# Time dependence of wave packets without eigenfunctions

In general, to obtain the time dependence of an arbitrary wave packet $\left| \phi(x)\right>$ in the Schödinger picture, we expand the wave packet in the energy eigenfunction basis $\left| \psi_n(x)\right>$

$$\left| \phi(x)\right> = \sum_\limits{n} \left< \psi_n(x) | \phi(x)\right> \left| \psi_n(x)\right>$$

Then we solve the time dependent equation:

$$\left| \phi(x, t)\right> = \sum_\limits{n} \left< \psi_n(x) | \phi(x)\right> \left| \psi_n(x)\right> e^{-iE_nt\hbar}$$

1. Can the time dependent equation be applied to the wave packet itself $\phi(x)$ without using the energy eigenfunctions?

2. 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?

$$\left[-\frac{\hbar}{2m}\nabla^2 + V(x)\right]\phi(x, t) = i\hbar\frac{\partial \phi}{\partial t}$$

• Of course you are right. It is the first thing your QM course should have taught you. Provided you have the propagator (Green's function) for the potential in the square bracket. So, for vanishing of harmonic potentials, the WP ones provided will do the trick. For general potentials, however, you normally use the eigenfunctions to produce the propagator! – Cosmas Zachos Nov 10 '16 at 22:31
• Related to 22639, and 242689 . – Cosmas Zachos Nov 10 '16 at 22:38

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).
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 $$\int \phi(x,\omega,t) \phi^*(x,\omega',t)\ dx = \delta(\omega-\omega').$$
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.
Now for the alternative approach. 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 $$f(x,t) = \int f(x',0) K(x-x',t)\ dx'.$$