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

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6.5 The wavefunction of a system of identical particles 183

where σ

and σ

are the projections of the spin of each of the electrons

onto the chosen axis. Together with the solution (6.82) of the corresponding

Schr¨odinger equation, three additional wavefunctions, which are also solutions

of the Schr¨odinger equation, may be obtained from Eq. (6.82) by permutations of

spatial variables r

↔ r

and spin variables σ

↔ σ

. These four functions give

us two linear combinations, which have the correct type of symmetry – two linear

combinations are made antisymmetric. These wavefunctions can be written in

the following form:

,σ

; r

,σ

) = ψ

, r

)ϕ

(σ

,σ

), (6.83)

,σ

; r

,σ

) = ψ

, r

)ϕ

(σ

,σ

). (6.84)

Here, the symmetric and antisymmetric parts of the coordinate dependence of

the wavefunction of the system of two free electrons are given by the expressions

, r

) = A



)

+ e

)



, (6.85)

, r

) = A



)

− e

)



, (6.86)

and the symmetric and antisymmetric parts of the spin dependence by

(σ

,σ

) = ϕ









, (6.87)

(σ

,σ

) = ϕ







−



. (6.88)

Thus, in the symmetric spin state the spins of the two electrons are parallel to

each other and the total projection of the spin of the system is equal to S

= 1.

In the antisymmetric spin state the spins of the two electrons are anti-parallel

to each other and S

= 0. The procedure for ﬁnding the correct wavefunctions

is sometimes called symmetrization. The operation of symmetrization must be

carried out in all cases of systems with identical particles.

If there are more than two identical particles, then the system has such states,

whose wavefunctions, for particles with integer spins, are symmetric with respect

to permutations of any pair of particles (i.e., the wavefunctions do not change their

sign), and, for particles with half-integer spins, are antisymmetric with respect

to the permutations of any pair of particles (i.e., the wavefunctions change their

sign to the opposite). One of the formulations of the Pauli exclusion principle

which relates to a system of electrons is that the total wavefunction of the system

must be antisymmetric, since electrons are particles with half-integer spin.This

formulation is closely connected with the fact that it is not possible to have more

than one electron occupying the same quantum state.

For complex particles (nuclei, atoms, and molecules) the following rule

approximately applies: if a particle consists of an odd number of fermions (elec-

trons, protons, or neutrons), then its spin is half-integer and it behaves as a

fermion. If the complex particle consists of an even number of fermions, then

184 Quantum states in atoms and molecules

its spin is integer and it behaves as a boson. Looking at a particular example,

individual helium atoms of types

and

are chemically indistinguish-

able. The atom

is a fermion since it contains two electrons and a nucleus,

which consists of two protons and one neutron. The atom

is a boson since

its nucleus contains one additional neutron. This difference substantially affects

the properties of a system consisting of a large number of the considered iso-

topes of helium. Liquid helium of type

at temperature T ≈ 2 K becomes a

superﬂuid, but liquid helium of type

does not exhibit such a property. This

is because the creation of a so-called Bose–Einstein condensate is possible only

in a system of bosons. Its main property is the transition of the entire ensemble

of particles to the lowest energy level when the phase-transition temperature is

reached, which for

is approximately equal to 2 K.

Example 6.6. Show that, if at some time a quantum system that consists of

identical particles is in the state described by the symmetric wavefunction 

then it will be described by the symmetric wavefunction for all subsequent times.

Reasoning. Let us write the time derivative of the wavefunction as

d

→



t

, (6.89)

where by 

we understand the change of the wavefunction during the time t.

Let us substitute expression (6.89) into the time-dependent Schr¨odinger equation

(2.182), which describes the initial state of the system:



H

t. (6.90)

Since the Hamiltonian

H is symmetric with respect to the system of coordinates of

the particles, the wavefunction

H

is also a symmetric function of coordinates.

Therefore, in the process of evolution of the wavefunction, which is deﬁned by

Eq. (6.90), its symmetry does not change. Let us note that the preservation of

the symmetry of the wavefunction is a universal property. The same argument

applies also to an antisymmetric wavefunction.

