The generalised uncertainty principle for any Hermitian operators $\hat{A}$ and $\hat{B}$ can be proven using the fact that the inner product of the following operator is non-negative: $$\hat{O} = \hat{\alpha} + i\lambda\hat{\beta},$$ where $\hat{\alpha} = \hat{A} - \langle\hat{A}\rangle$ and $\hat{\beta} = \hat{B} - \langle\hat{B}\rangle$. These are chosen such that $$(\Delta{A})^2 = \langle(\hat{A} - \langle\hat{A}\rangle)^2\rangle = \langle \hat{\alpha}^2\rangle$$ and $$(\Delta{B})^2 = \langle(\hat{B} - \langle\hat{B}\rangle)^2\rangle = \langle\hat{\beta}^2\rangle.$$ I will write the full proof below for reference.
We construct the operators $\hat{\alpha}$, $\hat{\beta}$, and $\hat{O}$ in order to relate the uncertainties $\Delta A$ and $\Delta B$ of measuring $\hat{A}$ and $\hat{B}$. We use the positive-definite property of the inner product in order to prove the generalised uncertainty principle. This seems like a mathematical trick. My question is: is there any (other) significance of the operator $\hat{O} = \hat{\alpha} + i\lambda\hat{\beta}$. Is there anything deeper to be said about the fact that it is this operator's inner product that gives us the uncertainty principle and not any other's. I understand the proof mathematically, but since the principle itself is so fundamental to nature, I was wondering what the importance of $\hat{O}$ is in nature, if that makes sense.
Full proof: $$\langle\hat{O}\psi|\hat{O}\psi\rangle = (\langle\hat{\alpha}\psi|+i\lambda\langle\hat{\beta}\psi|)\cdot(|\hat{\alpha}\psi\rangle+i\lambda|\hat{\beta}\psi\rangle)$$ $$= \langle\hat{\alpha}^2\rangle+\lambda^2\langle\hat{\beta}^2\rangle + i\lambda\langle[\hat{\alpha},\hat{\beta}]\rangle$$ because $\hat{A}$ and $\hat{B}$ are Hermitian, which implies that $\hat{\alpha}$ and $\hat{\beta}$ are Hermitian.
$$\langle[\hat{\alpha},\hat{\beta}]\rangle = \langle\psi|\hat{\alpha}\hat{\beta}|\psi\rangle - \langle\psi|\hat{\beta}\hat{\alpha}|\psi\rangle\\
= \langle\psi|\hat{\alpha}\hat{\beta}|\psi\rangle-\langle\psi|\hat{\alpha}\hat{\beta}|\psi\rangle^{*} \\
= z - z^{*} \in \mathbb{I}.$$
Define $ib = \langle[\hat{\alpha},\hat{\beta}]\rangle$, where $b \in \mathbb{R}$. Therefore,
$$\langle\hat{O}\psi|\hat{O}\psi\rangle = (\Delta{A})^2 + \lambda^2(\Delta{B})^2 -\lambda b \geq 0 ~.$$
This is a quadratic equation in $\lambda$. The positive definite condition of the quaratic implies that its discriminant must be $\leq0$, $$b^2 - 4 {{(\Delta{A})}^{2}} {{(\Delta{B})}^{2}} \leq 0$$ $${{(\Delta{A})}^{2}} {{(\Delta{B})}^{2}} \geq \frac{1}{4}b^2,$$ where $b = |ib| = |\langle[\hat{\alpha},\hat{\beta}]\rangle|$ and it can be shown that $[\hat{\alpha},\hat{\beta}]=[\hat{A},\hat{B}]$. Therefore, $${{(\Delta{A})}^{2}} {{(\Delta{B})}^{2}} \geq \frac{1}{4}|\langle[\hat{A},\hat{B}]\rangle|^2 .$$
