1000 Solved Problems in Modern Physics

126 2 Quantum Mechanics – I

This givesa=1,b=0forλ=1 anda=0,b=1forλ= 3

Hence the eigen states ofAare

(

1

0

)

and

(

0

1

)

(c) AsA=A†,Ais Hermitian and hence an observable.

2.76 (a) A general rule for commutators is

[A^2 ,B]=A[A,B]+[A,B]A

HereH=P^2 / 2 μ

[H,X]=(1/ 2 μ)[P^2 ,x]=(1/^2 μ)(P[P,x]+[P,x]P)

=(1/ 2 μ)2p/i=p/iμ

Therefore [x,H]=ip/μ

(b)[[x,H],x]=

[

iPx

μ

,x

]

=

(

i

μ

)

[Px,x]=

(

i

μ

)

(−i)=

^2

μ

2.77[A^2 ,B]=AAB−BAA=AAB−ABA+ABA−BAA

=A[A,B]+[A,B]A

2.78(σ.A)(σ.B)=(σxAx+σyAy+σzAz)(σxBx+σyBy+σzBz)

=AxBxσx^2 +AyByσy^2 +AzBzσz^2 +σxσyAxBy+σxσzAxBz

+σyσxAyBx+σyσzAyBz+σzσxAzBx+σzσyAzBy

=A.B+iσz(AxBy−AyBx)+iσx(AyBz−AzBy)

+iσy(AzBx−BzAx)

=A.B+i[σ.(A×B)]z+i[σ.(A×B)]x+i[σ.(A×B)]y

=A.B+iσ.(A×B)

where we have used the identities in simplifying:

σx^2 =σy^2 =σz^2 =^1

andσyσx=−σxσyetc.

2.79 (a)σyμ=σμy†as can be seen from the matrix elements ofσy. Thereforeσyis
Hermitian. It is the matrix of a Hermitian operator whose eigen values are
real.
(b) The eigen valuesλare found by setting
∣
∣
∣
∣

σy 11 −λσy 11

σy 21 σy 22 −λ

∣

∣

∣

∣=

∣

∣

∣

∣

−λ−i

i −λ

∣

∣

∣

∣=^0

λ^2 − 1 = 0 ,λ=± 1