6.6 The hydrogen molecule

Let us consider one more important example where the Pauli exclusion principle

considerably affects the formation of stationary quantum states. This is the system

of two atoms of hydrogen, which form a hydrogen molecule. The wavefunctions

of each of the electrons in their atoms substantially differ from zero only in a

small area close to their nucleus. If the atoms are separated from each other by

such a distance that they can be considered to be independent, the wavefunctions

of electrons do not overlap and for this reason it is meaningless to make the

total wavefunction of the system of two electrons antisymmetric. If the atoms are

brought together to a distance such that the wavefunctions of individual electrons

6.6 The hydrogen molecule 185

Figure 6.13 The scheme of

the hydrogen molecule.

The nuclei are designated

as a and b, the electrons

as 1 and 2.

overlap, then the quantum states of such a system of electrons must be described

by a total antisymmetric wavefunction.

Let us place the nuclei a and b of two hydrogen atoms such that they are

separated by a distance R (see Fig. 6.13). Let us use an adiabatic approximation,

whereby the kinetic energy of the nuclei can be neglected and we can consider

the nuclei as motionless. In the zeroth approximation we can consider the atoms

as isolated and therefore the interaction between them can be neglected. Electron

1 near nucleus a and electron 2 near nucleus b are described by the atomic

wavefunctions of the 1s-state given in Table 6.2. Let us denote these functions as

) ≡ ψ

(1), (6.91)

) ≡ ψ

(2). (6.92)

These functions satisfy the following equations:

(0)

(1) = E

(1), (6.93)

(0)

(2) = E

(2), (6.94)

where the Hamiltonians of the unperturbed atoms are as follows:

(0)

=−

∇

−

, (6.95)

(0)

=−

∇

−

. (6.96)

Here E

, which is deﬁned by Eq. (6.8), is the energy of the electron in the ground

state of the hydrogen atom. The total energy of the electron in the hydrogen

molecule in the zeroth approximation is equal to twice the energy of the electron

in the hydrogen atom, i.e.,

(0)

= 2E

=−

. (6.97)

If we consider that the atoms are not isolated we have to take into account all

the interactions in the hydrogen molecule: interaction of electron 1 with nucleus

186 Quantum states in atoms and molecules

b, interaction of electron 2 with nucleus a, and interaction of electron 1 with

electron 2. In this case the Schr¨odinger equation for the hydrogen molecule will

be the following:

[

(0)

W ]ψ(1, 2) = E

ψ(1, 2), (6.98)

where the unperturbed Hamiltonian,

(0)

, is deﬁned as

(0)

. (6.99)

The perturbation term,

W , is the potential energy of the interactions mentioned

above:

W = W(1, 2) =



−



. (6.100)

Since the nuclei are considered motionless, their interaction is not included in the

total Hamiltonian, and therefore we will take into account this term by adding

/R to the total energy.

The wavefunctions of the unperturbed Hamiltonian of the hydrogen molecule,

(0)

, can be constructed from two one-particle wavefunctions (6.3). In order for

the total wavefunction of the system of two electrons to be antisymmetric, the

radial part of this function must be symmetric in the case of the total spin equal

to S = 0 and antisymmetric in the case of the total spin equal to S = 1, i.e.,

(0)

(1, 2) = ψ

(1)ψ

(2) + ψ

(2)ψ

(1), (6.101)

(0)

(1, 2) = ψ

(1)ψ

(2) − ψ

(2)ψ

(1). (6.102)

These functions have the following property: after permutation of coordinates

(replacement r

↔ r

, i.e., 1 ↔ 2) the symmetric function stays the same and

the antisymmetric one changes its sign:

(0)

(1, 2) = ψ

(0)

(2, 1), (6.103)

(0)

(1, 2) =−ψ

(0)

(2, 1). (6.104)

Using the methods of perturbation theory for degenerate stationary states (see

Section 5.2), we can show that under the inﬂuence of perturbation,

W ,the

unperturbed degenerate level E

splits into two. The one of these levels with lower

energy, E

, corresponds to the symmetric wavefunction, ψ

(0)

