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\fancyhead[LO]{\textbf{Noncommutative Maccone-Pati Uncertainty Principles}}
\fancyhead[RE]{\textbf{K. Mahesh Krishna}}


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{\bf\large{{Noncommutative Maccone-Pati Uncertainty Principles}}}\\
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\textbf{K. Mahesh Krishna}\\
School of Mathematics and Natural Sciences\\
Chanakya University Global Campus\\
NH-648, Haraluru Village\\
Devanahalli Taluk, 	Bengaluru  North District\\
Karnataka State, 562 110, India \\
Email: kmaheshak@gmail.com\\

Date: \today
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\textbf{Abstract}: Let $\mathcal{E}$ be a  Hilbert C*-module  over a unital C*-algebra $\mathcal{A}$. 	Let  $A:	\mathcal{D}(A) \subseteq \mathcal{E} \to  \mathcal{E}$ and $B:	\mathcal{D}(B)\subseteq \mathcal{E}\to  \mathcal{E}$  be  possibly unbounded self-adjoint morphisms. Then for all $x \in \mathcal{D}(AB)\cap  \mathcal{D}(BA)$ with $\langle x, x \rangle =1$, we show that 
\begin{align*}
(1)	\quad\quad\quad 	d_x(A)^2+d_x(B)^2&\geq \mp \langle \{A,B\}x, x \rangle +\langle (A\pm B)x, y \rangle\langle y, (A\pm B)x \rangle\pm \{\langle Ax, x \rangle, \langle Bx, x \rangle\}, \\
&\quad \forall y \in \mathcal{E} \text{ satisfying } \|y\|\leq 1 \text{ and } \langle x,y \rangle =0
\end{align*}
\begin{align*}
(2)	\quad \quad \quad d_x(A)^2+d_x(B)^2&\geq \mp i\langle [A,B]x, x \rangle +\langle (A\pm iB)x, y \rangle\langle y, (A\pm iB)x \rangle \mp i[\langle Ax, x \rangle, \langle Bx, x \rangle], \\
	&\quad \forall y \in \mathcal{E} \text{ satisfying } \|y\|\leq 1 \text{ and } \langle x,y \rangle =0.
\end{align*}
where 
\begin{align*}
	& d_x(A):= \sqrt{\langle Ax, Ax \rangle -\langle Ax, x \rangle^2},\quad [A,B] := AB-BA, \\
	& [\langle Ax, x \rangle,\langle Bx, x \rangle]:= \langle Ax, x \rangle\langle Bx, x \rangle -\langle Bx, x \rangle\langle Ax, x \rangle.
\end{align*}
We call Inequalities (1) and (2) as noncommutative  Maccone-Pati uncertainty principles. They  reduce to   uncertainty principles derived by  Maccone and Pati [\textit{Phys. Rev. Lett., 2014}] whenever $\mathcal{A}=\mathbb{C}$.



\textbf{Keywords}:  Uncertainty Principle,  Hilbert C*-module.

\textbf{Mathematics Subject Classification (2020)}:  46L08.\\

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\section{Introduction}
Let $\mathcal{H}$ be a complex Hilbert space and $A$ be a possibly unbounded self-adjoint linear  operator defined on domain $\mathcal{D}(A)\subseteq \mathcal{H}$. For $h \in \mathcal{D}(A)$ with $\|h\|=1$, define the \textbf{uncertainty} of $A$ at the point $h$ as 
\begin{align*}
	\Delta _h(A)\coloneqq \|Ah-\langle Ah, h \rangle h \|=\sqrt{\|Ah\|^2-\langle Ah, h \rangle^2}. 
\end{align*}
In 1929, Robertson \cite{ROBERTSON} derived the following mathematical form of the uncertainty principle of Heisenberg derived in 1927 \cite{HEISENBERG}. Recall that for two linear operators $A:	\mathcal{D}(A)\subseteq \mathcal{H}\to  \mathcal{H}$ and $B:	\mathcal{D}(B)\subseteq \mathcal{H}\to  \mathcal{H}$, we define the commutator $[A,B] \coloneqq AB-BA$ and the anti-commutator $\{A,B\}\coloneqq AB+BA$.
\begin{theorem} \cite{ROBERTSON, HEISENBERG, VONNEUMANNBOOK, DEBNATHMIKUSINSKI}  (\textbf{Heisenberg-Robertson Uncertainty Principle})
	Let  $A:	\mathcal{D}(A)\subseteq \mathcal{H}\to  \mathcal{H}$ and $B:	\mathcal{D}(B)\subseteq \mathcal{H}\to  \mathcal{H}$  be self-adjoint linear operators. Then for all $h \in \mathcal{D}(AB)\cap  \mathcal{D}(BA)$ with $\|h\|=1$, we have 
	\begin{align}\label{HR}
		\frac{1}{2} \left(\Delta _h(A)^2+	\Delta _h(B)^2\right)\geq \frac{1}{4} \left(\Delta _h(A)+	\Delta _h(B)\right)^2 \geq  \Delta _h(A)	\Delta _h(B)   \geq  \frac{1}{2}|\langle [A,B]h, h \rangle |.
	\end{align}
\end{theorem}
In 1930,  Schrodinger improved Inequality (\ref{HR}) \cite{SCHRODINGER}. 
\begin{theorem} \cite{SCHRODINGER, BERTLMANNFRIIS}\label{SE}
	(\textbf{Heisenberg-Robertson-Schrodinger  Uncertainty Principle})
	Let  $A:	\mathcal{D}(A)\subseteq \mathcal{H}\to  \mathcal{H}$ and $B:	\mathcal{D}(B)\subseteq \mathcal{H}\to  \mathcal{H}$  be self-adjoint linear operators. Then for all $h \in \mathcal{D}(AB)\cap  \mathcal{D}(BA)$ with $\|h\|=1$, we have 
	\begin{align*}
		\Delta _h(A)	\Delta _h(B)    \geq |\langle Ah, Bh \rangle-\langle Ah, h \rangle \langle Bh, h \rangle|
		&=\frac{\sqrt{|\langle [A,B]h, h \rangle |^2+|\langle \{A,B\}h, h \rangle -2\langle Ah, h \rangle\langle Bh, h \rangle|^2}}{2}\\
			&=\frac{\sqrt{(\langle \{A,B\}h, h \rangle -2\langle Ah, h \rangle\langle Bh, h \rangle)^2-\langle [A,B]h, h \rangle ^2}}{2}.
	\end{align*}	
\end{theorem}
 In 2014,  Maccone and Pati derived the following  uncertainty principles which work for orthogonal vectors  \cite{MACCONEPATI}.
\begin{theorem}  \cite{MACCONEPATI} (\textbf{Maccone-Pati Uncertainty Principles}) \label{MP}
	Let  $A:	\mathcal{D}(A)\subseteq \mathcal{H}\to  \mathcal{H}$ and $B:	\mathcal{D}(B)\subseteq \mathcal{H}\to  \mathcal{H}$  be self-adjoint linear operators. Then for all $h \in \mathcal{D}(AB)\cap  \mathcal{D}(BA)$ with $\|h\|=1$, we have 
		\begin{enumerate}[\upshape(i)]
			\item 
			\begin{align*}
	\Delta _h(A)^2+	\Delta _h(B)^2 &\geq|\langle (A+B)h, k \rangle|^2-\langle \{A,B\}h, h \rangle+2\langle Ah, h \rangle\langle Bh, h \rangle, \\
	&\quad \forall k \in \mathcal{H} \text{ satisfying } \|k\|=1 \text{ and } \langle h, k \rangle =0.
