1000 Solved Problems in Modern Physics

1.3 Solutions 45

HXi=λiXi (1)

NowX ̄ι′HXi=X ̄′ιλiXi=λiX ̄ι′Xi (2)

is real and non-zero. Similarly the conjugate transpose

X ̄′ιHXi=λ ̄ιX ̄′ιXi (3)

Comparing (2) and (3),

λ ̄i=λi

Thusλiis real

1.28 The characteristic equation is given by

∣ ∣ ∣ ∣ ∣ ∣

1 −λ − 11

03 −λ − 1

002 −λ

∣ ∣ ∣ ∣ ∣ ∣

= 0

(1−λ)(3−λ)(2−λ)+ 0 + 0 =0(1)

orλ^3 − 6 λ^2 + 11 λ− 6 =0 (characteristic equation) (2)

The eigen values areλ 1 = 1 ,λ 2 =3, andλ 3 =2.

1.29 LetX=

(

x

y

)

AX=

(

− 10

01

)(

x

y

)

=

(

−x

−y

)

It produces reflection through the origin, that is inversion. A performs the

parity operation, Fig. 1.10a.

Fig. 1.10aParity operation
(inversion through origin)

BX=

(

01

10

)(

x

y

)

=

(

y

x

)

Here thexandycoordinates are interchanged. This is equivalent to a reflec-

tion about a line passing through origin atθ= 45 ◦,Fig.1.10b

CX=

(

20

02

)(

x

y

)

=

(

2 x

2 y

)