[NOTES/QM-12003] Propagator for Free Particle

Green function

The Green function \(G(x,t;x\Prime,t\Prime)\) for the time dependent Schrodinger equation \begin{equation}i\hbar \dd[\psi]{t} =- \frac{\hbar^2}{2m} \pp[\psi]{x} + (V(x) - E) \psi =0\end{equation}satisfies the equation\begin{align}i\hbar \pp{t}G(x,t;x\Prime,t\Prime)- H G(x,t;x\Prime, t\Prime)&= i\hbar\delta(x-x\Prime)\delta(t-t\Prime)\\\hbar \pp{t}G(x,t;x\Prime,t\Prime) +\frac{\hbar^2}{2m} \PP{x}G(x,t;x\Prime,t\Prime) &- (V(x) - E)G(x,t;x\Prime,t\Prime)\\&=\delta(x-x\Prime)\delta(t-t\Prime).\end{align}where the Green function satisfies the boundary condition\begin{equation} G(x,t;x\Prime,t\Prime)\Big|_{t=t\Prime}= \delta(x-x\Prime).\end{equation}

Given the initial wave function at time \(t_0\) as \(\psi(x, t_0)\), the wave function at time \(t>t_0\) is given by \begin{equation} \psi(x,t) = \int dx\Prime G(x,t;x\Prime, t_0)\psi(x\Prime,t_0).\end{equation}PropagatorIf we define propagator \(K(x,t;x_0,t_0)\) by equation\begin{equation} G(x,t;x_0,t_0)= \theta(t-t_0) K(x,t;x_0,t_0),\end{equation}then the propagator obeys\begin{equation} i\hbar \pp{t} =K(x,t;x_0t_0)=\big[- \frac{\hbar^2}{2m} \pp{x} + (V(x) - E) \big]K(x,t;x_0,t_0) =0\end{equation}and \begin{equation}\left. K(x,t;x_0,t_0)\right|_{t=t_0} = \delta(t-t_0).\end{equation}

Free particle propagator

Since the propagator is solution of the time dependent free particle Schrodinger equation has the form given by superposition of free particle solutions of definite energy. These solutions of time dependent free particle Schr\"{o}dinger equation are given by\begin{equation} \psi_p(x,t) = \frac{1}{\sqrt{2\pi\hbar}} e^{ipx/\hbar} e^{-iEt/\hbar},\qquad E=\frac{p^2}{2m}.\end{equation}Writing the propagator $K(x,t; x_0,t_0)$ as superposition of the above solutions we get\begin{equation} K (x,t; x_0,t_0) = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^\infty C(p) e^{ipx/\hbar} e^{-iE_pt/\hbar},\end{equation}where $C(p)$ are coefficients to be fixed and are determined by setting $t=t_0$.This gives\begin{eqnarray} \delta (x,t; x_0,t_0) &=& (\sqrt{2\pi\hbar})^{-1/2}\int_{-\infty}^\infty C(p) e^{ipx/\hbar} e^{-iE_pt_0/\hbar},\, dp.\end{eqnarray}where $E_p=\frac{p^2}{2m}$. The function $C(p)$ is determined bytaking inverse Fourier transform of the above equation. Thus we have\begin{eqnarray} C(p) &=& (\sqrt{2\pi\hbar})^{-1/2}\int_{-\infty}^\infty \delta(x-x_0) e^{-ipx/\hbar} e^{-iE_pt_0/\hbar},\, dx.\\ &=& (\sqrt{2\pi\hbar})^{-1/2} e^{-ipx_0/\hbar} e^{-iE_pt_0/\hbar}.\end{eqnarray}Therefore the expression for the propagator becomes\begin{equation} K(x,t; x_0,t_0) = \left( \frac{1}{2\pi\hbar}\right)\int_{-\infty}^\infty e^{ip(x-x_0)/\hbar} e^{iE_p(t-t_0)} dp. \end{equation}Substituting $E_p=p^2/2m$ and doing a Gaussian integral over $p$ gives the final answer for the free particle propagator\begin{equation} K(x,t ; x_0,t_0)= \left(\frac{m}{2\pi\hbar (t-t_0)}\right)^{1/2} \exp\left( -\frac{m(x-x_0)^2}{2i\hbar(t-t_0)} \right)\end{equation}The answer for three dimensions can be written down and is given by\begin{equation} K(\xbf,t ; \xbf_0,t_0)= \left(\frac{m}{2\pi\hbar (t-t_0)}\right)^{3/2} \exp\left( -\frac{m(\xbf-\xbf_0)^2}{2i\hbar(t-t_0)} \right).\end{equation}