			\end{align*}
			\item 
				\begin{align*}
				\Delta _h(A)^2+	\Delta _h(B)^2 &\geq|\langle (A-B)h, k \rangle|^2+\langle \{A,B\}h, h \rangle-2\langle Ah, h \rangle\langle Bh, h \rangle, \\
				&\quad \forall k \in \mathcal{H} \text{ satisfying } \|k\|=1 \text{ and } \langle h, k \rangle =0.
			\end{align*}
		\item
	\begin{align*}
		\Delta _h(A)^2+	\Delta _h(B)^2 \geq \frac{|\langle (A+B)h, k \rangle|^2+|\langle (A-B)h, k \rangle|^2}{2}, \quad \forall k \in \mathcal{H} \text{ satisfying } \|k\|=1 \text{ and } \langle h, k \rangle =0.
	\end{align*}		
\item 
	\begin{align*}
		\Delta _h(A)^2+	\Delta _h(B)^2 \geq-i\langle [A,B]h, h \rangle +|\langle (A+iB)h, k \rangle|^2, \quad \forall k \in \mathcal{H} \text{ satisfying } \|k\|=1 \text{ and } \langle h, k \rangle =0.
	\end{align*}
\item 
	\begin{align*}
	\Delta _h(A)^2+	\Delta _h(B)^2 \geq i\langle [A,B]h, h \rangle +|\langle (A-iB)h, k \rangle|^2, \quad \forall k \in \mathcal{H} \text{ satisfying } \|k\|=1 \text{ and } \langle h, k \rangle =0.
\end{align*}
\item 
	\begin{align*}
	\Delta _h(A)^2+	\Delta _h(B)^2 \geq \frac{|\langle (A+iB)h, k \rangle|^2+|\langle (A-iB)h, k \rangle|^2}{2}, \quad \forall k \in \mathcal{H} \text{ satisfying } \|k\|=1 \text{ and } \langle h, k \rangle =0.
\end{align*}		
\item 
\begin{align*}
	\Delta _h(A)\Delta _h(B)  \geq	\frac{-i\langle [A,B]h, h \rangle}{2-\left|\left\langle \left(\frac{A}{\Delta _h(A)}+i\frac{B}{\Delta _h(B)}\right)h, k\right\rangle \right|^2}, \quad \forall k \in \mathcal{H} \text{ satisfying } \|k\|=1 \text{ and } \langle h, k \rangle =0.
\end{align*}
\item 
\begin{align*}
	\Delta _h(A)\Delta _h(B)  \geq	\frac{i\langle [A,B]h, h \rangle}{2-\left|\left\langle \left(\frac{A}{\Delta _h(A)}-i\frac{B}{\Delta _h(B)}\right)h, k\right\rangle \right|^2}, \quad \forall k \in \mathcal{H} \text{ satisfying } \|k\|=1 \text{ and } \langle h, k \rangle =0.
\end{align*}
\item 
\begin{align*}
	\Delta _h(A)\Delta _h(B)  \geq	\frac{2\langle Ah, h \rangle\langle Bh, h \rangle-\langle \{A,B\}h, h \rangle}{2-\left|\left\langle \left(\frac{A}{\Delta _h(A)}+\frac{B}{\Delta _h(B)}\right)h, k\right\rangle \right|^2}, \quad \forall k \in \mathcal{H} \text{ satisfying } \|k\|=1 \text{ and } \langle h, k \rangle =0.
\end{align*}
\item 
\begin{align*}
	\Delta _h(A)\Delta _h(B)  \geq	\frac{-2\langle Ah, h \rangle\langle Bh, h \rangle+\langle \{A,B\}h, h \rangle}{2-\left|\left\langle \left(\frac{A}{\Delta _h(A)}-\frac{B}{\Delta _h(B)}\right)h, k\right\rangle \right|^2}, \quad \forall k \in \mathcal{H} \text{ satisfying } \|k\|=1 \text{ and } \langle h, k \rangle =0.
\end{align*}
	\end{enumerate}	
\end{theorem}

We  ask:  What is   noncommutative analogues of Theorems \ref{MP}? For this, we need the notion of noncommutative Hilbert spaces known as  Hilbert C*-modules which  are first introduced by Kaplansky \cite{KAPLANSKY} for modules over commutative C*-algebras and later developed for modules over arbitrary C*-algebras by Paschke  \cite{PASCHKE} and Rieffel \cite{RIEFFEL}. Idea behind Hilbert C*-modules is simple: replace the field of complex numbers in the theory of Hilbert spaces by C*-algebras. Now the positive numbers are replaced by positive elements in C*-algebras. Hilbert C*-modules play an essential role in operator algebras, K-theory, KK-theory, operator spaces, quantum group theory, Morita equivalence, AW*-algebras, operator K-theory, group representation theory etc. We refer monographs \cite{MANUILOVTROITSKY, WEGGEOLSEN, THOMSEN, LANCE} for the beautiful use of Hilbert C*-modules in these  areas. 
\begin{definition}\cite{KAPLANSKY, PASCHKE, RIEFFEL}
	Let $\mathcal{A}$ be a  unital C*-algebra. A left module 	 $\mathcal{E}$  over $\mathcal{A}$ is said to be a (left)  \textbf{Hilbert C*-module} if there exists a  map $ \langle \cdot, \cdot \rangle: \mathcal{E}\times \mathcal{E} \to \mathcal{A}$ such that the following conditions hold. 
	\begin{enumerate}[\upshape(i)]
		\item $\langle x, x \rangle  \geq 0$, $\forall x \in \mathcal{E}$. If $x \in  \mathcal{E}$ satisfies $\langle x, x \rangle  =0 $, then $x=0$.
		\item $\langle ax+y, z \rangle  =a\langle x, z \rangle+\langle y, z \rangle$, $\forall x,y,z \in \mathcal{E}$, $\forall a \in \mathcal{A}$.
		\item $\langle x, y \rangle=\langle y,x \rangle^*$, $\forall x,y \in \mathcal{E}$.
		\item $\mathcal{E}$ is complete w.r.t. the norm $\|x\|\coloneqq \sqrt{\|\langle x, x \rangle\|}$,  $\forall x \in \mathcal{E}$.
	\end{enumerate}
\end{definition}
In the paper,  we use the following noncommutative Cauchy-Schwarz inequality. 
\begin{theorem}\cite{PASCHKE, RIEFFEL} \label{NCS}
Let $\mathcal{E}$ be a  Hilbert C*-module. Then 
\begin{align*}
	\langle x, y \rangle 	\langle y, x \rangle \leq \|	\langle y, y \rangle \|	\langle x, x \rangle , \quad \forall x, y \in \mathcal{E}.
\end{align*}	
\end{theorem}
 Note that,  eventhough we use Theorem \ref{NCS},   there are other variations of noncommutative Cauchy-Schwarz inequality, see \cite{BELLOMONTE, FUJIIFUJIIMOSLEHIANSEO, ARAMBASICBAKICMOSLEHIAN, ILLSEVICVAROSANEC}. We derive noncommutative analogues of  Theorem \ref{MP} in Theorem \ref{NMP}.



