29.01.2013 Views

Heiss W.D. (ed.) Quantum dots.. a doorway to - tiera.ru

Heiss W.D. (ed.) Quantum dots.. a doorway to - tiera.ru

Heiss W.D. (ed.) Quantum dots.. a doorway to - tiera.ru

SHOW MORE
SHOW LESS

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

146 C.W.J. Beenakker<br />

Fig. 6. Ensemble averag<strong>ed</strong> density of states of a chaotic billiard coupl<strong>ed</strong> by a<br />

point contact <strong>to</strong> a superconduc<strong>to</strong>r, for several values of the transmission probability<br />

through the point contact. The energy is in units of the Thouless energy<br />

ET = NΓδ/4π. The solid curves are comput<strong>ed</strong> from (43) and (44), for modeindependent<br />

transmission probabilities Γ =1,0.5, 0.25, 0.1. The dash<strong>ed</strong> curve is<br />

the asymp<strong>to</strong>tic result (51) for Γ ≪ 1 Adapt<strong>ed</strong> from [22] (The definition of δ us<strong>ed</strong> in<br />

that paper differs from the one us<strong>ed</strong> here by a fac<strong>to</strong>r of two)<br />

When E ≫ Egap the density of states approaches the value 2/δ from above,<br />

twice the value in the isolat<strong>ed</strong> billiard. The doubling of the density of states<br />

occurs because electron and hole excitations are combin<strong>ed</strong> in the excitation<br />

spect<strong>ru</strong>m of the Andreev billiard, while in an isolat<strong>ed</strong> billiard electron and<br />

hole excitations are consider<strong>ed</strong> separately.<br />

A rather simple clos<strong>ed</strong>-form expression for 〈ρ(E)〉 exists in two limiting<br />

cases [22]. In the case Γ = 1 of a ballistic point contact one has<br />

〈ρ(E)〉 = ET<br />

√<br />

3<br />

�<br />

�<br />

Q+(E/ET ) − Q−(E/ET ) , (47)<br />

3Eδ<br />

�<br />

Q±(x) = 8 − 36x 2 ± 3x � 3x4 + 132x2 �1/3 − 48 , (48)<br />

E>Egap =2γ 5/2 ET =0.60 ET =<br />

0.30 �<br />

=0.048 Nδ ,<br />

τdwell<br />

(49)<br />

where γ = 1<br />

2 (√ 5 − 1) is the golden number. In this case the parameter ∆gap<br />

in (46) is given by<br />

∆gap = [(5 − 2 √ 5)δ 2 Egap/8π 2 ] 1/3 =0.068 N 1/3 δ. (50)<br />

In the opposite tunneling limit Γ ≪ 1 one finds

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!