Barrier Penetration and Quantum Tunnelling

BARRIER PENETRATION AND QUANTUM TUNNELLING

Barrier Penetration

If a particle with energy 'E' is incident on a thin energy barrier of height 'V', greater than ‘E', then there is a finite probability of the particle to penetrate the barrier. This phenomenon is called barrier penetration and this effect is called Tunnelling Effect (or) Quantum Tunnelling. According to classical mechanics, the probability of a particle to penetrate / tunnel the barrier is zero, but according to quantum mechanics it is finite.

Concept of Barrier Penetration

Let us consider a particle of kinetic energy 'E' which moves from the left (Region‒1) and strikes the potential barrier of height 'V' as shown in Fig. 7.10. If the kinetic energy (E) is lesser than the potential energy (V) (i.e) if E<V then, according to classical mechanics there is no chance for the particle to cross the potential barrier (V).

But, according to quantum mechanics, the particle has certain probability (few chances) to penetrate (or) cross the potential barrier (V) and comes out to Region‒3, by tunnelling the Region‒2.

Note: For wider and higher potential barrier, the chance of crossing the barrier through tunnelling process is very less.

Derivation / Proof

Let us consider such a particle with energy E<V incident from left side (Region‒1) and tunnel the Region‒2 of width 'l' [x=0 to l] and comes out as a transmitted wave in Region‒3, as shown in Fig. 7.10

Here, the potential energy V=0, on both sides of the barrier, which means, no forces will act upon the particle in Region‒1 and in Region‒3.

∴ The boundary conditions shall be written for various regions as

For Region‒1: When x <0; V=0

For Region‒2: When 0<x<1; V=V

For Region‒3: When x>1; V=0

Let ᴪ₁, ᴪ₂ and ᴪ₃ be the wavefunctions in Regions 1, 2 and 3 respectively.

Then, the Schroedinger's wave equations for all the three regions shall be written as

For Region‒1

d²ᴪ₁/dx² + 2m/ḧ² [E‒V] ᴪ₁ = 0

Since V=0 in Region‒1, we can write the above equation as

For Region‒2

d²ᴪ₂/dx² + 2m/ḧ² [V‒E] ᴪ₂ = 0

Since V>E in Region‒2, we can write the above equation as

For Region‒3

d²ᴪ₃/dx² + 2m/ḧ² [E‒V] ᴪ₃ = 0

Since V=0 in Region‒3, we can write the above equation as

∴ Equations (1), (2) and (3) shall be written as

The solutions for eqns (4), (5) and (6) shall be written as

For Region‒1: ᴪ₁ = Ae^iax + Be^‒iax       ………..(7)

For Region‒2: ᴪ₂ = Fe^βx + Ge^‒βx       ………..(8)

For Region‒3: ᴪ₃ = Ce^iax + De^‒iax       ………..(9)

Here A, B, C, D, F, G are the amplitudes of corresponding waves in various regions as shown in Fig. 7.10.

Let us discuss the behaviour of wave function and the amplitudes in each Region, viz.,

Region‒1

The wave function of the incident wave in Region‒1 shall be written from equation (7) as

 ᴪ₁ (Incident) = Ae^iαx                      ……………..(10)

Where 'A' is the amplitude of the incident wave in Region‒1

Since there are ample chances for the wave to get reflected within the Region‒1, due to higher potential barrier (or) larger width of the barrier, the wave function of the reflected wave in Region‒1 shall be written from equation (7) as

ᴪ₁ (Reflected) = Be^‒iαx                      ……………..(11)

Where 'B' is the amplitude of the reflected wave in Region‒1.

Region‒2

The wave function of the transmitted (or) tunnelling wave at Region‒2 shall be written from equation (8) as

ᴪ₂ = Fe^βx + Ge^‒βx                     ……………..(12)

Where

 β - is the wave number given by β = √[2m (V‒E)] / h

F -  is the amplitude of the barrier penetrating wave (or) tunnelling wave in Region‒2.

G - is the amplitude of the reflected wave at the boundary between Region‒1 and Region‒2.

From eqn (12) we can see that the exponents are real quantities, so the wave function ᴪ₂ will not oscillate and therefore does not represent a moving particle at Region‒2.

Thus, the particle can either penetrate (or) tunnel through Region‒2 and transmitted to Region‒3 (or) it shall be reflected back to Region‒1 itself. [Given by equation (11)].

Region‒3

The wave function of the transmitted wave in Region‒3 shall be written from equation (9) as

ᴪ₃ (Transmitted) = Ce^lax                ...(13)

Where 'C' is the amplitude of the transmitted wave in Region‒3.

In Region‒3, i.e., x>l, there can be only transmitted wave and there will not be any reflected wave and therefore the amplitude of the reflected wave in Region‒3 i.e., D=0

Thus, the wave function of the reflected wave in Region‒3 is also zero.

 ᴪ₃ (Reflected) = 0              ...(14)

The transmission and reflection co‒efficient shall be obtained as follows:

Transmission Co‒efficient

We know that the probability density is the square of the amplitude of that function. Therefore the barrier transmission co‒efficient (T) is the ratio between the square of the amplitude of the transmitted wave |C|² and the square of the amplitude of the incident wave |A|².

 ∴ The transmission co‒efficient T = |C|² / |A|² = 4√E√(E‒V)  /  [√E + √(E –V)]²      ...(15)

Equation (15) is also called as the "Penetrability" of the barrier.

Reflection Co‒efficient

Similarly, the reflection co‒efficient (R) for the barrier surface at x=0 is the ratio between the square of the amplitude of the reflected wave |B|² and the square of the amplitude of the incident wave |A|²

∴ The reflection co‒efficient R = |B|² / |A|² = [ √E‒√(E‒V)  /  [√E+√(E –V) ]²      ...(16)