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 .

unnamed

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}}>

 

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s