\section{Noncommutative  Maccone-Pati  Uncertainty Principles}

Let $\mathcal{E}$ be a  Hilbert C*-module  over a unital C*-algebra $\mathcal{A}$ and $A$ be a possibly unbounded self-adjoint morphism defined on domain $\mathcal{D}(A)\subseteq \mathcal{E}$. For $x \in \mathcal{D}(A)$ with $\langle x, x\rangle =1$, we define the noncommutative uncertainty of $A$ at the point $x$ as 
\begin{align*}
	\Delta _x(A)\coloneqq \|Ax-\langle Ax, x \rangle x \|=\sqrt{\|\langle Ax, Ax \rangle -\langle Ax, x \rangle^2\|}. 
\end{align*}
We  define 
\begin{align*}
	d_x(A)\coloneqq \sqrt{\langle Ax, Ax \rangle -\langle Ax, x \rangle^2}=\sqrt{\langle Ax -\langle Ax, x \rangle x, Ax -\langle Ax, x \rangle x\rangle}.
\end{align*}
Then $d_x(A)\in \mathcal{A}$ and 
\begin{align*}
	\Delta _x(A)=\|	d_x(A)\|.
\end{align*}
Given two elements $a, b \in \mathcal{A}$, we define $[a,b]\coloneqq ab-ba$ and $\{a,b\}\coloneqq ab+ba$. Similarly, for two morphisms $A:	\mathcal{D}(A)\subseteq \mathcal{E}\to  \mathcal{E}$ and $B:	\mathcal{D}(B)\subseteq \mathcal{E}\to  \mathcal{E}$, we define $[A,B] \coloneqq AB-BA$ and $\{A,B\}\coloneqq AB+BA$.



\begin{theorem} (\textbf{Noncommutative Maccone-Pati Uncertainty Principles}) \label{NMP}
Let $\mathcal{E}$ be a  Hilbert C*-module  over a unital C*-algebra $\mathcal{A}$. 	Let  $A:	\mathcal{D}(A)\subseteq \mathcal{E}\to  \mathcal{E}$ and $B:	\mathcal{D}(B)\subseteq \mathcal{E}\to  \mathcal{E}$  be self-adjoint morphisms. Then for all $x \in \mathcal{D}(AB)\cap  \mathcal{D}(BA)$ with $\langle x, x \rangle =1$, we have 	
	\begin{enumerate}[\upshape(i)]
		\item 
			\begin{align*}
			d_x(A)^2+d_x(B)^2&\geq -\langle \{A,B\}x, x \rangle +\langle (A+B)x, y \rangle\langle y, (A+B)x \rangle+\{\langle Ax, x \rangle, \langle Bx, x \rangle\}, \\
			&\quad \forall y \in \mathcal{E} \text{ satisfying } \|y\|\leq 1 \text{ and } \langle x,y \rangle =0.
		\end{align*}
	\item 	\begin{align*}
		\Delta_x(A)^2+\Delta_x(B)^2&\geq \|-\langle \{A,B\}x, x \rangle +\langle (A+B)x, y \rangle\langle y, (A+B)x \rangle+\{\langle Ax, x \rangle, \langle Bx, x \rangle\}\|, \\
		&\quad \forall y \in \mathcal{E} \text{ satisfying } \|y\|\leq 1 \text{ and } \langle x,y \rangle =0.
	\end{align*}
	\item 	\begin{align*}
		d_x(A)^2+d_x(B)^2&\geq \langle \{A,B\}x, x \rangle +\langle (A-B)x, y \rangle\langle y, (A-B)x \rangle-\{\langle Ax, x \rangle, \langle Bx, x \rangle\}, \\
		&\quad \forall y \in \mathcal{E} \text{ satisfying } \|y\|\leq 1 \text{ and } \langle x,y \rangle =0.
	\end{align*}
	\item 
		\begin{align*}
		\Delta_x(A)^2+\Delta_x(B)^2&\geq \|\langle \{A,B\}x, x \rangle +\langle (A-B)x, y \rangle\langle y, (A-B)x \rangle-\{\langle Ax, x \rangle, \langle Bx, x \rangle\}\|, \\
		&\quad \forall y \in \mathcal{E} \text{ satisfying } \|y\|\leq 1 \text{ and } \langle x,y \rangle =0.
	\end{align*}
		\item 
		\begin{align*}
		d_x(A)^2+d_x(B)^2&\geq  \frac{\langle (A+B)x, y \rangle\langle y, (A+B)x \rangle+\langle (A-B)x, y \rangle\langle y, (A-B)x \rangle}{2}, \\
		&\quad \forall y \in \mathcal{E} \text{ satisfying } \|y\|\leq 1 \text{ and } \langle x,y \rangle =0.
		\end{align*}
		\item 
		\begin{align*}
			\Delta _x(A)^2+\Delta _x(B)^2&\geq \frac{\|\langle (A+B)x, y \rangle\langle y, (A+B)x \rangle+\langle (A-B)x, y \rangle\langle y, (A-B)x \rangle\|}{2}, \\
			&\quad \forall y \in \mathcal{E} \text{ satisfying } \|y\|\leq 1 \text{ and } \langle x,y \rangle =0.
		\end{align*}
	\item 
	\begin{align*}
	d_x(A)^2+d_x(B)^2&\geq -i\langle [A,B]x, x \rangle +\langle (A+iB)x, y \rangle\langle y, (A+iB)x \rangle-i[\langle Ax, x \rangle, \langle Bx, x \rangle], \\
	&\quad \forall y \in \mathcal{E} \text{ satisfying } \|y\|\leq 1 \text{ and } \langle x,y \rangle =0.
\end{align*}
\item 
	\begin{align*}
	\Delta_x(A)^2+\Delta_x(B)^2&\geq \|-i\langle [A,B]x, x \rangle +\langle (A+iB)x, y \rangle\langle y, (A+iB)x \rangle-i[\langle Ax, x \rangle, \langle Bx, x \rangle]\|, \\
	&\quad \forall y \in \mathcal{E} \text{ satisfying } \|y\|\leq 1 \text{ and } \langle x,y \rangle =0.
\end{align*}
\item 	
\begin{align*}
	d_x(A)^2+d_x(B)^2&\geq i\langle [A,B]x, x \rangle +\langle (A-iB)x, y \rangle\langle y, (A-iB)x \rangle+i[\langle Ax, x \rangle, \langle Bx, x \rangle], \\
	&\quad \forall y \in \mathcal{E} \text{ satisfying } \|y\|\leq 1 \text{ and } \langle x,y \rangle =0.
\end{align*}
\item 	
\begin{align*}
	\Delta_x(A)^2+\Delta_x(B)^2&\geq \|i\langle [A,B]x, x \rangle +\langle (A-iB)x, y \rangle\langle y, (A-iB)x \rangle+i[\langle Ax, x \rangle, \langle Bx, x \rangle]\|, \\
	&\quad \forall y \in \mathcal{E} \text{ satisfying } \|y\|\leq 1 \text{ and } \langle x,y \rangle =0.
\end{align*}
\item 
\begin{align*}
		d_x(A)^2+d_x(B)^2&\geq \frac{\langle (A+iB)x, y \rangle\langle y, (A+iB)x \rangle+\langle (A-iB)x, y \rangle\langle y, (A-iB)x \rangle}{2}, \\
		&\quad \forall y \in \mathcal{E} \text{ satisfying } \|y\|\leq 1 \text{ and } \langle x,y \rangle =0.
