Momentum and wave vector representation

[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?

 

[jtbell]

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

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

 

[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?

 

[jtbell]

$\psi(x)=\frac{1}{\sqrt{2\pi}}\frac{1}{\hbar}\int^{\infty}_{-\infty}F(p)e^{i\frac{p}{\hbar}x}dp$

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,

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

(Your F(k) is my A(k).)

 

[LagrangeEuler]

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