Operators Matrices and Group Theory

Chapter 13
Operators, Matrices, and Group Theory
Exercise 13.3. Find the commutator [x 2 ,
EXERCISES
Exercise 13.1. Find the eigenfunctions and eigenvalues of
√
d
the operator i , where i = −1.
dx
i
df
= af
dx
[x 2 ,
2
d2
d2 2
2 d
]
f
=
x
f
−
(x f )
dx 2
dx 2
dx 2
2
d
2d f
2df
−
= x
2x f + x
dx 2
dx
dx
2
d2 f
df
2d f
−
x
−
2
f
−
2x
dx 2
dx
dx 2
df
= −2 f − 2x
dx
2 d
d
x 2 , 2 = −2 − 2x
dx
dx
= x2
Separate the variables:
df
df
a
dx =
= = −ia
dx
f
i
ln ( f ) = −iax + C
f = eC e−iax = Ae−iax
df
= −ia Ae−iax = −ia f
dx
The single eigenfunction is Ae−iax and the eigenvalue is
−ia. Since no boundary conditions were specified, the
constants A and a can take on any values.
Exercise 13.2. Find the operator equal to the operator
d2
product 2 x.
dx
df
d2 f
df
d2
d
df
x
+
f
=
x
+
x
f
=
+
dx 2
dx
dx
dx 2
dx
dx
= x
df
d2 f
+2
2
dx
dx
The operator equation is
d2
d2
d
x
=
x
+2
dx 2
dx 2
dx
Mathematics for Physical Chemistry. http://dx.doi.org/10.1016/B978-0-12-415809-2.00059-8
© 2013 Elsevier Inc. All rights reserved.
d2
].
dx 2
3 .
= x + d , find A
Exercise 13.4. If A
dx
d
d
3 = x + d
x+
x+
A
dx
dx
dx
d
d2
d
d
x2 +
x+x
+ 2
= x+
dx
dx
dx
dx
d2
d
d
d 2
d2
x + x2
+x 2 +
x + 2x
dx
dx
dx
dx
dx
3
d
d d
+ 3.
+ x
dx dx
dx
Exercise 13.5. Find an expression for B 2 if B = x(d/dx)
2
4
and find B f if f = bx .
d 2
d
df
B2 f = x
f =x
x
dx
dx
dx
2
df
d f
d2 f
df
+x 2 =x
+ x2 2
= x
dx
dx
dx
dx
2
d f
df
+ x 2 2 bx 4
B 2 (bx 4 ) = x
dx
dx
= x3 + x
= 4bx 4 + x 2 (4)(3)bx 2 = 16bx 4
e79
e80
Mathematics for Physical Chemistry
Exercise 13.6. Show that the solution in the previous
example satisfies the original equation.
d2 y
dy
+ 2y = 0
−3
2
dx
dx
d 2x
d2 2x
x
x
c
c
−
3
e
+
c
e
e
+
c
e
1
2
1
2
dx 2 dx
+2 c1 e2x + c2 e x
= 4c1 e2x + c2 e x − 3 2c1 e2x + c2 e x
+ 2 c1 e2x + c2 e x = 0
Exercise 13.7. Find the eigenfunction of the Hamiltonian
operator for motion in the x direction if V (x) = E 0 =
constant.
∂2
+
Ė
−
0 ψ = Eψ
2m ∂ x 2
2m ∂ 2ψ
E − E 0 ψ = −κ 2 ψ
=−
∂x2
where we let
2m
(E − E 0 )
If E E 0 the real solution is
κ2 =
ψreal = c1 sin (κ x) + c2 cos (κ x)
In order for ψreal to be an eigenfunction, either c1 or c2 has
to vanish. Another version of the solution is
ψcomplex = b1 eiκ x + b2 e−iκ x
In order for ψcomplex to be an eigenfunction, either b1 or b2
has to vanish.
Exercise 13.8. Show that the operator for the momentum
in Eq. (13.19) is hermitian.
We integrate by parts
∞ ∗ dψ
∗ ∞
∞ dχ ∗
dx = χ ψ −∞ −
ψdx
χ
i −∞
dx
i
i −∞ dx
∞
dχ ∗
ψdx
=−
i −∞ dx
The other side of the equation is, after taking the complex
conjugate of the operator
∞ dχ ∗
−
ψdx
i −∞ dx
which is the same expression.
Exercise 13.9. Write an equation similar to Eq. (13.20) for
the σ̂v operator whose symmetry element is the y-z plane.
σ̂v(yz) (x1 ,y1 ,z 1 ) = (− x1 ,y1 ,z 1 )
2(x) (1,2, − 3).
Exercise 13.10. Find C
2(z) (1,2, − 3) = (− 1, − 2, − 3)
C
Exercise 13.11. Find S2(y) (3,4,5).
