1000 Solved Problems in Modern Physics

3.3 Solutions 209

Fig. 3.24ψ 1 (x)forSHO

Substitute

Bxexp(−x^2 / 2 a)=ψ 1 anda=

(

mω

) 1 / 2

and rearrange to get

−

(

^2

2 m

)

d^2 ψ 1

dx^2

+

(

mω^2

2

)

ψ 1 =

(

3 ω

2

)

ψ 1

(c) E=^3  2 ω

(d) <px>=

∫

ψ 1 ∗

(

−i∂∂ψx^1

)

dx

=−iB^2

∫∞

−∞xexp

(

−x

2

a^2

)(

1 −x

2

a^2

)

dx

=zero (because integration over an odd function between symmetrical lim-

its is zero). This result is expected because half of the time the particle will

be pointing along positive direction and for the half of time in the negative

direction.

(e) The results of SHO are valuable for the analysis of vibrational spectra of

diatomic molecules, identification of unknown molecules, estimation of

force constants etc.

3.3.5 Hydrogen Atom...................................

3.64=<−e

2

r >=

∫

ψ∗Vψdτ=−e

2

πa 03

∞∫

0

(exp

(

−^2 ar 0

)

/r)4πr^2 dr

wheree=charge of electron

=−

4 e^2

a^30

∫∞

0

exp

(

−

2 r

a 0

)

rdr=−

e^2

a 0

Therefore<V>=−e

2

a 0

=

∫ [

ψ∗

(

−

^2

2 m

)

∇^2 ψ

]

dτ (1)

In polar coordinates (independent ofθandφ);