\end{align*}
\item 
\begin{align*}
	\Delta_x(A)^2+\Delta_x(B)^2&\geq \frac{\|\langle (A+iB)x, y \rangle\langle y, (A+iB)x \rangle+\langle (A-iB)x, y \rangle\langle y, (A-iB)x \rangle\|}{2}, \\
	&\quad \forall y \in \mathcal{E} \text{ satisfying } \|y\|\leq 1 \text{ and } \langle x,y \rangle =0.
\end{align*}
\end{enumerate}
\end{theorem}
\begin{proof}
	Let $x \in \mathcal{D}(AB)\cap  \mathcal{D}(BA)$ with $\langle x, x \rangle =1$. Let  	$y \in \mathcal{E}$ be such that $ \|y\|\leq 1 $ and $ \langle x,y \rangle =0.$
\begin{enumerate}[\upshape(i)]
		\item 
		\begin{align*}
				&\langle (A+B)x, y \rangle \langle y, (A+B)x \rangle\\
				&=\langle Ax-\langle Ax, x \rangle x+Bx-\langle Bx, x \rangle x, y \rangle\langle y,  Ax-\langle Ax, x \rangle x+Bx-\langle Bx, x \rangle x\rangle\\
				&\leq \| \langle y, y \rangle \|\langle  (Ax-\langle Ax, x \rangle x)+ (Bx-\langle Bx, x \rangle x), (Ax-\langle Ax, x \rangle x)+ (Bx-\langle Bx, x \rangle x)\rangle\\
				&\leq \langle Ax-\langle Ax, x \rangle x, Ax-\langle Ax, x \rangle x\rangle +\langle Bx-\langle Bx, x \rangle x, Bx-\langle Bx, x \rangle x\rangle \\
				&\quad +\langle Bx-\langle Bx, x \rangle x, Ax-\langle Ax, x \rangle x\rangle+\langle Ax-\langle Ax, x \rangle x, Bx-\langle Bx, x \rangle x\rangle\\
				&=d_x(A)^2+d_x(B)^2+\langle Bx, Ax \rangle -\langle Bx, x \rangle\langle Ax, x \rangle+\langle Ax, Bx \rangle -\langle Ax, x \rangle\langle Bx, x \rangle\\
				&=d_x(A)^2+d_x(B)^2+\langle \{A,B\}x, x \rangle -\{\langle Ax, x \rangle, \langle Bx, x \rangle\}.
			\end{align*}
		Therefore 
	\begin{align*}
	d_x(A)^2+d_x(B)^2\geq -\langle \{A,B\}x, x \rangle +\langle (A+B)x, y \rangle\langle y, (A+B)x \rangle+\{\langle Ax, x \rangle, \langle Bx, x \rangle\}.
\end{align*}
	\item Take norm on  (i).
	\item Similar to (i).
	\item Take norm on (iii).
	\item \begin{align*}
		&\langle (A+B)x, y \rangle \langle y, (A+B)x \rangle+\langle (A-B)x, y \rangle \langle y, (A-B)x \rangle=2\langle Ax, y \rangle \langle y, Ax \rangle +2\langle Bx, y \rangle \langle y, Bx \rangle\\
		&=2\langle Ax-\langle Ax, x \rangle x, y \rangle \langle y, Ax-\langle Ax, x \rangle x \rangle +2\langle Bx-\langle Bx, x \rangle x, y \rangle \langle y, Bx-\langle Bx, x \rangle x\rangle\\
		&\leq 2\|\langle y, y \rangle \|\langle Ax-\langle Ax,x \rangle x,  Ax-\langle Ax, x \rangle x \rangle + 2\|\langle y, y \rangle \|\langle Bx-\langle Bx,x \rangle x,  Bx-\langle Bx, x \rangle x \rangle\\
		&\leq 2d_x(A)^2+2d_x(B)^2.
	\end{align*}	
\item Take norm on (v).
\item 
\begin{align*}
	&\langle (A+iB)x, y \rangle\langle y, (A+iB)x \rangle=	\langle Ax+iBx, y \rangle\langle y, Ax+iBx \rangle\\
	&=	\langle Ax-\langle Ax, x \rangle x+i(Bx-\langle Bx, x \rangle x), y \rangle\langle y,  Ax-\langle Ax, x \rangle x+i(Bx-\langle Bx, x \rangle x)\rangle\\
	&\leq \| \langle y, y \rangle \|\langle  (Ax-\langle Ax, x \rangle x)+i (Bx-\langle Bx, x \rangle x), (Ax-\langle Ax, x \rangle x)+i (Bx-\langle Bx, x \rangle x)\rangle\\
	&\leq \langle Ax-\langle Ax, x \rangle x, Ax-\langle Ax, x \rangle x\rangle +\langle Bx-\langle Bx, x \rangle x, Bx-\langle Bx, x \rangle x\rangle \\
	&\quad +i\langle Bx-\langle Bx, x \rangle x, Ax-\langle Ax, x \rangle x\rangle-i\langle Ax-\langle Ax, x \rangle x, Bx-\langle Bx, x \rangle x\rangle\\
	&=d_x(A)^2+d_x(B)^2+i(\langle Bx, Ax \rangle -\langle Bx, x \rangle\langle Ax, x \rangle)-i(\langle Ax, Bx \rangle -\langle Ax, x \rangle\langle Bx, x \rangle)\\
	&=d_x(A)^2+d_x(B)^2+i\langle [A,B]x, x \rangle +i[\langle Ax, x \rangle, \langle Bx, x \rangle].
\end{align*}
Therefore 
\begin{align*}
	d_x(A)^2+d_x(B)^2\geq -i\langle [A,B]x, x \rangle +\langle (A+iB)x, y \rangle\langle y, (A+iB)x \rangle-i[\langle Ax, x \rangle, \langle Bx, x \rangle].
\end{align*}
\item Take norm on (vii).
\item Similar to  (vii).
\item Take norm on (ix).
\item Add (vii) and (ix).
\item Add (viii) and (ix), then take norm.
\end{enumerate} 	
\end{proof}
\begin{corollary}
Let $\mathcal{E}$ be a  Hilbert C*-module  over a unital commutative C*-algebra $\mathcal{A}$. 	Let  $A:\mathcal{D}(A)\to  \mathcal{E}$ and $B:	\mathcal{D}(B)\to  \mathcal{E}$  be self-adjoint morphisms. Then for all $x \in \mathcal{D}(AB)\cap  \mathcal{D}(BA)$ with $\langle x, x \rangle =1$, we have 
\begin{enumerate}[\upshape(i)]
\item If $\mathcal{A}$ is commutative, then
\begin{align*}
	d_x(A)^2+d_x(B)^2&\geq -i\langle [A,B]x, x \rangle +\langle (A+iB)x, y \rangle\langle y, (A+iB)x \rangle, \\
	&\quad \forall y \in \mathcal{E} \text{ satisfying } \|y\|\leq 1 \text{ and } \langle x,y \rangle =0.
\end{align*}
\item If $\mathcal{A}$ is commutative, then
\begin{align*}
	\Delta_x(A)^2+\Delta_x(B)^2&\geq \|-i\langle [A,B]x, x \rangle +\langle (A+iB)x, y \rangle\langle y, (A+iB)x \rangle\|, \\
	&\quad \forall y \in \mathcal{E} \text{ satisfying } \|y\|\leq 1 \text{ and } \langle x,y \rangle =0.
