QMGreensite_merged

5.5. GAUSSIANWAVEPACKETS 73

thisbecomes
f(p)=(4πa^2 )^1 /^4 e−a

(^2) (p−p 0 ) (^2) /2 ̄h 2
(5.67)
Then,substitutingf(p)into(5.44),wefindthewavefunctionatanylatertimet
ψ(x,t) = (4πa^2 )^1 /^4
∫ dp
2 π ̄h
e−a
(^2) (p−p 0 ) (^2) /2 ̄h 2
exp
[
i(px−
p^2
2 m
t)/ ̄h
]

=

(4πa^2 )^1 /^4

2 π ̄h

∫

dq exp

[

−

a^2

2 ̄h^2

q^2 +i

q+p 0

h ̄

x−i

(q+p 0 )^2

2 m ̄h

t

]

=

(4πa^2 )^1 /^4

2 π ̄h

exp[i(p 0 x−

p^20

2 m

t)/ ̄h]

×

∫

dq exp

[

−

(

a^2

2 ̄h^2

+

it

2 m ̄h

)

q^2 +i

(

x

̄h

−

p 0 t

m ̄h

)

q

]

=

(4πa^2 )^1 /^4

2 π ̄h



 π

a^2

2 ̄h^2 +

it

2 m ̄h





1 / 2

exp

{

−

(x−v 0 t)^2

2(a^2 +i ̄ht/m)

}

ei(p^0 x−

p^20

2 mt)/ ̄h(5.68)

where
v 0 ≡

p 0

m

(5.69)

Although(5.68)lookslikeacomplicatedexpression,itsmodulusisverysimple:

|ψ|^2 = ψ∗(x,t)ψ(x,t)

=

1

√

πa^2 (t)

exp

[

−

(x−v 0 t)^2

a^2 (t)

]

(5.70)

wherewehavedefined

a(t)≡

√

a^2 +

̄h^2

m^2 a^2

t^2 (5.71)

Notethatthemodulusofthewavefunctionremainsagaussianatalltimes,butthe
width a(t)of the gaussianincreases, i.e. the wavefunction ”spreadsout”as time
passes.
Wecanthencalculateand∆xatanytimet>0:

<x> =

1

√

πa^2 (t)

∫

dxxexp

[

−

(x−v 0 t)^2

a^2 (t)

]

=

1

√

πa^2 (t)

∫

dx′(x′+v 0 t)exp

[

−

x′^2

a^2 (t)

]

= v 0 t (5.72)

andalso

∆x^2 = <(x−<x>)^2 >=<(x−v 0 t)^2 >