S2(y) (3,4,5) = (− 3. − 4, − 5)
Exercise 13.12. List the symmetry elements of a uniform
cube centered at the origin with its faces perpendicular to
the coordinate axes.
The inversion center at the origin.
Three C4 axes coinciding with the coordinate axes.
Four C3 axes passing through opposite corners of the cube.
Four S6 axes coinciding with the C3 axes.
Six C2 axes connecting the midpoints of opposite edges.
Three mirror planes in the coordinate planes.
Six mirror planes passing through opposite edges.
Exercise 13.13. List the symmetry elements for
a. H2 O (bent)
E,C2(z) ,σx z ,σ yz
b. CO2 (linear)
E,i,σh ,C∞(z) ,∞C2 ,σh ,∞σv
Exercise 13.14. Find iψ2 px where i is the inversion
operator. Show that ψ2 px is an eigenfunction of the
inversion operator, and find its eigenvalue.
−(x 2 + y 2 + z 2 )1/2
iψ2 px = i x exp
2a0
2
−(x + y 2 + z 2 )1/2
= −ψ2 px
= −x exp
2a0
The eigenvalue is equal to −1.
Exercise 13.15. The potential energy of two charges Q 1
and Q 2 in a vacuum is
V=
Q1 Q2
4π 0 r12
where r12 is the distance between the charges and ε0
is a constant called the permittivity of a vacuum, equal
to 8.854187817 × 10−12 Fm−1 = 8.854187817 ×
10−12 C2 N−1 m−2 . The potential energy of a hydrogen
molecules is given by
V =
e2
e2
e2
e2
−
−
−
4π 0 r AB
4π 0 r1A
4π 0 r1B
4π 0 r2 A
e2
e2
−
+
4π 0 r2B
4π 0 r12
where A and B represent the nuclei and 1 and 2 represent
the electrons, and where the two indexes indicate the
CHAPTER | 13 Operators, Matrices, and Group Theory
two particles whose interparticle distance is denoted. If a
hydrogen molecule is placed so that the origin is midway
between the two nuclei and the nuclei are on the z axis, show
that the inversion operator ı̂ and the reflection operator σ̂h
do not change the potential energy if applied to the electrons
but not to the nuclei.
We use the fact that the origin is midway between the
nuclei. Under the inversion operation, electron 1 is now the
same distance from nucleus A as it was originally from
nucleus B, and the same is true of electron 2. Under the σ̂h
operation, the same is true. The potential energy function
is unchanged under each of these operations.
e81
A
BC
Exercise 13.16. Find the product
⎤
⎤⎡ ⎤ ⎡
2
0
1 0 2
⎥
⎥⎢ ⎥ ⎢
⎢
⎣ 0 −1 1 ⎦ ⎣ 3 ⎦ = ⎣ −2 ⎦
1
1
0 0 1
⎡
A(BC)
Exercise 13.17. Find the two matrix products
⎤
⎤
⎡
⎤⎡
1 3 2
1 3 2
1 2 3
⎥
⎥
⎢
⎥⎢
⎢
⎣ 3 2 1 ⎦ ⎣ 2 2 −1 ⎦ and ⎣ 2 2 −1 ⎦
−2 −1 −1
−2 1 −1
1 −1 2
⎤
⎡
1 2 3
⎥
⎢
×⎣3 2 1⎦
1 −1 2
⎡
The left factor in one product is equal to the right factor
in the other product, and vice versa. Are the two products
equal to each other?
⎡
⎤⎡
1 2 3
1
⎢
⎥⎢
⎣3 2 1⎦⎣ 2
1 −1 2
−2
⎤⎡
⎡
1
1 3 2
⎥⎢
⎢
⎣ 2 2 −1 ⎦ ⎣ 3
1
−2 −1 −1
⎤
3 2
⎥
2 −1 ⎦ =
1 −1
⎤
2 3
⎥
2 1⎦ =
−1 2
⎡
−1
⎢
⎣ 5
−5
⎡
12
⎢
⎣ 7
−6
⎤
10 −3
⎥
14 3 ⎦
3 1
⎤
6 10
⎥
9 6 ⎦.
−5 −9
The two products are not equal to each other.
