Chapter 13 Tunneling and Coulomb blockage

Chapter 13
Tunneling and Coulomb blockage

13.1 Tunneling

Modern technology allows to fabricate various structures involving tunneling barriers. One of the ways is a split-gate structure.

Fig 1. Split-gate structure allowing resonant tunneling.

Such a system can be considered as a specific example of series connection of to obstacles. The complex amplitude of the wave transmitted through the whole system is

$D=\frac{t_1t_2e^{i\phi}}{1-e^{2i\phi}r_2r_1^\prime} = \frac{t_1 t_2 e^{i\phi}}{1-e^{\theta}\sqrt{R_1R_2}} ,$ (1)

where $\theta = 2 \phi + {\rm arg}(r_2r_1^\prime)$ . It is clear that the transmittance

$T=\frac{T_1T_2}{1+R_1R_2 -2 \sqrt{R_1R_2}\cos \theta}$ (2)

is maximal at some specific value of $\theta$ where $\cos \theta =1$ , the maximal value being

$T_{\max}=\frac{T_1T_2}{(1-\sqrt{R_1R_2})^2} .$ (3)

This expression is specifically simple at T₁, T₂ << 1,

$T_{\max}=\frac{4T_1T_2}{(T_1 + T_2)^2} .$ (4)

Thus we observe that two low-transparent barriers in series can have a unit transmittance if they have the same partial transparencies, T₁ = T₂ = T. The reason of this fact in quantum interference in the region between the barriers which makes wave functions near the barriers very large to overcome low transmittance of each barrier.

An important point is that the phase $\theta$ gained in the system is a function of the electron energy. Thus near a particular value E^(r) defined by the equality

$\cos \theta (E^{(r)}) = 0 \rightarrow \theta (E^{(r)}_k) = 2\pi k$

one can expand $\cos \theta$ as

$1 - \frac{1}{2} \left(\frac{\partial \theta}{\partial E} \right)^2 \left(E-E^{(r)} \right)^2 .$

Thus at low transmittance we arrive at a very simple formula of a Breit-Wigner type,

T $\approx$ $\frac{T_1T_2}{(T_1 + T_2)^2/4 + (\theta')^2\left(E-E^{(r)} \right)^2 }$
= $\frac{\Gamma_1 \Gamma_2}{(\Gamma_1 + \Gamma_2)^2/4 +\left(E-E^{(r)} \right)^2} .$ (5)

Here we denote $\theta' \equiv (\partial \theta/\partial E)_{E-E^{(r)}}$ and introduce $\Gamma_i = T_i/|\theta'|$ .

The physical meaning of the quantities $\Gamma_i$ is transparent. Let us assume that all the phase shift is due to ballistic motion of an electron between the barriers. Then,

$\theta =2ka = 2ah^{-1}\sqrt{2mE} \rightarrow \theta' =\frac{a}{\hbar}\sqrt{\frac{2m}{E}}=\frac{2a}{\hbar v}$

where v is the electron velocity. As a result, the quantity $\Gamma_i$ can be rewritten as $\Gamma=\hbar \nu_a T_i ,$ where $\nu_a=v/2d$ is the frequency of oscillations inside the inter-barrier region, the so-called attempt frequency. Thus it is clear that $\Gamma_i$ are the escape rates through i-th barrier.

To specify the transition amplitudes let us consider a 1D model for a particle in a well between two barriers. We are interested in the scattering problem shown in Fig. 2.

Fig 2. On the resonant tunneling in a double-barrier structure.

To find a transmission probability one has to match the wave functions and their gradients at the interfaces 1-4 between the regions A-C. They have the following form

$\begin{array}{ll} e^{ikx}+re^{-ikx}& {\rm in the regionA};\\ a_1e^{\kappa_B x}+ a_2 e^{-\kappa_B x}& {\rm in theregion B;}\\ b_1 e^{ikx}+b_2e^{-ikx}& {\rm in the region C};\\ c_1e^{\kappa_D x}+ c_2 e^{-\kappa_D x}& {\rm in the region D;}\\ te^{ikx}& {\rm in the region E} \end{array} .$

Here

$k =\hbar^{-1}\sqrt{2mE} , \kappa_i =\sqrt{\kappa_{0i}^2-k^2} , \kappa_{0i} =\hbar^{-1}\sqrt{2mU_i} .$

