# What is the energy of a Gaussian wave packet?

Suppose we have a potential barrier situation, that is $V(x)$ is zero everywhere except on the interval $[-a,a]$, where it is equal to some $V_0 > 0$. Introduce some Gaussian shaped wave packet to the left of the barrier, moving right.

What is the energy of the packet (i.e. of the system described by this wavefunction) at each instant of time?

Well, the wavefunction $\psi(x,t)$ is not an energy eigenstate, so the question is asking about the expected value of the energy, I suppose. Does that just mean carrying out the calculation

$$\langle E(t) \rangle = \int_{-\infty}^\infty dx \ \langle \psi_t | x \rangle \langle x | H \psi_t \rangle \quad ?$$

• It seems correct. Commented Dec 23, 2014 at 22:31
• Note that $E(t)$ is constant. The wavefunction is only a Gaussian at $t=0$, after which the shape is not so easy to compute. Commented Dec 23, 2014 at 22:38
• The expected value of kinetic energy should be easy and intuitive. The expected value of $V(x)$ is harder, and will be expressible in terms of the error function. Commented Dec 23, 2014 at 22:40
• The way I imagined this, it is still more or less a gaussian until it collides with the barrier, then some weird stuff happens for a while, and then its two almost-gaussians travelling in opposite direction. Commented Dec 23, 2014 at 22:41
• Yes, this is correct. They are not perfect Gaussians, though, as far as I know. Commented Dec 23, 2014 at 22:42

As Nemis L. pointed out, the expectation value $\langle H\rangle$ is constant, because of Ehrenfest's theorem: $$\frac{d}{dt} \langle H \rangle = \frac{1}{i \hbar} \langle [ H,H ] \rangle = 0.$$ The other way of seeing this is that the state can be written as a superposition of orthogonal energy eigenstates.