[proofplan] Let $\hbar>0$, let $Q:\mathcal S(\mathbb R)\to L^2(\mathbb R)$ be the position operator defined by $(Qf)(x)=xf(x)$, and let $P:\mathcal S(\mathbb R)\to L^2(\mathbb R)$ be the momentum operator defined by $(Pf)(x)=-i\hbar f'(x)$. Center these operators by subtracting their expectations in the state $\psi$. Cauchy-Schwarz bounds the product of the centered norms below by the absolute value of their inner product, and the imaginary part of that inner product is controlled by the commutator $[Q,P]=i\hbar I$. [/proofplan]

[step:Center the two observables and apply Cauchy-Schwarz]Let $Q:\mathcal S(\mathbb R)\to L^2(\mathbb R)$ denote the position operator defined by $(Qf)(x)=xf(x)$, and let $P:\mathcal S(\mathbb R)\to L^2(\mathbb R)$ denote the momentum operator defined by $(Pf)(x)=-i\hbar f'(x)$, where $f\in\mathcal S(\mathbb R)$. Let $I:L^2(\mathbb R)\to L^2(\mathbb R)$ denote the identity operator. Let \begin{align*} \mu_Q=(Q\psi,\psi)_{L^2} \end{align*} and \begin{align*} \mu_P=(P\psi,\psi)_{L^2}. \end{align*} Since $Q$ and $P$ are symmetric on $\mathcal S(\mathbb R)$, the scalars $\mu_Q$ and $\mu_P$ are real. Define \begin{align*} u=(Q-\mu_Q I)\psi \end{align*} and \begin{align*} v=(P-\mu_P I)\psi. \end{align*} Then $\|u\|_{L^2}=\Delta_\psi Q$ and $\|v\|_{L^2}=\Delta_\psi P$. By the Cauchy-Schwarz inequality in $L^2(\mathbb R)$, \begin{align*} \Delta_\psi Q\,\Delta_\psi P=\|u\|_{L^2}\|v\|_{L^2}\geq |(u,v)_{L^2}|. \end{align*} Since the absolute value of a complex number dominates the absolute value of its imaginary part, \begin{align*} \Delta_\psi Q\,\Delta_\psi P\geq |\operatorname{Im}(u,v)_{L^2}|. \end{align*}[/step]

[guided]Let $Q:\mathcal S(\mathbb R)\to L^2(\mathbb R)$ be defined by $(Qf)(x)=xf(x)$, let $P:\mathcal S(\mathbb R)\to L^2(\mathbb R)$ be defined by $(Pf)(x)=-i\hbar f'(x)$, and let $I:L^2(\mathbb R)\to L^2(\mathbb R)$ be the identity operator. The uncertainty product is the product of the norms of the centered position and momentum vectors. Define \begin{align*} \mu_Q=(Q\psi,\psi)_{L^2} \end{align*} and \begin{align*} \mu_P=(P\psi,\psi)_{L^2}. \end{align*} Because $Q$ and $P$ are symmetric on $\mathcal S(\mathbb R)$, both expectation values are real. Set \begin{align*} u=(Q-\mu_Q I)\psi \end{align*} and \begin{align*} v=(P-\mu_P I)\psi. \end{align*} By definition, $\|u\|_{L^2}=\Delta_\psi Q$ and $\|v\|_{L^2}=\Delta_\psi P$. The Cauchy-Schwarz inequality gives \begin{align*} \Delta_\psi Q\,\Delta_\psi P=\|u\|_{L^2}\|v\|_{L^2}\geq |(u,v)_{L^2}|. \end{align*} For any complex number $z$, one has $|z|\geq |\operatorname{Im}z|$. Applying this to $z=(u,v)_{L^2}$ gives \begin{align*} \Delta_\psi Q\,\Delta_\psi P\geq |\operatorname{Im}(u,v)_{L^2}|. \end{align*} Thus it remains only to compute the imaginary part of the centered inner product.[/guided]

