1. Zero bias Conductance

2. 1. Side-coupled device
1.1. The Model
t
H = εLk c† cLk +
Lk √ 0 d† cLk + H.c.
k
2N k
t
+ εRk c† cRk +
Rk √ 0 d† cRk + H.c.
k
2N k
+ V c† f0 + H.c + Kf0 f0
d
†
+ εd c† cd + U nd↑ nd↓ .
d
(1.1)
We consider εF ≡ 0, εLk = εk − ∆µ and εRk = εk + ∆µ. The chemical potential dif-
ference between the reservoirs is 2∆µ and ∆µ = e∆φ/2 for an applied bias ∆φ. The
zero bias conductance occurs for the regime ∆φ → 0. Using the canonical even-odd
transformation
1
cLk = √ (cek + cok ) (1.2)
2
1
cRk = √ (cek − cok ) (1.3)
2
we derive a free Hamiltonian
Ho = εk c† cok
ok (1.4)
k
and an interacting one in terms of a renormalized hybridization
√ t0 †
He = εk c† cek +
ek 2√ d cek + V c† f0 + H.c.
d
k
2N kσ
+ †
Kf0 f0 + εd c† cd + U nd↑ nd↓ .
d (1.5)
Note that
H = Ho + He , (1.6)
which satisfies the property
[Ho , He ] = 0. (1.7)

3. 2. Linear response theory
We consider the Hamiltonian splitted as
H =H+H (2.1)
and the time evolution of an eigenstate |Φn in the interaction picture
t
i
|Φn = 1− H (τ ) dτ |Ψn , (2.2)
−∞
where |Ψn is an eigenstate of the Hamiltonian H. The Hamiltonian H is treated as a
perturbation. So the current J through the system is determined by the thermal average
of the current operator IL (t) over the states |Ψn . Therefore, we have
t
i
J =− IL (t) , H (τ ) dτ. (2.3)
−∞
The time evolution for the operators is
H H
IL (t) = exp i t IL exp −i t (2.4)
H H
H (t) = exp i t H exp −i t . (2.5)
The perturbative Hamiltonian is
H = −∆µ c† cLk − c† cRk = −∆µ
Lk Rk c† cok + c† cek .
ek ok (2.6)
k k

4. 3. The Conductance
We want to derive the conductance expressed in the form
J 2e2 ∂fF (ε)
G= = dεT (ε, β) − (3.1)
∆φ h ∂ε
where T (ε, β) is the transmission coefficient. Let us first calculate the current
ie
IL = − c† cLk , H
Lk (3.2)
k
ie t0
IL = − √ c† d − d† cLk
Lk (3.3)
2N k
ie t0
IL = − √ √ c† d − d† cek + c† d − d† cok
ek ok . (3.4)
2 2N k
The substitution of Eqs.(2.4), (2.5), (2.6) and (3.4) in Eq.(2.3) leads to
√ e +∞
J = − 2 ∆µ dτ F (t − τ ) (3.5)
−∞
with the correlator
i t
F (t − τ ) = − θ (t − τ ) √0 c† (t) d (t) , c† (τ ) coq (τ )
ok eq . (3.6)
kq
2N
The integral of Eq.(3.5) results in
+∞ +∞
i
dτ F (t − τ ) = − Z−1 exp (−βEn ) θ (t − τ ) dτ
−∞ kq mn −∞
i t0 †
× exp − (En − Em ) (τ − t) Ψn | √ cok d |Ψm
2N
× Ψm | c† coq |Ψn
eq
+∞
i
− − Z−1 exp (−βEn ) θ (t − τ ) dτ
kq mn −∞
i
× exp − (Em − En ) (τ − t) Ψn | c† coq |Ψm
eq
t0 †
× Ψm | √ cok d |Ψn . (3.7)
2N
Now we exchange n ↔ m in the second term of (3.7) and introduce an infinitesimal
parameter η in the exponentials,
+∞
(exp (−βEn ) − exp (−βEm ))
dτ F (t − τ ) = limη→0+ Z−1
−∞ En − Em + iη
kq mn
t0 †
× Ψn | √ cok d |Ψm Ψm | c† coq |Ψn .
eq (3.8)
2N

5. We see that the term Ψm | c† coq |Ψn mixtures the conduction bands.
eq
We eliminate such mixture calculating
Ψm | c† coq , H |Ψn
eq = Ψm | c† coq H − Hc† coq |Ψn
eq e
= (En − Em ) Ψm | c† coq |Ψn ,
eq (3.9)
1 √ t0
Ψm | c† coq |Ψn
eq = − 2√ Ψm | d† coq |Ψn .
(En − Em ) 2N
(3.10)
The combination of Eq.(3.10) with the limit
exp (−βEn ) − exp (−βEm )
limEm →En = −β exp (−βEn ) (3.11)
En − Em
in Eq.(3.8) leads to
e t
J = −2 π∆µβZ−1 exp (−βEn ) δ (En − Em ) Ψn | √ 0 c† d |Ψm
ok
mn k
2N
t
× Ψm | √ 0 d† coq |Ψn . (3.12)
q 2N
The property of Eq.(1.7) allows to write
|Ψm = |r |s (3.13)
Z−1 exp (−βEn ) δ (En − Em ) = Z−1 Z−1 exp (−βEr ) exp (−βEs )
e o
e o
e o e o
× δ (Er + Es − Er − Es )
e o e e o o
= exp (−βEr ) exp (−βEs ) δ (ε + Er − Er ) δ (ε + Es − Es ) dε
1 o o e e
= [exp (−βEs ) + exp (−βEs )] [exp (−βEr ) + exp (−βEr )]
β
∂fF (ε) e e o o
× − δ (ε + Er − Er ) δ (ε + Es − Es ) dε, (3.14)
∂ε
where |r and |s are even and odd eigenstates, respectively. Applying them on Eq.(3.12),
we finally derive the conductance
2e2 d|d† ε ∂fF (ε)
G= πΓ dε − − (3.15)
h π ∂ε
2
Nq
where Γ = π t
√0
N
1
ρ and ρ = − π kq c† |coq
ok . Using t
√0
N
= πρ , we
ε=εF
recover Eq.(5) of the manuscript.