Exercise 13.18. Show that the properties of Eqs. (13.45)
and (13.46) are obeyed by the particular matrices
⎤
⎡
⎤
1 2 3
0 2 2
⎥
⎢
⎢
⎥
= ⎣ 4 5 6 ⎦ B = ⎣ −3 1 2 ⎦
7 8 9
1 −2 −3
⎤
⎡
1 0 1
⎥
⎢
× C = ⎣ 0 3 −2 ⎦
2 7 −7
⎤
⎤⎡
⎡
1 0 1
0 2 2
⎥
⎥⎢
⎢
= ⎣ −3 1 2 ⎦ ⎣ 0 3 −2 ⎦
2 7 −7
1 −2 −3
⎤
⎡
4 20 −18
⎥
⎢
= ⎣ 1 17 −19 ⎦
−5 −27 26
⎤
⎤⎡
⎡
4 20 −18
1 2 3
⎥
⎥⎢
⎢
= ⎣ 4 5 6 ⎦ ⎣ 1 17 −19 ⎦
−5 −27 26
7 8 9
⎡
⎤
−9 −27 22
⎢
⎥
= ⎣ −9 3 −11 ⎦
−9 33 −44
⎤
⎤ ⎡
⎤⎡
⎡
−3 −2 −3
0 2 2
1 2 3
⎥
⎥ ⎢
⎥⎢
⎢
= ⎣ 4 5 6 ⎦ ⎣ −3 1 2 ⎦ = ⎣ −9 1 0 ⎦
−15 4 3
1 −2 −3
7 8 9
⎤
⎤⎡
⎡
1 0 1
−3 −2 −3
⎥
⎥⎢
⎢
= ⎣ −9 1 0 ⎦ ⎣ 0 3 −2 ⎦
2 7 −7
−15 4 3
⎤
⎡
−9 −27 22
⎥
⎢
= ⎣ −9 3 −11 ⎦ .
−9 33 −44
⎡
AB
(AB)C
⎤⎞
⎤ ⎡
⎤ ⎛⎡
1 0 1
0 2 2
1 2 3
⎥⎟
⎥ ⎢
⎥ ⎜⎢
⎢
⎣ 4 5 6 ⎦ ⎝⎣ −3 1 2 ⎦ + ⎣ 0 3 −2 ⎦⎠
2 7 −7
1 −2 −3
7 8 9
⎤
⎡
4 25 −27
⎥
⎢
= ⎣ 7 58 −48 ⎦
10 91 −69
⎤⎡
⎤ ⎡
⎤⎡
⎡
⎤
1 0 1
1 2 3
0 2 2
1 2 3
⎥⎢
⎥ ⎢
⎥⎢
⎢
⎥
⎣ 4 5 6 ⎦ ⎣ −3 1 2 ⎦ + ⎣ 4 5 6 ⎦ ⎣ 0 3 −2 ⎦
7 8 9
1 −2 −3
7 8 9
2 7 −7
⎤
⎤ ⎡
⎤ ⎡
⎡
4 25 −27
7 27 −24
−3 −2 −3
⎥
⎥ ⎢
⎥ ⎢
⎢
= ⎣ −9 1 0 ⎦ + ⎣ 16 57 −48 ⎦ = ⎣ 7 58 −48 ⎦
10 91 −69
25 87 −72
−15 4 3
⎡
e82
Exercise 13.19. Show by
that
⎤⎡
⎡
a11 a12 a13
1 0 0 0
⎥⎢
⎢
⎢ 0 1 0 0 ⎥ ⎢ a21 a22 a31
⎥⎢
⎢
⎣ 0 0 1 0 ⎦ ⎣ a31 a31 a31
0 0 0 1
a41 a41 a31
Mathematics for Physical Chemistry
explicit matrix multiplication
⎤ ⎡
a11
a14
⎥ ⎢
a41 ⎥ ⎢ a21
⎥=⎢
a41 ⎦ ⎣ a31
a41
a41
a12
a21
a31
a41
a13
a31
a31
a31
⎤
a14
⎥
a41 ⎥
⎥.
a41 ⎦
a41
:Each element is produces as a single term since the other
terms in the same contain a factor zero.
Exercise 13.20. Show that AA−1 = E and that A−1 A = E
for the matrices of the preceding example.
