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4 Coulomb blockade

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4.5 Cotunneling 89<br />

In the metallic dots we can assume that the distribution over internal<br />

single-particle states is equilibrium, and obtain<br />

Γ (2)n<br />

in<br />

2π<br />

=<br />

¯h<br />

<br />

k∈L,q∈R αβ<br />

<br />

1<br />

Eβ + ∆E + +<br />

n − Ek<br />

1<br />

Eq − ∆E + n−1 − Eα<br />

2<br />

|Vαq| 2 |Vβk| 2<br />

× f 0 L(Ek) 1 − f 0 R(Eq) 1 − f 0 S(Eβ) f 0 S(Eα)δ(Ek + Eα − Eq − Eβ). (4.67)<br />

The two terms in this expression correspond to two physical processes. The<br />

first one describes the jump (virtual!) of the extra electron from the state k in<br />

the left lead into the state β in the system, and then the other electron jumps<br />

through the second junction from the state α into the state q in the right<br />

lead (Fig. 4.13b). In the second process (Fig. 4.13d) first the electron jumps<br />

from the system into the right lead, and then the second electron jumps from<br />

the left lead. All other matrix elements of the tunneling Hamiltonians in the<br />

expression (4.63) are zero. The distribution functions describe as usually the<br />

probabilities of necessary occupied or empty states.<br />

In the wide-band limit with energy independent matrix elements we arrive<br />

at<br />

Γ (2)n<br />

in<br />

=¯hGLGR<br />

2πe4 <br />

dEk dEq dEα dEβf 0 L(Ek) 1 − f 0 S(Eβ) f 0 S(Eα) 1 − f 0 R(Eq) <br />

<br />

1<br />

×<br />

Eβ + ∆E + 1<br />

+<br />

n − Ek Eq − ∆E + 2 δ(Ek + Eα − Eq − Eβ).<br />

n−1 − Eα<br />

(4.68)<br />

At zero temperature the integrals in this expression can be performed<br />

analytically [31, 24].<br />

The current is determined by the balance between forward and backward<br />

tunneling rates<br />

Jin = ¯hGLGR<br />

12πe 3<br />

<br />

J(V )=e<br />

∆E + n<br />

Γ (2)n<br />

in<br />

∆E + n−1<br />

(V ) − Γ (2)n<br />

in<br />

<br />

(−V ) . (4.69)<br />

At finite temperatures and small voltage V ≪ ∆E + n ,∆E + n−1 , the analytical<br />

solution for the current is<br />

<br />

1<br />

−<br />

1<br />

2 (eV 2 2<br />

) +(2πT) V. (4.70)<br />

The linear conductance can be estimated as<br />

<br />

T<br />

2 . (4.71)<br />

G ∼ h<br />

GLGR<br />

e2 EC

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