- #1

sspitz

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## Homework Statement

Check the energy-time uncertainty principle for:

[tex]\Psi(x,0)=Ae^{-ax^2}e^{ilx}[/tex]

using the observable x.

## Homework Equations

[tex]

\Delta{E}\Delta{t}\geq\hbar/2\\

\Delta{E}=\sigma_H=\sqrt{<H^2>-<H>^2}\\

\Delta{t}=\frac{\sigma_x}{\frac{d<x>}{dt}}

[/tex]

## The Attempt at a Solution

First I find the wavefunction with time dependence

[tex]

\Psi(x,t)=\frac{1}{\sqrt{2\pi}}\int{\phi(k)e^{ikx-i\frac{\hbar k^{2}t}{2m}}dk}\\

\phi(k)=\frac{1}{\sqrt{2\pi}}\int{\Psi(x,0)e^{-ikx}dx}

[/tex]

This gives me a fairly complicated expression

[tex]

\Psi(x,t)=\frac{Aexp(\frac{-ax^{2}+lix-\frac{tl^{2}i\hbar}{2m}}{w^{2}})}{w}\\

A^{2}=\sqrt{\frac{2a}{\pi}},w^{2}=1+\frac{\hbar2ita}{m}

[/tex]

Just finding <H> (not to mention <H^2> is giving me a lot of trouble. Is grinding out the derivatives and integrals the only way to do this problem?

The only other options I can think of:

1: <H> and <H^2> should not change in time (?), so I can just use the wavefunction at t=0 to find them? (But, still have to use the wavefunction with time to get the std dev of x)

2: Considering [tex] \frac{e^{ikx}}{\sqrt{2\pi}} [/tex] are like orthonormal eigenfunctions for H, could I say:

[tex]

<H>=\int{\phi(k)^{2}E(k)dk}

[/tex]