Mitin V.V., Sementsov D.I., Vagidov N.Z. Quantum mechanics for nanostructures

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3.5 Tunneling 93

E,U

E < U

E > U

Incident

electron

Incident

electron

Figure 3.8 Reﬂection of an

electron from a potential

step with U = U

through the barrier becomes equal to zero, and the reﬂection coefﬁcient becomes

equal to unity: R = 1. The case with ﬁnite U

and width L →∞corresponds to

the total reﬂection of the electron from the so-called potential step with a ﬁnite

height of the barrier U

(see Fig. 3.8). In this case a phase shift δ between the

wavefunctions of the incident and reﬂected waves occurs:

√

iδ

. (3.161)

Here the phase shift δ is deﬁned by the expression

δ = arctan



2kκ

− κ



= arctan





E(U

− E)

2E − U



. (3.162)

We see that at E = U

the phase shift is equal to zero, at E = U

/2 the phase

shift is equal to δ = π/2, and at E  U

the phase shift is δ → π . In the case

of reﬂection from an inﬁnitely high barrier (U

→∞) the phase shift between

incident and reﬂected wavefunctions of the electron is equal to δ = π as in

the case of the reﬂection of an electromagnetic wave from an optically denser

medium.

It is interesting to study the probability of ﬁnding an electron in the region

x > 0. In spite of the fact that E < U

and that the reﬂection coefﬁcient of an

electron from the potential step is equal to unity, the probability of ﬁnding an

electron in the region x > 0 is not equal to zero! The electron wavefunction,

(x), in this region is equal to

(x ) = A

−κx

, (3.163)

where

k − iκ

. (3.164)

94 Layered nanostructures

Therefore, |ψ

= 0. The effective penetration depth of the electron under the

potential step can be estimated by a quantity l

eff

= κ

−1

. At this distance from the

boundary x = 0 the probability of ﬁnding the electron decreases by a factor of

in comparison with its magnitude at x = 0. The probability that the electron

penetrates the potential step in this case is deﬁned by the transmission coefﬁcient,

D, through the boundary x = 0:

D =

4kκ

+ κ

. (3.165)

For the reﬂection coefﬁcient in this case we obtain the expression

R =



k + iκ

k − iκ



= 1. (3.166)

Thus, the electron penetrates through the boundary of the potential step into

the region x > 0, but then it returns back into the region with x < 0, reﬂecting

effectively from the potential step with reﬂection coefﬁcient R = 1. Because of

such penetration there occurs a phase shift δ between the two wavefunctions

which correspond to reﬂected and incident waves.

3.5.2 Electron motion above the barrier

Let us write the general expressions for the electron reﬂection and transmission

coefﬁcients in the case of electron motion with energy E > U

above the barrier

with ﬁnite width L (see Fig. 3.7):

R =



1 +

4E(E − U

)

sin

(κ L)



−1

, (3.167)

D =



1 +

sin

(κ L)

4E(E − U

)



−1

, (3.168)

where κ =

√

(E −U

)/h

. In the case of electron motion above the potential

well (Eqs. (3.129) and (3.131)) as well as in the case of motion above the potential

barrier (Eqs. (3.152) and (3.153)) the change in electron energy leads to oscilla-

tions of both coefﬁcients, R and D. The maxima of the transmission coefﬁcient,

D (and minima of reﬂection coefﬁcient, R), are deﬁned by the condition

L = π n, n = 1, 2, 3,... (3.169)

It follows from the last condition that

= U

. (3.170)

The transmission and reﬂection coefﬁcients are D = 1 and R = 0, and we are

dealing with the so-called resonant interaction of an electron with the potential

barrier. We note that the quantity E

= E

−U

coincides with the expression

3.5 Tunneling 95

3π/2π

π/2

2π

0.5

0.1

0.2

0.5

0.01

Figure 3.9 The

dependence of the

electron’s transmission

coefﬁcient, D,onthe

magnitude of the

potential barrier, U

,and

κ L. The six curves

correspond to E

0, 1, 0.5, 0.2, 0.1, and 0.01.

Here E

= E

−U

and

is the electron’s ﬁrst

energy level.

(3.44) for the energy of the nth level of an electron localized inside of the potential

well with inﬁnite barriers.