\end{align*}
\end{enumerate}	
\end{corollary}
We also have following uncertainty principles. 
\begin{theorem}(\textbf{Noncommutative Maccone-Pati Uncertainty Principles})
Let $\mathcal{E}$ be a  Hilbert C*-module  over a unital C*-algebra $\mathcal{A}$. 	Let  $A:	\mathcal{D}(A)\to  \mathcal{E}$ and $B:	\mathcal{D}(B)\to  \mathcal{E}$  be self-adjoint morphisms. Then for all $x \in \mathcal{D}(AB)\cap  \mathcal{D}(BA)$ with $\langle x, x \rangle =1$, we have 	
\begin{enumerate}[\upshape(i)]
	\item 	\begin{align*}
		\Delta_x(A)^2+\Delta_x(B)^2&\geq   \langle y, (A+B)x \rangle\langle (A+B)x, y \rangle-\|\langle \{A,B\}x, x \rangle -\{\langle Ax, x \rangle, \langle Bx, x \rangle\}\|, \\
		&\quad \forall y \in \mathcal{E} \text{ satisfying } \|y\| \leq1 \text{ and } \langle x,y \rangle =0.
	\end{align*}
	\item 
	\begin{align*}
		\Delta_x(A)^2+\Delta_x(B)^2&\geq \langle y, (A-B)x \rangle\langle (A-B)x, y \rangle-\|\langle \{A,B\}x, x \rangle -\{\langle Ax, x \rangle, \langle Bx, x \rangle\}\|\\
		&\quad \forall y \in \mathcal{E} \text{ satisfying } \|y\| \leq1 \text{ and } \langle x,y \rangle =0.
	\end{align*}
	\item 	\begin{align*}
		\Delta_x(A)^2+\Delta_x(B)^2&\geq  \frac{\|\langle y, (A+B)x \rangle\langle (A+B)x, y \rangle+\langle (A-B)x, y \rangle\langle y, (A-B)x \rangle\|}{2},\\
		&\quad \forall y \in \mathcal{E} \text{ satisfying } \|y\| \leq1 \text{ and } \langle x,y \rangle =0.
	\end{align*}
	\item 
	\begin{align*}
		\Delta_x(A)^2+\Delta_x(B)^2&\geq \langle y, (A+iB)x \rangle\langle (A+iB)x, y \rangle-\|\langle [A,B]x, x \rangle +[\langle Ax, x \rangle, \langle Bx, x \rangle]\|,\\
		&\quad \forall y \in \mathcal{E} \text{ satisfying } \|y\| \leq1 \text{ and } \langle x,y \rangle =0.
	\end{align*}
	\item 
	\begin{align*}
		\Delta_x(A)^2+\Delta_x(B)^2&\geq  \langle y, (A-iB)x \rangle\langle (A-iB)x, y \rangle-\|\langle [A,B]x, x \rangle +[\langle Ax, x \rangle, \langle Bx, x \rangle]\|,\\
		&\quad \forall y \in \mathcal{E} \text{ satisfying } \|y\| \leq 1 \text{ and } \langle x,y \rangle =0.
	\end{align*} 
\end{enumerate}	
\end{theorem}
\begin{proof}
		Let $x \in \mathcal{D}(AB)\cap  \mathcal{D}(BA)$ with $\langle x, x \rangle =1$. Let  	$y \in \mathcal{E}$ be such that $ \|y\|\leq 1 $ and $ \langle x,y \rangle =0.$
\begin{enumerate}[\upshape(i)]
	\item 	\begin{align*}
		&\langle y, (A+B)x \rangle \langle (A+B)x, y \rangle\\
		&=\langle y, Ax-\langle Ax, x \rangle x+(Bx-\langle Bx, x \rangle x) \rangle\langle Ax-\langle Ax, x \rangle x+(Bx-\langle Bx, x \rangle x), y\rangle\\
		&\leq \|\langle  (Ax-\langle Ax, x \rangle x)+ (Bx-\langle Bx, x \rangle x), (Ax-\langle Ax, x \rangle x)+ (Bx-\langle Bx, x \rangle x)\rangle\|\langle y, y \rangle\\
		&\leq \|\langle  (Ax-\langle Ax, x \rangle x)+ (Bx-\langle Bx, x \rangle x), (Ax-\langle Ax, x \rangle x)+ (Bx-\langle Bx, x \rangle x)\rangle\|\|\langle y, y \rangle\|\\
		&\leq \|\langle Ax-\langle Ax, x \rangle x, Ax-\langle Ax, x \rangle x\rangle +\langle Bx-\langle Bx, x \rangle x, Bx-\langle Bx, x \rangle x\rangle \\
		&\quad +\langle Bx-\langle Bx, x \rangle x, Ax-\langle Ax, x \rangle x\rangle+\langle Ax-\langle Ax, x \rangle x, Bx-\langle Bx, x \rangle x\rangle\|\\
		&=\|d_x(A)^2+d_x(B)^2+(\langle Bx, Ax \rangle -\langle Bx, x \rangle\langle Ax, x \rangle)+\langle Ax, Bx \rangle -\langle Ax, x \rangle\langle Bx, x \rangle\|\\
		&\leq \|d_x(A)^2+d_x(B)^2\|+\|\langle \{A,B\}x, x \rangle -\{\langle Ax, x \rangle, \langle Bx, x \rangle\}\|.
	\end{align*}
	Therefore 
	\begin{align*}
	\Delta_x(A)^2+\Delta_x(B)^2\geq	\|d_x(A)^2+d_x(B)^2\|\geq \langle y, (A+B)x \rangle\langle (A+B)x, y \rangle-\|\langle \{A,B\}x, x \rangle -\{\langle Ax, x \rangle, \langle Bx, x \rangle\}\|.
	\end{align*}
	\item Similar to (i).
\item \begin{align*}
	&\langle y, (A+B)x \rangle \langle  (A+B)x, y \rangle+\langle y, (A-B)x \rangle \langle  (A-B)x, y \rangle=2\langle y, Ax \rangle \langle Ax, y \rangle +2\langle y, Bx \rangle \langle Bx, y \rangle\\
	&=2\langle y, Ax-\langle Ax, x \rangle x\rangle \langle Ax-\langle Ax, x \rangle x, y \rangle +2\langle y, Bx-\langle Bx, x \rangle x \rangle \langle Bx-\langle Bx, x \rangle x, y \rangle\\
	&\leq 2 \|\langle Ax-\langle Ax,x \rangle x,  Ax-\langle Ax, x \rangle x \rangle \|\langle y, y \rangle+ 2\|\langle Bx-\langle Bx,x \rangle x,  Bx-\langle Bx, x \rangle x \rangle\|\langle y, y \rangle\\
	&\leq 2\Delta_x(A)^2+2\Delta_x(B)^2.