Mathematica and other software packages can find a
matrix product with a single command. Using the Scientific
Work Place software
⎡
⎤
3
1 1
−
⎤
⎡
⎤⎢
⎡
4
2 4 ⎥
⎥
1
0
0
2 1 0 ⎢
1⎥
⎥
⎥⎢ 1
⎢
⎥ ⎢
AA−1 = ⎣ 1 2 1 ⎦ ⎢
⎢ −2 1 −2 ⎥ = ⎣ 0 1 0 ⎦
⎢
⎥
0 0 1
0 1 2 ⎣ 1
1 3 ⎦
−
4
2 4
⎡ 3
1 1 ⎤
−
⎤
⎤ ⎡
⎡
⎢ 4
2 4 ⎥
⎢
⎥ 2 1 0
1 0 0
⎢
⎥
1
1 ⎥⎢
⎥
⎥ ⎢
A−1 A = ⎢
⎢ −2 1 −2 ⎥ ⎣ 1 2 1 ⎦ = ⎣ 0 1 0 ⎦
⎢
⎥
0 0 1
⎣
⎦ 0 1 2
1
1 3
−
4
2 4
Exercise 13.21. Use Mathematica or another software
package to verify the inverse found in the preceding
example. Using the Scientific Work Place software
⎡ 3
1 1 ⎤
−
⎤−1 ⎢ 4
⎡
2 4 ⎥
⎢
⎥
2 1 0
⎢
1
1⎥
⎥
⎢
⎢
1 − ⎥
⎣1 2 1⎦ = ⎢−
2⎥
⎢ 2
⎥
0 1 2
⎣
⎦
1
1 3
−
4
2 4
Exercise 13.22. Find the inverse of the matrix
12
A=
3 4
Using the Scientific Work Place software,
⎤
−1 ⎡
−2 1
1 2
A−1 =
=⎣ 3
1⎦
3 4
−
2
2
Check this
⎤ ⎡
−2 1
1 2 ⎣
1
0
3
1⎦=
3 4
0 1
−
2
2
Exercise 13.23. Expand the following determinant by
minors:
3 2 0
7 5
−1 5 −2
7 −1 5 = 3 2 4
3 4
2 3 4
= 3(− 4 − 15) − 2(28 − 10) = −93
Exercise 13.24. a. Find the value of the determinant
3 4 5 2 1 6 = 47
3 −5 10 b. Interchange the first and second columns and find the
value of the resulting determinant.
4 3 5 1 2 6 = −47
−5 3 10 c. Replace the second column by the sum of the first
and second columns and find the value of the resulting
determinant.
7 4 5 3 1 6 = 47
−2 −5 10 d. Replace the second column by the first, thus making
two identical columns, and find the value of the
resulting determinant.
3 3 5 2 2 6 = 0
3 3 10 Exercise 13.25. Obtain the inverse of the following matrix
by hand. Then use Mathematica to verify your answer.
⎡
1
⎢
⎣3
1
⎡
1
⎢
⎣3
1
⎤
⎤−1 ⎡
−2 0 3
3 0
⎥
⎢
⎥
1 0 −1 ⎥
0 4⎦ = ⎢
⎣ 3 1 9⎦
−
2 0
2 4⎤ 4
⎤⎡
⎤
⎡
3 0 ⎢ −2 0 3 ⎥
1 0 0
⎥ 1 0 −1 ⎥ ⎢
⎥
0 4⎦⎢
⎣ 3 1 9 ⎦ = ⎣0 1 0⎦
−
2 0
0 0 1
2 4 4
Exercise 13.26. Obtain the multiplication table for the C2v
point group and show that it satisfies the conditions to be a
group.
CHAPTER | 13 Operators, Matrices, and Group Theory
e83
Think of the H2 O molecule, which possesses all of the
symmetry operators in this group. Place the molecule in
the y − z plane with the rotation axis on the z axis. Note
that in this group, each element is its own inverse. For the
other operators, one inspects the action of the right-most
operator, followed by the action of the left-most operator.
If the result is ambiguous, you need to use the fact that a
reflection changes a right-handed system to a left-handed
system while the rotation does not. For example, σv(yz)
followed by σv(x z) exchanges the hydrogens, but changes
the handedness of a coordinate system, so the result is the
same as σv(x z) .
x = x
y = y
z = −z
1 0 0
σ̂h ↔ 0 1 0
0 0 −1
Exercise 13.28. By transcribing Table 13.1 with appropriate changes in symbols, generate the multiplication table for
the matrices in Eq. (13.65).
$
'
E
C2
σv(yz)
σv(xz)
E
E
2
C
σv(yz)
σv(xz)
2
C
2
C
E
σv(xz)
σv(yz)
2
C
E
σv(yz)
σv(yz)
σv(xz)
E
σv(xz)
σv(xz)
σv(xz)
2
C
&
c. Find the matrix equivalent to σ̂h .
E
A
B
C
D
F
%
These operators form a group because (1) each product
is a member of the group; (2) the group does include the
identity operator; (3) because each of the members is its
own inverse; and (4) multiplication is associative.
Exercise 13.27.
2 (z).
a. Find the matrix equivalent to C
x = −x
y = −y
z = z
−1 0 0
2 (z) ↔ 0 −1 0
C
0 0 1
b. Find the matrix equivalent to S3 (z).
1
x = cos (2π/3)x − sin (2π/3)y = − x
2
1√
−
3 y
2
1√
1
3 x− y
y = sin (2π/3)x + cos (2π/3)y =
2
2
z = z.