The transmission amplitude is given by the quantity t while the reflection amplitude – by the quantity r. In fact we have 8 equations for 8 unknowns (r,t,a_i,b_i,c_i), so after some tedious algebra we can find everything. For a single barrier one would get

$T(E) =\frac{4k^2 \kappa^2}{\kappa_0^4 \sinh^2(\kappa d)+4 k^2 \kappa^2} \approx \frac{k^2 \kappa^2}{\kappa_0^4}e^{-2\kappa d} .$

Here d is the barrier's thickness. So the transparency exponentially decays with increase of the product $\kappa d$ . The calculations for a double-barrier structure is tedious, so we consider a simplified model of the potential

$U(x)= U_0d [\delta (x) +\delta (x-a)] .$

In this case we have 3 regions,

$\begin{array} {ll} e^{ikx} + re^{-ikx} & x<0 . \\ A \sin kx + B \cos kx & 0 <x < a , \\ t e^{ik(x-a)} & x>a \end{array}$ (6)

The matching conditions for the derivatives at the $\delta$ -functional barrier has the form

$\psi'(x_0+0) - \psi'(x_0-0) = \kappa^2 d \psi (x_0) .$ (7)

Here $\kappa^2 = 2mU_0/\hbar^2$ . One can prove it by integration of the Schrödinger equation

$(\hbar^2/2m)\nabla^2 \psi + U_0d \delta(x-x_0) \psi = E \psi$

around the point x₀. Thus we get the following matching conditions

B = 1+r ,
kA –ik(1–r) = $\kappa^2 a (1+r) ,$
A sin ka + B cos ka = t ,
ikt –k(Acos ka –B sin ka) = $t\kappa^2 a .$

First one can easily see that there is a solution with zero reflectance, r = 0. Substituting r = 0 we get the following requirement for the set of equation to be consistent

$k=k_0 , \tan k_0a = - \frac{2k_0}{\kappa^2 d} .$ (8)

We immediately observe that at that k |t| = 1 (total transmission). At strong enough barrier, $\kappa d \gg 1$ , this condition means

$k_0a = \pi (2s+1) , s= 0, \pm 1, ..$

Physically, that means that an electron gains the phase $2\pi s$ during its round trip (cf. with optical interferometer). Thus two barriers in series can have perfect transparency even if the transparency of a single barrier is exponentially small. The physical reason is quantum interference.

The condition (8) defines the energy

$E_0 =\frac{\hbar^2 k_0^2}{2m}$

where the transparency is maximal. Near the peak one can expand all the quantities in powers of

$k-k_0 \approx \frac{E-E_0}{(\partial E/ \partial k)_{k_0}} \approx k_0 \frac{E-E_0}{2E_0} .$

The result for a general case can be expressed in the Breit-Wigner form

$T(E) = \frac{\Gamma_L \Gamma_R}{(E-E_0)^2 + \frac{1}{4}(\Gamma_L +\Gamma_R)^2} .$

Here $\Gamma_{L(R)}/\hbar$ are the escape rates for the electron inside the well to the left(right) lead. They are given by the attempt frequency $v_0/2a = \hbar k_0/2ma$ times the transparency of a given barrier.

Of course, if voltage across the system is zero the total number of electrons passing along opposite directions is the same, and the current is absent. However, in a biased system we obtain the situation shown in Fig. 3.

Fig 3. Negative differential conductance in double-barrier resonant-tunneling structure.

Negative differential conductance, $dJ/dV \le 0$ , allows one to make a generator. One can also control the system moving the level E₀ with respect to the Fermi level by the gate voltage. In this way, one can make a transistor.

Commercial resonant tunneling transistors are more complicated than the principle scheme described above. A schematic diagram of a real device is shown in Fig. 4. In this device resonant tunneling takes place through localized states in the barrier.

Fig 4. Schematic diagram of a Si MOSFET with a split gate (a), which creates a potential barrier in the inversion layer (b). In the right panel oscillations in the conductance as a function of gate voltage at 0.5 K are shown. They are attributed to resonant tunneling through localized states in the barrier. A second trace is shown for a magnetic field of 6 T. From T. E. Kopley et al., Phys. Rev. Lett. 61, 1654 (1988).