(1, 2), and the other

one, with higher energy, E

, corresponds to the antisymmetric wavefunction,

(0)

(1, 2). The expressions for the energies of the symmetric and antisymmetric

states have the following forms:

= 2E

+ W

(R), (6.105)

= 2E

+ W

(R), (6.106)

6.6 The hydrogen molecule 187

where

Q + A

1 + F

(6.107)

and

Q − A

1 − F

. (6.108)

Here we introduced the following quantities:

Q =



|ψ

(1)|

W (1, 2)|ψ

(2)|

, (6.109)

A =



∗

(1)ψ

(2)W (1, 2)ψ

∗

(2)ψ

(1)dV

, (6.110)

F =



∗

(1)ψ

(1)dV



∗

(2)ψ

(2)dV

. (6.111)

The quantity Q is called the Coulomb integral since it includes the energy of

the electrostatic interaction of the electron clouds with each other and with the

atomic nuclei. The electron clouds have the following electric charge density:

(1) =−e|ψ

(1)|

, (6.112)

(2) =−e|ψ

(2)|

. (6.113)

The quantity A, which is called the exchange integral, is closely connected with

the wave nature of the electron. In this integral there are two “exchange charge

densities”:

(1) =−eψ

∗

(1)ψ

(1), (6.114)

(2) =−eψ

∗

(2)ψ

(2), (6.115)

which can be interpreted as if each of the electrons belonged to nucleus a and

nucleus b simultaneously.

The quantity F is called the overlap integral and it is a measure of the

overlap of the wavefunctions of atoms (i.e., overlap of electron clouds) at a given

distance between them, R. All the above-mentioned quantities are functions of

this distance, R, and consequently the corrections to the energy, E

(Eqs. (6.107)

and (6.108)), also depend on R.

In analyzing the expressions (6.105)–(6.111) for the energy of the system of

electrons E

and E

we emphasize that, when atoms approach each other,

the energy level E

= 2E

deﬁned by Eq. (6.97) splits into two sublevels.

The magnitude of the splitting depends on the distance between nuclei and is

equal to

E(R) = E

(R) − E

(R) = W

(R) − W

(R) =−2

A − QF

1 − F

. (6.116)

Under the assumption of the adiabatic approximation, the energy of the electrons

plays a role as one of the components of the potential energy of interaction of the

188 Quantum states in atoms and molecules

−

Figure 6.14 Tw o

dependences of the

potential energy U on R:

(1) the dependence with

the correction W

(total

spin equal to S = 1) and

(2) the dependence with

the correction W

(total

spin equal to S = 0). Here

= U(R) − 2E

and

= 0.74

atoms. The total potential energy of interaction of the hydrogen atoms, including

the energy of the Coulomb repulsion of the nuclei, is deﬁned by the following

expression:

U (R) = 2E



(R),

(R).

(6.117)

The total potential energy of interaction of the hydrogen atoms strongly

depends on the total spin of the system of electrons. Figure 6.14 shows two

dependences of the potential energy U(R): (1) the dependence that corresponds

to the correction W

and to the total spin S = 1 and (2) the dependence that

corresponds to the correction W

and to the total spin S = 0. The monotonic

decrease of energy with increasing R in the state with parallel spins means that

atoms repel each other. Thus, it is impossible to form the hydrogen molecule

with S = 1. The minimum potential energy for the hydrogen molecule, U (R

) =

−4.73 eV, which occurs at R

= 0.74

A ≈ 1.4r

, occurs only for the state with

opposite spins. Therefore, for this state (S = 0) the bonding between hydrogen

atoms, which is called covalent bonding, occurs. This type of bonding plays a

signiﬁcant role in the formation of molecules as well as of numerous crystals.

Example 6.7. Estimate the oscillation frequency of the nuclei of the hydrogen

molecule in the ground state and the dissociation energy of the hydrogen molecule

(i.e., the energy required for the breakup of the hydrogen molecule into separate

atoms).