⎤
⎡
√
−1/2 − 3/2 0
⎥
⎢√
S3 (z) ↔ ⎣ 3/2 −1/2 0 ⎦
0
0
1
E
E
A
B
C
D
F
A
A
B
E
D
D
C
B
B
E
A
F
F
D
C
C
F
D
E
C
A
D
D
C
F
A
E
B
F
F
D
C
B
A
E
Exercise 13.29. Verify several of the entries in the
multiplication table by matrix multiplication of the matrices
in Eq. (13.65).
⎡
⎤⎡
⎤
√
√
−1/2 − 3/2 0
−1/2
3/2 0
⎢√
⎥⎢ √
⎥
AB = ⎣ 3/2 −1/2 0 ⎦ ⎣ − 3/2 −1/2 0 ⎦
0
0
1
0
0 1
⎤
⎡
1 0 0
⎥
⎢
= ⎣0 1 0⎦ = E
0 0 1
⎡
⎤⎡
⎤
√
√
−1/2 − 3/2 0
1/2 − 3/2 0
⎢√
⎥⎢ √
⎥
AD = ⎣ 3/2 −1/2 0 ⎦ ⎣ − 3/2 −1/2 0 ⎦
0
0
1
0
0
1
⎤
⎡ 1 1√
30
⎥
⎢ 2 2
⎥
⎢ √
1
1
⎥
⎢
= ⎢
3 − 0⎥ = C
⎦
⎣2
2
0 1
⎤⎡
⎤
√
√
1/2
1/2 − 3/2 0
3/2 0
⎢√
⎥⎢ √
⎥
CD = ⎣ 3/2 −1/2 0 ⎦ ⎣ − 3/2 −1/2 0 ⎦
0
0 1
0
0
1
⎡ 1
⎤
1√
− −
3 0
⎢ 2
⎥
2
⎢ √
⎥
1
1
⎥
= ⎢
⎢
0⎥ = A
3 −
⎣2
⎦
2
⎡
0
0
0
1
e84
Exercise 13.30. Show by matrix multiplication that two
matrices with a 2 by 2 block and two 1 by 1 blocks produce
another matrix with a 2 by 2 block and two 1 by 1 blocks
when multiplied together.
⎡
⎤⎡
⎤ ⎡
⎤
a b 0 0
α β 0 0
aα + bγ aβ + bδ 0 0
⎢
⎥⎢
⎥ ⎢
⎥
⎢ c d 0 0 ⎥ ⎢ γ δ 0 0 ⎥ ⎢ cα + dγ cβ + dδ 0 0 ⎥
⎢
⎥⎢
⎥=⎢
⎥
⎣ 0 0 e 0 ⎦⎣ 0 0 ε 0 ⎦ ⎣
0
0
εe 0 ⎦
0 0 0 f
0 0 0 φ
0
0
0 fφ
Exercise 13.31. Pick a few pairs of 2 by 2 submatrices
from Eq. (13.65) and show that they multiply in the same
way as the 3 by 3 matrices.
√
√
3/2
−1/2
1 0
−1/2 − 3/2
√
√
=
− 3/2 −1/2
3/2 −1/2
0 1
⎡ 1
1√ ⎤
√
√
3
− −
3/2
1/2 − 3/2
1/2
⎥
⎢
2
√
√
= ⎣ √2
⎦
1
1
− 3/2 −1/2
3/2 −1/2
3 −
2
2
⎡ 1 1√ ⎤
√
√
3
−1/2 − 3/2
1/2 − 3/2
⎥
⎢ 2 2
√
√
=⎣ √
⎦.
1
1
− 3/2 −1/2
3/2 −1/2
3 −
2
2
Exercise 13.32. Show that the 1 by 1 matrices (scalars) in
Eq. (13.67) obey the same multiplication table as does the
group of symmetry operators.
Mathematics for Physical Chemistry
= cos (x) f + sin (x)
df
dx
df
= cos (x) f
− sin (x)
dx d
, sin (x) = cos (x)
dx
b.
d2
,x ;
dx 2
d2
,x
dx 2
d2
d2
[x f ] − x 2 f
2
dx
dx
d
df
d2
=
x
+ f −x 2 f
dx
dx
dx
f =
df
d2
d2
df
+x 2 f +
−x 2 f
dx
dx
dx
dx
df
= 2
dx
2 d
d
,x = 2
dx 2
dx
=
3. The components of the angular momentum correspond
to the quantum mechanical operators:
∂
∂
∂
∂
y
−z
, Ly =
z
−x
,
Lx =
i
∂z
∂y
i
∂x
∂z
∂
∂
−y
Lz =
x
.
i
∂y
∂x
2 ↔ 1,
↔ 1 C
3 ↔ 1 C
Since the elements E
3
the product of any two of these will yield +1. Since
σ̂a ↔ −1 σ̂b ↔ −1 σ̂c ↔ −1, the product of any two of
these will yield 1. The product of any of the first three with
These operators do
not commute with each other. Find
any of the second three will yield −1. the multiplication
Ly .
