Download QUANTUM CHEMISTRY AND GROUP THEORY(2) M.Sc. DEGREE

QUANTUM CHEMISTRY AND GROUP THEORY(2)
M.Sc. DEGREE (C.S.S), Chemistry (1st Sem)
̂ of spherical polar co-ordinates.
1. Write equation for 𝐿̂2 and 𝐿𝑧
𝜕
Lz = -iħ𝜕∅
1 𝜕
∂
1
𝐿̂2 = -ħ2 [ sinθ 𝜕𝜃 (sinθ ∂θ) + Sin2θ
𝜕2
]
𝜕𝜙2
2. Explain spin postulates
Postulates of spin by Goudsmith
Uhienbeck and Goudsmith ,in 1925, proposed that in addition to the orbital angular
momentum an electron possessed an intrinsic angular momentum with quantum number s
= ½ and which could exist in two states with ms = ½ denoted as ∝ and ms =(-) ½ denoted
as 𝛽
Since the inherent spin angular momentum of a microscopic particle has no analog in
classical mechanics, we cannot construct operators for spin. Therefore we simply use
symbols for the spin operators without giving an explicit form for them.
Analogous to the orbital-angular momentum operators L2 , Lx , Ly, Lz we have the spin
angular momentum operators S2, Sx, Sy, Sz, which are postulated to be linear and
hermitian
S2 is the operator for the square of the magnitude of the total spin angular momentum of a
particle. S =√𝑠(𝑠 + 1) ħ and Sz = msħ the z component of the spin angular momentum.
1
1
s=2 and ms = ± 2
The spin angular momentum operators obey the same commutation relations as the
orbital angular momentum oprators. S2 commute with any of the components, Sx, Sy, Sz,
such that
[ S2, Sx] = [S2, Sy] = [S2, Sz] = 0 and the components, Sx, Sy, Sz, do not
commute each other.
3. Prove that a cyclic group is always Abelian
In certain groups ,every two elements may commute. Such groups are called abelian. All
pure rotational groups are abelian.
Sometimes all the elements of a group can be derived from one basic element. Such
group is referred to as cyclic group.
In C3 point group, the elements are E, C3, C32
C3 is the basic element here and the other elements can be obtained as C32 = C3 xC3
and E = C3 x C3 x C3
E x C3 = C3 x E
C32 x C3 = C32 x C3
E x C32 = C32 x E
4. Derive a general expression for the matrix representation of 𝜎𝑣̂ using the basis (x,y,z).
What is the character of this matrix.
Reflection through a plane 𝜎𝑥𝑧 changes the coordinate point (x, y, z) as x =x1, -y = y1
and z =z1. Hence the matrix is given as
1 0 0
𝜎𝑥𝑧 = [0 −1 0]
0 0 1
Character = 1
Reflection through a plane 𝜎𝑦𝑧 changes the coordinate point (x, y, z) as -x =x1 ,y = y1
and z =z1. Hence the matrix is given as
−1 0
𝜎𝑦𝑧 = [ 0 1
0 0
0
0]
1
Character = 1
5. State great Orthogonality theorem. What are the consequences of the theorem.
The GOT is used to generate the limited set of irreducible representations that are
allowed for each group. GOT results in a number of important relationships among the
characters of representations and among representations themselves.
Important rules derived from GOT are
Rule 1: The number of irreducible representations (IRs) of a group is equal to the number
of classes in the group.
Rule 2 : In a given representation the characters of all the elements of the same class will
be identical.
Rule 3: The sum of the squares of the dimensions of the IRs of a group is equal to the
order of the group. If l1, l2, l3,….. = ∑ li2 = h the order
Rule 4: The sum of the squares of the characters in a given IR is equal to the order of the
group. ∑[𝜆𝑖(𝑅)]2 = h
If different classes occur in a group the above equation takes the form, ∑gc[𝜆𝑖(𝑅𝑐)]2 = h
Where R is a class of operations Rc and gc is the order of the class.
Rule 5: Any two different IRs are orthogonal. ie. The characters of any two IRs of a
group are orthogonal means ∑gc λi (Rc) λj (Rc) = 0.where i≠j
6. Set up the Schrodinger equation and find eigen values and eigen function for a particle
moving in a ring.
Let us consider a circle of radius r and the particle is free to move along the
circumference of the circle. The mass of the particle is m. We have to find allowed
energy levels and associated wave functions by solving the time independent Schrodinger
equation.
ħ2
𝜕2
𝜕2
If the Cartesian coordinates are x and y, then the kinetic energy = -2𝑚 ( 𝜕𝑥2 + 𝜕𝑦2 ). But it
is difficult to use x and y Cartesian coordinates in such a system. So we have to use
plane polar co ordinates. Instead of x and y we can use r and ϕ , where r is the radius of
the circle and ϕ is the angle makes between the line joining the position of the particle
and origin with x axis.
