Noncommutative Maccone-Pati Uncertainty Principles

1. Introduction

Let $\mathcal{H}$ be a complex Hilbert space and A be a possibly unbounded self-adjoint linear operator defined on domain $\mathcal{D}\left(A\right)\subseteq \mathcal{H}$. For $h\in \mathcal{D}\left(A\right)$ with $\parallel h\parallel =1$, define the uncertainty of A at the point h as

$$\begin{array}{c}\hfill {\Delta }_h\left(A\right)\parallel Ah-\langle Ah,h\rangle h\parallel =\sqrt{{\parallel Ah\parallel }^2-{\langle Ah,h\rangle }^2}.\end{array}$$

In 1929, Robertson [1] derived the following mathematical form of the uncertainty principle of Heisenberg derived in 1927 [2]. Recall that for two linear operators $A:\mathcal{D}\left(A\right)\subseteq \mathcal{H}\to \mathcal{H}$ and $B:\mathcal{D}\left(B\right)\subseteq \mathcal{H}\to \mathcal{H}$, we define the commutator $[A,B]AB-BA$ and the anti-commutator $\{A,B\}AB+BA$.

Theorem 1.

[1,2,3,4] (Heisenberg-Robertson Uncertainty Principle) Let $A:\mathcal{D}\left(A\right)\subseteq \mathcal{H}\to \mathcal{H}$ and $B:\mathcal{D}\left(B\right)\subseteq \mathcal{H}\to \mathcal{H}$ be self-adjoint linear operators. Then for all $h\in \mathcal{D}\left(AB\right)\cap \mathcal{D}\left(BA\right)$ with $\parallel h\parallel =1$, we have

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{2}}\left({\Delta }_h{\left(A\right)}^2+{\Delta }_h{\left(B\right)}^2\right)\ge {\displaystyle \frac{1}{4}}{\left({\Delta }_h\left(A\right)+{\Delta }_h\left(B\right)\right)}^2\ge {\Delta }_h\left(A\right){\Delta }_h\left(B\right)\ge {\displaystyle \frac{1}{2}}\left|\langle [A,B]h,h\rangle \right|.\end{array}$$

(1)

In 1930, Schrodinger improved Inequality (1) [5].

Theorem 2.

[5,6] (Heisenberg-Robertson-Schrodinger Uncertainty Principle) Let $A:\mathcal{D}\left(A\right)\subseteq \mathcal{H}\to \mathcal{H}$ and $B:\mathcal{D}\left(B\right)\subseteq \mathcal{H}\to \mathcal{H}$ be self-adjoint linear operators. Then for all $h\in \mathcal{D}\left(AB\right)\cap \mathcal{D}\left(BA\right)$ with $\parallel h\parallel =1$, we have

$$\begin{array}{cc}\hfill {\Delta }_h\left(A\right){\Delta }_h\left(B\right)\ge |\langle Ah,Bh\rangle -\langle Ah,h\rangle \langle Bh,h\rangle |& ={\displaystyle \frac{\sqrt{{\left|\langle [A,B]h,h\rangle \right|}^2+{|\langle \{A,B\}h,h\rangle -2\langle Ah,h\rangle \langle Bh,h\rangle |}^2}}{2}}\hfill \\ \hfill & ={\displaystyle \frac{\sqrt{{(\langle \{A,B\}h,h\rangle -2\langle Ah,h\rangle \langle Bh,h\rangle )}^2-{\langle [A,B]h,h\rangle }^2}}{2}}.\hfill \end{array}$$

In 2014, Maccone and Pati derived the following uncertainty principles which work for orthogonal vectors [7].

Theorem 3.

[7] (Maccone-Pati Uncertainty Principles) Let $A:\mathcal{D}\left(A\right)\subseteq \mathcal{H}\to \mathcal{H}$ and $B:\mathcal{D}\left(B\right)\subseteq \mathcal{H}\to \mathcal{H}$ be self-adjoint linear operators. Then for all $h\in \mathcal{D}\left(AB\right)\cap \mathcal{D}\left(BA\right)$ with $\parallel h\parallel =1$, we have

(i)

$$\begin{array}{cc}\hfill {\Delta }_h{\left(A\right)}^2+{\Delta }_h{\left(B\right)}^2& \ge {\left|\langle (A+B)h,k\rangle \right|}^2-\langle \{A,B\}h,h\rangle +2\langle Ah,h\rangle \langle Bh,h\rangle ,\hfill \\ \hfill & \forall k\in \mathcal{H}\mathrm{satisfying}\parallel k\parallel =1\mathrm{and}\langle h,k\rangle =0.\hfill \end{array}$$

