[proofplan]
We subtract the expectation values from $A$ and $B$ so that the variances become ordinary squared norms of centered vectors. The [Cauchy-Schwarz inequality](/theorems/432) then gives a lower bound for $\Delta_\psi A\,\Delta_\psi B$ in terms of the [inner product](/page/Inner%20Product) of the centered vectors. The imaginary part of that inner product is exactly a constant multiple of the expectation of the commutator, and the scalar shifts do not change the commutator.
[/proofplan]
[step:Center the observables without changing the commutator]
Define the centered operators
\begin{align*}
\widetilde{A}: H &\to H
\end{align*}
by
\begin{align*}
\widetilde{A} := A-\mathbb{E}_\psi[A]I
\end{align*}
and
\begin{align*}
\widetilde{B}: H &\to H
\end{align*}
by
\begin{align*}
\widetilde{B} := B-\mathbb{E}_\psi[B]I.
\end{align*}
Since $A$ and $B$ are self-adjoint and $\|\psi\|_H=1$, the scalars $\mathbb{E}_\psi[A]$ and $\mathbb{E}_\psi[B]$ are real. Hence $\widetilde{A}$ and $\widetilde{B}$ are self-adjoint.
By definition of the standard deviations,
\begin{align*}
\Delta_\psi A = \|\widetilde{A}\psi\|_H
\end{align*}
and
\begin{align*}
\Delta_\psi B = \|\widetilde{B}\psi\|_H.
\end{align*}
Because scalar multiples of the identity commute with every operator,
\begin{align*}
[\widetilde{A},\widetilde{B}] = [A,B].
\end{align*}
[/step]
[step:Apply Cauchy-Schwarz to the centered state vectors]
Apply the finite-dimensional Cauchy-Schwarz inequality (citing a result not yet in the wiki: Cauchy-Schwarz inequality) in the [Hilbert space](/page/Hilbert%20Space) $H$ to the vectors $\widetilde{A}\psi$ and $\widetilde{B}\psi$. This gives
\begin{align*}
|(\widetilde{A}\psi,\widetilde{B}\psi)_H| \le \|\widetilde{A}\psi\|_H\|\widetilde{B}\psi\|_H.
\end{align*}
Using the identities from the previous step, we obtain
\begin{align*}
\Delta_\psi A\,\Delta_\psi B \ge |(\widetilde{A}\psi,\widetilde{B}\psi)_H|.
\end{align*}
[guided]
The purpose of centering is that the uncertainty $\Delta_\psi A$ is exactly the norm of the centered vector $\widetilde{A}\psi$, and similarly for $B$. Thus the product of uncertainties is a product of two Hilbert-space norms:
\begin{align*}
\Delta_\psi A\,\Delta_\psi B = \|\widetilde{A}\psi\|_H\|\widetilde{B}\psi\|_H.
\end{align*}
We now use the finite-dimensional Cauchy-Schwarz inequality (citing a result not yet in the wiki: Cauchy-Schwarz inequality). Its hypotheses are satisfied because $H$ is a complex Hilbert space and $\widetilde{A}\psi,\widetilde{B}\psi \in H$. Applying it to these two vectors gives
\begin{align*}
|(\widetilde{A}\psi,\widetilde{B}\psi)_H| \le \|\widetilde{A}\psi\|_H\|\widetilde{B}\psi\|_H.
\end{align*}
Therefore
\begin{align*}
\Delta_\psi A\,\Delta_\psi B \ge |(\widetilde{A}\psi,\widetilde{B}\psi)_H|.
\end{align*}
This is the only place where the metric structure of the Hilbert space enters. The rest of the proof identifies the imaginary part of the scalar $(\widetilde{A}\psi,\widetilde{B}\psi)_H$ with the commutator expectation.
[/guided]
[/step]
[step:Identify the imaginary part with the commutator expectation]
Let
\begin{align*}
z := (\widetilde{A}\psi,\widetilde{B}\psi)_H \in \mathbb{C}.
\end{align*}
Since $\widetilde{A}$ and $\widetilde{B}$ are self-adjoint and the inner product is linear in the first argument,
\begin{align*}
(\widetilde{A}\widetilde{B}\psi,\psi)_H = (\widetilde{B}\psi,\widetilde{A}\psi)_H = \overline{z}.
\end{align*}
Similarly,
\begin{align*}
(\widetilde{B}\widetilde{A}\psi,\psi)_H = (\widetilde{A}\psi,\widetilde{B}\psi)_H = z.
\end{align*}
Thus
\begin{align*}
([\widetilde{A},\widetilde{B}]\psi,\psi)_H = \overline{z}-z = -2i\,\operatorname{Im} z.
\end{align*}
Equivalently,
\begin{align*}
\operatorname{Im} z = \frac{i}{2}([\widetilde{A},\widetilde{B}]\psi,\psi)_H.
\end{align*}
Using $[\widetilde{A},\widetilde{B}]=[A,B]$, we get
\begin{align*}
|\operatorname{Im} z| = \frac{1}{2}|([A,B]\psi,\psi)_H|.
\end{align*}
[/step]
[step:Compare a complex number with its imaginary part]
For every complex number $w \in \mathbb{C}$, $|\operatorname{Im} w| \le |w|$. Applying this to $w=z$ gives
\begin{align*}
|z| \ge |\operatorname{Im} z|.
\end{align*}
Combining this with the Cauchy-Schwarz estimate and the commutator identity,
\begin{align*}
\Delta_\psi A\,\Delta_\psi B \ge |z|.
\end{align*}
Hence
\begin{align*}
\Delta_\psi A\,\Delta_\psi B \ge |\operatorname{Im} z| = \frac{1}{2}|([A,B]\psi,\psi)_H|.
\end{align*}
This is the desired Robertson uncertainty inequality.
[/step]