Since the wavenumber is equal to κ = 2π/λ

, the condition (3.169) means

that the width of the potential barrier, L, contains an integer number of de Broglie

half-wavelengths, λ

/2. Near values of the wavenumber of



n +



(3.171)

we have maxima of the reﬂection coefﬁcient and minima of the transmission

coefﬁcient. The existence of oscillations in the coefﬁcients R and D is the result

of the interference of waves reﬂected from the potential steps at the boundaries of

the rectangular potential barrier. The dependences of the transmission coefﬁcient,

D,onκ L found for six values of the ratio E

, namely 0, 1, 0.5, 0.2, 0.1, and

0.01, where E

= π

/(2m

), are shown in Fig. 3.9. We see that an increase

of the barrier height or an increase of the barrier width results in a decrease of the

amplitude of oscillations of the transmission coefﬁcient, D. When the electron

energy, E, increases the transmission coefﬁcient, D, quickly increases, tending

to unity, whereas the amplitude of oscillations decreases. The locations of the

maxima correspond to the resonance values of the energy, E

(Eq. (3.170)).

The corresponding expressions for the reﬂection and transmission coefﬁcients

at E > U

for the potential step shown in Fig. 3.8 have the forms

R =



k − κ

k + κ





√

E −

√

E −U

√

E +

√

E −U



, (3.172)

D =

4kκ

(k + κ)



E(E −U

)

(

√

E +

√

E −U

)

. (3.173)

96 Layered nanostructures

E/ U

0.5

123

Figure 3.10 The

dependence of an

electron’s reﬂection

coefﬁcient, R,onthe

barrier height, U

Figure 3.10 shows the dependence of the reﬂection coefﬁcient, R,onthetotal

energy of an electron, E, for a given height of the potential step, U

. We see that

even when an electron moves above the potential step (E /U

> 1) the electron

still undergoes reﬂection. At E = U

reﬂection stays the same as at E < U

, i.e.,

R = 1.

Example 3.7. An electron traveling from the negative part of the x-axis is

incident on a rectangular potential step. The height of the potential barrier U

greater than the total energy of the electron E (see Fig. 3.8). Find the probability

of ﬁnding the electron “under the barrier” up to the distance a from the boundary

x = 0.

Reasoning. The probability density of an electron’s penetration under the barrier

depends on the coordinate x and can be written as

|ψ

(x )|

+ κ

)

−2κx

. (3.174)

The probability density, |ψ

(x)|

, of ﬁnding the electron at x > 0 exponentially

decreases with increasing x. If the distance from the barrier at which we can

ﬁnd the electron is equal to a, then the total probability of ﬁnding the electron

between x = 0 and x = a is equal to

P =



|ψ

(x )|

dx =

+ κ

)κ



1 −e

−2κa



. (3.175)

Thus, with increasing a the total probability, P, of ﬁnding the electron at 0 ≤

x ≤ a increases, and at a →∞this probability tends to its limit:

max

+ κ

)κ

. (3.176)

3.5 Tunneling 97

Example 3.8. A rectangular potential barrier (Fig. 3.7) has a width L = 0.15

nm. Find the barrier height U

and the electron energy E for which the electron’s

total probability of transmission through the barrier is equal to P = 0.4 and the

coefﬁcient D

from Eq. (3.156) is equal to unity.

Reasoning. Considered from the physical point of view, the probability P of

electron transmission through the potential barrier is equal to the transmission

coefﬁcient D, which is deﬁned by Eq. (3.156):

P = D = D

−2L

√

−E)/h

. (3.177)

From Eq. (3.157) and the condition D

= 1 we can ﬁnd the ratio γ = E/U

16γ (1 −γ ) = 1. (3.178)

On solving this quadratic equation, we ﬁnd two possible values of γ :

γ =

√

. (3.179)

Let us take the logarithm of Eq. (3.177). As a result we obtain

ln P =−



− E). (3.180)

From the last equation, taking into account E = γ U

, we ﬁnd the expression for

1 −γ

(ln P)

. (3.181)

After the substitution of the given parameters we obtain two values of U

= 8.51 × 10

−19

J = 5.31 eV,

= 0.61 × 10

−19

J = 0.38 eV.

The corresponding values of the electron’s energy are

= 7.94 ×10

−19

J = 4.95 eV,

= 0.4 ×10

−19

J = 0.25 eV.

