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

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5.6 Problems 153

is subjected to a perturbation of the following form:

W (x) =



, b ≤ x ≤ L − b,

0, x < b, x > L − b.

Find the energy of the electron in stationary states, taking into account

corrections arising from the ﬁrst-order perturbation theory.

Problem 5.4. Find the energy of the ground state of a one-dimensional harmonic

oscillator that is subjected to an anharmonic potential perturbation W (x) = γ x

Problem 5.5. A charged one-dimensional oscillator is placed in a perturbation

homogeneous electric ﬁeld directed along the axis of oscillations. Find the

energy of the oscillator, taking into account the ﬁrst two orders of correction of

perturbation theory. The charge of the particle is positive and equal to e and its

mass is equal to m.

Problem 5.6. The electron is in a spherically-symmetric potential well (4.41):

U (r) =



0, r ≤ a,

∞, r > a.

The unperturbed magnitudes of the electron energy in the well are E

. Find, in

the ﬁrst order of perturbation theory, the electron energy and wavefunction in the

magnetic ﬁeld directed along the z-axis. The perturbation operator

V is deﬁned

V =

∂

∂θ

where θ is the polar angle in the plane perpendicular to the z-axis. Hint: use the

spherical coordinate system.

Problem 5.7. An electron, which is in a one-dimensional potential U(x) = βx

(a linear harmonic oscillator), is placed in a homogeneous electric ﬁeld. The

electric ﬁeld is directed along the axis of oscillations and changes with time as

E(t) = E

−t

/τ

Before the electric ﬁeld is turned on (t →−∞) the electron was in one of its

unperturbed states. Find, in the ﬁrst order of perturbation theory, the probability

of electron excitation to the higher states at t →∞.

Problem 5.8. The motion of a plane harmonic oscillator takes place in the

xy-plane in the following potential:

U (x , y) =

β(x

+ y

)

Find, in the ﬁrst order of perturbation theory, the splitting of the ﬁrst excited

energy level of the oscillator under the perturbation potential W = γ xy.

154 Approximate methods of ﬁnding quantum states

Problem 5.9. Find, in the quasiclassical approximation, the probability of electron

transmission, P, through the potential barrier

U (x ) =



0, x < 0,

−x /b

, x ≥ 0,

for an electron whose energy is E < U

Chapter 6

Quantum states in atoms and molecules

The main characteristics of atoms and molecules are their structure and their

energy spectrum. Under the term structure of an arbitrary particle we usually

understand the size and the distribution of its mass and its charge in space. In

quantum physics such a distribution is deﬁned by the square of the modulus of

the wavefunction of a particle. The particle itself may consist of a system of

other particles that are bound by a certain type of coupling. For example, an atom

consists of a nucleus and a system of interacting electrons, a molecule consists

of a system of interacting atoms, and so on. If we consider an atom, the nucleus

of the atom is assumed to be at rest (the so-called adiabatic approximation). This

assumption can be made because the nucleus has a much larger mass than that of

an electron (a proton’s mass is 1836 times larger than the mass of an electron).

Then, the square of the modulus of the wavefunction, |ψ(r

, r

,...,r

,for

the system of electrons deﬁnes the probability density of ﬁnding the jth electron

at the point r

. Graphically it is very convenient to depict |ψ |

in the form

of an electron cloud, which can be considered as an averaged distribution of

matter added to the mass of the nucleus located at the center of an atom. Since

electrons have not only mass, m

, but also the electric charge, −e, the electron

cloud should be considered as the averaged density of the negative electric charge,

=−e|ψ(r

, r

,...,r

, which compensates for the nucleus’ positive charge.

The current chapter is devoted to the study of the wavefunction and the geometry

of the electron cloud, as well as the energy spectra of the simplest atoms and

molecules.

6.1 The hydrogen atom

6.1.1 An electron in a Coulomb potential

As has already been shown, the de Broglie wavelength of an electron in an atom is

comparable to the atomic size. Therefore, we cannot neglect the wave properties

of the electron and we have to describe its behavior in an atom on the basis of

the Schr¨odinger equation.

155

156 Quantum states in atoms and molecules

Figure 6.1 Quantization of

an electron in the

potential well U (r).

