Free Particle

Has no definite energy. Because $p=\hslash k$, you can replace the below equations of $k$ with $\frac{p}{\hslash }$

Topics

• Phase Velocity
• Group Velocity

$\Psi (x,t)=\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }\tilde{\Psi }(k)e^{i(kx-\omega t)}\mathrm{d}k$

• Wave function for free particle
• $\omega =\frac{\hslash k^2}{2m}$
• $e^{i(kx-\omega t)}$ can be rewritten as $e^{ikx}{e}^{-iEt/\hslash }=e^{ikx}{e}^{-i\hslash k^{2}t/2m}$ because $E=\frac{{\hslash }^2{k}^2}{2m}$
• Analogous to $\Psi (\vec{r},t)=\sum _{n=1}^{\infty }{c}_n{\psi }_n(\vec{r})e^{-i{E}_{n}t/\hslash }$, where $c_n$ is something like $\tilde{\Psi }(k)$

$\tilde{\Psi }(k)=\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }\Psi (x,0)e^{-ikx}\mathrm{d}x$

• Analogous to $c_n=\int {\psi }_n^{\ast }(x)\Psi (x,0)\mathrm{d}x$, or the wave function weighting coefficient
• $\tilde{\Psi }(k)$ is the Fourier transform of $\Psi (x,t)$
• Solved by Plancherel’s Theorem

$\Psi (x,0)=\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }\tilde{\Psi }(k)e^{ikx}\mathrm{d}k$

$\frac{1}{2\pi }{\int }_{-\infty }^{\infty }{e}^{ikx}{e}^{-i{k}^{\prime }x}\mathrm{d}x=\delta (k-k^{\prime })$

• Dirac orthonormality condition
• $\delta$ is the delta function