dω nF (ω) E0 A˜B (q, ω) 2π

(8)

1

If we add these three contributions and set E0 = 0 for the symmetric case, one obtain, hHi = 2

XZ k

dω nF (ω) ω A(k, ω) 2π

(9)

So we can sum over k to obtain the spectral function of the local Green function, hHi = 2M

Z

dω nF (ω) ω A(ω) 2π

(10)

So the energy divided by the number of site M is given by E = 2

Z

dω nF (ω) ω N (ω)

(11)

and the specific heat is CV =

dE dT

(12)

Just an ordinary derivative. In fact we have computed hH − µN i. But N is constant, and also µ. So this gives the correct value of CV . Next, if we assume that the anticommuatator of L is zero, the we obtain the mean value of H1 as a sum over Matsubara frequencies, hH1 i = −

2X Σ(k, iωn ) G(k, iωn ) β k,n

(13)

We then need the value of the fermions 1 1X hc†k,↑ ck,↑ i = + G(k, iωn ) 2 β n

(14)

So we have the contribution hH0 i = 2

X

εk hc†k,↑ ck,↑ i

(15)

k

From NCA we obtain E = −0.33 for T = 0.1,−0.36 for T = 0.06, E = −0.37 for T = 0.04.

2