Momentum Basis

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

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

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