The expression $\Delta_{\psi} A$ denotes the uncertainty of a quantum mechanical observable $A$ in a state with normalised wavefunction $\psi$. Prove that the Heisenberg uncertainty principle

$$\left(\Delta_{\psi} x\right)\left(\Delta_{\psi} p\right) \geqslant \frac{\hbar}{2}$$

holds for all normalised wavefunctions $\psi(x)$ of one spatial dimension.

[You may quote Schwarz's inequality without proof.]

A Gaussian wavepacket evolves so that at time $t$ its wavefunction is

$$\psi(x, t)=(2 \pi)^{-\frac{1}{4}}(1+i \hbar t)^{-\frac{1}{2}} \exp \left(-\frac{x^{2}}{4(1+i \hbar t)}\right)$$

Calculate the uncertainties $\Delta_{\psi} x$ and $\Delta_{\psi} p$ at each time $t$, and hence verify explicitly that the uncertainty principle holds at each time $t$.

[You may quote without proof the results that if $\operatorname{Re}(a)>0$ then

$$\int_{-\infty}^{\infty} \exp \left(-\frac{x^{2}}{a^{*}}\right) x^{2} \exp \left(-\frac{x^{2}}{a}\right) d x=\frac{1}{4}\left(\frac{\pi}{2}\right)^{\frac{1}{2}} \frac{|a|^{3}}{(\operatorname{Re}(a))^{\frac{3}{2}}}$$

and

$$\left.\int_{-\infty}^{\infty}\left(\frac{d}{d x} \exp \left(-\frac{x^{2}}{a^{*}}\right)\right)\left(\frac{d}{d x} \exp \left(-\frac{x^{2}}{a}\right)\right) d x=\left(\frac{\pi}{2}\right)^{\frac{1}{2}} \frac{|a|}{(\operatorname{Re}(a))^{\frac{3}{2}}} \cdot\right]$$