1000 Solved Problems in Modern Physics

3.3 Solutions 203

Similarly<Δpx>^2 =<p^2 >

(^1) /
2 mω
(^2) <x (^2) >·

(

1

2 m

)

<px>^2 ≤

1

4

(

ω

2

) 2

[

<x^2 ><px^2 >

] 1 / 2

≤



2

orΔx.Δpx≤ 2

Now compare this result with the uncertainty principle

Δx·Δpx≤



2

We conclude thatΔx.Δpx≥ 2. Obviously the zero point energy could not

have been lower than 2 ωwithout violating the uncertainty principle.

3.55 The probability distribution for the quantum mechanical simple harmonic
oscillator (S.H.O) is

P(x)=|ψn|^2 =

αexp(−ξ^2 )Hn^2 (ξ)

√

π 2 nn!

(1)

ξ=αx;α^4 =mk/^2

Stirling approximation gives

n!→(2nπ)^1 /^2 nne−n (2)

Furthermore the asymptotic expression for Hermite function is

Hn(ξ)(forn→∞)→ 2 n+^1

(n/ 2 e)

n 2

√

2 cosβ

exp(nβ^2 ) cos

[

(2n+^1 / 2 )β−

nπ

2

]

(3)

where sinβ=ξ/

√

2 n (4)

Using (2) and (3) in (1)

P(x)→ 2 αexp(−ξ^2 )exp(2nβ^2 )

cos^2

[(

2 n+^12

)

β−n 2 π

]

π

√

2 ncosβ

But<cos^2

[(

2 n+^12

)

β−n 2 π

]

>=^12

ThereforeP(x)=

αexp(−ξ^2 )exp(2nβ^2 )

π

√

2 ncosβ

(5)

Fig. 3.23Probability distribution of quantum mechanical oscillator and classical oscillator