Fourier Transforms, Momentum and Position

1. Jun 9, 2009

r16

In quantum mechanics, why does the Fourier transform

$$f(x) = \int_{-\infty}^\infty F(k) e^{ikx}dk$$
represent position and

$$F(k) = \int_{-\infty}^\infty f(x) e^{-ikx} dx$$
represent momentum?

2. Jun 9, 2009

malawi_glenn

you must have something given to you, e.g. given that f(x) is the position wavefunction, then it's fourier transform is the momenum wavefunction. It just follows since p = - i hbar d/dx

<x'|p|p'> = -i h d/dx' <x'|p'>

p'<x'|p'> = -i h d/dx' <x'|p'>

solve this differential equation

<x'|p'> = N exp { i p' x' /h }

now look at

psi_a(x') = <x'|a> = (insert completeness realtion ) = integral dp' <x'|p'><p'|a>

psi_a(x') = N integral dp' exp { i p' x' /h } phi_a(p')

determine the normalization N using the delta function:

delta(x' - x'') = N^2 integral dp' exp { i p'( x' - x'') /h }

we get N = 1/sqrt(2 pi h)

3. Jun 10, 2009

r16

I'm only starting out in Quantum Mechanics (chapter 2 of the griffiths book) and I am not familiar with the notation

I'm sure ill get to it later on in the book. Until then, could you explain it?

4. Jun 10, 2009

malawi_glenn

you might want to look up "bra - ket" notation or "dirac notation" in Griffiths book, then I can explain if you don't understand, but basically:

p|p'> = "p operator on p-eigenstate with momentum p' " = p'|p'> (I denote operator with the letter and eigenvalues with prime"

since p = -i h d/dx, we can do the same operation on the right hand side, but with <x'|p'> as just an arbitrary function of x' and p'

The basic idea is that, without getting too much into math behind it;

|a'> is a vector in hilbert space, it denotes the state with quantun number a'

a|a'> = a'|a>

ok?

these are called "kets"

now the dual vector, called "bra":

<a'|

we can "think" of this as the ket's beeing column vectors and bra's as row vectors:

<a''|a'> is then a number

ok, this was assuming that a',a'' are discrete quantum numbers

now, for x and p, which are continuous, we can use the same notation, but we can not imagine/represent them as discrete vectors as we do in introductory linear algebra.