1000 Solved Problems in Modern Physics

(Grace) #1

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


dx

)

dx


dψ∗
dx

dx+


x

(

dψ∗
dx

)(


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


=−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

)


+

∫ ((


^2

2 m

)

∇^2 ψ+Vψ

)

∂ψ
∂x

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