\end{align*}
\item 
\begin{align*}
	&\langle y, (A+iB)x \rangle\langle (A+iB)x, y \rangle=	\langle y, Ax+iBx \rangle\langle Ax+iBx, y \rangle\\
	&=	\langle y, Ax-\langle Ax, x \rangle x+i(Bx-\langle Bx, x \rangle x) \rangle\langle Ax-\langle Ax, x \rangle x+i(Bx-\langle Bx, x \rangle x), y\rangle\\
	&\leq  \|\langle  (Ax-\langle Ax, x \rangle x)+i (Bx-\langle Bx, x \rangle x), (Ax-\langle Ax, x \rangle x)+i (Bx-\langle Bx, x \rangle x)\rangle\| \langle y, y \rangle\\
	&\leq \|\langle Ax-\langle Ax, x \rangle x, Ax-\langle Ax, x \rangle x\rangle +\langle Bx-\langle Bx, x \rangle x, Bx-\langle Bx, x \rangle x\rangle \\
	&\quad +i\langle Bx-\langle Bx, x \rangle x, Ax-\langle Ax, x \rangle x\rangle-i\langle Ax-\langle Ax, x \rangle x, Bx-\langle Bx, x \rangle x\rangle\|\\
	&=\|d_x(A)^2+d_x(B)^2+i(\langle Bx, Ax \rangle -\langle Bx, x \rangle\langle Ax, x \rangle)-i(\langle Ax, Bx \rangle -\langle Ax, x \rangle\langle Bx, x \rangle)\|\\
	&=\|d_x(A)^2+d_x(B)^2\|+\|(\langle Bx, Ax \rangle -\langle Bx, x \rangle\langle Ax, x \rangle)-(\langle Ax, Bx \rangle -\langle Ax, x \rangle\langle Bx, x \rangle)\|\\
	&=\|d_x(A)^2+d_x(B)^2\|+\|\langle [A,B]x, x \rangle +[\langle Ax, x \rangle, \langle Bx, x \rangle]\|.
\end{align*}
Therefore 
\begin{align*}
	&\Delta_x(A)^2+\Delta_x(B)^2\geq \|d_x(A)^2+d_x(B)^2\|\\
	&\geq \langle y, (A+iB)x \rangle\langle (A+iB)x, y \rangle-\|\langle [A,B]x, x \rangle +[\langle Ax, x \rangle, \langle Bx, x \rangle]\|.
\end{align*}
\item Similar to (iv).
\end{enumerate}		
\end{proof}
\begin{theorem}
Let $\mathcal{E}$ be a  Hilbert C*-module  over a unital commutative C*-algebra $\mathcal{A}$. 	Let  $A:\mathcal{D}(A)\to  \mathcal{E}$ and $B:	\mathcal{D}(B)\to  \mathcal{E}$  be self-adjoint morphisms. Then for all $x \in \mathcal{D}(AB)\cap  \mathcal{D}(BA)$ with $\langle x, x \rangle =1$, we have 	
\begin{enumerate}[\upshape(i)]
	 \item If  $d_x(A)$, $d_x(B)$ are invertible, then 
	\begin{align*}
		d_x(A)d_x(B)&\geq \frac{-i\langle [A,B]x, x \rangle}{2-\left\langle \left(\frac{A}{d_x(A)}+\frac{iB}{d_x(B)}\right)x, y\right \rangle \left\langle y,  \left(\frac{A}{d_x(A)}+\frac{iB}{d_x(B)}\right)x\right \rangle},\\
		&\quad \forall y \in \mathcal{E} \text{ satisfying } \|y\|\leq 1 \text{ and } \langle x,y \rangle =0.
	\end{align*}
	\item 
	\begin{align*}
		\Delta_x(A)\Delta_x(B)&\geq \frac{-i\langle [A,B]x, x \rangle}{2-\left\langle \left(\frac{A}{\Delta_x(A)}+\frac{iB}{\Delta_x(B)}\right)x, y\right \rangle \left\langle y,  \left(\frac{A}{\Delta_x(A)}+\frac{iB}{\Delta_x(B)}\right)x\right \rangle},\\
		&\quad \forall y \in \mathcal{E} \text{ satisfying } \|y\|\leq 1 \text{ and } \langle x,y \rangle =0.
	\end{align*}
	\item If $d_x(A)$, $d_x(B)$ are invertible, then 
	\begin{align*}
		d_x(A)d_x(B)&\geq \frac{i\langle [A,B]x, x \rangle}{2-\left\langle \left(\frac{A}{d_x(A)}-\frac{iB}{d_x(B)}\right)x, y\right \rangle \left\langle y,  \left(\frac{A}{d_x(A)}-\frac{iB}{d_x(B)}\right)x\right \rangle},\\
		&\quad \forall y \in \mathcal{E} \text{ satisfying } \|y\|\leq 1 \text{ and } \langle x,y \rangle =0.
	\end{align*}
\item 
	\begin{align*}
	\Delta_x(A)\Delta_x(B)&\geq \frac{i\langle [A,B]x, x \rangle}{2-\left\langle \left(\frac{A}{\Delta_x(A)}-\frac{iB}{\Delta_x(B)}\right)x, y\right \rangle \left\langle y,  \left(\frac{A}{\Delta_x(A)}-\frac{iB}{\Delta_x(B)}\right)x\right \rangle},\\
	&\quad \forall y \in \mathcal{E} \text{ satisfying } \|y\|\leq 1 \text{ and } \langle x,y \rangle =0.
\end{align*}
 \item If  $d_x(A)$, $d_x(B)$ are invertible, then 
\begin{align*}
	d_x(A)d_x(B)&\geq \frac{2\langle Ax, x \rangle\langle Bx, x \rangle -\langle \{A,B\}x, x \rangle}{2-\left\langle \left(\frac{A}{d_x(A)}+\frac{B}{d_x(B)}\right)x, y\right \rangle \left\langle y,  \left(\frac{A}{d_x(A)}+\frac{B}{d_x(B)}\right)x\right \rangle},\\
	&\quad \forall y \in \mathcal{E} \text{ satisfying } \|y\|\leq 1 \text{ and } \langle x,y \rangle =0.
\end{align*}
\item 
\begin{align*}
	\Delta_x(A)\Delta_x(B)&\geq \frac{2\langle Ax, x \rangle\langle Bx, x \rangle -\langle \{A,B\}x, x \rangle}{2-\left\langle \left(\frac{A}{\Delta_x(A)}+\frac{B}{\Delta_x(B)}\right)x, y\right \rangle \left\langle y,  \left(\frac{A}{\Delta_x(A)}+\frac{B}{\Delta_x(B)}\right)x\right \rangle},\\
	&\quad \forall y \in \mathcal{E} \text{ satisfying } \|y\|\leq 1 \text{ and } \langle x,y \rangle =0.
\end{align*}
\item If  $d_x(A)$, $d_x(B)$ are invertible, then 
\begin{align*}
	d_x(A)d_x(B)&\geq \frac{-2\langle Ax, x \rangle\langle Bx, x \rangle +\langle \{A,B\}x, x \rangle}{2-\left\langle \left(\frac{A}{d_x(A)}-\frac{B}{d_x(B)}\right)x, y\right \rangle \left\langle y,  \left(\frac{A}{d_x(A)}-\frac{B}{d_x(B)}\right)x\right \rangle},\\
	&\quad \forall y \in \mathcal{E} \text{ satisfying } \|y\|\leq 1 \text{ and } \langle x,y \rangle =0.