There exist also transistors with two quantum wells where electrons pass through the resonant levels in two quantum wells from the emitter to collector if the levels are aligned. The condition of alignment is controlled by the collector-base voltage, while the number of electrons from emitter is controlled by the base-emitter voltage.

13.2 Coulomb blockade

Now let us discuss a specific role of Coulomb interaction in a mesoscopic system. Consider a system with a dot created by a split-gate system (see above).

If one transfers the charge Q from the source to the grain the change in the energy of the system is

$\Delta E = QV_G + \frac{Q^2}{2C} .$

Here the first item is the work by the source of the gate voltage while the second one is the energy of Coulomb repulsion at the grain. We describe it by the effective capacitance C to take into account polarization of the electrodes. The graph of this function is the parabola with the minimum at

Q = Q₀ = –CV_G ,

So it can be tuned by the gate voltage V_G. Now let us remember that the charge is transferred by the electrons with the charge –e. Then, the energy as a function of the number n of electrons at the grain is

$\Delta E (n)= -neV_G + \frac{n^2e^2}{2C} .$

Now let us estimate the difference

$\Delta E (n+1) - \Delta E (n)=-eV_G +n\frac{e^2}{C} .$

We observe that at certain values of V_G,

$V_{Gn} =n \frac{e}{C} ,$ (9)

the difference vanishes. It means that only at that values of the gate voltage resonant transfer is possible. Otherwise one has to pay for the transfer that means that only inelastic processes can contribute. As a result, at
$k_B T \le \frac{e^2}{2C}$

the linear conductance is exponentially small if the condition (9) is met. This phenomenon is called the Coulomb blockade of conductance.

As a result of the Coulomb blockade, electron tunnel one-by-one, and the conductance vs. gate voltage dependence is a set of sharp peaks. That fact allows one to create a so-called single-electron transistor (SET) which is now the most sensitive electrometer. Such a device (as was recently demonstrated) can work at room temperature provided the capacitance (size!) is small enough.

Coulomb blockade as a physical phenomenon has been predicted by Kulik and Shekhter [26]. There are very good reviews [13,14,15] about single-change effects which cover both principal and applied aspects. Below we shall review the simplest variant of the theory, so called ''orthodox model''.

A simple theory of single charge tunneling

For simplicity, let us ignore discrete character of energy spectrum of the grain and assume that its state is fully characterized by the number n of excess electrons with respect to an electrically neutral situation. To calculate the energy of the systems let us employ the equivalent circuit shown in Fig. 5.

Fig 5. Equivalent circuit for a single-electron transistor. The gate voltage, V_g, is coupled to the grain via the gate capacitance, C_g. The voltages V_e and V_c of emitter and collector are counted from the ground.

The left (emitter) and right (collector) tunnel junctions are modeled by partial resistances and capacitances.

The charge conservation requires that

–ne = Q_e +Q_c + Q_g
= C_e(V_e –U) +C_c(V_c –U)+C_g(V_g –U) , (10)

where U is the potential of the grain. The effective charge of the grain is hence

$Q=CU=ne +\sum_{i=e,c,g} C_i V_i , C\equiv \sum_i C_i .$

This charge consists of 4 contributions, the charge of excess electrons and the charges induced by the electrodes. Thus, the electrostatic energy of the grain is

$E_n=\frac{Q^2}{2C} = \frac{(ne)^2}{2C} + \frac{ne}{C}\sum_{i} C_i V_i + \frac{1}{2C}\left(\sum_{i} C_iV_i \right)^2 .$ (11)

The last item is not important because it is n-independent. In the stationary case, the currents through both junctions are the same. Here we shall concentrate on this case. In the non-stationary situation, an electric charge can be accumulated at the grain, and the currents are different. To organize a transport of one electron one has to transfer it first from emitter to grain and then from grain to collector. The energy cost for the first transition,

$E_{n+1} - E_n=\frac{(2n+1)e^2}{2C}+\frac{e}{C}\sum_{i} C_i V_i$ (12)

must be less than the voltage drop eV_e. In this way we come to the criterion

$E_n -E_{n + 1} + eV_e \ge 0 .$ (13)

In a similar way, to organize the transport from grain to collector we need

$E_{n+1} -E_{n} - eV_c \ge 0 .$ (14)

