Lecture 12

03/30/2023, Th.

Today

Landauer formula
1D localization

Reading

Girvin & Yang ch. 10.2-10.4

1. Beyond Boltzmann: quantum effects

$\rho(B)$ $B$ increases
Cyclotron resonance
Localizaiton: weak, strong and many-body
$\sigma=e^2\frac{\partial n}{\partial \mu}D$
- Band insulator
- Mott insulator
- Anderson insulator: MIT by Anderson transition is a quantum phenomenon

Plan

Landauer formula (leads to 1d localization, and later QHE edges)
Diffusion, and weak localization
Anderson localization

2. Landauer formula

$L\times W$ strip, with two leads L and R. The 2DEG is metallic. and We assume the device is perfect: electrons can move in the trip free of scattering (elastic or inelastic) and without any decoherence (退相干). In other words,

\begin{matrix} (1) & L ≫ ℓ_{ϕ} \end{matrix}

$\ell_\phi$ is coherence length of electrons. The question: what is the device conductance?

The electron states in a confined electron gas

\begin{matrix} (2) & \begin{aligned} ψ_{n k} (x, y) = \sqrt{\frac{2}{L w}} \sin (\frac{n π y}{w}) e^{j k x} \\ ε_{n k} = ε_{k} + \underset{E_{n}}{\underset{⏟}{\frac{ℏ^{2}}{2 m} {(\frac{n π}{W})}^{2}}} \end{aligned} \end{matrix}

$k$ $n=1,2,3\cdots$ labels the subbands (子能带) due to confinement in the y-direction.

We also impose a voltage by having different chemical potentials, say

\begin{matrix} (3) & μ_{L} > μ_{R} \end{matrix}

in the two leads, so the voltage is given in the relation

\begin{matrix} (4) & - e V = μ_{L} - μ_{R} . \end{matrix}

$\mu_L$ ;
$\mu_R$ ;

We now compute the current

\begin{matrix} \begin{matrix} \begin{aligned} I & = - 2 e \sum_{n = 1}^{\infty} \int \frac{d k}{2 π} v_{n k} [f^{0} (ε_{n k} - μ_{L}) θ (k) + f^{0} (ε_{n k} - μ_{R}) θ (- k)] \\ = - 2 e \sum_{n = 1}^{\infty} \int_{0}^{\infty} \frac{d k}{2 π} \frac{\partial ε_{k}}{ℏ \partial k} [f^{0} (ε_{k} + E_{n} - μ_{L}) - f^{0} (ε_{k} + E_{n} - μ_{R})] \\ = - 2 \frac{e}{h} \sum_{n} \int_{0}^{\infty} d ε [f^{0} (ε + E_{n} - μ_{L}) - f^{0} (ε + E_{n} - μ_{R})] \\ [T = 0] & = - 2 \frac{e}{h} \sum_{n} \int_{E_{n}}^{\infty} d ε [θ (μ_{L} - ε) - θ (μ_{R} - ε)] \end{aligned} \end{matrix} \end{matrix}

The above integral is

$\mu<E_n$ $n$ th channel (subband) is closed.
$-2\frac{e}{h}(\mu_L-\mu_R)=2\frac{e^2}{h}V$

$G$

\begin{matrix} (5) & I = G V \end{matrix}

$2e^2/h$ $G$ . And

\begin{matrix} (6) & G = N \frac{2 e^{2}}{h} \end{matrix}

where the number of open channels is

\begin{matrix} (7) & N = \sum_{n = 1}^{\infty} θ (μ - E_{n}) . \end{matrix}

$e^2/h$ $h/e^2$

\begin{matrix} (8) & \frac{e^{2}}{h} \approx \frac{1}{25812.807 Ω} \end{matrix}

The derivation above is independent of the details of the band structure. So, so long as the wire is perfect, the conductance is determined by the number of open channels, without regard to the shape of the bands.

The paradox: a perfect wire has finite conductance, and therefore finite resistance. This means there must be dissipation.
$\mu_L/\mu_R$ .
The electron from the left reservoir must dissipate energy in the right reservoir;
The device resistance (i.e. global) of the ideal wire is really the contact resistance.

Now we turn to an imperfect wire. Suppose there is one scatterer in the wire.

