1000 Solved Problems in Modern Physics

(Grace) #1

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

)
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