130_notes.dvi

5.6.4 Fourier Transform of Gaussian*

We wish toFourier transform the Gaussian wave packetin (momentum) k-spaceA(k) =
( 2 α
π

) 1 / 4

e−α(k−k^0 )

2

to getf(x) in position space. The Fourier Transform formula is

f(x) =

1

√

2 π

(

2 α

π

) 1 / 4 ∫∞

−∞

e−α(k−k^0 )

2

eikxdk.

Now we will transform the integral a few times to get to the standard definite integral of a Gaussian
for which we know the answer. First,
k′=k−k 0

which does nothing really sincedk′=dk.

f(x) =

( α

2 π^3

) 1 / 4

eik^0 x

∫∞

−∞

e−α(k−k^0 )

2

ei(k−k^0 )xdk

f(x) =

( α

2 π^3

) 1 / 4

eik^0 x

∫∞

−∞

e−αk

′ 2

eik

′x

dk′

Now we want tocomplete the squarein the exponent inside the integral. We plan a term like
e−αk

′′ 2

so we define

k′′=k′−

ix

2 α

.

Againdk′′=dk′=dk. Lets write out the planned exponent to see what we are missing.

−α

(

k′−

ix

2 α

) 2

=−αk′^2 +ik′x+

x^2

4 α

We need to multiply bye−
x 42 α
to cancel the extra term in the completed square.

f(x) =

( α

2 π^3

) 1 / 4

eik^0 x

∫∞

−∞

e−α(k

′−ix

2 α)

2

e−

x 42 α

dk′

That term can be pulled outside the integral since it doesn’t depend onk.

f(x) =

(α

2 π^3

) 1 / 4

eik^0 xe−

x 42 α

∫∞

−∞

e−αk

′′ 2

dk′′

So now we have the standard Gaussian integral which just gives us

√π

α.

f(x) =

( α

2 π^3

) 1 / 4 √π

α

eik^0 xe−

x 42 α

f(x) =

(

1

2 πα

) 1 / 4

eik^0 xe−

x 42 α