the commutator L x ,
table is
'
$
L x ,
Ly f
σ̂a
σ̂b
σ̂c
E
C3
C32
∂
∂ ∂
∂
y
−z
z
−x
f
=
1
1
1
−1
−1
−1
E
i
∂z
∂y i
∂x
∂z
1
1
1
−1
−1
−1
C3
∂
∂ ∂
∂
2
z
−
x
y
−
z
f
−
1
1
1
−1
−1
−1
C3
i
∂x
∂z i
∂z
∂y
−1
−1
−1
1
1
1
σ̂a
∂ ∂f
∂ ∂f
∂ ∂f
2
y z
−y x
−z z
=
−
−1
−1
−1
1
1
1
σ̂b
∂z ∂ x
∂z ∂z
∂ y ∂x
−1
−1
−1
1
1
1
σ̂c
∂ ∂f
∂ ∂f
∂ ∂f
+z x
−z y
+z z
&
%
∂ y ∂z
∂ x ∂z
∂x ∂ y
∂ ∂f
∂ ∂f
+x y
−x z
∂z ∂ y
∂z ∂z
2
PROBLEMS
∂f
∂2 f
∂2 f
∂ f
+y
− x y 2 − z2
= − 2 yz
∂z∂ x
∂z
∂z
∂ y∂ x
1. Find the following commutators, where Dx = d/dx:
2f
2f
2f
d
∂
∂
∂
a. dx , sin (x) ;
− zy
+ z2
+ zx
∂
y∂z
∂
x∂z
∂ x∂ y
d
d
df
2
2
∂ f
∂f
∂ f
, sin (x) f =
[sin (x) f ] − sin (x)
−
x
z
−
x
+zy
dx
dx
dx
∂z 2
∂y
∂z∂ y
CHAPTER | 13 Operators, Matrices, and Group Theory
We can now apply Euler’s reciprocity relation to cancel
all of the terms but two:
∂f
∂f
2
−x
L x , L y f = − y
∂z
∂y
∂f
∂f
2
−y
= i
Lz
= x
∂y
∂z
5. In quantum mechanics, the expectation value of a
mechanical quantity is given by
! ∗
dx
ψ Aψ
,
A = ! ∗
ψ ψ dx
is the operator for the mechanical quantity
where A
and ψ is the wave function for the state of the system.
The integrals are over all permitted values of the
coordinates of the system. The expectation value is
defined as the prediction of the mean of a large number
of measurements of the mechanical quantity, given that
the system is in the state corresponding to ψ prior to
each measurement.
For a particle moving in the x direction only and
confined to a region on the x axis from x = 0 to x = a,
the integrals are single integrals from 0 to a and p̂x is
given by (/i)∂/∂ x. The normalized wave function is
"
πx 2
sin
ψ=
a
a
Normalization means that the integral in the
denominator of the expectation value expression is
equal to unity.
a. Show that this wave function is normalized. We
let u = π x/a
2 a π 2 2 a 2 πx dx =
sin
sin u du
a 0
a
aπ 0
π
sin 2x 2 x
2π
=
−
=1
=
π 2
4
π 2
0
b. Find the expectation value of x.
πx πx 2 a
x sin
dx
sin
x =
a 0
a
a
πx 2 a
=
dx
x sin2
a 0
a
2 a 2 π
=
u sin2 u du
a π
0
π
2
2a x
x sin (2x) cos 2x = 2
−
−
π
4
4
8
0
2
1 1
a
2a π
− +
=
= 2
π
4
8 8
2
e85
c. The operator corresponding to px is i ddx .
Find the expectation value of px .
a
πx πx d
2
px =
sin
dx
sin
i a 0
a dx
a
πx πx 2 π a
=
cos
dx
sin
ia a 0
a
a
2 π a π
=
sin (u) cos (u)du
ia a π 0
π
2 sin (u) =
=0
ia 2 0
d. Find the expectation value of px2 .
a
π x d2
πx 2
22
px = −
dx
sin
sin
a 0
a dx 2
a
a
πx πx d
2π
= −
cos
dx
sin
aa 0
a dx
a
2 π 2 a 2 π x = 2
dx
sin
a a
a
0
2 π 2 a π 2
= 2
sin (u)du
a a
π 0
π
sin
2x x
2π
= 2 2
−
a
2
4
0
2π π 2 π 2
h2
=
= 2 2
=
a
2
a2
4a 2
7. If x is an ordinary variable, the Maclaurin series for
1/(1 − x) is
1
= 1 + x2 + x3 + x4 + · · · .
1−x
If X is some operator, show that the series
1+ X+
X2 + X3 + X4 + · · ·
is the inverse of the operator 1 − X.
