Conductance in terms of cross section

We consider scattering of a single-quasiparticle wavefunction from the general 2-terminal system described in the Introduction (see Fig. 7.1a). The Hamiltonian is ${\mathcal{H}} = -(\hbar^2/2m) \nabla^2 + V{({\mathbf r})}$, for a quasiparticle mass $m$. The elastic scattering potential $V{({\mathbf r})}$ completely defines the system. We imagine a monochromatic unit plane wave $\psi_{{\mbox{\tiny I}}} = e^{i{\bf k}\cdot {\bf r}}$ incident from the free-space left-hand region^7.2. The wavevector is ${\bf k}\equiv (k,\phi)$ in polar coordinates, $\phi$ being the angle of incidence. The free-space wavevector magnitude is taken as $k = k_{{\mbox{\tiny F}}}$ (corresponding to a total energy $E = \hbar^2 k^2 / 2m$ equal to the Fermi energy), unless stated otherwise.

We are at liberty to choose our definition of the `unscattered' wave $\psi_{{\mbox{\tiny0}}}$. We take it to be the wavefunction which would result from reflection of the incident wave off a wall uniform in the $y$ direction. We can imagine creating such a wall by replacing the `system box' shown in Fig. 7.1a by the surrounding $y$-invariant wall profile. Note that $\psi_{{\mbox{\tiny0}}}$ exists only on the left side. In the left free-space region it is

$$\psi_{{\mbox{\tiny0}}} \; = \; e^{i(k_x x + k_y y)} - e^{i(-k_x x + k_y y + \gamma_{\bf k})}$$ (7.1)

where the first term is $\psi_{{\mbox{\tiny I}}}$, and the angle-dependent reflection phase $\gamma_{\bf k}$ of the second term depends on both $(k,\phi)$ and the wall profile^7.3. Upon introduction of our true system potential, the full wavefunction becomes

$$\psi \; \equiv \; \psi_{{\mbox{\tiny0}}} + \psi_{{\mbox{\tiny R}}} + \psi_{{\mbox{\tiny T}}} ,$$ (7.2)

where the change in reflected wave $\psi_{{\mbox{\tiny R}}}$ exists only on the left side, and the new transmitted wave $\psi_{{\mbox{\tiny T}}}$ exists only on the right. These scattered waves have the asymptotic ($r > L$ and $kr \gg 1$) forms of 2D scattering theory (see [128], and Chapter 7 of [174]) ^7.4,

$$\psi_{{\mbox{\tiny R}}} = f_{{\mbox{\tiny R}}}(\theta)\fra... ...T}}} = f_{{\mbox{\tiny T}}}(\theta')\frac{e^{ikr}}{\sqrt{r}}.$$ (7.3)

See Fig. 7.1a for definitions of $\theta$ and $\theta'$.

The transmission cross section $\sigma_{{\mbox{\tiny T}}}(k,\phi)$ is the ratio of $\Gamma_{{\mbox{\tiny T}}}$, the transmitted particle flux (number per unit time), to $j_{{\mbox{\tiny I}}}$, the incident particle flux per unit length normal to the incident beam:

$$\sigma_{{\mbox{\tiny T}}}(k,\phi) \; \equiv \; \frac{\Gamma_{{\mbox{\tiny T}}}}{j_{{\mbox{\tiny I}}}}.$$ (7.4)

Physically, $\sigma_{{\mbox{\tiny T}}}(k,\phi)$ is the length required of an aperture oriented normal to the incident beam in order to transmit an equivalent flux of classical particles. (Note that $\sigma_{{\mbox{\tiny T}}}(k,\phi)$ is proportional to the injection distribution [20] which can be measured in mesoscopic systems [181]). It depends on the incident angle because $V{({\mathbf r})}$ has no radial symmetry. $j_{{\mbox{\tiny I}}}$ is the magnitude of the incoming probability flux density vector ${\bf j} \equiv (\hbar/m) {\mbox{\tiny Im}} [ \psi_{{\mbox{\tiny I}}}^* \mbox{\boldmath$\nabla$}\psi_{{\mbox{\tiny I}}} ]$, which for a unit wave gives $j_{{\mbox{\tiny I}}} = v$, the particle speed. The transmitted flux is defined as

$$\Gamma_{{\mbox{\tiny T}}}\; \equiv \; \int \! dl \: \hat{\b... ...iny T}}}^* \mbox{\boldmath$\nabla$}\psi_{{\mbox{\tiny T}}} ] ,$$ (7.5)

where the line integral encloses the entire transmitted wave, and the (rightwards-pointing) surface normal is $\hat{\bf n}$. Applying this and (7.4) to the asymptotic form gives

$$\sigma_{{\mbox{\tiny T}}}(k,\phi) \; = \; \int_{-\pi/2}^{\pi/2} d\theta' \, \vert f_{{\mbox{\tiny T}}}(\theta')\vert^2 ,$$ (7.6)

familiar from scattering theory apart from the restriction to the right half-plane. There is a corresponding form

$$\sigma_{{\mbox{\tiny R}}}(k,\phi) \; = \; \int_{-\pi/2}^{\pi/2} d\theta \, \vert f_{{\mbox{\tiny R}}}(\theta)\vert^2 ,$$ (7.7)

for the reflective cross section (removal from the unscattered wave without being transmitted).