\end{align*}
\item 
\begin{align*}
	\Delta_x(A)\Delta_x(B)&\geq \frac{-2\langle Ax, x \rangle\langle Bx, x \rangle +\langle \{A,B\}x, x \rangle}{2-\left\langle \left(\frac{A}{\Delta_x(A)}-\frac{B}{\Delta_x(B)}\right)x, y\right \rangle \left\langle y,  \left(\frac{A}{\Delta_x(A)}-\frac{B}{\Delta_x(B)}\right)x\right \rangle},\\
	&\quad \forall y \in \mathcal{E} \text{ satisfying } \|y\|\leq 1 \text{ and } \langle x,y \rangle =0.
\end{align*}
\end{enumerate}
\end{theorem}
\begin{proof}
		Let $x \in \mathcal{D}(AB)\cap  \mathcal{D}(BA)$ with $\langle x, x \rangle =1$. Let  	$y \in \mathcal{E}$ be such that $ \|y\|\leq 1 $ and $ \langle x,y \rangle =0.$
	\begin{enumerate}[\upshape(i)]
		\item Since   $\mathcal{A}$ is  commutative and $d_x(A)$, $d_x(B)$ are invertible, 
		\begin{align*}
			&\left\langle \left(\frac{A}{d_x(A)}+\frac{iB}{d_x(B)}\right)x, y\right \rangle \left\langle y,  \left(\frac{A}{d_x(A)}+\frac{iB}{d_x(B)}\right)x\right \rangle\\
			&=\left\langle \frac{Ax-\langle Ax, x \rangle x}{d_x(A)}+\frac{i(Bx-\langle Bx, x \rangle x)}{d_x(B)}, y\right \rangle \left\langle y,  \frac{Ax-\langle Ax, x \rangle x}{d_x(A)}+\frac{i(Bx-\langle Bx, x \rangle x)}{d_x(B)}\right \rangle\\
			&\leq \|\langle y, y \rangle \|\left\langle \frac{Ax-\langle Ax, x \rangle x}{d_x(A)}+\frac{i(Bx-\langle Bx, x \rangle x)}{d_x(B)}, \frac{Ax-\langle Ax, x \rangle x}{d_x(A)}+\frac{i(Bx-\langle Bx, x \rangle x)}{d_x(B)}\right \rangle\\
			&\leq \left\langle \frac{Ax-\langle Ax, x \rangle x}{d_x(A)},  \frac{Ax-\langle Ax, x \rangle x}{d_x(A)}\right \rangle+\left\langle \frac{Bx-\langle Bx, x \rangle x}{d_x(B)},  \frac{Bx-\langle Bx, x \rangle x}{d_x(B)}\right \rangle\\
			&\quad +i\left\langle \frac{Bx-\langle Bx, x \rangle x}{d_x(B)},  \frac{Ax-\langle Ax, x \rangle x}{d_x(A)}\right \rangle-i\left\langle \frac{Ax-\langle Ax, x \rangle x}{d_x(A)},  \frac{Bx-\langle Bx, x \rangle x}{d_x(B)}\right \rangle\\
			&=1+1+\frac{i\langle Bx, Ax \rangle-i\langle Bx, x \rangle\langle Ax, x \rangle}{d_x(A)d_x(B)}+\frac{-i\langle Ax, Bx \rangle+i\langle Ax, x \rangle\langle Bx, x \rangle}{d_x(A)d_x(B)}\\
			&=2+\frac{i\langle [A,B]x, x \rangle}{d_x(A)d_x(B)}.
		\end{align*}
		Therefore 
		\begin{align*}
			&\frac{-i\langle [A,B]x, x \rangle}{d_x(A)d_x(B)}\leq 2-\left\langle \left(\frac{A}{d_x(A)}+\frac{iB}{d_x(B)}\right)x, y\right \rangle \left\langle y,  \left(\frac{A}{d_x(A)}+\frac{iB}{d_x(B)}\right)x\right \rangle\\
			&\implies \frac{-i\langle [A,B]x, x \rangle}{2-\left\langle \left(\frac{A}{d_x(A)}+\frac{iB}{d_x(B)}\right)x, y\right \rangle \left\langle y,  \left(\frac{A}{d_x(A)}+\frac{iB}{d_x(B)}\right)x\right \rangle}\leq  d_x(A)d_x(B).
		\end{align*}
		\item Since  $\mathcal{A}$ is commutative, 
		\begin{align*}
			&\left\langle \left(\frac{A}{\Delta_x(A)}+\frac{iB}{\Delta_x(B)}\right)x, y\right \rangle \left\langle y,  \left(\frac{A}{\Delta_x(A)}+\frac{iB}{\Delta_x(B)}\right)x\right \rangle\\
			&=\left\langle \frac{Ax-\langle Ax, x \rangle x}{\Delta_x(A)}+\frac{i(Bx-\langle Bx, x \rangle x)}{\Delta_x(B)}, y\right \rangle \left\langle y,  \frac{Ax-\langle Ax, x \rangle x}{\Delta_x(A)}+\frac{i(Bx-\langle Bx, x \rangle x)}{\Delta_x(B)}\right \rangle\\
			&\leq \|\langle y, y \rangle \|\left\langle \frac{Ax-\langle Ax, x \rangle x}{\Delta_x(A)}+\frac{i(Bx-\langle Bx, x \rangle x)}{\Delta_x(B)}, \frac{Ax-\langle Ax, x \rangle x}{\Delta_x(A)}+\frac{i(Bx-\langle Bx, x \rangle x)}{\Delta_x(B)}\right \rangle\\
			&\leq \left\langle \frac{Ax-\langle Ax, x \rangle x}{\Delta_x(A)},  \frac{Ax-\langle Ax, x \rangle x}{\Delta_x(A)}\right \rangle+\left\langle \frac{Bx-\langle Bx, x \rangle x}{\Delta_x(B)},  \frac{Bx-\langle Bx, x \rangle x}{\Delta_x(B)}\right \rangle\\
			&\quad +i\left\langle \frac{Bx-\langle Bx, x \rangle x}{\Delta_x(B)},  \frac{Ax-\langle Ax, x \rangle x}{\Delta_x(A)}\right \rangle-i\left\langle \frac{Ax-\langle Ax, x \rangle x}{\Delta_x(A)},  \frac{Bx-\langle Bx, x \rangle x}{\Delta_x(B)}\right \rangle\\
			&=1+1+\frac{i\langle Bx, Ax \rangle-i\langle Bx, x \rangle\langle Ax, x \rangle}{\Delta_x(A)\Delta_x(B)}+\frac{-i\langle Ax, Bx \rangle+i\langle Ax, x \rangle\langle Bx, x \rangle}{\Delta_x(A)\Delta_x(B)}\\
			&=2+\frac{i\langle [A,B]x, x \rangle}{\Delta_x(A)\Delta_x(B)}.
		\end{align*}
		Therefore 
		\begin{align*}
			&\frac{-i\langle [A,B]x, x \rangle}{\Delta_x(A)\Delta_x(B)}\leq 2-\left\langle \left(\frac{A}{\Delta_x(A)}+\frac{iB}{\Delta_x(B)}\right)x, y\right \rangle \left\langle y,  \left(\frac{A}{\Delta_x(A)}+\frac{iB}{\Delta_x(B)}\right)x\right \rangle\\
			&\implies \frac{-i\langle [A,B]x, x \rangle}{2-\left\langle \left(\frac{A}{\Delta_x(A)}+\frac{iB}{\Delta_x(B)}\right)x, y\right \rangle \left\langle y,  \left(\frac{A}{\Delta_x(A)}+\frac{iB}{\Delta_x(B)}\right)x\right \rangle}\leq  \Delta_x(A)\Delta_x(B).