The hydrogen atom is the simplest atomic system, where the conﬁned electron

with electric charge, −e, moves in the spherically-symmetric electrostatic ﬁeld

of a positively charged nucleus (proton) (the charge of the proton is +e). The

potential energy of an electron in such a system is deﬁned by its Coulomb

interaction with the nucleus and is equal to

U (r) =−

, (6.1)

where r is the distance between the electron and the nucleus. The nucleus, in

the ﬁrst order of approximation, can be considered as a point (the proton size

≈ 10

−15

m while the distance between the electron and the proton is of the

order of 10

−10

m =1

A). The motion of an electron in the potential (6.1) can

be considered to occur in a spherically-symmetric well of unlimited size (see

Fig. 6.1). However, the size of the atom itself, i.e., the average diameter of an

electron orbit in a stationary state, is ﬁnite and is of the order of 10

−10

For the description of possible electron quantum states in an atom we will

ﬁrst consider the hydrogen atom (H) and assume that its nucleus is motionless.

Because of the large difference between proton and electron masses, m

and m

= 1836), their center of mass coincides with the proton’s center of mass.

The stationary Schr

odinger equation (4.26) in the case of the potential deﬁned

by Eq. (6.1) can be written as

∇

ψ(r) +



E +



ψ(r) = 0. (6.2)

Since the potential within which the electron is conﬁned is spherically-symmetric,

it is convenient to seek the solution of Eq. (6.2) in spherical coordinates, whose

6.1 The hydrogen atom 157

origin (r = 0) coincides with the center of mass of the nucleus. In such a coordi-

nate system the electron wavefunction will depend on the distance of the electron

from the atom’s center, r , as well as on the azimuthal and polar angles, ϕ and θ,

respectively.

We will write the wavefunction for Eq. (6.2) in the form of a product of

two functions, as we have done already in the case of electron motion in a

spherically-symmetric potential well (see Eq. (4.27)):

ψ(r,ϕ,θ) = X(r)Y (ϕ, θ). (6.3)

According to Eq. (4.28), for a spherically-symmetric potential the function

(ϕ, θ) is the eigenfunction of the operator

with the eigenvalues of this

operator equal to l(l + 1) h

, where l is the orbital quantum number. This quan-

tum number can have values l = 0, 1, 2,... and the orbital magnetic quantum

number is equal to m = 0, ±1,...,±l. The forms of several functions Y

(ϕ, θ)

for l = 0, 1, 2 and m = 0, ±1, ±2 are given by the relations (4.37).

Let us substitute the wavefunction (6.3) into Eq. (6.2) and separate variables.

As a result we ﬁnd the following equation for the radial function, X(r):







E −

l(l + 1) h



X = 0. (6.4)

The solution of this equation depends on two quantum numbers, l and n, and can

be written as

(ρ) = ρ

−ρ/n

n−l−1



j=0

, (6.5)

where we introduced the variable ρ = r/r

, with

= 0.53 × 10

−10

m (6.6)

the radius of the ﬁrst Bohr orbit in the hydrogen atom for the ground state. The

distinctive feature of Eq. (6.5) is that it has a ﬁnite number of terms a

.Only

in this case will the exponential factor in Eq. (6.5) provide the vanishing of the

square of the modulus of the wavefunction at inﬁnity, i.e., at ρ →∞.

The principal quantum number, n, which deﬁnes the energy state of an electron

in an atom, can take on the values n = 1, 2, 3,...The total energy of the electron

is quantized (see Fig. 6.1) and is equal to

=−

, (6.7)

where E

is deﬁned as

=−

=−13.6eV. (6.8)

Here, E

is the energy of the electron ground state in the hydrogen atom (i.e., the

electron energy for the ﬁrst Bohr orbit). The magnitude |E

|=13.6eVisthe

158 Quantum states in atoms and molecules

ionization energy of the hydrogen atom. This energy is required to transfer the

electron from the ground energy state, E

(shown schematically in Fig. 6.1), to

the continuous energy spectrum (denoted by “free electron” in Fig. 6.1).

The coefﬁcients a

of the power series (6.5)for j > 0 can be found from the

following relation (we will not show here how to derive this relation because of

the complexity of the derivation):

j+1

= 2a

√

( j +l + 1) −1

( j + l + 2)( j + l + 1) −l(l + 1)

. (6.9)

Here,  = E/E

is a positive dimensionless parameter for the conﬁned states of

an electron, i.e., for the region of the discrete spectrum with E < 0. Relation

(6.9) is called recursive since it allows us to ﬁnd the coefﬁcient a

knowing the

coefﬁcient a

, to ﬁnd the coefﬁcient a

knowing the coefﬁcient a

, and so on.

