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Oct 29, 20231 min read

Similar to the wave functions generated from the free particle, except using $p$ instead of $k$

$Φ (p, t) = \frac{1}{2 π ℏ} \int_{- \infty}^{\infty} e^{- i p x /ℏ} Ψ (x, t) d x$

$\displaystyle \Phi(p,t)=\frac{1}{\sqrt{ 2\pi \hbar }}\int_{-\infty}^{\infty}e^{-ipx/\hbar}\Psi(x,t) \, \mathrm{d}x$
$\displaystyle \Psi(x,t)=\frac{1}{\sqrt{ 2\pi \hbar }}\int_{-\infty}^{\infty} e^{ipx/\hbar}\Phi(p,t) \, \mathrm{d}p$