We will calculate the conductance by assuming the chemical potential is slightly higher on the left side than the right, and as is usual[20,55] consider only the left-to-right transport of the states in this narrow energy range. We take the left region to be a large ($\gg l_\phi$) closed region of area $A$ containing single-particle states, and find their decay rate through the QPC into the right side. Semiclassically each single-particle state occupies a phase-space volume $h^d$, where we have $d=2$. Therefore the phase-space density in the 2DEG Fermi sea is $2/h^2$ where the factor of 2 comes from the spin degeneracy. We can project this density onto momentum space in order to find the effective number of plane-wave states impinging on the wall^7.5: this corresponds to a uniform density of states in ${\bf k}$-space given by

$$\rho(k,\phi)\, k dk \, d\phi \; = \; \frac{A}{2 \pi^2} \, k dk \, d\phi .$$ (7.8)

Each state has an amplitude $A^{-1/2}$ due to the requirement of unity area normalisation in the left region, so has incoming flux density $j_{{\mbox{\tiny I}}} = v/A$. Substituting this into (7.4) gives the decay rate of a state $i$ as

$$\Gamma_{{\mbox{\tiny T}}}^{(i)} \; = \; \frac{v}{A} \sigma_{{\mbox{\tiny T}}}(k_i, \phi_i) .$$ (7.9)

We can now sum the decay rates of all the left-hand states in a given wavevector range $k_{{\mbox{\tiny F}}}$ to $k_{{\mbox{\tiny F}}}+ \delta k$, to get the current

$$\delta I = e \sum_{i} \Gamma_{{\mbox{\tiny T}}}^{(i)} = \frac{ev}{A} \int_{-\pi/2}^{\pi/2} \!\! d\phi \int_{k_{{\mbox{\ti... ...}}+ \delta k} \!\!\!\! kdk \, \rho(k,\phi) \, \sigma_{{\mbox{\tiny T}}}(k,\phi)$$

$$= \frac{ev \, k_{{\mbox{\tiny F}}}\delta k}{2\pi^2} \int_{-\pi/2}^{\pi/2} \! \! d\phi \, \sigma_{{\mbox{\tiny T}}}(k_{{\mbox{\tiny F}}},\phi) ,$$ (7.10)

where the last step incorporated the linear-response assumption that $\sigma_{{\mbox{\tiny T}}}$ is constant over the range $\delta k$.

When a potential difference $\delta V$ is applied across the QPC, the energy range carrying current is $\delta E = e \, \delta V$, which we can equate with $\hbar v \, \delta k$ using the dispersion relation. This can be used with (7.10) to write the conductance

$$G \; \equiv \; \frac{\delta I}{\delta V} = \frac{2e^2}{h} \cdot \frac{1}{\lambda_{{\mbox{\tiny F}}}} \int_{-\pi/2}^{\pi/2} \! \! d\phi \, \sigma_{{\mbox{\tiny T}}}(k_{{\mbox{\tiny F}}},\phi)$$ (7.11)

$$= \frac{2e^2}{h} \cdot \frac{k_{{\mbox{\tiny F}}}}{2} \langle \sigma_{{\mbox{\tiny T}}}\rangle_\phi ,$$ (7.12)

where the particle wavelength is $\lambda_{{\mbox{\tiny F}}}\equiv 2\pi/k_{{\mbox{\tiny F}}}$. The latter form is written in terms of the angle-averaged cross section at the Fermi energy. The weighting of this average is uniform because of the ergodic assumption that incoming states are uniformly distributed in angle.

Eq.(7.1) is a key result of this chapter. Like the Landauer formula, it directly connects conductance and scattering. In a scattering measurement from the left side, $\sigma_{{\mbox{\tiny T}}}$ appears to be the QPC's inelastic cross section (since the transmitted waves never return to this side). In a current measurement the corresponding conductance is given by (7.1). An independent verification is provided by the result (L.4) of Appendix L, when combined with the Landauer formula Eq.(7.14). Our derivation was for temperature $T=0$, but it applies at a finite $T$ as long as $\sigma_{{\mbox{\tiny T}}}$ does not change significantly over the energy range $k_{{\mbox{\tiny B}}} T$. This can be seen by generalizing the above to include integration over the Fermi distribution.

In the limit where a QPC is adiabatic, its conductance is known to be quantized [193,20,65]: $G = (2e^2/h)N$ where $N$ is the integer number of open channels at the Fermi energy. Looking at (7.11), this corresponds to quantization of the angular integral of the cross section in units of $\lambda_{{\mbox{\tiny F}}}$.