# Momentum and wave vector representation

1. Mar 11, 2013

### LagrangeEuler

$$p=\hbar k$$
So $dp=\hbar dk$
How to define Fourier transform from momentum to coordinate space and from wave vector to coordinate space? I'm confused. Is there one way to do it or more equivalent ways?

2. Mar 11, 2013

### Staff: Mentor

As far as I know, the only way to do it that preserves normalization is:

$$\psi(x) = \frac{1}{\sqrt{2 \pi \hbar}} \int^{+\infty}_{-\infty} {\phi(p) e^{ipx/\hbar} dp} \\ \phi(p) = \frac{1}{\sqrt{2 \pi \hbar}} \int^{+\infty}_{-\infty} {\psi(x) e^{-ipx/\hbar} dx}$$

or

$$\psi(x) = \frac{1}{\sqrt{2 \pi}} \int^{+\infty}_{-\infty} {A(k) e^{ikx} dk} \\ A(k) = \frac{1}{\sqrt{2 \pi}} \int^{+\infty}_{-\infty} {\psi(x) e^{-ikx} dx}$$

3. Mar 12, 2013

### LagrangeEuler

Thanks for your answer. And when you want to go from momentum to wave vector representation
For example
$\psi(x)=\frac{1}{\sqrt{2\pi}}\int^{\infty}_{-\infty}F(k)e^{ikx}dk$
$p=\hbar k$ so $dp=\hbar dk$
and
$\psi(x)=\frac{1}{\sqrt{2\pi}}\frac{1}{\hbar}\int^{\infty}_{-\infty}F(p)e^{i\frac{p}{\hbar}x}dp$
Right?

4. Mar 12, 2013

### Staff: Mentor

You will find that, using this equation, if F(p) is normalized, that is

$$\int^{+\infty}_{-\infty} {F^*F dp} = 1$$

then $\psi(x)$ is not, that is

$$\int^{+\infty}_{-\infty} {\psi^*\psi dx} \ne 1$$

whereas for my version (in the first pair of my previous post), if $\phi(p)$ is normalized, then so is $\psi(x)$, and vice versa. The extra factor of $\sqrt{\hbar}$ makes the difference. In my version,

$$\phi(p) = \frac{1}{\sqrt{\hbar}} A(k) = \frac{1}{\sqrt{\hbar}} A \left( \frac{p}{\hbar} \right)$$

But why? Why is $\phi(p)=\frac{1}{\sqrt{\hbar}}A(k)$?