Fourier Transform and Basis Change

Classical Mechanics Level 3

A quantum-mechanical state has a position wavefunction given by

$\large \psi(x) = \pi^{-1/4} e^{-\frac{x^2}{2}}.$

What is the wavefunction $\psi(p)$ of the same state written in the momentum basis?

\pi^{-1/4} e^{-\frac{p^2}{2}}

\pi^{-1/4} \hbar^{-1/2} e^{-\frac{p^2}{2\hbar^2}}

\frac{e^{ipx/\hbar}}{2\pi \hbar}

\delta(p)

1 solution

Matt DeCross
May 10, 2016

This wavefunction is a Gaussian, so its Fourier transform (i.e., the same state in the momentum basis) will also be a Gaussian. This leaves two possible solutions. Computing the integral explicitly to compute $\psi(p)$ by completing the square in the exponent yields:

$\psi(p) = \int_{-\infty}^{\infty} \frac{ e^{ipx / \hbar}}{\sqrt{2\pi \hbar}} \pi^{-1/4} e^{-x^2/2} dx = \pi^{-1/4} \hbar^{-1/2} e^{-p^2/(2\hbar^2)},$

as claimed.

In numerator, $e^{\frac{-ipx}{\hslash}}$ for this is inverse transform you wrote.

lovro cupic - 1 year, 10 months ago

Fourier Transform and Basis Change

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