Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 35, Problem 11P
To determine

The uncertainty product ΔxΔpx .

Expert Solution & Answer
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Answer to Problem 11P

The uncertainty product ΔxΔpx is h2 .

Explanation of Solution

Formula used:

The expression for ground state of harmonic motion is given by,

  x=ψ0*(x)xψ0(x)dx

The wave function for ground state is,

  ψ0(x)=A0emω0x22h

Calculation:

The expression for ground state of harmonic motion is calculated as,

  x= ψ 0 *( x)x ψ 0( x)dx= A 0 e m ω 0 x 2 2h x A 0 e m ω 0 x 2 2h dx=( A 0)2 x e ( 2m ω 0 x 2 2h )dx=( A 0)2 x e ( m ω 0 x 2 h )dx

Let,

  mω0x2h=k

Then,

  ( m ω 0 h)(2xdx)=dkxdx=h2mω0dkdx=(h 2m ω 0 ) m ω 0 h 2 dkk= h 2 4m ω 0 dkk

For the limits,

  x,kx,k

  x=( A 0)2 ( kh m ω 0 ) e k h 2 4m ω 0 dkk=( A 0)2( h 2 4m ω 0 ) e kdk=( A 0)2( h 2 4m ω 0 )[ e k 1]=0

The expectation value for x2 for ground state is,

  x2= ψ 0 *( x) x 2 ψ 0( x)dx=20 A 0 e m ω 0 x 2 2h x 2 A 0 e m ω 0 x 2 2h dx=2( A 0)20 x 2 e ( 2m ω 0 x 2 2h )dx=2( A 0)2 x 2 e ( m ω 0 x 2 h )dx

For the limits,

  x0,k0x,k

  x2=2( A 0)20( kh m ω 0 ) e k h 2 4m ω 0 dkk=( m ω 0 hπ)12( h m ω 0 )320 k e kdk=( m ω 0 hπ)12( h m ω 0 )320 k 3 2 1 e kdk

By definition of Gamma function,

  0 e k k 3 2 1dk=Γ(32)=12π

Then,

  x2=( m ω 0 hπ)12( h m ω 0 )32π2=( m ω 0 hπ)12( h m ω 0 )12(h m ω 0 )π2=h2mω0

The uncertainty in position is calculated as,

  Δx= ( xx ) 2=( x 2 2xx+ x 2 )=( x 2 2x0+ 0 2 )= x 2

Further simplify the above,

  Δx=h2mω0   ...... (1)

The uncertainty in momentum is calculated as,

  Δpx= ( p x p x ) 2=( p x 2 2 p x p x + p x 2 )=( p x 2 2 p x 0+ 0 2 )= p x 2

Further simplify the above,

  Δpx=12mhω0   ...... (2)

Multiply equation (1) and (2).

  ΔxΔpx=h 2m ω 0 12mhω0=h2

Conclusion:

Therefore, the product ΔxΔpx is h2 .

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