Normalization of the Fourier transform

[doggydan42]

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?

 

[eys_physics]

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

 

[doggydan42]

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.

 

[eys_physics]

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.

In your formula it should be $\sin\theta$ instead of $\sin\phi$.

 

[doggydan42]

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.

In your formula it should be $\sin\theta$ instead of $\sin\phi$.

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.

 

[eys_physics]

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

 

[vela]

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

The OP is using the mathematics convention. Physicists typically use $\theta$ to be the polar angle and $\phi$ for the azimuthal angle.