Electron Transport through a Molecular Conductor with Center-of-Mass Motion

The linear conductance of a molecular conductor oscillating between two metallic leads is investigated numerically both for Hubbard interacting and noninteracting electrons. The molecule-leads tunneling barriers depend on the molecule displacement from its equilibrium position. The results present an interesting interference which leads to a conductance dip at the electron-hole symmetry point that could be experimentally observable. It is shown that this dip is caused by the destructive interference between the purely electronic and phonon-assisted tunneling channels, which are found to carry opposite phases. When an internal vibrational mode is also active, the electron-hole symmetry is broken but a Fano-like interference is still observed.

Figure 1

A schematic of the system studied in this Letter. The molecule can oscillate between the two leads, thus modulating the tunneling barriers.

Figure 2

(a) NRG and (b) $\mathrm{ED}+\mathrm{DE}$ results for $G$ as a function of $V_g$ in the Kondo regime for $\alpha =0$ and increasing $\lambda$ ($\lambda /{\omega }_0=0.0$, 0.4, 0.8, and 1.2).

Figure 3

(a) $G$ vs $V_g$ in the Kondo regime for $\alpha =0$ (solid line) and $\alpha =0.4$ (dashed line), $\lambda =0.2$ in both cases. In the first case, the usual Kondo peak is obtained. In the second case, a conductance dip is obtained. (b) Average occupation for the same set of parameters. Note that the charging behavior is almost the same in both cases. (c) $G$ vs $V_g$ in the absence of Coulomb repulsion ($U=0$), for $\alpha =0$ (solid line) and $\alpha =0.4$ (dashed line), $\lambda =0.1$ in both cases. The dip is also obtained in this case; thus the physical mechanism behind this effect does not depend on the electron-electron interaction.

Figure 4

$G$ as a function of $V_g$ in the Kondo regime for $\alpha =0.4$ and $\lambda =0.15$, 0.20, and 0.25 (dashed, dotted, and solid lines, respectively). The dip becomes more pronounced as $\lambda$ increases. The inset shows the convergence of the results with the maximum number of phonons ($N_{\mathrm{ph}}$). Note that the qualitative effect of the conductance cancellation is preserved all the way down to $N_{\mathrm{ph}}=1$.

Figure 5

(a) A schematic representation of the two conductance channels, the purely electronic tunneling represented by the two upper arrows and the “phonon-assisted tunneling” channel represented by the two lower arrows. (b) Partial conductance when only one of the channels is active. The dotted line shows the conductance of the first channel $G_{\mathrm{E}}$, while the dashed line shows the conductance of the second channel $G_{\mathrm{I}}$. (c) $G$ when both channels are active. (d) Phase difference $\Delta \Phi$ of the two channels. Note that for $V_g=-U/2$, $\Delta \Phi =\pi$ and $G_{\mathrm{E}}=G_{\mathrm{I}}=2e^2/h$, thus leading to a perfect cancellation in the overall conductance. ($\lambda =0.2$, $\alpha =0.4$.)

Figure 6

The solid line shows $G$ when a breathing vibrational mode is active (no c.m. motion). The particle-hole symmetry is broken as expected and no conductance dip is obtained. The dashed line shows $G$ when both breathing and c.m. vibrational modes are active. The combined effects of the two modes lead to a Fano-like interference. The breathing mode parameters used are ${\lambda }^{\prime }=0.2$, ${\omega }_0^{\prime }=0.3$, and ${\alpha }^{\prime }=0.3$.

Figure 7

(a) Convergence of $G$ with the size of the exactly solved cluster $L$. The solid line, dashed line, and the circles show the results obtained using $L=3$, 7, and 11, respectively. In the three cases, $N_{\mathrm{ph}}=3$ was used. (b) The real and (c) imaginary parts of the isolated cluster Green function $g_{\mathrm{lr}}$ that propagates an electron from the left to the right ends of the cluster ($L=3$ and $V_g=-U/2$). Note that both parts are equal to zero at the Fermi level (located at $\omega =0.0$). Thus, the origin of the conductance dip can be traced back to the exactly solved cluster. $\lambda =0.2$ and $\alpha =0.4$ in the three figures.