The inequalities (14) and (13) provide the relations between V_e,V_c and V_g to make the current possible. For simplicity let us consider a symmetric system, where

$G_e=G_c=G, C_e=C_c \approx C/2 (C_g \ll C), V_e=-V_c=V_b/2$

where V_b is bias voltage. Then we get the criterion,

$V_b \ge (2n+1)|e|/C - 2(C_g/C)V_g .$

We observe that there is a threshold voltage which is necessary to exceed to organize transport. This is a manifestation of Coulomb blockade. It is important that the threshold linearly depends on the gate voltage which makes it possible to create a transistor. Of course, the above considerations are applicable at zero temperature.

The current through the emitter-grain transition we get

$I =e \sum_n p_n \left[\Gamma_{e \rightarrow g} - \Gamma_{g \rightarrow e} \right] .$ (15)

Here p_n is the stationary probability to find n excess electrons at the grain. It can be determined from the balance equation,

$p_{n-1}\Gamma_{n-1}^n +p_{n+1}\Gamma_{n+1}^n- \left(\Gamma^{n-1}_n +\Gamma_{n+1}^n \right)p_n=0 .$ (16)

Here

$\Gamma_{n-1}^n$ = $\Gamma_{e \rightarrow g}(n-1) +\Gamma_{c \rightarrow g}(n-1) ;$ (17)
$\Gamma_{n+1}^n$ = $\Gamma_{g \rightarrow e}(n+1) +\Gamma_{g \rightarrow c}(n+1) .$ (18)

The proper tunneling rates can be calculated from the golden rule expressions using tunneling transmittance as perturbations. To do that, let us write down the Hamiltonian as

${\cal H}_0$ = ${\cal H}_e + {\cal H}_g + {\cal H}_{\rm ch} + {\cal H}_{\rm bath} ;$
${\cal H}_{e,c}$ = $\sum_{{\bf k} \sigma} \epsilon_{\bf k} c^{†}_{{\bf k} \sigma} c_{{\bf k} \sigma} ,$
${\cal H}_{g}$ = $\sum_{{\bf q} \sigma} \epsilon_{\bf q} c^{†}_{{\bf q} \sigma} c_{{\bf q} \sigma} ,$
${\cal H}_{\rm ch}$ = $({\hat n} - Q_0)/2C , {\hat n} = \sum_{{\bf q} \sigma} c^{†}_{{\bf q} \sigma}c_{{\bf q} \sigma} -N^+ .$

Here ${\cal H}_{\rm bath}$ is the Hamiltonian for the thermal bath. We assume that emitter and collector electrodes can have different chemical potentials. N⁺ is the number of positively charged ions in the grain. To describe tunneling we introduce the tunneling Hamiltonian between, say, emitter and grain as

${\cal H}_{e \leftrightarrow g}=\sum_{{\bf k},{\bf q}, \sigma} T_{{\bf k} {\bf q}}c^{†}_{{\bf k} \sigma} c_{{\bf q} \sigma} + {\rm h.c.}$

Applying the golden rule we obtain

$\Gamma_{e \rightarrow g} (n) = \frac{G_e}{e^2}\int_{-\infty}^{\infty} d \epsilon_k \int_{-\infty}^{\infty} d \epsilon_q f_e (\epsilon_k) [ 1- f_g (\epsilon_q)] \delta(E_{n+1} - E_n -eV_e) .$

Here we have introduced the tunneling conductance of e–g junction as

$G_e=(4 \pi e^2/\hbar) g_e(\epsilon_F) g_g(\epsilon_F) {\cal V}_e {\cal V}_g \langle |T_{{\bf k} {\bf q}}|^2 \rangle .$

along the Landauer formula, ${\cal V}_{e,g}$ being the volumes of the lead and grain, respectively. In this way one arrives at the expressions

$\Gamma_{e \rightarrow g}(n,V_e)$ = $\Gamma_{g \rightarrow e}(-n,-V_e)= \frac{2G_e}{e^2}{\cal F}(\Delta_{+,e}) ;$ (19)
$\Gamma_{g \rightarrow c}(n, V_c)$ = $\Gamma_{c \rightarrow g}(-n, -V_c)= \frac{2G_c}{e^2}{\cal F}( \Delta_{-,c}) .$ (20)