(ii)

$$\begin{array}{cc}\hfill {\Delta }_h{\left(A\right)}^2+{\Delta }_h{\left(B\right)}^2& \ge {\left|\langle (A-B)h,k\rangle \right|}^2+\langle \{A,B\}h,h\rangle -2\langle Ah,h\rangle \langle Bh,h\rangle ,\hfill \\ \hfill & \forall k\in \mathcal{H}\mathrm{satisfying}\parallel k\parallel =1\mathrm{and}\langle h,k\rangle =0.\hfill \end{array}$$

(iii)

$$\begin{array}{c}\hfill {\Delta }_h{\left(A\right)}^2+{\Delta }_h{\left(B\right)}^2\ge {\displaystyle \frac{{\left|\langle (A+B)h,k\rangle \right|}^2+{\left|\langle (A-B)h,k\rangle \right|}^2}{2}},\forall k\in \mathcal{H}\mathrm{satisfying}\parallel k\parallel =1\mathrm{and}\langle h,k\rangle =0.\end{array}$$

(iv)

$$\begin{array}{c}\hfill {\Delta }_h{\left(A\right)}^2+{\Delta }_h{\left(B\right)}^2\ge -i\langle [A,B]h,h\rangle +{\left|\langle (A+iB)h,k\rangle \right|}^2,\forall k\in \mathcal{H}\mathrm{satisfying}\parallel k\parallel =1\mathrm{and}\langle h,k\rangle =0.\end{array}$$

(v)

$$\begin{array}{c}\hfill {\Delta }_h{\left(A\right)}^2+{\Delta }_h{\left(B\right)}^2\ge i\langle [A,B]h,h\rangle +{\left|\langle (A-iB)h,k\rangle \right|}^2,\forall k\in \mathcal{H}\mathrm{satisfying}\parallel k\parallel =1\mathrm{and}\langle h,k\rangle =0.\end{array}$$

(vi)

$$\begin{array}{c}\hfill {\Delta }_h{\left(A\right)}^2+{\Delta }_h{\left(B\right)}^2\ge {\displaystyle \frac{{\left|\langle (A+iB)h,k\rangle \right|}^2+{\left|\langle (A-iB)h,k\rangle \right|}^2}{2}},\forall k\in \mathcal{H}\mathrm{satisfying}\parallel k\parallel =1\mathrm{and}\langle h,k\rangle =0.\end{array}$$

(vii)

$$\begin{array}{c}\hfill {\Delta }_h\left(A\right){\Delta }_h\left(B\right)\ge {\displaystyle \frac{-i\langle [A,B]h,h\rangle }{2-{\left|〈\left(\frac{A}{{\Delta }_h\left(A\right)}+i\frac{B}{{\Delta }_h\left(B\right)}\right)h,k〉\right|}^2}},\forall k\in \mathcal{H}\mathrm{satisfying}\parallel k\parallel =1\mathrm{and}\langle h,k\rangle =0.\end{array}$$

(viii)

$$\begin{array}{c}\hfill {\Delta }_h\left(A\right){\Delta }_h\left(B\right)\ge {\displaystyle \frac{i\langle [A,B]h,h\rangle }{2-{\left|〈\left(\frac{A}{{\Delta }_h\left(A\right)}-i\frac{B}{{\Delta }_h\left(B\right)}\right)h,k〉\right|}^2}},\forall k\in \mathcal{H}\mathrm{satisfying}\parallel k\parallel =1\mathrm{and}\langle h,k\rangle =0.\end{array}$$

(ix)

$$\begin{array}{c}\hfill {\Delta }_h\left(A\right){\Delta }_h\left(B\right)\ge {\displaystyle \frac{2\langle Ah,h\rangle \langle Bh,h\rangle -\langle \{A,B\}h,h\rangle }{2-{\left|〈\left(\frac{A}{{\Delta }_h\left(A\right)}+\frac{B}{{\Delta }_h\left(B\right)}\right)h,k〉\right|}^2}},\forall k\in \mathcal{H}\mathrm{satisfying}\parallel k\parallel =1\mathrm{and}\langle h,k\rangle =0.\end{array}$$

