 Solution

For a normalized state and an observable $A$, define $⟨A⟩=⟨\psi ,A\psi ⟩$ and ${\Delta }_{\psi }A=\Vert (A-⟨A⟩)\psi \Vert$. Suppose first that the position and momentum variances are finite and the state has the regularity needed for the usual position-momentum quadratic forms. Put $a=⟨x⟩$, $b=⟨p⟩$, $u=(x-a)\psi$ and $v=(-i\hslash {\partial }_x-b)\psi$. The Cauchy-Schwarz inequality givesThe term involving $b$ has zero imaginary part. Integration by parts givesConsequently $\Im ⟨u,v⟩=\hslash /2$, andThis proves the Heisenberg uncertainty relation. For smooth rapidly decaying states the boundary terms vanish directly. The same proof extends to $x\psi \in L^2$ and weak derivative ${\psi }^{\prime }\in L^2$ using smooth cutoffs and taking their radius to infinity; the necessary mixed products are integrable by Cauchy-Schwarz. Normalization alone need not give finite variances or an operator-domain momentum expectation. A state with infinite spread is interpreted through the corresponding extended bound, rather than assigning a finite uncertainty to an undefined expression.

For the specified Gaussian wave packet, write $s=\hslash t$ and $a_t=1+is$. Its probability density isIt is a normalized centered normal distribution with variance $1+s^2$. Symmetry gives $⟨x⟩=0$, and henceAlso ${\partial }_x\psi =-x\psi /(2a_t)$, so $p\psi =i\hslash x\psi /(2a_t)$ and $⟨p⟩=0$ by oddness. Integration by parts, or directly taking the squared norm, givesThusThe packet saturates the bound at $t=0$ and has a larger position-momentum uncertainty product at every nonzero time. Its spatial spreading does not change the momentum distribution.