Here

${\cal F}(\epsilon)=\frac{\epsilon}{1+\exp(-\epsilon/kT)} \rightarrow \epsilon \Theta (\epsilon) {\rm at} T \rightarrow 0 ,$

while

$\Delta_{\pm,\mu} (n)= E_n -E_{n \pm 1} \pm eV_\mu = \frac{1}{C} \left[\frac{e^2}{2} \mp en \mp e\sum_iC_iV_i \right] \pm eV_\mu$

is the energy cost of transition. The temperature-dependent factor arise from the Fermi occupation factor for the initial and final states, physically they describe thermal activation over Coulomb barrier. The results of calculation of current-voltage curves for a symmetric transistor structure are shown in Fig. 6.

Fig 6. The current of a symmetric transistor as a function of gate and bias voltage at T = 0 (from the book [5]).

At low temperatures and low bias voltages, VC/e <1, only two charge states play a role. At larger bias voltage, more charge states are involved. To illustrate this fact, a similar plot is made for symmetrically biased transistor, V_e = –V_g = V/2, for different values of Q₀, Fig. 7.

Fig 7. The current of asymmetric transistor, G_e = 10G_c, as a function of bias voltage at T = 0 and different Q₀.e = 0, 0.25 and 1 (from the book [5]). At Q₀ = 0 the Coulomb blockade is pronounced, while at Q₀/e = 0.5 the current-voltage curve is linear at small bias voltage. The curves of such type are called the Coulomb staircase.

Cotunneling processes

As we have seen, at low temperature the sequential tunneling can be exponentially suppressed by Coulomb blockade. In this case, a higher-order tunneling process transferring electron charge coherently through two junctions can take place. For such processes the excess electron charge at the grain exists only virtually.

A standard next-order perturbation theory yields the rate

$\Gamma_{i\rightarrow f}= \frac{2\pi}{\hbar} \left|\sum_{\psi} \frac{\langle i|{\cal H}_{\rm int}|\psi \rangle \langle \psi |{\cal H}_{\rm int}|i \rangle}{E_\psi -E_i} \right|^2 \delta(E_i -E_f) .$

Two features are important.

There are 2 channel which add coherently: (i) $e\rightarrow g, g\rightarrow c$ with the energy cost $\Delta_{-,e}(n+1)$ , and (ii) $g \rightarrow c, e\rightarrow g$ with the energy cost $\Delta_{+,c}(n-1)$ .
The leads have macroscopic number of electrons. Therefore, with overwhelming probability the outgoing electron will come from a different state than the one which the incoming electron occupies. Hence, after the process an electron-hole excitation is left in the grain.

Transitions involving different excitations are added incoherently, the result being

$\Gamma_{\rm cot}$ = $\frac{\hbar G_eG_c}{2 \pi e^4}\int_e d \epsilon_k \int_g d \epsilon_q \int_g d \epsilon_{q'}\int_c d \epsilon_{k'} f(\epsilon_k)[1-f(\epsilon_q)]f(\epsilon_{q'})[1-f(\epsilon_{k'})]$
$\times \left[\frac{1}{\Delta_{-,e}(n+1)} + \frac{1}{\Delta_{+,c}(n-1)}\right]^2 \delta(eV + \epsilon_k -\epsilon_q + \epsilon_{q'} - \epsilon_{k'} ) .$

At T = 0 the integrals can be done explicitly, and one obtains

$\Gamma_{\rm cot}=\frac{\hbar G_eG_c}{12 \pi e} \left[\frac{1}{\Delta_{-,e}(n+1)} + \frac{1}{\Delta_{+,c}(n-1)}\right]^2V^3 {\rm for} eV \ll \Delta_i .$

As a result, the current appears proportional to V³ that was observed experimentally. The situation is not that simple for the degenerate case when $\Delta_i=0$ . In that case the integrals are divergent and the divergence must be removed by a finite life time of a state. A detailed treatment of that case is presented in the book [5].

There is also a process when an electron tunnels through the system leaving no excitations in the grain. The probability of such elastic cotunneling has a small factor $(g_g {\cal V}_g)^{-1}$ . However, it leads to the current, proportional to V, thus it can be important at very low bias voltage.

