# Normalization of the Fourier transform

## Homework Statement

The fourier transfrom of the wave function is given by
$$\Phi(p) = \frac{N}{(1+\frac{a_0^2p^2}{\hbar^2})^2}$$
where ##p:=|\vec{p}|## in 3 dimensions.
Find N, choosing N to be a positive real number.

## Homework Equations

$$\int d^3\vec{p}|\Phi(p)|^2=1$$
, over all p in the 3 dimensions.

## The Attempt at a Solution

First finding the complex conjugate,
$$\Phi^*(p) = \frac{N}{(1+\frac{a_0^2 p^2}{\hbar^2})^2}$$
So,
$$|\Phi(p)|^2 = \frac{N^2}{(1+\frac{a_0^2 p^2}{\hbar^2})^4}$$
So,
$$\frac{1}{N^2} = \int d^3 \vec{p}\frac{1}{(1+\frac{a_0^2 p^2}{\hbar^2})^4}$$

How would I change ##d^3\vec{p}## to be a triple integral, one of which is over dp?

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By using spherical coordinates for ##\vec{p}## ...

By using spherical coordinates for ##\vec{p}## ...
I thought of using spherical coordinates for p, but would they be the same as for x.
So would it be that the integral, ignoring the function of integration, becomes?
$$\int d^3\vec{p}= \int_{0}^{\pi}\int_{0}^{2\pi}\int_{0}^{\infty}p^2sin(\phi)dpd\theta d\phi$$?

If the above is true, why can the same be written for p as was for x.

It doesn't matter if you have ##p## or ##x##. You can write write any three-dimensional vector ##\vec{a}## as
$$\vec{a}=a_x e_x + a_y e_y + a_z e_z$$,
where ##e_x##, ##e_y## and ##e_z## are the cartesian basis vectors.
You can then transform to spherical coordinates, i.e. instead writing ##\vec{a}## in terms of ##e_r##, ##e_\theta## and ##e_\phi## . It is not important what ##\vec{a}## represent.

It doesn't matter if you have ##p## or ##x##. You can write write any three-dimensional vector ##\vec{a}## as
$$\vec{a}=a_x e_x + a_y e_y + a_z e_z$$,
where ##e_x##, ##e_y## and ##e_z## are the cartesian basis vectors.
You can then transform to spherical coordinates, i.e. instead writing ##\vec{a}## in terms of ##e_r##, ##e_\theta## and ##e_\phi## . It is not important what ##\vec{a}## represent.

Why would you use ##sin(\theta)##?
Using ##sin(\theta)##,
$$\int_{0}^{2\pi}sin(\theta)d\theta = [-cos(\theta)]_0^{2\pi} = -1-(-1) = 0$$
Also, http://mathworld.wolfram.com/SphericalCoordinates.html. ##sin(\phi)## is used where ##\phi## ranges from 0 to ##\pi##, which is what I have in my integral.

Yes, you are correct. I thought the range for ##\phi## was ##[0,2\pi]## and ##[0,\pi]## for ##\theta##.

vela
Staff Emeritus