1− X 1+ X+
X2 + X3 + X4 + · · ·
=1+ X+
X2 + X3 + X4 + · · ·
− X+
X2 + X3 + X4 + · · · = 1
9. Find the result of each operation on the given point
(represented by Cartesian coordinates):
√
3(z) (1,1,1) = − 1 , 1 3.1
a. C
2 2
b. S4(z) (1,1,1) = (1, − 1, − 1)
11. Find the result of each operation on the given point
(represented by Cartesian coordinates):
2(z) (− 1, − 1, − 1)
2(z) ı̂(1,1,1) = C
a. C
= (1, − 1, − 1)
2(z) (1,1,1) = ı̂(− 1, − 1,1) = (1,1, − 1)
b. ı̂ C
e86
Mathematics for Physical Chemistry
13. Find the 3 by 3 matrix that is equivalent in its action
to each of the symmetry operators:
8 (x): Let α = π/8 ↔ 45◦
a. C
x = x
1
y = cos (α)y + sin (α)z = √ (y + z)
2
1
z = sin (α)y + cos (α)z = √ (y + z)
2
1 0 0
1
0
8 (x) ↔ 0 √
C
2
1
0 0 √
2
b. S6 (x): Let α = π/3 ↔ 60◦
√
3
1
y
x = cos (α)x − sin (α)y = x −
2
2
√
1
3
x+ y
y = sin (α)x + cos (α)y =
2
2
z = −z.
√
1
3
−
0
2
√2
8 (x) ↔ 3 1
C
0
2
2
0
0 −1
15. Give the function that results if the given symmetry
operator operates on the given function for each of the
following:
a. ı̂(x + y + z 2 ) = (− x − y + z 2 )
b. S4(x) (x + y + z) = x + z − y
17. Find the matrix products. Use Mathematica to check
your result.
⎤
⎤ ⎡
⎡
1 2 3
1 2 3
⎢ 0 3 −4 ⎥ ⎢ 0 3 −4 ⎥
⎥
⎥ ⎢
⎢
a. 3 2 1 4 ⎢
⎥
⎥=⎢
⎣ 1 −2 1 ⎦ ⎣ 1 −2 1 ⎦
3 1 0
3 1 0
⎤
⎡
⎤
⎡
17
1 2 3 ⎡ ⎤
⎥
⎢
⎥ 2
⎢
⎢ 0 3 −4 ⎥ ⎢ ⎥ ⎢ −3 ⎥
b. ⎢
⎥
⎥⎣3⎦ = ⎢
⎣ −1 ⎦
⎣ 1 −2 1 ⎦
3
9
3 1 0
⎤
⎡
1 4 −7 3
6 3 −1 ⎢
⎥
c.
⎣ 2 5 8 −2 ⎦
7 4 −2
3 6 −9 1
9 33 −9 11
=
9 36 1 11
19. Show that A(B + C) = AB + AC for the example
matrices in the previous problem.
⎤
⎤ ⎡
⎡
0 3 1
3 1 4
⎥
⎥ ⎢
⎢
B + C = ⎣ −2 0 1 ⎦ + ⎣ −4 2 3 ⎦
3 1 −2
3 2 1
⎤
⎡
3 4 5
⎥
⎢
= ⎣ −6 2 4 ⎦
6 3 −1
⎤
⎤⎡
⎡
3 4 5
0 1 2
⎥
⎥⎢
⎢
A(B + C) = ⎣ 3 1 −4 ⎦ ⎣ −6 2 4 ⎦
6 3 −1
2 3 1
⎤
⎡
6 8 2
⎥
⎢
= ⎣ −21 2 23 ⎦
−6 17 21
⎤
⎤⎡
⎡
3 1 4
0 1 2
⎥
⎥⎢
⎢
AB = ⎣ 3 1 −4 ⎦ ⎣ −2 0 1 ⎦
3 2 1
2 3 1
⎤
⎡
4 4 3
⎥
⎢
= ⎣ −5 −5 9 ⎦
3 4 12
⎤
⎤⎡
⎡
0 3 1
0 1 2
⎥
⎥⎢
⎢
AC = ⎣ 3 1 −4 ⎦ ⎣ −4 2 3 ⎦
3 1 −2
2 3 1
⎤
⎡
2 4 −1
⎥
⎢
= ⎣ −16 7 14 ⎦
−9 13 9
⎤
⎡
4 4 3
⎥
⎢
AB + AC = ⎣ −5 −5 9 ⎦
3 4 12
⎤
⎤ ⎡
⎡
6 8 2
2 4 −1
⎥
⎥ ⎢
⎢
+ ⎣ −16 7 14 ⎦ = ⎣ −21 2 23 ⎦
−6 17 21
−9 13 9
21. Test the following matrices for singularity. Find the
inverses of any that are nonsingular. Multiply the
original matrix by its inverse to check your work. Use
Mathematica to check your work.
