Example: Wave packet

We consider the wave function (wave packet, see above)

$$\Psi(x)= \frac{1}{\sqrt{\sqrt{\pi a^2}}} \exp{\left(-\frac{x^2}{2a^2}\right)}.$$ (70)

1. We calculate

$$\langle x \rangle =\int_{-\infty}^{\infty} dx \Psi^*({x},t) x \Ps... ...t_{-\infty}^{\infty} dx \Psi^*({x},t) \frac{\hbar \partial_x}{i} \Psi({x},t)=0.$$ (71)

2. We calculate (see problem sheet 1)

$$\langle p^2\rangle = {{\hbar^2}\over{2a^2}},\quad \langle x^2\rangle = {{a^2}\over 2}.$$ (72)

By this we obtain for the mean square deviations of the position $x$ and the momentum $p$,

$${\Delta p^2 = {\langle p^2\rangle - \langle p\rangle^2} = {\langl... ...gle x^2\rangle - \langle x\rangle^2} = {\langle x^2\rangle } = {a^2\over{2}}. }$$

The product of the two is just

$$\Delta x^2 \cdot\Delta p^2 ={\hbar^2\over 4}.$$

The particular case of our wave packet fulfills the Heisenberg uncertainty relation

$$\Delta x^2 \cdot\Delta p^2 \ge {\hbar^2\over 4}.$$ (73)

with the $=$-sign. We will later prove that there are Heisenberg uncertainty relations for arbitrary pairs of operators and not only for $x$ and $p$.