1000 Solved Problems in Modern Physics

128 2 Quantum Mechanics – I

2.3.8 Uncertainty Principle...............................

2.82ΔxΔpx∼/ 2

P=



2 x

E=

p^2

2 m

+ 1 / 2 mω^2 x^2

=

^2

8 mx^2

+ 1 / 2 mω^2 x^2

The ground state energy is obtained by setting∂∂Ex= 0

∂E

∂x

=−

^2

4 mx^3

+mω^2 x= 0

whencex^2 = 2 mω

∴ E= 1 / 4 ω+ 1 / 4 ω=^12 ω

2.83 IfEandpare to be measured simultaneously their operators must commute.
Now
H=−^2 ∇^2 / 2 m+Vandp=−i∇
[H,p]=[−^2 ∇^2 / 2 m+V,−i∇]
=i^3 ∇^2 ∇/ 2 m−iV∇−i^3 ∇∇^2 / 2 m+i∇V

The first and the third term on the RHS get cancelled because∇^2 ∇=∇∇^2.

Therefore

[H,P]=−i(V∇−∇V)

IfV = constant, the commutator vanishes. To put it differently energy

and momentum can be measured with arbitrary precision only for unbound

particles.

2.84 Consider the motion of a particle along x-direction.
The uncertaintyΔxis defined as

(Δx)^2 =<(x−<x>)^2 >=<x^2 >− 2 <x><x>+<x>^2

=<x^2 >−<x>^2 (1)

Similarly

(ΔPx)^2 =<Px^2 >−<Px>^2 (2)