Lecture 06 Quantum Tunneling (1D-case)

ECE440 Nanoelectronics

Quantum Tunneling (1D Case)0

,          0, 0,

In classical physics, bouncedpenetrating

In quantum physics, penetrating probability depends on , and .

When 0, or ,

0 ħ 0 (Schrodinger Equation)

2ħ 0, 2

ħ

2 0

In 0 region, (reflection)

In a region, (transmission)

Quantum Tunneling (1D Case)When

ħ (Schrodinger Equation)

2

ħ 0,

α2 0

0 ,

α2

ħ

at 0, 0 0 and | |

11 α

12 1 α 1 α12 1 α 1 α

1

(Classical forbidden region)

2 1 α

2 1 α

since and | |

Quantum Tunneling (1D Case)

α α

2

Combining 1 and 2 , we obtain

1 α 1 α 1 α

1 α 1 α 1 α

Quantum Tunneling (1D Case)1 1 

1 α1 α2 α

1 α sin α 2 α cos α

| |4 α

α sin 2 α 4 α cos 2 α4 α

α sin 2 α 4 α

1α

4 α s 2α

11

1sin 2α

| | 1 | |α sin 2α

α sin 2α 4 α

Quantum Tunneling (1D Case)

If α ≫ 1, sin α ~ 12 ≫ 1

16 αα .

16exp

2ħ 2

Quantum reflection, transmission and tunneling effects 

We considered the simplest potential in the form of a rectangular barrier, as shown in the Figure below:

for 2( )

0 for 2bV z L

V zz L

where Vb, the barrier height, is greater than zero, and L is the barrier width.

L2 z

Vbiinir it

-L2 0

ε

There exist two distinct energy intervals, where the behavior of particle isexpected to exhibit different characteristics depending on which energyinterval the particle’s energy is in: 0 < ε < Vb and ε > Vb. In classical physics,for the first energy interval, obviously the reflection coefficient is R = 1, whilethe transmission coefficient is T = 0; for the second energy interval R = 0 andT = 1.

In wave mechanics we have to solve the equation 

with the potential

2 2

2 ( ) ( ) 02

d V z zm dz

for 2( )

0 for 2bV z L

V zz L

.

At the left of the barrier we introduce the incident wave with a unit

magnitude, eikz, and a reflected wave, re-ikz. At the right of the barrier there is

only a transmitted wave, teikz. Then, the exponential factors are

0 02 2 ( )and

bm m V

k

For ε < Vb, κ is a real number, and for ε > Vb, κ is imaginary number. The

parameters r, a, b, and t are still arbitrary functions of ε, which we find by

matching these solutions and their derivatives at the walls of the well, z =

± L/2.

The fluxes at z→ ± ∞ can be calculated in terms of r and t :2and .ti t

The solutions are

2( ) 2 2.

2

ikz ikz

z z

ikz

e re z Lz ae be L z L

te z L

21 , in ri i r

Omitting the procedure of calculation of r and t, we write down the obvious relationship: 

1R T

and two different results for the two energy intervals are as follows: 

2 2 2 22

1

1 sinh 2b

kk

T VL

2 2 2 22

1

1 sin 2b

k KkK

T VKL

In the latter formulas we used K2 = -κ2. 

Let us consider first the case of an energy less than the barrier height, i.e., ε

< Vb. Classical results can be obtained formally at the limit of zero Planck’s

constant, ћ . Indeed, when ћ → 0, we obtain

sinh 2κL→ ∞, and T→ 0 . That is, no transmission through the barrier occurs,

as expected. In reality, this classical result is realized for high and wide barriers

( Vb, L→ ∞ ).

However, at finite values of these parameters, we always obtain a finiteprobability of particle transmission through the barrier, which is, as wediscussed previously, the tunneling effect. For κL >> 1, sinh 2κL = ½ e2κL thesecond term in the denominator

predominates and we can approximate the formula as2 2

42 2 2

16( )

LkT ek

Thus, the tunneling effect is determined primarily by the exponential factor.The probability of finding the particle under the barrier is also exponentiallysmall.

02 ( ),

bm V

2 2 2 22

1

1 sinh 2kk

TL

For the second case, corresponding to an energy greater than the barrier

height, when classical physics gives strictly T = 1, from the quantum‐

mechanical equation

we find that T reaches 1 only at sin 2KL = 0, i.e., at certain “resonant"

energies, when 2KL = πn with n being an integer. Otherwise, T < 1 with

minima at 2KL = (n + ½) π. All of this means that there exists a reflection of

the particle even at large energies.

2 2 2 22

1 ,1 sin 2

bk KkK

T VKL

Figure Dependences of transmission coefficient, T, on electron incident energy, ε, for differentthicknesses of the barrier, L: (a) L = 10 Å and (b) L = 20 Å. Height of the barrier Vb= 0.3 eV.

Oscillations in T with energy for ε > Vb are explained by “overbarrier reflection” of theparticles.

Barrier height

10 Å0.

3 eV

(a)

ε (eV)

T

(b)

Barrier height

20 Å

0.3

eVTε (eV)

0.0 0.3 0.6 0.9 1.2 1.5 1.8 2.1 2.40.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0.0 0.3 0.6 0.9 1.2 1.5 1.8 2.1 2.40.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

2 2 2 22

1

1 sinh 2b

kk

T VL

2 2 2 2

2

1

1 sin 2b

k KkK

T VKL

WKB (Wentzel, Kramers, and Brillouin) Method

WKB Method (cont’d)

WKB Method (cont’d)