**Quantum Mechanics and Momentum**

Part 3: The Fourier Transform, Dirac Delta, and Plancherel’s Theorem

Consider the Fourier transform in one dimension. There are several common conventions for defining it; here, we will use the unitary angular form frequently used in physics: for function *f*(*t*) the Fourier transform is

,

and the inverse transform is

These convert between the time domain and the angular frequency domain. For transforms on a function of a position variable *f*(*x*), we have

,

and

which convert between the position domain and the (angular) wavenumber domain.

For an *n* dimensional position space with position vector we find the transform of a function to be in the *n* dimensional wavenumber space with wavenumber vector

via

and

Now, consider the Fourier transform of the Dirac delta “function”:

and by examination of this, and the inverse transform, we obtain

.

(Note that this is not a mathematically rigorous equality, but is valid under integration with a function: .)

This relation is useful for proving a number of relations, one of which follows.

For a valid wavefunction of one spatial dimension *ψ*(*x*), let us define its Fourier transform *φ*(*k*).

Then

The complex conjugate of this is

where ^{*} indicates the complex conjugate.

Thus, let us now find :

This is known as Plancherel’s Theorem (or in some physics and engineering sources as Parseval’s Theorem), and we see that if the (position space) wavefunction *ψ*(*x*) is normalized, , then the corresponding “wavenumber space” wavefunction *φ*(*k*) is also normalized.

The reason we want to use such a Fourier transformed function goes back to our de Broglie relations, particularly . Using this, we see that our “wavenumber domain,” via this equation, represents the momentum space basis for our quantum state.

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Tags: Fourier Transform, Friday Physics, Momentum Space, Parseval's Theorem, physics, Plancherel's Theorem, Quantum Mechanics

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November 21, 2008 at 12:11 am |

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October 26, 2010 at 11:29 am |

Cool! But how can one combine ω and k in the Fourier Transform and in what conditions? I would like to “derive” Schödinger equation but I need to justify why to use ψ(x,t) that contains both ω and k. So via expection value calculations Schödinger equation can be “derived”.

October 27, 2010 at 12:54 pm |

One can have both ω and k if your function depends on both time t and space x, as each of the former is the variable for the Fourier transform of the corresponding latter variable. I touch on the time transform, and its relation to the time-dependent Schrödinger equation in part 6 of this series.