(x)

$$\begin{array}{c}\hfill {\Delta }_h\left(A\right){\Delta }_h\left(B\right)\ge {\displaystyle \frac{-2\langle Ah,h\rangle \langle Bh,h\rangle +\langle \{A,B\}h,h\rangle }{2-{\left|〈\left(\frac{A}{{\Delta }_h\left(A\right)}-\frac{B}{{\Delta }_h\left(B\right)}\right)h,k〉\right|}^2}},\forall k\in \mathcal{H}\mathrm{satisfying}\parallel k\parallel =1\mathrm{and}\langle h,k\rangle =0.\end{array}$$

Recently we derived noncommutative version of Theorem 2 in [8]. We therefore ask: What is noncommutative analogues of Theorems 3? For this, we need the notion of noncommutative Hilbert spaces known as Hilbert C*-modules which are first introduced by Kaplansky [9] for modules over commutative C*-algebras and later developed for modules over arbitrary C*-algebras by Paschke [10] and Rieffel [11].

Definition 1.

[9,10,11] Let $\mathcal{A}$ be a unital C*-algebra. A left module $\mathcal{E}$ over $\mathcal{A}$ is said to be a (left)Hilbert C*-moduleif there exists a map $\langle ·,·\rangle :\mathcal{E}\times \mathcal{E}\to \mathcal{A}$ such that the following conditions hold.

(i)

$\langle x,x\rangle \ge 0$, $\forall x\in \mathcal{E}$. If $x\in \mathcal{E}$ satisfies $\langle x,x\rangle =0$, then $x=0$.

(ii)

$\langle ax+y,z\rangle =a\langle x,z\rangle +\langle y,z\rangle$, $\forall x,y,z\in \mathcal{E}$, $\forall a\in \mathcal{A}$.

(iii)

$\langle x,y\rangle ={\langle y,x\rangle }^{*}$, $\forall x,y\in \mathcal{E}$.

(iv)

$\mathcal{E}$ is complete w.r.t. the norm $\parallel x\parallel \sqrt{\parallel \langle x,x\rangle \parallel }$, $\forall x\in \mathcal{E}$.

In the paper, use the following noncommutative Cauchy-Schwarz inequality.

Theorem 4.

[10,11] Let $\mathcal{E}$ be a Hilbert C*-module. Then

$$\begin{array}{c}\hfill \langle x,y\rangle \langle y,x\rangle \le \parallel \langle y,y\rangle \parallel \langle x,x\rangle ,\forall x,y\in \mathcal{E}.\end{array}$$

We derive noncommutative analogues of Theorem 3 in Theorem 5.

2. 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 $\mathcal{D}\left(A\right)\subseteq \mathcal{E}$. For $x\in \mathcal{D}\left(A\right)$ with $\langle x,x\rangle =1$, we define the noncommutative uncertainty of A at the point x as

$$\begin{array}{c}\hfill {\Delta }_x\left(A\right)\parallel Ax-\langle Ax,x\rangle x\parallel =\sqrt{{\parallel \langle Ax,Ax\rangle -\langle Ax,x\rangle }^2\parallel }.\end{array}$$

We define

$$\begin{array}{c}\hfill d_x\left(A\right)\sqrt{\langle Ax,Ax\rangle -{\langle Ax,x\rangle }^2}=\sqrt{\langle Ax-\langle Ax,x\rangle x,Ax-\langle Ax,x\rangle x\rangle }.\end{array}$$

Then $d_x\left(A\right)\in \mathcal{A}$ and

$$\begin{array}{c}\hfill {\Delta }_x\left(A\right)=\parallel d_x\left(A\right)\parallel .\end{array}$$

Given two elements $a,b\in \mathcal{A}$, we define $[a,b]ab-ba$ and $\{a,b\}ab+ba$. Similarly, for two morphisms $A:\mathcal{D}\left(A\right)\subseteq \mathcal{E}\to \mathcal{E}$ and $B:\mathcal{D}\left(B\right)\subseteq \mathcal{E}\to \mathcal{E}$, we define $[A,B]AB-BA$ and $\{A,B\}AB+BA$.

