# Minimum-uncertainty wave function -- contradiction?

1. Dec 22, 2015

### Happiness

The following equation (5.122) is the minimum-uncertainty wave function, which is a Gaussian wave packet. Since it is Gaussian in $x$, we may get $\Delta x$ directly from the $\sigma$ of the Gaussian distribution: $(\Delta x)^2=\frac{\hbar^2}{2(\Delta p_x)^2}$. We have $\Delta x\Delta p_x=\frac{\hbar}{\sqrt2}$, which contradicts the fact that the wave function has minimum uncertainty, ie., $\Delta x\Delta p_x=\frac{\hbar}{2}$.

Derivation of (5.122):

Gaussian distribution:

2. Dec 22, 2015

### blue_leaf77

You should first calculate $|\psi(x)|^2$ and then extract $\Delta x$ from there, not from $\psi(x)$