Concluding remarks

There are many experiments where Coulomb-blockaded devices are investigated. Probably most interesting are the devices where tunneling takes place through a small quantum dot with discrete spectrum. An example of such device is shown in Fig. 8.

Fig 8. (a) A typical structure of a quantum dot. The depleted (shaded) areas are controlled by electrodes 1-4, C, and F. Electrode C also controls the electrostatic potential in the dot. (b) a model of a quantum dot. From [7].

The linear conductance of such a structure as a function of the gate electrode C is shown in Fig. 9

Fig 9. Conductance of a quantum dot vs. the voltage of gate electrode C. From L. P. Kouwenhoven et al., Z. Phys. B 85, 367 (1991).

An important point is that at present time the devices can be fabricated so small that the criterion $kT \le e^2/C$ can be satisfied at room temperatures. Now several room temperature operating Coulomb blockade devices were reported. Among them are devices consisting of large molecules with the probes attached in different ways. This is probably a starting point for new extremely important field - molecular electronics. Such devices are extremely promising both because they are able to operate at room temperatures and because they will allow high integration. This is one of important trends. Another one concerns with single-electron devices which include superconducting parts. There is a lot of interesting physics regarding transport in such systems.

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T	$\approx$	$\frac{T_1T_2}{(T_1 + T_2)^2/4 + (\theta')^2\left(E-E^{(r)} \right)^2 }$
	=	$\frac{\Gamma_1 \Gamma_2}{(\Gamma_1 + \Gamma_2)^2/4 +\left(E-E^{(r)} \right)^2} .$	(5)

B	=	1+r ,
kA –ik(1–r)	=	$\kappa^2 a (1+r) ,$
A sin ka + B cos ka	=	t ,
ikt –k(Acos ka –B sin ka)	=	$t\kappa^2 a .$

–ne	=	Q_e +Q_c + Q_g
	=	C_e(V_e –U) +C_c(V_c –U)+C_g(V_g –U) ,	(10)

$\Gamma_{n-1}^n$	=	$\Gamma_{e \rightarrow g}(n-1) +\Gamma_{c \rightarrow g}(n-1) ;$	(17)
$\Gamma_{n+1}^n$	=	$\Gamma_{g \rightarrow e}(n+1) +\Gamma_{g \rightarrow c}(n+1) .$	(18)

${\cal H}_0$	=	${\cal H}_e + {\cal H}_g + {\cal H}_{\rm ch} + {\cal H}_{\rm bath} ;$
${\cal H}_{e,c}$	=	$\sum_{{\bf k} \sigma} \epsilon_{\bf k} c^{†}_{{\bf k} \sigma} c_{{\bf k} \sigma} ,$
${\cal H}_{g}$	=	$\sum_{{\bf q} \sigma} \epsilon_{\bf q} c^{†}_{{\bf q} \sigma} c_{{\bf q} \sigma} ,$
${\cal H}_{\rm ch}$	=	$({\hat n} - Q_0)/2C , {\hat n} = \sum_{{\bf q} \sigma} c^{†}_{{\bf q} \sigma}c_{{\bf q} \sigma} -N^+ .$

$\Gamma_{e \rightarrow g}(n,V_e)$	=	$\Gamma_{g \rightarrow e}(-n,-V_e)= \frac{2G_e}{e^2}{\cal F}(\Delta_{+,e}) ;$	(19)
$\Gamma_{g \rightarrow c}(n, V_c)$	=	$\Gamma_{c \rightarrow g}(-n, -V_c)= \frac{2G_c}{e^2}{\cal F}( \Delta_{-,c}) .$	(20)

$\Gamma_{\rm cot}$	=	$\frac{\hbar G_eG_c}{2 \pi e^4}\int_e d \epsilon_k \int_g d \epsilon_q \int_g d \epsilon_{q'}\int_c d \epsilon_{k'} f(\epsilon_k)[1-f(\epsilon_q)]f(\epsilon_{q'})[1-f(\epsilon_{k'})]$
		$\times \left[\frac{1}{\Delta_{-,e}(n+1)} + \frac{1}{\Delta_{+,c}(n-1)}\right]^2 \delta(eV + \epsilon_k -\epsilon_q + \epsilon_{q'} - \epsilon_{k'} ) .$