A potential barrier is a region in which the potential energy of a particle exceeds the total energy.
Let us consider a potential barrier defined by V(x) = 0 for x < 0
= V for 0 ≤ ≤x L
= 0 for x > L ...(4.2.1)
The potential barrier is sketched in the Fig. (4.2.1). Assume that a particle moving from left to right encounters the potential barrier of height V and width L on its path.
In terms of classical mechanics the behavior may be predicted as follows:
(i) If the energy of the particle is greater than the height of the barrier (E > V), the particle passes over the barrier without any hindrance. Inside the barrier, the velocity of the particle diminishes and beyond it the particle acquires its initial value.
(ii) If E < V, the particle is reflected from the barrier and is unable to penetrate through the barrier.
Quantum mechanical treatment of this problem predicts different results. If E > V, there is a finite probability that the particle will be reflected from the barrier. If E < V, there is a finite probability that the particle will penetrate through the barrier and will be found on the other side of the barrier.
Thus the quantum mechanics allows the particle to leak through the barrier. This phenomenon is called tunnel effect. This is a purely quantum mechanical result having no classical analogue. Thus by this mechanism the alpha-particles are emitted by radioactive nuclei, although the potential barrier is such that classically they cannot be able to surmount it.
Fig. 4.2.1 One dimensional potential barrier
In the Fig. (4.2.1) the potential barrier divides the space into three regions I, II and III. The Schrửdinger wave equations in these regions are:
Region I
2 1
2 2 1
2 E 0
d m
dx
ψ + ψ =
D or
2 1 2
1 1
2 0
ψ + ψ =
d k
dx ...(4.2.2)
where 1 2
2 Em k =
D ...(4.2.3)
Region II
2 2
2 2 2
2 (E V)
d m 0
dx
ψ + − ψ =
D or
2 2 2
2 2
2 0
d k
dx
ψ + ′ ψ = ...(4.2.4)
where 2 2
2m(E V)
k′ = −
D ...(4.2.5)
Here k'2 is imaginary and therefore we can write 2 2 2 2
2 (V E)
, m
k′ =ik k = −
D ...(4.2.6)
Region III
2 3
2 2 3
2 E 0
d m
dx
ψ + ψ =
D
or
2 3 2
3 3
2 0
ψ + ψ =
d k
dx ...(4.2.7)
where 3 2
2 Em k =
D = k1 ...(4.2.8)
Solutions of Eqns. (4.2.2), (4.2.4) and (4.2.7) are
ψ1( )x =A exp (ik x1 )+B exp (−ik x1 ) ...(4.2.9)
ψ2( )x =C exp (−ik x2′ )+D exp (ik x2′ )
2 2
C exp (k x) Dexp ( k x)
= + − ...(4.2.10)
ψ3( )x =F exp (ik x1 )+G exp (−ik x1 ) ...(4.2.11a) The term exp (ik1x) corresponds to a wave propagating in the positive direction of x-axis and exp (– ik1x) to a wave propagating in opposite direction. In region III there must be only one wave that has penetrated through the barrier and is propagating from left to right. We must, therefore, assume G = 0.
The wave function y3 (x) then becomes
ψ3( )x =F exp (ik x1 ) ...(4.2.11b) To find out other coefficients we use the following boundary conditions that wave functions must satisfy.
At x = 0,
y1(0) = y2(0) ...(4.2.12)
1 2
0 0
= =
ψ ψ
=
x x
d d
dx dx ...(4.2.13)
At x = L,
y2(L) = y3(L) ...(4.2.14)
2 3
x L x L
d d
dx = dx =
ψ
ψ
=
...(4.2.15)
These boundary conditions lead to following equations:
A + B = C + D ...(4.2.16)
ik1A – ik1B = k2C – k2D ...(4.2.17)
C exp ( L)k2 +D exp (−k2L)=F exp (ik1L) ...(4.2.18)
2C exp ( L)2 2Dexp ( 2L) 1F exp ( 1L)
k k −k −k =ik ik ...(4.2.19)
Here we are interested in transmission coefficient or transmission probability T, and reflection coefficient or reflection probability R. T is defined as the ratio of the current density associated with transmitted beam and that associated with incident beam. Similarly, R is defined as the ratio of current density associated with transmitted beam to that associated with incident beam.
2 2
trans 1
2 2
inci 1
J ( / ) | F | | F | F F
T J ( / )| A | | A | A A
k m k m
∗
= = = = D
D ...(4.2.20)
2 2
ref 1
2 2
inc 1
J ( / ) | B | | B | B B R J ( / )| A | | A | A A
k m k m
∗
= = = = D
D ...(4.2.21)
The conservation of energy demands that
R + T = 1 ...(4.2.22)
From Eqns. (4.2.16) to (4.2.19) the expressions for
2 2
2 2
| F | | B |
| A | and | A |
come out to be
2 2 2
1 2 2
2 2 2 2 2 2 2 2
2 1 2 1 2 2
16 exp (2 L)
| F |
| A | ( ) [1 exp (2 L)] 4 [1 exp (2 L)]
k k k
k k k k k k
= − − + + ...(4.2.23)
2 2 2 2
2
2 1 2
2 2 2 2 2 2 2 2
2 1 2 1 2 2
( ) [1 exp (2 L)]
| B |
| A | ( ) [1 exp (2 L)] 4 [1 exp (2 L)]
k k k
k k k k k k
+ −
= − − + + ...(4.2.24)
The expression for T after simplification becomes
2 2 2
1 2
2 2 2 2 2 2 2
2 1 2 1 2
4
| F | T
| A | ( ) sin L 4
k k
k k h k k k
= =
+ + ...(4.2.25)
Substituting the expressions for k1 and k2, we have
2 2 1 2
2 2
2
V sin L
T 1 1
4E (V E)
V sin L
1 4E(V E)
h k h k
−
= + = + −
−
...(4.2.26)
The expression for coefficient of reflection comes out to be
2 2 2
2
2 1
2 2 2
2 2 2 1 2
2 1 2
2
( )
| B |
R | A | 4
( )
sin L
k k
k k k k
h k
= = +
+ + ...(4.2.27)
Substituting the expressions for k1 and k2, we obtain
2 1
2 2
2 2
2 2
V 4E (V E)
R 1
4E (V E) V sin L
V sin L
h k h k
− −
= + − = + ...(4.2.28)
Classical limit is obtained by setting h ® 0. In this limit k2 ® ¥ and k1 ® ¥. This implies that T ® 0 and R ® 1. The probability of transmission becomes zero and that for reflection is unity. This is the classical prediction.