Example 3.9. Find the electron transmission probability through the potential

barrier of the following proﬁle:

U (x ) =











0, x < −a,

(1 + x /a), −a ≤ x ≤ 0,

(1 − x /b), 0 < x ≤ b,

0, x > b,

(3.182)

where a and b are positive numbers. Assume that the electron is incident on the

barrier from the negative part of the x-axis and that the electron’s energy is less

than U

(E < U

98 Layered nanostructures

with energy E

Incident electron

Figure 3.11 An electron’s

transmission through the

triangular potential barrier

deﬁned by Eq. (3.182).

Reasoning. Equation (3.156) for the probability of an electron’s transmission

through the rectangular potential barrier can be generalized for the case of a

potential barrier of arbitrary form. If the potential in the barrier region changes

slowly, i.e., its change is small at distances comparable to the de Broglie wave-

length, the potential of an arbitrary proﬁle can be presented as a sum of a large

number of rectangular potentials (see Fig. 3.11). Then for the transmission coef-

ﬁcient the following expression is valid:

D = D

−2

√



√

U (x)−Edx/h

, (3.183)

where x

and x

are the points of the electron’s penetration under the poten-

tial barrier (see Fig. 3.11) and are deﬁned from the solution of the following

equations:

(

1 + x /a

)

= E, x

=−

(

1 − E/U

)

(

1 − x /b

)

= E, x

(

1 − E/U

)

b. (3.184)

Thus, to ﬁnd D it is necessary to ﬁnd the following integral:

I =





U (x ) − E dx =





x − E dx +





−

x − E dx.

(3.185)

In order to evaluate the integrals on the right-hand side of Eq. (3.185)itis

necessary to change variables: t =

√

x/a − E in the ﬁrst integral and

t =

√

−U

x/a − E in the second integral. As a result of integration and

taking into account the expressions for x

and x

, we obtain

I =

− E)

3/2

− E)

3/2

a + b

− E)

3/2

. (3.186)

3.5 Tunneling 99

b =

(e|E|)U

metal vacuum

Figure 3.12 The triangular

potential barrier at the

metal–vacuum interface

under an applied electric

ﬁeld, E.

On substituting the last expression into Eq. (3.183) we obtain

D = D

−4

√

(a+b)(U

−E)

3/2

/(3h

)

. (3.187)

In practice very often a potential with one of the parameters, a or b,having

a value of zero is encountered. For example, at a = 0 the potential has the form

shown in Fig. 3.12. Such a potential is common for a metal–vacuum interface to

which an electric ﬁeld, E, is applied. Without the electric ﬁeld the interface of the

metal with vacuum is a rectangular potential step of ﬁnite height, i.e., U(x) = 0

at x < 0 and U(x) = U

at x ≥ 0. If an external electric ﬁeld E is applied along

the electron’s direction of motion (perpendicular to the interface), then at x > 0

the electron’s potential energy becomes equal to

U (x ) = U

− e|E|x, (3.188)

and the coordinate for which U (b) = 0is

x = b =

e|E|

For this potential barrier the transmission probability for the electron is deﬁned

by the expression

D = D

−4

√

−E)

3/2

/(3eh

|E|)

. (3.189)

It follows from the last equation that with increasing electric ﬁeld the probability

of an electron’s transmission through the barrier rapidly increases. This is because

of the decrease of the barrier width.

100 Layered nanostructures

E,U

U = ∞

Figure 3.13 Awellwith

the potential proﬁle

(3.190).

Example 3.10. Find the dependence on energy of the probability of an electron

escaping a potential well through a potential barrier with the following proﬁle:

U (x ) =







∞, x < 0,

0, 0 ≤ x ≤ L ,

α/x, x > L .

(3.190)

The graph of the potential well deﬁned by Eq. (3.190) is shown in Fig. 3.13.

Reasoning. The probability of the electron escaping from the potential well is

deﬁned by its transmission coefﬁcient, D. Expression (3.156) was obtained for a

barrier of rectangular form. The main difference of the present problem from the

potential barrier with proﬁle (3.141) is that the barrier width, d, now depends on

the energy, E , and, as follows from Eq. (3.190), it is equal to

d(E) = x

(E) − L =

− L , (3.191)

where the coordinate x

is the solution of the equation E = α/x. To calculate

the probability of an electron escaping from the well as a result of transmission

through the barrier, we have to use expression (3.183):

