# A note on Wave-functions and Fourier Transforms

In quantum mechanics you can denote the wave-function in the position or the momentum basis. Written in the momentum basis, it would look something like: $|\psi(x)> = a_0 |p_0> + a_1 |p_1> + \hdots$ $|\psi(x)> = \sum\limits_{n} a_n |p_n>$

But momentum is a continuous variable and it varies from $-\infty$ to $\infty$. Therefore changing to the integral representation we get that: $|\psi(x)> = \int\limits_{-\infty}^{\infty} dp \ a_n(p) |p>$

But $a_n(p)$ is just the projection of the momentum vector on the wavefunction: $|\psi(x)> = \int\limits_{-\infty}^{\infty} dp |p>$

We are also aware from our knowledge of Fourier Transform* that the wave function written in momentum space is given as : $|\psi(x)> = \frac{1}{\sqrt{2 \pi \hbar}} \int\limits_{-\infty}^{\infty} dp \ \tilde{\psi}(p) |e^{\frac{ipx}{\hbar}}>$

Comparing both the above equations if we take the momentum basis as $|p> = e^{\frac{ipx}{\hbar}}$, then: $ = \frac{1}{\sqrt{2 \pi \hbar}} \tilde{\psi}(p)$

We can perform a similar analysis by expanding the wavefunction about the position basis and get $ = \frac{1}{\sqrt{2 \pi \hbar}} \psi(x)$

** Where does the $\frac{1}{\sqrt{2 \pi \hbar}}$ in the Fourier Transform come from ?

We know from Fourier transform is defined as follows: $|\psi(x)> = \frac{1}{\sqrt{2 \pi}} \int\limits_{-\infty}^{\infty} dk \ \psi_{1}(k) |e^{ikx}>$

Plugging in $p = \hbar k$ and rewriting the above equation we get, $|\psi(x)> = A \int\limits_{-\infty}^{\infty} dp \ \tilde{\psi}(p) |e^{\frac{ipx}{\hbar}}>$

We find that from $\int\limits_{-\infty}^{\infty} dx <\psi(x)| \psi(x)> = 1$

that the normalization constant is not $\frac{1}{\sqrt{2 \pi}}$ but $\frac{1}{\sqrt{2 \pi \hbar}}$. Therefore, $|\psi(x)> = \frac{1}{\sqrt{2 \pi \hbar}} \int\limits_{-\infty}^{\infty} dp \ \tilde{\psi}(p) |e^{\frac{ipx}{\hbar}}>$