Time dependence of wave packets without eigenfunctions - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-07-17T20:30:49Z https://physics.stackexchange.com/feeds/question/292020 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/292020 0 Time dependence of wave packets without eigenfunctions spoofd https://physics.stackexchange.com/users/100271 2016-11-10T21:47:22Z 2019-02-28T07:02:40Z <p>In general, to obtain the time dependence of an arbitrary wave packet $\left| \phi(x)\right&gt;$ in the Schödinger picture, we expand the wave packet in the energy eigenfunction basis $\left| \psi_n(x)\right&gt;$</p> <p>$$\left| \phi(x)\right&gt; = \sum_\limits{n} \left&lt; \psi_n(x) | \phi(x)\right&gt; \left| \psi_n(x)\right&gt;$$</p> <p>Then we solve the time dependent equation: </p> <p>$$\left| \phi(x, t)\right&gt; = \sum_\limits{n} \left&lt; \psi_n(x) | \phi(x)\right&gt; \left| \psi_n(x)\right&gt; e^{-iE_nt\hbar}$$</p> <ol> <li><p>Can the time dependent equation be applied to the wave packet itself $\phi(x)$ without using the energy eigenfunctions?</p></li> <li><p>I think it might be possible to do so by using the time dependent Schödinger equation considering the wave packet as a function of $x$ and $t$ (equation below). Am I right?</p></li> </ol> <p>$$\left[-\frac{\hbar}{2m}\nabla^2 + V(x)\right]\phi(x, t) = i\hbar\frac{\partial \phi}{\partial t}$$</p> https://physics.stackexchange.com/questions/292020/-/299798#299798 1 Answer by flippiefanus for Time dependence of wave packets without eigenfunctions flippiefanus https://physics.stackexchange.com/users/55261 2016-12-20T10:56:00Z 2016-12-20T10:56:00Z <p>Here is a possible suggestion how such an approach might work. The mechanism follows from rather general considerations and does not require quantum mechanics. So I'll use a more general mathematical notation.</p> <p>Imagine I want to know the time evolution $f(x,t)$ when I only have knowledge of the initial condition $f(x,0)$. The assumption is that there is some dynamics that uniquely fixes $f(x,t)$, given $f(x,0)$. One can express the dynamics by some linear operator (linear equation of motion).</p> <p>One way, as you pointed out, is to expand the solution in terms of the eigen-functions (let's briefly review it) $$f(x,t) = \int F(\omega) \phi(x,\omega,t)\ d\omega .$$ Here $\phi(x,\omega,t)$ are the eigen-functions and they are indexed by $\omega$ (associated with the energy). These eigen-functions are also orthogonal in the sense that $$\int \phi(x,\omega,t) \phi^*(x,\omega',t)\ dx = \delta(\omega-\omega').$$</p> <p>This now allows us to obtain the spectrum for $f(x,0)$ at $t=0$ using $$F(\omega) = \int f(x,0) \phi^*(x,\omega,0)\ dx .$$ Then we can substitute $F(\omega)$ into the original expansion to get a general expression for $f(x,t)$ that is valid for all time.</p> <p><strong>Now for the alternative approach.</strong> Imagine we do the substitution, just mentioned and then change the order of integration $$f(x,t) = \int f(x',0) \int \phi^*(x',\omega,0) \phi(x,\omega,t)\ d\omega \ dx'.$$ The inner integral now represents a Green function or propagator for the process $$K(x,x',t) = \int \phi^*(x',\omega,0) \phi(x,\omega,t)\ d\omega.$$ If the dynamics is translation invariant in $x$, we'll get $$K(x-x',t) = \int \phi^*(x',\omega,0) \phi(x,\omega,t)\ d\omega.$$ If we substitute this back we obtain a convolution integral $$f(x,t) = \int f(x',0) K(x-x',t)\ dx'.$$</p> <p>So this provides an alternative approach. However, one needs to know the Green function, or alternatively derive it from knowledge of the eigen-functions.</p>