130_notes.dvi

xis|f(x)|^2 dx. The probability for a particle to have wave number in regiondkaround some value
ofkis|A(k)|^2 dk. (Remember thatp= ̄hkso the momentum distribution is very closely related. We
work withkfor a while for economy of notation.)

5.2 Two Examples of Localized Wave Packets

Lets now trytwo examples of a wave packet localized inkand properly normalized att= 0.

1. A “square” packet:A(k) =√^1 afork 0 −a 2 < k < k 0 +a 2 and 0 elsewhere.


2. A Gaussian packet:A(k) =

( 2 α

π

) 1 / 4

e−α(k−k^0 )

2

.

These are both localized in momentum aboutp= ̄hk 0.

Check the normalization of (1).

∫∞

−∞

|A(k)|^2 dk=

1

a

k∫ 0 +a 2

k 0 −a 2

dk=

1

a

a= 1

Check the normalization of (2) using the result for a definite integral of a Gaussian (See section

5.6.3)

∞∫

−∞

dx e−ax

2

=

√π

a.

∫∞

−∞

|A(k)|^2 dk=

√

2 α

π

∫∞

−∞

e−^2 α(k−k^0 )

2

dk=

√

2 α

π

√

π

2 α

= 1

So now we take the Fourier Transform of (1) right here.

f(x) =

1

√

2 π

∫∞

−∞

A(k)eikxdk=

1

√

2 π

1

√

a

k∫ 0 +a 2

k 0 −a 2

eikxdk

f(x) =

1

√

2 πa

1

ix

[

eikx

]k 0 +a 2

k 0 −a 2 =

1

√

2 πa

1

ix

eik^0 x

[

eiax/^2 −e−iax/^2

]

f(x) =

1

√

2 πa

1

ix

eik^0 x

[

2 isin

(ax

2

)]

=

√

a

2 π

eik^0 x

2 sin

(ax

2

)

ax

Note that

2 sin(ax 2 )
ax is equal to 1 atx= 0 and that it decreases from there. If you square this,
it should remind you of a single slit diffraction pattern! In fact, the single slit gives us a square
localization in position space and the F.T. is thissin(xx)function.

The Fourier Transform of a Gaussian (See section 5.6.4) wave packetA(k) =

( 2 α

π

) 1 / 4

e−α(k−k^0 )

2

is

f(x) =

(

1

2 πα

) 1 / 4

eik^0 xe−

x 42 α