Reasoning. The adiabatic approximation can be used for the solution of the

problem of oscillations of nuclei in the hydrogen molecule. To ﬁnd the wave-

function of the nuclei (R) in a given quantum state we will take the energy of

6.6 The hydrogen molecule 189

the electrons, E

(R), as the potential energy of the interaction of nuclei. Let us

consider the motion of the nuclei in the system of their center of mass. Then, the

wavefunction of the nuclei, (a, b), will depend only on the relative distance,

R, between them and deﬁne their relative motion under the inﬂuence of internal

forces. To ﬁnd this function it is necessary to solve the following Schr¨odinger

equation:



−

+U (R)



(R) = E(R), (6.118)

where the reduced mass of two identical nuclei, m

, is equal to the half-mass of

the proton,

+ m

, (6.119)

and E is the total energy of the hydrogen molecule, taking into account the motion

of the nuclei. The magnitude U (R) is the total potential energy of the nuclei and

it is deﬁned by Eq. (6.117). It includes the energy of the Coulomb interaction of

the nuclei and the energy of the electrons, E

(R), in the given quantum state. To

ﬁnd the frequency of the small-amplitude oscillations of the nuclei let us expand

the function U (R) near the equilibrium position of the nuclei R

in series over the

displacement R − R

and let us keep only the ﬁrst two terms of the expansion:

U (R) = U (R

) +

(

R − R

)

, (6.120)

where

β =





R=R

. (6.121)

Taking into account the relation of the elastic constant with frequency, we get

ω =









R=R

. (6.122)

The expression for the energy U(R) can be rewritten in the form

U (R) = U (R

) +

(

R − R

)

. (6.123)

On choosing R

as the origin of coordinates and U (R

) as the reference point for

energy, i.e., taking R

= 0 and U(R

) = 0, we get the equation



−



(R) = E(R), (6.124)

which coincides with the one-dimensional Schr¨odinger equation for an electron

in the parabolic potential (4.74). On solving this equation we get the quantized

harmonic oscillator states for the system of the nuclei, whose energy according

190 Quantum states in atoms and molecules

to Eq. (4.88) is given by the expression

osc

= h



n +



, (6.125)

where n is the quantum number of the oscillator. The ground state of the hydrogen

molecule corresponds to the energy of the n = 0 oscillator state:

osc

= 0.27 eV. (6.126)

This value of oscillatory motion energy was established from measurements of

the optical spectra of the hydrogen molecule. From Eq. (6.126) it follows that

the oscillations of the nuclei happen with frequency ω = 8.6 ×10

−1

.The

oscillatory motion of the nuclei splits the electron levels into closely spaced

oscillatory sublevels. Then, the energy levels of the hydrogen molecule are given

by the expression

E(R) = E

) +

+ h



n +



. (6.127)

The energy of dissociation, E

dis

, of the hydrogen molecule is deﬁned by the

minimal energy necessary for dividing the hydrogen molecule into two individual

atoms:

dis



U (R

) +



=|−4.73 +0.27| eV = 4.46 eV. (6.128)

We note that a more precise solution of the problem for the hydrogen molecule

requires the taking into account of two more factors – rotational motion of

the molecule and the fact that the nuclei are identical. Under the inﬂuence of

rotational motion the oscillatory energy levels split into a system of rotatory

sublevels. The distance between these sublevels is of the order of 8 × 10

−3

eV.

6.7 Summary

1. The electron state in the hydrogen atom is deﬁned by the set of four quantum numbers:

the principal quantum number, n, orbital quantum number, l, orbital magnetic quantum

number m, and spin magnetic quantum number m

. The electron energy spectrum of

the hydrogen atom, which is deﬁned only by the principal quantum number, n,is

discrete and all quantized levels are in the region of negative energy values.

2. One of the main principles of quantum physics is the Pauli exclusion principle, accord-

ingtowhichaparticular state, which is characterized by a given set of quantum num-

bers (in the case of an atom the quantum numbers are n, l, m, m

), may be occupied

by only one electron.