The coefﬁcient a

is found from the normalization of the total wavefunction of

the hydrogen atom’s ground state, i.e.,



∞



2π



100

(r,ϕ,θ)

sin θ dr dϕ dθ = 1. (6.10)

It follows from Eq. (6.5) that the last term of the power series has the index

j = n − l − 1. Then, the coefﬁcient a

j+1

with j = n − l − 1 must be equal

to zero. According to Eq. (6.9), a

j+1

is equal to zero if the numerator in Eq. (6.9)

is equal to zero:

√

( j +l + 1) −1 = 0, or n

√

 − 1 = 0. (6.11)

Thus,

√

 = 1, i.e., n



E/E

= 1. (6.12)

From the last expression we get Eq. (6.7). The solution of Eq. (6.4) gives us the

eigenfunctions, X, and the corresponding eigenvalues, E. Equation (6.5)shows

that eigenfunctions X

depend on l for any particular n. In the general case

each speciﬁc eigenvalue, E

, must correspond to a certain eigefunction, X

However, the solution of Eq. (6.4) gives a different result and the eigenvalue

E depends solely on n and does not depend on l at all (see Eq. (6.7)). Since

the total energy, E

, does not depend on the orbital quantum number, l, the total

energy, E

, corresponds to several radial wavefunctions X

, i.e., the energy level

is degenerate. We have considered so far an individual hydrogen atom and

determined the energy levels for a single electron. If this hydrogen atom were to

interact with other atoms or were placed in an external electric or magnetic ﬁeld

then this degeneracy would be lifted and the total energy E

would depend on

the quantum number l as well.

For a given principal quantum number, n, the orbital quantum number, l, can

take on the following values:

l = 0, 1, 2,...,n − 1.

6.1 The hydrogen atom 159

Table 6.1. Notation of different quantum states

Quantum number l 012345

Symbol of state s p d f g h

The orbital magnetic quantum number, m, for each value of l can take on 2l + 1

values, and the total energy E

depends only on the principal quantum number,

n. Thus, all the electron states with n > 1 are degenerate, i.e., more than one

eigenfunction corresponds to the nth energy state. The order of degeneracy, g,is

deﬁned by the number of different electron states for a given energy E

g =

n−1



l=0

(2l + 1) = 1 +3 +5 +···+(2n − 1) = n

. (6.13)

Thus, for the state described by the wavefunction ψ

100

the order of degeneracy is

equal to unity and therefore the energy E

is non-degenerate. For the energy E

g is equal to four and corresponds to four different electron states in the atom

described by the wavefunctions

200

,ψ

210

,ψ

21−1

,ψ

21+1

As will be shown later, the order of degeneracy of the nth energy level is equal

to g = 2n

because the electron has intrinsic angular momentum (spin), which

can take on two values (we will talk about spin in Section 6.3).

6.1.2 Symbols deﬁning states and probability density

Different states of electrons in atoms, which correspond to different orbital

quantum numbers, l, are denoted by lower-case Latin letters. Table 6.1 shows

the correspondence of numbers and letters. When considering energy states we

usually talk about s-states (or s-electrons) for l = 0, p-states (or p-electrons) for

l = 1, and so on. In this notation of a quantum state the value of the principal

quantum number, n, is given before the symbol of the state with a given orbital

quantum number, l. Let us write the ﬁrst four groups of states for the hydrogen

atom: (1) 1s; (2) 2s, 2p; (3) 3s, 3p, 3d; and (4) 4s, 4p, 4d, 4f (see Table 6.1). Thus,

an electron in state 4f has n = 4 and l = 3. The wavefunctions ψ

nlm

(r,ϕ,θ)for

the ﬁrst two groups are given in Table 6.2.

The probability distribution of an electron’s location in an atom in the corre-

sponding stationary state is deﬁned by the magnitude |ψ

nlm

. Its spatial distri-

bution tells us about the form of the electron cloud.

