Coulomb blockade and non-Fermi-liquid behavior in quantum dots

The non-Fermi-liquid properties of an ultrasmall quantum dot coupled to a lead and to a quantum box are investigated. Tuning the ratio of the tunneling amplitudes to the lead and box, we find a line of two-channel Kondo fixed points for arbitrary Coulomb repulsion on the dot, governing the transition between two distinct Fermi-liquid regimes. The Fermi liquids are characterized by different values of the conductance. For an asymmetric dot, spin and charge degrees of freedom are entangled: a continuous transition from a spin to a charge two-channel Kondo effect evolves. The crossover temperature to the two-channel Kondo effect is greatly enhanced away from the local-moment regime, making this exotic effect accessible in realistic quantum-dot devices.

Figure 1

The magnetic susceptibility of the dot vs $T$, for ${\Gamma }_L=E_C=0.1D$, $N_B=0$, and different values of $U=-2{ϵ}_d$. Here ${\Gamma }_B$ is tuned for each value of $U$ to the two-channel point ${\Gamma }_B^{2\mathrm{CK}}$. The Kondo temperature $T_K$ is defined in Eq. (2). For all $U$, ranging from the local-moment regime $U⪢{\Gamma }_L$, to the strongly mixed-valent regime $U=0$, to negative $U$ (not shown), there exists a two-channel point ${\Gamma }_B^{2\mathrm{CK}}$ where $\chi \left(T\right)$ diverges logarithmically as $T\to 0$, and the finite-size spectrum converges to the conventional two-channel fixed point. Inset: $T_K$ vs $U$. For $U⪢{\Gamma }_L$, $E_C$ (local-moment regime), $T_K$ decreases exponentially with $U$. For $U<{\Gamma }_L$ (mixed-valent regime), it exceeds $E_C$.

Figure 2

$1∕T_K$ vs $N_B$ for $U={ϵ}_d=0$ and ${\Gamma }_L∕D=0.1$. Here $E_C∕D$ equals 0.1 (circles) and 0.01 (pluses). For each combined value of $N_B$ and $E_C$, the coupling ${\Gamma }_B$ is tuned to the two-channel point ${\Gamma }_B^{2\mathrm{CK}}$. As a function of $N_B$, ${\Gamma }_B^{2\mathrm{CK}}\left(N_B\right)$ varies by less than 0.2% for each fixed value of $E_C$. The ratio $T_K\left(0\right)∕T_K\left(N_B\right)$ is well described by ${\cos }^2\left(\pi N_B\right)$ (solid line). Inset: The box charge $⟨\widehat{N}⟩$ vs $N_B$ for ${\Gamma }_L=E_C=0.1D$ and $U={ϵ}_d=0$. Here ${\Gamma }_B∕{\Gamma }_L$ equals 1 (squares), 1.72 (filled $\text{circles}+\text{solid}$ line), and 3.24 (crosses). The Coulomb staircase is completely washed out for ${\Gamma }_B={\Gamma }_B^{2\mathrm{CK}}\left(0\right)=1.72{\Gamma }_L$, but is gradually recovered upon departure from ${\Gamma }_B^{2\mathrm{CK}}\left(0\right)$.

Figure 3

(a) The two-channel line ${\Gamma }_B^{2\mathrm{CK}}∕{\Gamma }_L$ vs ${ϵ}_d$ for $U∕D=2$, ${\Gamma }_L=E_C=0.1D$, and $N_B=0$. Dashed line: The local-moment estimate of Ref. 5. (b) The charge curve $⟨\widehat{N}⟩$ vs $N_B$, for ${ϵ}_d∕D=-1$, $-1.581$, $-1.733$, $-1.843$, and $-1.935$ (solid, dashed, dotted, dot-dashed, and long-dashed line, respectively). For each value of ${ϵ}_d$, ${\Gamma }_B$ is tuned to the corresponding $N_B=0$ value of ${\Gamma }_B^{2\mathrm{CK}}$, depicted in panel (a).

Figure 4

(a) The capacitance of the box and (b) the magnetic susceptibility of the dot, for $U∕D=2$, ${\Gamma }_L=E_C=0.1D$, $N_B=0$, and different values of ${ϵ}_d∕D$, as specified in the legends. For each ${ϵ}_d$, ${\Gamma }_B$ is tuned to the corresponding $N_B=0$ value of ${\Gamma }_B^{2\mathrm{CK}}$. Both $\chi \left(T\right)$ and $C\left(T\right)$ diverge logarithmically as $T\to 0$, but with different Kondo scales $T_K^{\mathrm{sp}}$ and $T_K^{\mathrm{ch}}$, extracted from the slopes of their $\ln \left(T\right)$ diverging terms. In going from ${ϵ}_d∕D=-1.581\text{to}-1.935$, $k_B{T}_K^{\mathrm{sp}}∕D$ takes the values 0.0022, 0.0054, 0.0457, 0.187, and 1.72, while $k_B{T}_K^{\mathrm{ch}}∕D$ equals 1.035, 0.36, 0.035, 0.009, and 0.00138, respectively.

Figure 5

The zero-temperature conductance of the two-lead device proposed in Ref. 5, for $U∕D=1$, $E_C∕D=0.1$, $N_B=0$, and different values of ${ϵ}_d$. The tunneling rates ${\Gamma }_l$ and ${\Gamma }_r$ are kept fixed in all curves, with ${\Gamma }_l+{\Gamma }_r=0.1D$. Here $G_{0}h∕2e^2$ equals $4{\Gamma }_l{\Gamma }_r∕{({\Gamma }_l+{\Gamma }_r)}^2$.