# Fourier transform. Impulse representation.

1. Mar 9, 2014

### LagrangeEuler

$\varphi(p)=\frac{1}{\sqrt{2\pi\hbar}}\int^{\infty}_{-\infty}dx\psi(x)e^{-\frac{ipx}{\hbar}}$. This $\hbar$ looks strange here for me. Does it holds identity
$\int^{\infty}_{-\infty}|\varphi(p)|^2dp=\int^{\infty}_{-\infty}|\psi(x)|^2dx=1$???
I'm don't think so because this $\hbar$. So state in impulse space is not normalized.

2. Mar 9, 2014

### Staff: Mentor

Yes, it does. The $\hbar$ in the transform is required in order to make this true. You may be thinking of the standard Fourier transform:
$$A(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}{\psi(x)e^{-ikx}dx}$$
Changing variables from k to $p = \hbar k$ introduces the extra $\hbar$ in the constant factor.