⎤
⎡
3 2 −1
⎥
⎢
a. ⎣ −4 6 3 ⎦
7 2 −1
3 2 −1 −4 6 3 = 48
7 2 −1 CHAPTER | 13 Operators, Matrices, and Group Theory
e87
⎡
Not singular
⎤−1
⎡
3 2 −1
⎥
⎢
⎣ −4 6 3 ⎦
7 2 −1
⎡
1
⎢
4
3 2 −1 ⎢
⎥ ⎢ 17
⎢
⎣ −4 6 3 ⎦ ⎢
⎢ 48
7 2 −1 ⎣ 25
−
24
⎤
⎡
0 2 3
⎥
⎢
b. ⎣ 1 0 1 ⎦
2 0 1
0 2
1 0
2 0
⎤
⎡
−
⎡
1
⎢ 4
⎢
⎢ 17
⎢
⎢ 48
⎣ 25
−
24
−
0
1
12
1
6
1
4
5
−
48
13
24
0
1
12
1
6
1
4
5
−
48
13
24
⎤
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎤
⎡
⎥
1 0 0
⎥
⎥
⎥ ⎢
⎥ = ⎣0 1 0⎦.
⎥
⎦
0 0 1
3 1 = 2
1
⎤−1 ⎡
0 −1
0 2 3
⎢1
⎥
⎢
⎣1 0 1⎦ = ⎢
⎣ 2 −3
2 0 1
0 2
Check:
⎤⎡
⎡
0 2 3 ⎢
⎥
⎢
⎣1 0 1⎦⎢
⎣
2 0 1
⎤
1
3 ⎥
⎥
2 ⎦
−1
⎤ ⎡
⎤
0 −1 1
1 0 0
⎥
1
3 ⎥ ⎢
⎥
= ⎣0 1 0⎦.
−3
⎦
2
2
0 0 1
0 2 −1
23. The H2 O molecule belongs to the point group C2v ,
C
2 , σ̂a ,
which contains the symmetry operators E,
and σ̂b , where the C2 axis passes through the oxygen
nucleus and midway between the two hydrogen nuclei,
and where the σa mirror plane contains the three nuclei
and the σb mirror plane is perpendicular to the σa
mirror plane.
a. Find the 3 by 3 matrix that is equivalent to each
symmetry operator.
⎤
⎡
1 0 0
⎥
↔ ⎢
E
⎣0 1 0⎦
0 0 1
⎤
0 0
⎥
−1 0 ⎦
0 1
⎤
0 0
⎥
1 0⎦
0 1
⎤
0 0
⎥
−1 0 ⎦
0 1
b. Show that the matrices obtained in part (a)
have the same multiplication table as the
symmetry operators, and that they form a group.
The multiplication table for the group was to
be obtained in an exercise. The multiplication
table is
E
E
C2
Not singular
⎡
−1
2 ↔ ⎢
C
⎣ 0
0
⎡
1
⎢
σ̂a ↔ ⎣ 0
0
⎡
−1
⎢
σ̂b ↔ ⎣ 0
0
E
C2
σv(yz) σv(yz)
σv(x z) σv(x z)
2 C
σv(yz) σv(x z)
2 σv(yz) σv(x z)
C
σv(yz)
E σv(x z) 2
σv(x z) E
C
σv(yz) C2
E
σv(x z) =σ̂b . We perform
where σv(yz) = σ̂a and a few of the multiplications.
⎤
⎤⎡
⎡
−1 0 0
1 0 0
⎥
⎥⎢
⎢
σ̂a σ̂b ↔ ⎣ 0 1 0 ⎦ ⎣ 0 −1 0 ⎦
0 0 1
0 0 1
⎤
⎡
−1 0 0
⎥
⎢
2
= ⎣ 0 −1 0 ⎦ ↔ C
0
0 1
⎤
⎤⎡
−1 0 0
−1 0 0
2
⎥
⎥⎢
2 ↔ ⎢
C
⎣ 0 −1 0 ⎦ ⎣ 0 −1 0 ⎦
0 0 1
0 0 1
⎤
⎡
1 0 0
⎥
⎢
= ⎣0 1 0⎦ ↔ E
0 0 1
⎤
⎤⎡
⎡
−1 0 0
−1 0 0
⎥
⎥⎢
⎢
2
C
σv(x z) ↔ ⎣ 0 −1 0 ⎦ ⎣ 0 −1 0 ⎦
0 0 1
0 0 1
⎤
⎡
1 0 0
⎥
⎢
σv(yz)
= ⎣0 1 0⎦ ↔
0 0 1
⎡