D = D

−2



√

(

α/x−E

)

dx/h

, (3.192)

where the constant D

is close to unity and the upper limit of integration is the

coordinate x

. As a result of integration, we obtain the expression

D ≈ D

−2α

√



arccos

√

E/U

−

√

(E/U

)

(

1−E/U

)



, (3.193)

where U

<α/L. In the case of a sufﬁciently high potential barrier, when U



E, this expression, taking into account that arccos(0) = π/2, transforms into the

3.7 Problems 101

simpler expression

D ≈ D

−πα

√

/E/h

. (3.194)

From the last expression it follows that with increasing electron energy, E,the

transmission coefﬁcient, D, increases. At E = U

, as follows from Eq. (3.193)

and from arccos(1) = 0, the transmission coefﬁcient, D, is equal to unity.

3.6 Summary

1. The wavefunction of a free electron can be described by a de Broglie plane wave,

whose group velocity coincides with the electron velocity. The energy spectrum of

such an electron is continuous.

2. In the case of electron motion in a limited space the main important property of such

motion is the discreteness of the energy spectrum, i.e., quantization of electron energy.

All quantum states of an electron and the corresponding energy states in the case of

one-dimensional motion can be enumerated by one quantum number, n.

3. In a potential well with inﬁnite barriers there is an inﬁnite number of energy levels

that correspond to the stationary quantum states of an electron in the well. The energy

of corresponding levels is proportional to n

and the distance between levels is pro-

portional to 2n + 1. If we place the center of coordinates at the center of the well,

then the wavefunctions of an electron for quantum states with odd numbers of n are

symmetric functions of coordinate, and those for quantum states with even numbers

of n are antisymmetric functions of coordinate.

4. Higher electron energy corresponds to higher wavenumber. The lowest energy level

with n = 1 is called the ground state and it corresponds to the minimal electron energy.

However, this state is not the state of rest.

5. In conﬁned quantum states the probability density of ﬁnding an electron, |ψ|

,has

maxima in the region of the potential well and exponentially decreases outside of the

well. This means that an electron can be outside of the well with a non-zero probability.

6. In a symmetric potential well with ﬁnite barriers there is always at least one energy

level, i.e., the energy level E

, which is lower than the barrier height U

7. An electron, just like any other quantum particle, can tunnel through a potential barrier

whose height is greater than the total energy of the electron. The probability of a particle

tunneling through a barrier exponentially decreases with increasing width and height

of the barrier and with increasing mass of the particle. The total energy of the particle

does not change during tunneling through the barrier.

3.7 Problems

Problem 3.1. Using the procedure of calculations of average values of physical

quantities, show that for a one-dimensional motion the operator of coordinate is

coordinate itself.

102 Layered nanostructures

Problem 3.2. A particle with mass m is moving in the region −L/2 ≤ x ≤ L/2.

The state of the particle is described by the wavefunction

ψ(x) = A exp





, (3.195)

where p is the particle’s momentum. Find the normalization constant, A, and

the average value of the particle’s kinetic energy, K .

Problem 3.3. An electron is in a one-dimensional rectangular potential well with

barriers of inﬁnite height. Find the well width if the energy difference between

the ﬁfth and fourth quantum states, E

− E

, is equal to the electron’s average

thermal energy at room temperature.

Problem 3.4. An electron is in a one-dimensional rectangular potential well

with barriers of inﬁnite height. The width of the well is equal to L = 5nm.

Find the wavelengths of photons emitted during electronic transitions from the

excited states with quantum numbers n = 2, λ

, and n = 3, λ

, to the ground

state with n = 1. (Answer: λ

≈ 1.15 µm and λ

≈ 0.43 µm.)

Problem 3.5. An electron is in a one-dimensional rectangular potential well with

barriers of inﬁnite height. Find the normalized wavefunctions and energy levels

if the width of the potential well is equal to L and the wavefunction is symmetric

with respect to the coordinate origin placed at the center of the well (see Fig. 3.14).

Problem 3.6. An electron is in a one-dimensional rectangular potential well with

barriers of inﬁnite height and with well width equal to L as shown in Fig. 3.14.

Find the average values of the coordinate, the momentum projection, and the

squares of these magnitudes for an electron that is in quantum states described

by symmetric wavefunctions.

L/2)=

∞

U(L/2)= ∞

Figure 3.14 A rectangular

symmetric potential well

with respect to the

coordinate origin with

barriers of inﬁnite height.