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 $\infty$ .

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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:

$|\psi(x)> = \int\limits_{-\infty}^{\infty} dp <p|\psi(x)>|p>$

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:

$<p|\psi(x)> = \frac{1}{\sqrt{2 \pi \hbar}} \tilde{\psi}(p)$

We can perform a similar analysis by expanding the wavefunction about the position basis and get

$<x| \tilde{\psi}(p)> = \frac{1}{\sqrt{2 \pi \hbar}} \psi(x)$

** 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,

$|\psi(x)> = \frac{1}{\sqrt{2 \pi \hbar}} \int\limits_{-\infty}^{\infty} dp \ \tilde{\psi}(p) |e^{\frac{ipx}{\hbar}}>$

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A note on Wave-functions and Fourier Transforms

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