1000 Solved Problems in Modern Physics

2.3 Solutions 129

The precise statement of the Heisenberg uncertainty principle is

ΔPxΔx≥/ 2

ΔPyΔy≥/2(3)

ΔPzΔz≥/ 2

Consider the integral, a function of a real parameterλ

I(λ)=

∫∞

−∞

dx|(x−<x>)ψ+iλ(−i∂ψ/∂x−<Px>ψ|^2 (4)

By definition,I(λ≥0). Expanding (4)

I(λ)=

∫∞

−∞

dxψ∗(x−<x>)^2 ψ+λ

∫∞

−∞

dx

(

ψ∗

∂ψ

∂x+ψ

∂ψ∗

∂x

)

(x−<x>)

+λ^2 ^2

∫∞

−∞

(

∂ψ∗

∂x

)

∂ψ

∂x−iλ

(^2) 
∫∞
−∞
dx[ψ
∂ψ∗
∂x +λ
(^2) 2
∫∞
−∞
dxψ∗ψ (5)
The term in the second line can be written as
∫∞
−∞
dx

∂

∂x

(ψ∗ψ)(x−<x>)=[(x−<x>)ψ∗ψ]∞−∞−

∫∞

−∞

dxψ∗ψ=− 1

because it is expected thatψ→0. Sufficiently fast asx→±∞so that the

integrated term is zero. Similarly the third term can be re-written as

^2

∫∞

−∞

dx

(

∂ψ∗

∂x

)(

∂ψ

∂x

)

=^2

[

ψ∗

∂ψ

∂x

]∞

−∞

+

∫∞

−∞

dxψ∗(−^2 ∂^2 ψ/∂x^2 )=<Px^2 >

In term (4) rewrite

−i

∫∞

−∞

dx

∂ψ∗

∂x

ψ=−i

[

ψ∗ψ

]∞

−∞+

∫∞

−∞

dxψ∗i

∂ψ

∂x

=−<Px>

So, the full term (4) becomes− 2 <Px>^2.

Collecting all the terms

I(λ)=(Δx)^2 −λ+(ΔPx)^2 λ^2 ≥ 0

Denoting I(λ) = aλ^2 +bλ+c, the condition I(λ ≥ 0) is satisfied if

b^2 − 4 ac≥0.

Thus,^2 −4(Δx)^2 (ΔPx)^2 ≤0, and therefore

ΔPxΔx≥/ 2

2.85ΔxΔp=

cΔP≈cp=

c

Δx

=

197 .3MeV−fm

0. 529 × 10 −^10 m

= 372. 97 × 10 −^5 MeV= 3 ,730 eV

T=c^2 p^2 /mc^2 =(3,730)^2 / 0. 511 × 106 = 13 .61 eV