ñòð. 45 |

useful in a variety of applications, but it will also establish the relation between the

Fourier transform and the closure relation introduced in section (10.1). Consider the delta

function centered at x = x0 :

f(x) = (x x0 ) : (11.29)

;

Problem a: Show that the Fourier transform F(k) of this function is given by:

F(k) = 21 e;ikx0 : (11.30)

Problem b: Show that this implies that the Fourier transform of the delta function (x)

centered at x = 0 is a constant. Determine this constant.

Problem c: Use expression (11.27) to show that

1 Z 1 eik(x;x0 ) dk :

(x x0 ) = 2 (11.31)

;1

;

Problem d: Use a similar analysis to derive that

1 Z 1 e;i(k;k0 )x dx :

(k k0 ) = 2 (11.32)

;1

;

These expressions are very useful in a number of applications. Again, there is close analogy

between this expression and the projection of vectors introduced in section (10.1). To

establish this connection let use de ne the following basis functions:

uk (x) 1 eikx (11.33)

p

2

and use the inner product de ned in (11.20) with the integration limits extending from

to

;1 1.

Problem e: Show that expression (11.32) implies that

(uk uk0 ) = (k k0 ) : (11.34)

;

Why does this imply that the functions uk (x) form an orthonormal set?

CHAPTER 11. FOURIER ANALYSIS

136

Problem f: Use (11.31) to derive that:

Z1

uk (x)uk (x0 ) dk = (x x0 ) : (11.35)

;1

;

This expression is the counterpart of the closure relation (10.13) introduced in section

(10.1) for nite-dimensional vector spaces. Note that the delta function (x x0 ) plays

;

the role of the identity operator I with components Iij = ij in equation (10.13) and that

R1

P

the summation N over the basis vectors is replaced by an integration ;1 dk over the

i=1

basis functions. Both di erences are due to the fact that we are dealing in this section

with an in nitely-dimensional function space rather than a nite-dimensional vector space.

Also note that in (11.35) the complex conjugate is taken of uk (x0 ). The reason for this is

that for complex unit vectors n the transpose in the completeness relation (10.13) should

^

be replaced by the Hermitian conjugate. This involves taking the complex conjugate as

well as taking the transpose.

11.5 Changing the sign and scale factor

In the Fourier transform (11.27) from the wave number domain (k) to the position domain

(x), the exponent has a plus sign exp (+ikx) and the coe cient multiplying the integral

is given by 1. In other texts on Fourier transforms you may encounter a di erent sign

of the exponent and di erent scale factors are sometimes used in the Fourier transform.

For example, the exponent in the Fourier transform from the wave number domain to the

position domain may have a minus sign exp (;ikx) and there may be a scale factor such

p

as 1= 2 that di ers from 1. It turns out that there is a freedom in choosing the sign of

the exponential of the Fourier transform as well as in the scaling of the Fourier transform.

We will rst study the e ect of a scaling parameter on the Fourier transform.

˜

Problem a: Let the function F(k) de ned in (11.28) be related to a new function F(k)

˜

by a scaling with a scale factor C: F(k) = C F(k). Use the expressions (11.27) and

(11.28) to show that: Z1

˜

F(k)eikxdk

f(x) = C (11.36)

;1

1 Z 1 f(x)e;ikxdx :

˜

F(k) = 2 C (11.37)

;1

These expressions are completely equivalent to the original Fourier transform pair (11.27)

and (11.28). The constant C is completely arbitrary. This implies that one may take any

multiplication constant for the Fourier transform the only restriction is that the product

of the coe cients for Fourier transform and the backward transform is equal to 1=2 .

Problem b: Show this last statement.

In the literature, notably in quantum mechanics, one often encounters the Fourier trans-

form pair using the value C = 1= 2 . This leads to the Fourier transform pair:

p

1 Z 1 F (k)eikx dk

˜

f(x) = (11.38)

2 ;1

p

11.5. CHANGING THE SIGN AND SCALE FACTOR 137

1 Z 1 f(x)e;ikx dx :

˜

F(k) = (11.39)

2 ;1

p

This normalization not only has the advantage that the multiplication factors for the

p

forward and backward are identical (1= 2 ), but the constants are also identical to the

constant used in (11.33) to create a set of orthonormal functions.

Next we will investigate a change in the sign of the exponent in the Fourier transform.

˜ ˜

To do this, we will use the function F(k) de ned by: F(k) = F(;k).

Problem c: Change the integration variable k in (11.27) to and show that the Fourier

;k

transform pair (11.27) and (11.28) is equivalent to:

Z1

F(k)e;ikx dk

˜

f(x) = (11.40)

;1

1 Z 1 f(x)eikx dx :

˜

F(k) = 2 (11.41)

;1

Note that these expressions only di er from earlier expression by the sign of the exponent.

ñòð. 45 |