Taking into account that an electron possesses not only mass but also charge,

the parameter

(r) =−e

nlm

(6.14)

can be interpreted as the electron charge density. The spatial motion of an electron

is equivalent to the existence of a current. The density of this current can be found

160 Quantum states in atoms and molecules

Table 6.2. The wavefunctions ψ

nlm

(r,ϕ,θ) for the ﬁrst two

groups of states (here, Z is the atomic number)

Wavefunction Expression State

100

√





3/2

−ρ

200

√

2π





3/2

(2 −ρ)e

−ρ/2

210

√

2π





3/2

ρe

−ρ/2

cos θ 2p

21−1

√





3/2

ρe

−ρ/2

sin θ e

−iϕ

21+1

√





3/2

ρe

−ρ/2

sin θ e

iϕ

048121620

(a.u.)

r/r

0.1

0.2

0.3

0.4

0.5

Figure 6.2 Distributions of

the radial functions r

for the 1s, 2p, and 3d

quantum states of the

hydrogen atom.

if we multiply the probability current density (2.190) in a given state by the charge

of the electron:

=−e

(ψ

nlm

∇ψ

∗

nlm

− ψ

∗

nlm

∇ψ

nlm

). (6.15)

In the 1s-state the orbital quantum number, l, and angular momentum, L,are

equal to zero. Therefore, the electron cloud in this state is spherically symmetric.

From the classical point of view this corresponds to electron motion only along

the radial direction – along r. The electron would have to cross the region

occupied by the nucleus, which is impossible in classical physics. To make this

result of quantum theory more understandable, let us consider the probability

of ﬁnding the electron within the volume dV :

dP =|ψ

nlm

dV.

6.1 The hydrogen atom 161

1s 2p, m = 0 2p, m = ±1

Figure 6.3 Apictorial

representation of the

electron clouds, |ψ

nlm

for the 1s and 2p states in

the hydrogen atom.

Let us choose a spherical shell of radius r, width dr, and volume dV = 4πr

dr.

Then, for the 1s-shell the magnitude

dP = 4πr

|ψ

100

dr (6.16)

deﬁnes the probability of ﬁnding an electron in this shell. The probability density,

dP/dr, is the probability of ﬁnding an electron in the spherical shell around the

radius r with width equal to unity. The radial dependences of r

of the three

states 1s, 2p, and 3d for a hydrogen atom are shown in Fig. 6.2. We see that the

maximum probability density for the quantum state 1s corresponds to r = r

which coincides with the radius of the ﬁrst Bohr orbit of the hydrogen atom. As

r goes to zero the probability density, dP/dr, also tends to zero, which means

that the electron is absent from the region occupied by the nucleus. The electron

clouds, |ψ

nlm

, corresponding to the 1s and 2p states in the hydrogen atom are

shown in Fig. 6.3. Note that |ψ

21±1

has a doughnut shape and Fig. 6.3 shows it

from the top.

Example 6.1. For the ground state of the hydrogen atom ﬁnd the average, mean-

square, and most probable distance between the electron and the nucleus.

Reasoning. The wavefunction of the ground state (n = 1, l = 0, m = 0) for the

hydrogen atom (Z = 1) is given by the expression

100

(r) =



πr

−r/r

(6.17)

(see Table 6.2). The average distance of an electron from the nucleus according to

the quantum-mechanical deﬁnition of the average value of the physical magnitude

(see Eq. (2.173)) can be obtained as follows:







∞

100

4πr

dr =



∞

−2r/r

dr. (6.18)

162 Quantum states in atoms and molecules

In order to calculate the integral, let us change variables from r to x:

x =

. (6.19)

As a result the integral takes the form











∞

−x

dx. (6.20)

Taking into account that



∞

−x

dx = n!, (6.21)

we get the ﬁnal result





. (6.22)

In order to ﬁnd the mean-square distance between the electron and the nucleus,

let us ﬁnd ﬁrst the following magnitude:







∞

100

4πr

dr =



∞

−2r/r

dr = 3r

. (6.23)

Thus, the mean-square distance, r

, is equal to







√

. (6.24)

Note that the integral in Eq. (6.23) was evaluated by using the same procedure

as for Eq. (6.18). To ﬁnd the most probable distance, r

, of an electron from the

nucleus, it is necessary to equate the derivative of the probability density deﬁned

by Eq. (6.16) to zero:







4πr

πr

−2r/r



−2r/r



1 −



= 0. (6.25)

From Eq. (6.25) it follows that r

= r

. This distance corresponds to the maxi-

mum of the probability density, dP/dr.

6.1.3 Orbital magnetic moment and orbital magnetic

quantum number

Since an electron is a charged particle, the current, I

, must be connected with

the electron motion around the atomic nucleus. The motion of an electron with

velocity v along a circular orbit of radius r with period T is equivalent to a

circular current, whose magnitude is given by the expression

=−

2πr

, (6.26)

where v is the orbital velocity of the electron. The minus sign in Eq. (6.26) indi-

cates that the electron velocity and current have opposite directions because