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On this page right at the top they mention two sets of fourier transform. First set is connection between $x$ (position) and $k$ (wave vector) space:

$$ \begin{split} f(x) &= \frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^{\infty} A(k) e^{ikx} dk\\ A(k) &= \frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^{\infty} f(x) e^{ikx} dx \end{split} $$

while the second set is connection between $x$ (position) and $p$ (momentum):

$$ \begin{split} \psi(x) &= \frac{1}{\sqrt{2\pi \hbar}} \int\limits_{-\infty}^{\infty} \phi(p) e^{i\frac{p}{\hbar}x} dp\\ \phi(p) &= \frac{1}{\sqrt{2\pi \hbar}} \int\limits_{-\infty}^{\infty} \psi(x) e^{-i\frac{p}{\hbar}x} dx\\ \end{split} $$


Q1: How do i derive the second set out of first one?

I know De Broglie relation $p = k \hbar$. Hence from $\exp[\pm ikx]$ in the first set we get $\exp \left[\pm i \frac{p}{\hbar} x\right]$ in the second set of equations. This is clear to me. What i dont know is how do we get from $1/\sqrt{2\pi}$ in the first set to $1/\sqrt{2 \pi \hbar}$ in the second set. Where does a $\hbar$ come from?

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The quick answer is that changing variables from $k$ to $p=\hbar k$ introduces a factor of $\hbar$ since $dp=\hbar\,dk$. This is split symmetrically into two factors of $\sqrt{\hbar}$ by adjusting the normalization. –  Emilio Pisanty Mar 3 '13 at 0:18
    
Is this a simple integration trick? –  71GA Mar 3 '13 at 7:35
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I would hesitate to call it a trick: it's just a change in the variable of integration. You can match your two sets of formulae with the identifications $f=\psi$ and $A=\sqrt{\hbar}\phi$, and changing variables from $k$ to $p=k/\hbar$, with $dp=\frac1\hbar dk$. –  Emilio Pisanty Mar 3 '13 at 12:03

1 Answer 1

up vote 3 down vote accepted

As far as where you put things like the $2 \pi$ and the $\hbar$ in the Fourier transform or Inverse Fourier transform, it doesn't really matter. What really matters is that the operations are the inverse of each other. For example:

$\phi(p) = \frac{1}{\sqrt{2 \pi \hbar}} \int^{\infty}_{-\infty} dx \hspace{2mm} \psi(x) e^{-i \frac{p}{\hbar} x} $

$ =\frac{1}{2 \pi \hbar} \int^{\infty}_{-\infty}dx \int^{\infty}_{-\infty}dq \hspace{2mm}\phi (q) e^{ i \frac{q}{\hbar} x } e^{-i \frac{p}{\hbar} x}$

$=\frac{1}{2 \pi \hbar} \int^{\infty}_{-\infty} dx \int^{\infty}_{-\infty}dq \hspace{2mm}\phi (q) e^{ i \frac{(q-p)}{\hbar} x } $

$ =\frac{1}{2 \pi \hbar} \int^{\infty}_{-\infty} dq\hspace{2mm} \phi (q) 2 \pi \hbar\delta(q - p)$

$= \int^{\infty}_{-\infty} dq \hspace{2mm}\phi (q) \delta(q - p) $

$ = \phi(p)$

This is what matters. Now if you really want to `derive' the second set of relations from the first, simply set $p\rightarrow \frac{p}{\hbar}$ and $A(k)\rightarrow A(k) \sqrt{\hbar}$. This gives

$f(x) = \frac{1}{\sqrt{2 \pi}} \int^{\infty}_{-\infty}A(k) \sqrt{\hbar} e^{i\frac{k}{\hbar} x} \frac{dk}{\hbar} =\frac{1}{\sqrt{2 \pi \hbar}} \int^{\infty}_{-\infty}A(k) e^{i\frac{k}{\hbar} x} dk $

and

$A(k ) \sqrt{\hbar} = \frac{1}{\sqrt{2 \pi}} \int^{\infty}_{-\infty}f(x) e^{i \frac{k }{\hbar}x} dx \rightarrow A(k ) = \frac{1}{\sqrt{2 \pi\hbar}} \int^{\infty}_{-\infty}f(x) e^{i \frac{k }{\hbar}x} dx$

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