1000 Solved Problems in Modern Physics

164 3 Quantum Mechanics – II

The first term on the RHS is zero at both the limits.

−

∫

∂^2 ψ∗

∂x^2

xψdx=−

∫

dψ∗

dx

(

ψ+x

dψ

dx

)

dx

=ψ

dψ∗

dx

dx+

∫

x

(

dψ∗

dx

)(

dψ

dx

)

dx (8)

Substituting (7) and (8) in (5), the terms underlined vanish together.

d<x>

dt

=

(

i

2 m

)(

−

∫

ψ∗

∂

∂x

dx+

∫

ψ

dψ∗

dx

dx

)

=

1

2 m

(∫

ψ∗

(

−i

∂

∂x

)

ψdx+

∫

ψ

(

i

∂

∂x

)

ψ∗dx

)

(9)

Now the operator forPxis−i∂∂x. The first term on RHS of (9) is the

average value of the momentumPx, the second term must represent the

average value ofPx∗.Butpxbeing real,Px∗=Px. Therefore

d<x>

dt

=

1

m

<Px> (10)

Thus (10) is similar to classical equationx=p/m

Equation (10) can be interpreted by saying that if the “position” and

“momentum” vectors of a wave packet are regarded as the average or

expectation values of these quantities, then the classical and quantum

motions will agree.

(b)

d<Px>

dt

=

d

dt

∫

ψ∗

(

−i

∂

∂x

)

ψdτ=−i

d

dt

∫

ψ∗

∂ψ

∂x

dτ

=−i

∫

dψ∗

dt

∂ψ

∂x

dτ+

∫

ψ∗

∂

∂x

∂ψ

∂t

dτ (11)

Nowi

∂ψ

∂t

=−

(

^2

2 m

)

∇^2 ψ+Vψ

−

i∂ψ∗

∂t

=−

(

^2

2 m

)

∇^2 ψ∗+Vψ∗ (12)

Using (12) in (11)

d

dt

<Px>=−

∫

ψ∗

∂

∂x

(

−

(

^2

2 m

)

∇^2 ψ+Vψ

)

dτ

+

∫ ((

−

^2

2 m

)

∇^2 ψ+Vψ

)

∂ψ

∂x

dτ