$r_n$ $t_n$ .
$\begin{matrix} (9) & | r_{n} |^{2} + | t_{n} |^{2} = 1 \end{matrix}$
ignore band hopping, for the moment.

Then

\begin{matrix} (10) & I = - \frac{2 e}{h} \sum_{n = 1}^{\infty} \int_{E_{n}}^{\infty} d ε [(1 - {| r_{n} |}^{2}) θ (μ_{L} - ε) - {| t_{n} |}^{2} θ (μ_{R} - ε)] \end{matrix}

$\mu_R<\varepsilon<\mu_L$ $1-|r_n|^2=|t_n|^2$ . So

\begin{matrix} (11) & G = T \frac{2 e^{2}}{h} \end{matrix}

where

\begin{matrix} (12) & T = \sum_{n = 1}^{\infty} | t_{n} |^{2} θ (μ_{R} - E_{n}) = N \overset{―}{T} \end{matrix}

$T$ : total transmission probability.
$\overline T$ : average transmission probability of an open channel.
$N$ : number of open channels.

$n$ $n'$ th channel, with the amplitude

\begin{matrix} (13) & t_{n^{'} n} \end{matrix}

$[t_{n'n}]$ is called the transmission matrix.

3. Multiterminal device

$\pi$ $\mu_i$ , controlled by external circuits.

$j$ is

\begin{matrix} (14) & \begin{matrix} I_{j} = - \frac{2 e}{h} \sum_{k} (T_{j k} μ_{k} - T_{k j} μ_{j}) \\ T_{j k} = \sum_{n, n^{'} = 1}^{N} {| t_{j k}^{n n^{'}} |}^{2} = Tr (t_{j k}^{†} t_{j k}) \end{matrix} \end{matrix}

$j,k$ $n,n'$ $T$ is a real and symmetric matrix.

Suppose all reservoirs have the same chemical potentials, then all currents vanish. We have sum rule (求和规则)

\begin{matrix} (15) & \sum_{k} T_{j k} = \sum_{k} T_{k j} \end{matrix}

Then

\begin{matrix} (16) & I_{j} = - \frac{2 e}{h} \sum_{k} T_{j k} (μ_{k} - μ_{j}) . \end{matrix}

If a terminal has no net current in or out, it is called a voltage probe. To make a voltage probe, we need

\begin{matrix} (17) & μ_{j} = \frac{\sum_{k \neq j} T_{j k} μ_{k}}{\sum_{k \neq j} T_{j k}} \end{matrix}

Now for the four-terminal device, if we float terminals 2 and 3, and there is a current flowing between terminal 1 and 4, this is the so-called four-terminal measurement. Then the voltage between 2 and 3 is

\begin{matrix} (18) & - e V_{23} = μ_{2} - μ_{3} . \end{matrix}

$V_{23}$ $V_{23}$ can have a non-trivial dependence on the global geometry.

More generally speaking, adding contacts to the system, even in the form of voltage probes (which supposedly do not affect the circuit classically), inevitably changes the system and results of measurements involving other contacts. This is yet another example of the fact that in a quantum world any measurement inevitably affects the objects being measured.

4. Two scatterers

Suppose there are two scatterers in a two-terminal device.

like Fabry-Perot cavity
$T=|D|^2$

Let's do the accounting

\begin{matrix} (19) & \begin{aligned} B & = t_{1} + C r_{1}^{'}, \\ C e^{- i ϕ} & = B e^{i ϕ} r_{2}, \\ D & = B e^{i ϕ} t_{2}, \end{aligned} \end{matrix}

where where the primed variables are the transmission and reflection amplitudes for right incidence (which have the same magnitude as their left-incidence counterparts but possibly different phases). The phase picked up by the traveling wave is

\begin{matrix} (20) & ϕ = k_{F} L_{12} \end{matrix}

Solving, we find

\begin{matrix} (21) & D = \frac{t_{1} t_{2} e^{i ϕ}}{1 - r_{2} r_{1}^{'} e^{2 i ϕ}} \end{matrix}

\begin{matrix} (22) & T = \frac{T_{1} T_{2}}{1 + R_{1} R_{2} - 2 \sqrt{R_{1} R_{2}} \cos (2 ϕ + \arg (r_{2} r_{1}^{'}))} \end{matrix}

$T_1,T_2$ $L_{12}$ $T$ $G=Te^2/h$ $e^2/h$ .