When the barrier height and width both are large, k2L >> 1 and sin h k2L đ ẵ exp (k2L).
Under this approximation 1 can be neglected in the expression for T. Thus
2(12 2 )2 2
4E(V E) 16E E
T 1 exp ( 2 L),
V V
V exp ( L)
k k
−
= = − −
where 2 2
2m(V E)
k = −
D ...(4.2.29a)
The exponential term in Eqn. (4.2.29a) is more dominant term than the coefficient 16E E
V 1 V
−
and the latter is assumed to be equal to 1 in most of the application. With this approximation the expression for transmission probability becomes
T ≅exp(−2 Lk2 ) ...(4.2.29b) The quantum mechanical analysis of potential barrier problem shows that the particle has finite probability of getting transmitted through the barrier even its energy is less than the height of the barrier.
The transmission probability of the particle depends on (i) width L of the barrier and (ii) the difference (V – E). This dependence of T on the width of the barrier and energy of the incident particle is displayed in the table.
E V L
2=
k 2
2 (V E)m −
D 2 Lk2 T
1 eV 4 eV 0.1 nm 0.886 × 1010 m–1 1.772 0.17
1 eV 4 eV 0.2 nm 0.886 × 1010 m–1 3.544 0.03
2 eV 4 eV 0.1 nm 0.724 × 1010 m–1 1.448 0.23
Notice that when the width of the barrier is doubled, the transmission probability decreases by nearly 6 times whereas when the energy of the incident particle is doubled, the transmission probability increases only by a factor of nearly 1.3 times. So the transmission probability strongly depends on the width of the barrier.
Fig. 4.2.2 Wave functions in the three regions
Fig. 4.2.3 A potential barrier of varying width
For potential barrier of variable width, as shown in the Fig. (4.2.3), the transmission probability is given by
2
1
T exp 2 2 (V E)
x
x
m dx
≅ − −
D∫ ...(4.2.30)
The emission of alpha-particle from radioactive nuclei, the passage of electron through potential barrier in tunnel diode and the crossing of electron through classically forbidden region between two superconductors are the well-known examples of tunneling phenomenon.
Case 2: E > V In this case 2
2
2m(V E)
k = −
D becomes imaginary. Let = β β = −
2 D2
, where 2m(E V).
k i The expressions for R and T become
1
2 2
2 2
1 4E (E V)
R 1
4E (E V) V sin L
1 V sin L
− −
= + − = + β β
...(4.2.31)
2 2 1
2 2
1 V sin L
T 1
4E(E V) V sin L
1 4E(E V)
β −
= + β = + −
−
...(4.2.32)
(a) When E đ V, bđ 0, sin bL ằbL and in this limit
2
2
T 1 1 VL
2
= m + D
...(4.2.33) (b) Eqn. (4.2.32) shows that when E increases above V, transmission probability T becomes oscillatory due to presence of sin bL. The barrier becomes transparent (T = 1) when
β = πL n , n=1, 2, 3,...
or 2 (E2 V) 2 2 2 m L
− = πn D
or L 2 nλ
= (4.2.34)
where
2 (E V) h λ = m
− = de Broglie wavelength of the particle.
Thus, when the width of the barrier is integral multiple of half the wavelength of the particle, the barrier becomes transparent. This phenomenon is called resonance scattering. Resonances are obtained for the values of E given by
2 2
2
2 (E V) L
nπ m −
β = = D
or
2 2 2
E V 1 2
2 VL n
m
π
= +
D ...(4.2.35)
Minimum value of T is obtained when
sin L 1 orβ = β =L (2n+ π1) / 2, n=0,1, 2, 3,...
or
2 2 2 2
(2 1) E V 1
8 VL n
m
+ π
= +
D ...(4.2.36)
For this value of E, T is minimum.
or
1
min
T 1 1
4E E 1
V V
−
= + −
...(4.2.37)
(c) When E decreases below V, T decreases monotonically. When k2L >> 1, sin h k2L = ẵ exp (k2L). In this case the expression for T becomes
2 2 1
V sin 2L
T 1
4E(V E) h k −
= + − (omitting 1).
2(12 2 )2 2
4E(V E) 16E E
1 exp ( 2 L).
V V
V exp L
k k
−
≅ = − − ...(4.2.38)
The variation of T with E/V and that of T with increasing thickness of barrier L are shown in the Fig. 4.2.4.
Fig. 4.2.4 Variation of T with E/V
Fig. 4.2.5 Variation of T with thickness L of barrier. Appearance of transmission resonances