[step:Compute the commutator contribution]The real shifts $\mu_Q I$ and $\mu_P I$ commute with all operators, so \begin{align*} [(Q-\mu_Q I),(P-\mu_P I)]=[Q,P]. \end{align*} For $\psi\in\mathcal S(\mathbb R)$, \begin{align*} (QP\psi)(x)=-i\hbar x\psi'(x) \end{align*} and \begin{align*} (PQ\psi)(x)=-i\hbar(\psi(x)+x\psi'(x)). \end{align*} Therefore \begin{align*} ([Q,P]\psi)(x)=i\hbar\psi(x). \end{align*} Equivalently, $[Q,P]\psi=i\hbar\psi$. Since $\|\psi\|_{L^2}=1$, \begin{align*} ([Q,P]\psi,\psi)_{L^2}=i\hbar. \end{align*} The space $\mathcal S(\mathbb R)$ is invariant under $Q$, $P$, $Q-\mu_Q I$, and $P-\mu_P I$, so all compositions below are defined on $\psi$. For symmetric operators $S$ and $T$ on a common invariant domain and a vector in that domain, with the $L^2$ inner product linear in the first variable, \begin{align*} 2i\,\operatorname{Im}(S\psi,T\psi)_{L^2}=-(([S,T]\psi,\psi)_{L^2}). \end{align*} Applying this identity with $S=Q-\mu_Q I$ and $T=P-\mu_P I$ yields \begin{align*} |\operatorname{Im}(u,v)_{L^2}|=\frac{\hbar}{2}. \end{align*} Combining this with the previous step gives \begin{align*} \Delta_\psi Q\,\Delta_\psi P\geq \frac{\hbar}{2}. \end{align*}[/step]

[guided]Because $\mu_Q$ and $\mu_P$ are real scalars, the scalar multiples $\mu_Q I$ and $\mu_P I$ commute with $Q$ and $P$. Therefore the centered commutator is the same as the original commutator: \begin{align*} [(Q-\mu_Q I),(P-\mu_P I)]=[Q,P]. \end{align*} We now compute $[Q,P]$ on the Schwartz function $\psi$. The position-after-momentum expression is \begin{align*} (QP\psi)(x)=-i\hbar x\psi'(x). \end{align*} The momentum-after-position expression uses the product rule: \begin{align*} (PQ\psi)(x)=-i\hbar(\psi(x)+x\psi'(x)). \end{align*} Subtracting these expressions gives \begin{align*} ([Q,P]\psi)(x)=i\hbar\psi(x). \end{align*} Thus $[Q,P]\psi=i\hbar\psi$. Since $\|\psi\|_{L^2}=1$, this implies \begin{align*} ([Q,P]\psi,\psi)_{L^2}=i\hbar. \end{align*} The space $\mathcal S(\mathbb R)$ is invariant under multiplication by $x$, differentiation, and scalar shifts. Therefore it is invariant under $Q$, $P$, $Q-\mu_Q I$, and $P-\mu_P I$, and all compositions used in the commutator are defined on $\psi$. For symmetric operators $S$ and $T$ on a common invariant domain, with the $L^2$ inner product linear in the first variable, the identity \begin{align*} 2i\,\operatorname{Im}(S\psi,T\psi)_{L^2}=-(([S,T]\psi,\psi)_{L^2}) \end{align*} follows by expanding the commutator and using symmetry. With $S=Q-\mu_Q I$ and $T=P-\mu_P I$, we obtain \begin{align*} |\operatorname{Im}(u,v)_{L^2}|=\frac{\hbar}{2}. \end{align*} The previous step already proved \begin{align*} \Delta_\psi Q\,\Delta_\psi P\geq |\operatorname{Im}(u,v)_{L^2}|. \end{align*} Combining the two displays gives \begin{align*} \Delta_\psi Q\,\Delta_\psi P\geq \frac{\hbar}{2}. \end{align*}[/guided]