This huge change depending on the separation between scatterers is a result of interference of waves from multiple reflection in the cavity.

$G$ is somewhat nonlocal, and can fluctuate strongly, with multiple disordered impurities in a wire. This is called universal conductance fluctuation (UCF, 普适电导涨落). 10.4 in Girvin and Yang has detailed explanations.

To better understand the interference phenomenon, let us consider the dimensionless resistance

\begin{matrix} (23) & Z = \frac{R}{T} = \frac{1}{T} - 1 \end{matrix}

$h/2e^2T$ $h/2e^2$ $Zh/2e^2$ is the resistance of the wire.

For the two-impurity case

\begin{matrix} (24) & Z = \frac{R}{T} = \frac{R_{1} + R_{2} - 2 \sqrt{R_{1} R_{2}} \cos (2 ϕ + \arg (r_{2} r_{1}^{'}))}{T_{1} T_{2}} \end{matrix}

We find

\begin{matrix} (25) & Z \neq Z_{1} + Z_{2} = \frac{R_{1}}{T_{1}} + \frac{R_{2}}{T_{2}} \end{matrix}

Quantum resistances connected in series (串联) do not add up!

The average resistance,

\begin{matrix} (26) & ⟨ Z ⟩ = \frac{R_{1} + R_{2}}{T_{1} T_{2}} > Z_{1} + Z_{2} \end{matrix}

on average, the resistance of two impurities is bigger than the sum of the resistances of individual impurities!

5. Many scatterers：1D localization

Landauer 1970: a sequence of nearly transparent (透明) scatterers,

\begin{matrix} (27) & T \approx 1, R ≪ 1 \end{matrix}

\begin{matrix} (28) & ⟨ Z_{n + 1} ⟩ \approx \frac{R_{n} + R}{T_{n}} = ⟨ Z_{n} ⟩ + \frac{R}{T_{n}} \end{matrix}

$R$ $n$ . So

\begin{matrix} (29) & \frac{d Z}{d n} = R (Z + 1) \end{matrix}

$Z$ $n$ , or the wire length if the impurity density is fixed. Integrating, we find

\begin{matrix} (30) & Z + 1 \sim e^{R n} \end{matrix}

which points squarely to localization as the wire length increases.

There is a bug in the above analysis:

$Z$ is not exactly narrow, due to the universal conductance fluctuation discussed in the previous section. So statistically, it is not very good quantity to use.
$\sim n$ , i.e., extensive or additive.
$\begin{matrix} (31) & \log (Z + 1) = \log (1 / T) = R n \end{matrix}$
$(1/T)$ $\eqref{eq:lec12-T12}$ using the following identity

\begin{matrix} (32) & \begin{aligned} \int_{0}^{2 π} d θ \log (a - \cos θ) = 2 π \log \frac{a + \sqrt{a^{2} - 1}}{2}, \\ \Rightarrow ⟨ \log T_{12} ⟩ & = \log (T_{1} T_{2}) - ⟨ \log (1 + R_{1} R_{2} + 2 \sqrt{R_{1} R_{2}} \cos θ) ⟩ \\ = \log (T_{1} T_{2}) - \log \frac{1 + R_{1} R_{2} + \sqrt{(1 + R_{1} R_{2})^{2} - 4 R_{1} R_{2}}}{2} \\ = \log T_{1} + \log T_{2} \end{aligned} \end{matrix}

If the two scatterers are identical

\begin{matrix} (33) & ⟨ \log T_{12} ⟩ = 2 \log T . \end{matrix}

$\log T$ is indeed additive and therefore extensive, and we have

\begin{matrix} (34) & ⟨ \log T_{n} ⟩ = n \log T \end{matrix}

Therefore

\begin{matrix} (35) & ⟨ \log (1 + Z_{n}) ⟩ = - n \log T \approx n Z \end{matrix}

$Z=R/T\approx R$ $\eqref{eq:lec12-RG}$ predicts.

This is the 1D localization. That is, waves cannot propagate in a 1D media with random disorder, due to interreference from multiple reflections. It can also be rigorously proved, in the case of electrons in 1D. Mott and Twose (1961) showed that all eigenstates of 1D electron system get localized with disorder (at low temperatures of course in order for the quantum interference effects to be significant).