		\end{align*}
	\item Similar to (i).
	\item Similar to (ii).
	\item Since   $\mathcal{A}$ is  commutative and $d_x(A)$, $d_x(B)$ are invertible, 
	\begin{align*}
		&\left\langle \left(\frac{A}{d_x(A)}+\frac{B}{d_x(B)}\right)x, y\right \rangle \left\langle y,  \left(\frac{A}{d_x(A)}+\frac{B}{d_x(B)}\right)x\right \rangle\\
		&=\left\langle \frac{Ax-\langle Ax, x \rangle x}{d_x(A)}+\frac{Bx-\langle Bx, x \rangle x}{d_x(B)}, y\right \rangle \left\langle y,  \frac{Ax-\langle Ax, x \rangle x}{d_x(A)}+\frac{Bx-\langle Bx, x \rangle x}{d_x(B)}\right \rangle\\
		&\leq \|\langle y, y \rangle \|\left\langle \frac{Ax-\langle Ax, x \rangle x}{d_x(A)}+\frac{Bx-\langle Bx, x \rangle x}{d_x(B)}, \frac{Ax-\langle Ax, x \rangle x}{d_x(A)}+\frac{Bx-\langle Bx, x \rangle x}{d_x(B)}\right \rangle\\
		&\leq \left\langle \frac{Ax-\langle Ax, x \rangle x}{d_x(A)},  \frac{Ax-\langle Ax, x \rangle x}{d_x(A)}\right \rangle+\left\langle \frac{Bx-\langle Bx, x \rangle x}{d_x(B)},  \frac{Bx-\langle Bx, x \rangle x}{d_x(B)}\right \rangle\\
		&\quad +\left\langle \frac{Bx-\langle Bx, x \rangle x}{d_x(B)},  \frac{Ax-\langle Ax, x \rangle x}{d_x(A)}\right \rangle+\left\langle \frac{Ax-\langle Ax, x \rangle x}{d_x(A)},  \frac{Bx-\langle Bx, x \rangle x}{d_x(B)}\right \rangle\\
		&=1+1+\frac{\langle Bx, Ax \rangle-\langle Bx, x \rangle\langle Ax, x \rangle}{d_x(A)d_x(B)}+\frac{\langle Ax, Bx \rangle-\langle Ax, x \rangle\langle Bx, x \rangle}{d_x(A)d_x(B)}\\
		&=2+\frac{-2\langle Ax, x \rangle\langle Bx, x \rangle +\langle \{A,B\}x, x \rangle}{d_x(A)d_x(B)}.
	\end{align*}
	Therefore 
	\begin{align*}
	&-\frac{-2\langle Ax, x \rangle\langle Bx, x \rangle +\langle \{A,B\}x, x \rangle}{d_x(A)d_x(B)}\leq 2-\left\langle \left(\frac{A}{d_x(A)}+\frac{B}{d_x(B)}\right)x, y\right \rangle \left\langle y,  \left(\frac{A}{d_x(A)}+\frac{B}{d_x(B)}\right)x\right \rangle\\	
	&\implies \frac{2\langle Ax, x \rangle\langle Bx, x \rangle -\langle \{A,B\}x, x \rangle}{2-\left\langle \left(\frac{A}{d_x(A)}+\frac{B}{d_x(B)}\right)x, y\right \rangle \left\langle y,  \left(\frac{A}{d_x(A)}+\frac{B}{d_x(B)}\right)x\right \rangle}\leq d_x(A)d_x(B).
	\end{align*}
	\item Since  $\mathcal{A}$ is commutative, 
	\begin{align*}
		&\left\langle \left(\frac{A}{\Delta_x(A)}+\frac{B}{\Delta_x(B)}\right)x, y\right \rangle \left\langle y,  \left(\frac{A}{\Delta_x(A)}+\frac{B}{\Delta_x(B)}\right)x\right \rangle\\
		&=\left\langle \frac{Ax-\langle Ax, x \rangle x}{\Delta_x(A)}+\frac{Bx-\langle Bx, x \rangle x}{\Delta_x(B)}, y\right \rangle \left\langle y,  \frac{Ax-\langle Ax, x \rangle x}{\Delta_x(A)}+\frac{Bx-\langle Bx, x \rangle x}{\Delta_x(B)}\right \rangle\\
		&\leq \|\langle y, y \rangle \|\left\langle \frac{Ax-\langle Ax, x \rangle x}{\Delta_x(A)}+\frac{Bx-\langle Bx, x \rangle x}{\Delta_x(B)}, \frac{Ax-\langle Ax, x \rangle x}{\Delta_x(A)}+\frac{Bx-\langle Bx, x \rangle x}{\Delta_x(B)}\right \rangle\\
		&\leq \left\langle \frac{Ax-\langle Ax, x \rangle x}{\Delta_x(A)},  \frac{Ax-\langle Ax, x \rangle x}{\Delta_x(A)}\right \rangle+\left\langle \frac{Bx-\langle Bx, x \rangle x}{\Delta_x(B)},  \frac{Bx-\langle Bx, x \rangle x}{\Delta_x(B)}\right \rangle\\
		&\quad +\left\langle \frac{Bx-\langle Bx, x \rangle x}{\Delta_x(B)},  \frac{Ax-\langle Ax, x \rangle x}{\Delta_x(A)}\right \rangle+\left\langle \frac{Ax-\langle Ax, x \rangle x}{\Delta_x(A)},  \frac{Bx-\langle Bx, x \rangle x}{\Delta_x(B)}\right \rangle\\
		&=1+1+\frac{\langle Bx, Ax \rangle-\langle Bx, x \rangle\langle Ax, x \rangle}{\Delta_x(A)\Delta_x(B)}+\frac{\langle Ax, Bx \rangle-\langle Ax, x \rangle\langle Bx, x \rangle}{\Delta_x(A)\Delta_x(B)}\\
		&=2+\frac{2\langle Ax, x \rangle\langle Bx, x \rangle +\langle \{A,B\}x, x \rangle}{\Delta_x(A)\Delta_x(B)}.
	\end{align*}
	Therefore 
\begin{align*}
	&-\frac{-2\langle Ax, x \rangle\langle Bx, x \rangle +\langle \{A,B\}x, x \rangle}{\Delta_x(A)\Delta_x(B)}\leq 2-\left\langle \left(\frac{A}{\Delta_x(A)}+\frac{B}{\Delta_x(B)}\right)x, y\right \rangle \left\langle y,  \left(\frac{A}{\Delta_x(A)}+\frac{B}{\Delta_x(B)}\right)x\right \rangle\\	
	&\implies \frac{2\langle Ax, x \rangle\langle Bx, x \rangle -\langle \{A,B\}x, x \rangle}{2-\left\langle \left(\frac{A}{\Delta_x(A)}+\frac{B}{\Delta_x(B)}\right)x, y\right \rangle \left\langle y,  \left(\frac{A}{\Delta_x(A)}+\frac{B}{\Delta_x(B)}\right)x\right \rangle}\leq \Delta_x(A)\Delta_x(B).
\end{align*}
	\item Similar to (v).
	\item Similar to (vi).
	\end{enumerate}
\end{proof}

\section{Acknowledgments}
The author thanks  anonymous reviewer for his/her constructive feedback, which significantly improved the final manuscript.

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