In plane polar coordinate system ,
𝜕2
𝜕2
( 𝜕𝑥2 + 𝜕𝑦2 ) =
𝜕2
𝜕𝑟2
1 𝜕
1 𝜕2
+𝑟 𝜕𝑟 + 𝑟2 𝜕𝜙2
-
ϕ = 0 and ϕ = 2𝜋 refers to same point. (r,ϕ) is same as (r, 𝜙 + 2𝜋).
𝜕2
𝜕2
Since r is fixed ( 𝜕𝑥2 + 𝜕𝑦2 ) =
1 𝜕2
𝑟2 𝜕𝜙2
ħ2 1 𝜕2
So kinetic energy is -2𝑚 𝑟2 𝜕𝜙2
We assume potential energy is constant in the system and for the sake of convenience it is
taken as zero. Moment of inertia of the particle mr2 = I
̂ = - ħ2 𝜕2
So 𝐻
2𝐼 𝜕𝜙2
The Hamiltonian is independent of time, so we can find stationary state.
̂Φ = E Φ
𝐻
Φ is a function of angle ϕ
ħ2 𝜕2
- 2𝐼 𝜕𝜙2Φ(ϕ) = E Φ(ϕ)
Since the wave function depend only ϕ ,no need partial differentiation.
ħ2 𝑑2
- 2𝐼 𝑑∅2 Φ(ϕ) = E Φ(ϕ)
𝑑2
Φ(ϕ) =
𝑑∅2
Let
𝑑2
𝑑∅2
−2𝐼𝐸
−2𝐼𝐸
ħ2
Φ(ϕ)
= m2 l
ħ2
Φ(ϕ) = - m2l Φ(ϕ)
Then Φ(ϕ) = A 𝑒 𝑖m𝑙𝜙
Since ϕ and (ϕ +2𝜋) refers to the same point, Φ is same at ϕ and (ϕ +2𝜋), otherwise the
wave function become multiple valued.
So Φ(ϕ) = Φ(2π + ϕ)
A 𝑒 𝑖m𝑙𝜙 = A 𝑒 𝑖m𝑙(𝜙+2𝜋)
1 = 𝑒 𝑖m𝑙.2𝜋
Since ml have to satisfy this equation, the possible values of ml is 0, ±1, ±2, ±3,…
We have
2𝐼𝐸
ħ2
= m2 l
So Energy of the particle, E = m2l (ħ2/2I)
2I
2
E0 = 0, E1 = ( ħ /2I), E2 = (4ħ2/2I) , E3 = (9 ħ2/2I) , …
Ground state energy is zero, it is non-degenerate while all other levels are doubly
degenerate. This degeneracy results from the fact that the particle may circulate in either
sense and therefore be in states of different angular momentum. The absence of
degeneracy in the ground state is because there is then no angular momentum and so the
question of its direction does not arise.
Evaluation of A
Φ(ϕ) = A 𝑒 𝑖m𝑙𝜙
Using the normalization condition,
2𝜋
∫0 A 𝑒 −𝑖m𝑙𝜙 A 𝑒 𝑖m𝑙𝜙 dϕ = 1
2𝜋
A2 ∫0 𝑑𝜙 = 2𝜋 A2 = 1 Hence A = √(1/2𝜋)
Φ(0) = √(1/2𝜋) , is independent of the value of ϕ. When the particle is
in a state of definite angular momentum its distribution is completely uniform.
The wave function corresponding to a definite state of angular momentum mlħ is
Φml (ϕ) = √1/2𝜋 𝑒 𝑖mlϕ = √1/2𝜋 (cos mlϕ + i sin mlϕ). Since the wave functions are
complex (unless ml = 0), they can be represented by plotting only the cosine component
and remembering that there is also a sine component of exactly the same shape but
displaced by a quarter of a wavelength.
Φ(0)
ϕ(±1)
The auxiliary function representation if ϕ(±1) and ϕ(±2)
7. Derive a general expression for the matrix form of rotation operation in the basis of
(x,y,z).