3. Considering the system of identical particles, we have to take into account that in

a quantum description the particles are distinguishable. Systems of particles with

integer spin (bosons) must be described by symmetric wavefunctions, whereas systems

6.8 Problems 191

of particles with half-integer spin (fermions) must be described by antisymmetric

wavefunctions. Since electrons have spin s = 1/2, the wavefunction of the electron

system must be antisymmetric.

4. All electrons of a many-electron atom with the same principal quantum number, n,

form a shell, which has 2n

states. Shells consist of subshells, which differ by the orbital

quantum number l. The number of different states in a subshell with different quantum

numbers m and m

is equal to 2(2l + 1). Completely ﬁlled shells and subshells have

the resulting quantum numbers of the orbital, spin, and total moment equal to zero:

L = 0, S = 0, and J = 0.

5. The order of ﬁlling by electrons of the atom’s subshells is deﬁned by Hund’s rules:

(1) the electron conﬁguration has the minimal energy in the state with the largest spin

S and with the largest (for such S) L; (2) for a less than half-ﬁlled subshell the quantum

number J is equal to J =|L − S|, whereas for a more than half-ﬁlled subshell J is

equal to J = L + S.

6. The covalent bonding between hydrogen atoms is caused by the exchange interaction,

and this type of bonding occurs for the state with anti-parallel spins (S = 0). This

type of bonding plays an important role in the formation of molecules and numerous

crystals.

6.8 Problems

Problem 6.1. Using Bohr’s model of the hydrogen atom and his postulates, ﬁnd

the radii of electron stationary orbits, r

, electron velocities, v

, orbital periods,

, the electron energy on the corresponding orbit, and the circular frequency of

a photon emitted during electron transition from the nth to the mth orbit. Find

the radius and velocity, r

and v

, for the ﬁrst electron orbit. Find the angular

frequency, ω

, of the electron transition from the second to the ﬁrst orbit.

(Answer: r

≈ 0.53

A, v

≈ 2.19 × 10

−1

, and ω

≈ 1.53 × 10

−1

Problem 6.2. Find the minimum, λ

min

, and maximum, λ

max

, wavelength of the

hydrogen spectral lines in the visible range of the spectrum. What is the minimum

speed, v

min

, of electrons incident on a hydrogen atom to excite these spectral

lines? (Answer: λ

min

≈ 365 nm, λ

max

≈ 656 nm, and v

min

≈ 8.2 × 10

−1

Problem 6.3. The electron in the hydrogen atom is in the ground state described

by the following wavefunction:

(r) =



πr



−1/2

−r/r

. (6.129)

Find the average electrostatic potential at the distance r from the nucleus.

Problem 6.4. Find the spectrum of the vibrational levels of a diatomic molecule.

For the diatomic CO molecule, estimate the distance between vibrational energy

levels, E

n+1,n

, and the emission wavelength, λ, during the corresponding

transitions. The carbon atom mass m

= 1.99 × 10

−26

kg, the oxygen atom

192 Quantum states in atoms and molecules

mass m

= 2.66 × 10

−26

kg, and the force constant β = 190 kg s

−2

.(Answer:

E

n+1,n

≈ 8.44 × 10

−2

eV and λ ≈ 24 µm. Hint: use Eq. (6.122).)

Problem 6.5. Find for the ﬁrst-order perturbation theory the energy, E

,ofthe

ground state of the helium, He, atom. Use as the perturbation the energy of

interaction between electrons. (Answer: E

≈−18.7 eV.)

Problem 6.6. In the alkaline atoms the nucleus with charge eZ and the ﬁrst Z − 1

electrons form a core with positive charge equal to the elementary charge, e.A

valence electron rotates in the ﬁeld of this core. Find the energy spectrum of the

valence electron if its potential energy is deﬁned by the following expression:

U (r) =−



1 +



, (6.130)

where we can consider C/r

 1(r

is the Bohr radius).

Problem 6.7. Find the angular momentum for electrons in an atom in 3s, 4d, and

5g states and the maximum number of electrons in an atom with the same sets

of quantum numbers n, l, and m.