The point P (x, y, z) changes to P1(x1,y1,z1) during rotation through θ where θ = 3600/n ,
i.e. rotation through a Cn. The distance OP is equal to r and PQ and P1Q1 are y and y1
respectively. Similarly OQ and OQ1 are x and x1 respectively. During the Cn operation,
OP changes to OP1 rotates through an angle θ
∠POQ = ϕ and ∠P1OQ1 = (θ +ϕ),
𝑂𝑄
𝑃𝑄
𝑦
sin ϕ = 𝑂𝑃 = 𝑟 ,
y = r sinϕ
𝑥
cos ϕ = 𝑂𝑃 = 𝑟 , x = r cos ϕ
P1Q1
sin (θ +ϕ) = sinθcosϕ + cosθsin ϕ =
OP1
=
y1
𝑟
r sinθcosϕ + r cosθsin ϕ = y1
x sinθ + y cosθ = y1 (since rcosϕ = x and rsinϕ =y)
OQ1
cos(θ + ϕ) = cosθcosϕ –sinθsinϕ = OP1 =
𝑥1
𝑟
r cosθcosϕ – r sinθsinϕ = x
x cosθ – y sinθ = x1
As the Cn rotation is about the z axis, z = z1
These expressions may be written as simultaneous equations
x cosθ - y sinθ +0z = x1
x sinθ + y cos θ + oz = y1
0x + 0y + z =z1
The corresponding matrix product would be
1
cosθ
[ sinθ
0
−sinθ 0 𝑥
cosθ 0] 𝑦
0
1 𝑧
x1
= y1
z1
Hence the general matrix for Cn is
cosθ
Cn = [ sinθ
0
−sinθ 0
cosθ 0]
0
1
Substituting appropriate values for θ, the matrix for any specific rotational operation
can be obtained from the general matrix.
8. Outline the essential postulates of quantum mechanics.
Significance of Postulates
The development of quantum mechanics is based on some basic postulates which can
neither be derived nor be proved. These postulates provide a convenient framework for
summarizing the basic concepts of quantum mechanics. The quantum mechanical
postulates have been extensively tested since they were proposed. The predicted
outcome is always found to be in agreement with the experimental observation. Hence
these postulates are universally accepted even though they cannot be deduced or proved.
Postulate 1
The state of a quantum mechanical system at any instant is completely specified by a
state function or wave function. Wave function is a function of position and time. It can
be expressed as Ψ (x, y, z, t) for a particle in 3D box. The probability of finding a particle
in the volume element dτ is Ψ *(x, y, z, t) Ψ (x, y, z, t) dτ at time t.
The wave function must satisfy certain mathematical conditions because of this
probabilistic interpretation. For the case of a single particle, the probability of finding it
somewhere is 1, so that we have the normalization condition
∫ Ψ*(x, y, z, t) .Ψ (x, y, z, t) dτ = 1
The wave function must also be single valued, continuous and finite to become
acceptable or well behaved.
Postulate 2
To every observable in classical mechanics there corresponds a linear Hermitian operator
in quantum mechanics.
Any measurable dynamical variable is called an observable. The classical mechanical
expressions for these observables can be written in terms of position co ordinates and
momentum coordinates of the particles that comprise the system.
Examples are, variable dynamics such as position, momentum, kinetic energy, potential
energy
To find the corresponding operator of an observable, write down the classical
mechanical expression for it in terms of Cartesian coordinates and corresponding linear
momentum components (px, py, …..) and replace each coordinate x by X and each
𝜕
momentum component px by the operator, Px = -iħ𝜕𝑥
Examples:
Postulate 3
Any measurement of an observable will lead to an eigen value of the associated operator.
AΨn = aΨn Where a is the eigenvalue and Ψ is a well-behaved eigenfunction. The operators
corresponding to these observable properties are Hermitian and the eigenvalues obtained are real.
Postulate 4
The expectation value of an observable is given by ∫Ψ*AΨ.dτ = <A> when Ψ is normalized. For
example, if we want to measure the energy of each member of a collection of similarly prepared
systems, each described by ,Ψ, then the average of the observed values is given by the above
equation.
Postulate 5
The wave function Ψ(x ,t) obeys the time-dependent Schrodinger equation.
𝜕
iħ
𝜕𝑡
Ψn(x, t)
̂ Ψn(x, t)
=𝐻
Let Ψn(x, t) = ψn(x).F(t) Where ψ & F are acceptable wave functions.
Solving the above equation we obtain, the time dependent part F(t) = e -iEt/ħ and
̂ ψn(x) = E ψn(x). ie the eigen value of hamiltonian operator is energy of the system. So we have
𝐻
Ψn(x ,t) = ψn(x). e -iEt/ħ , ∫ Ψn*(x ,t), Ψn(x ,t) = ∫ ψn *(x). e -iEt/ħ. ψn(x). e -iEt/ħ dx
= ∫ ψn *(x). ψn(x). dx = 1 for a normalized wave function . Let = ψ0, ψ1 , ψ2, ψ3,…….. be the
different eigenfunctions of the Hamiltonian , each of them will have its own associated eigen
value E0, E1, E2, E3, …… these solutions are referred to as stationary states for the system
because the probability ∫ ψn *(x). ψn(x). dx is explicitly time independent.