Theorem 5.(Noncommutative Maccone-Pati Uncertainty Principles) Let $\mathcal{E}$ be a Hilbert C*-module over a unital C*-algebra $\mathcal{A}$. Let $A:\mathcal{D}\left(A\right)\subseteq \mathcal{E}\to \mathcal{E}$ and $B:\mathcal{D}\left(B\right)\subseteq \mathcal{E}\to \mathcal{E}$ be self-adjoint morphisms. Then for all $x\in \mathcal{D}\left(AB\right)\cap \mathcal{D}\left(BA\right)$ with $\langle x,x\rangle =1$, we have

(i)

$$\begin{array}{cc}\hfill d_x{\left(A\right)}^2+d_x{\left(B\right)}^2& \ge -\langle \{A,B\}x,x\rangle +\langle (A+B)x,y\rangle \langle y,(A+B)x\rangle +\{\langle Ax,x\rangle ,\langle Bx,x\rangle \},\hfill \\ \hfill & \forall y\in \mathcal{E}\mathrm{satisfying}\parallel y\parallel \le 1\mathrm{and}\langle x,y\rangle =0.\hfill \end{array}$$

(ii)

$$\begin{array}{cc}\hfill {\Delta }_x{\left(A\right)}^2+{\Delta }_x{\left(B\right)}^2& \ge \parallel -\langle \{A,B\}x,x\rangle +\langle (A+B)x,y\rangle \langle y,(A+B)x\rangle +\{\langle Ax,x\rangle ,\langle Bx,x\rangle \}\parallel ,\hfill \\ \hfill & \forall y\in \mathcal{E}\mathrm{satisfying}\parallel y\parallel \le 1\mathrm{and}\langle x,y\rangle =0.\hfill \end{array}$$

(iii)

$$\begin{array}{cc}\hfill d_x{\left(A\right)}^2+d_x{\left(B\right)}^2& \ge \langle \{A,B\}x,x\rangle +\langle (A-B)x,y\rangle \langle y,(A-B)x\rangle -\{\langle Ax,x\rangle ,\langle Bx,x\rangle \},\hfill \\ \hfill & \forall y\in \mathcal{E}\mathrm{satisfying}\parallel y\parallel \le 1\mathrm{and}\langle x,y\rangle =0.\hfill \end{array}$$

(iv)

$$\begin{array}{cc}\hfill {\Delta }_x{\left(A\right)}^2+{\Delta }_x{\left(B\right)}^2& \ge \parallel \langle \{A,B\}x,x\rangle +\langle (A-B)x,y\rangle \langle y,(A-B)x\rangle -\{\langle Ax,x\rangle ,\langle Bx,x\rangle \}\parallel ,\hfill \\ \hfill & \forall y\in \mathcal{E}\mathrm{satisfying}\parallel y\parallel \le 1\mathrm{and}\langle x,y\rangle =0.\hfill \end{array}$$

(v)

$$\begin{array}{cc}\hfill d_x{\left(A\right)}^2+d_x{\left(B\right)}^2& \ge {\displaystyle \frac{\langle (A+B)x,y\rangle \langle y,(A+B)x\rangle +\langle (A-B)x,y\rangle \langle y,(A-B)x\rangle }{2}},\hfill \\ \hfill & \forall y\in \mathcal{E}\mathrm{satisfying}\parallel y\parallel \le 1\mathrm{and}\langle x,y\rangle =0.\hfill \end{array}$$

(vi)

$$\begin{array}{cc}\hfill {\Delta }_x{\left(A\right)}^2+{\Delta }_x{\left(B\right)}^2& \ge {\displaystyle \frac{\parallel \langle (A+B)x,y\rangle \langle y,(A+B)x\rangle +\langle (A-B)x,y\rangle \langle y,(A-B)x\rangle \parallel }{2}},\hfill \\ \hfill & \forall y\in \mathcal{E}\mathrm{satisfying}\parallel y\parallel \le 1\mathrm{and}\langle x,y\rangle =0.\hfill \end{array}$$

(vii)

$$\begin{array}{cc}\hfill d_x{\left(A\right)}^2+d_x{\left(B\right)}^2& \ge -i\langle [A,B]x,x\rangle +\langle (A+iB)x,y\rangle \langle y,(A+iB)x\rangle -i[\langle Ax,x\rangle ,\langle Bx,x\rangle ],\hfill \\ \hfill & \forall y\in \mathcal{E}\mathrm{satisfying}\parallel y\parallel \le 1\mathrm{and}\langle x,y\rangle =0.\hfill \end{array}$$

(viii)

$$\begin{array}{cc}\hfill {\Delta }_x{\left(A\right)}^2+{\Delta }_x{\left(B\right)}^2& \ge \parallel -i\langle [A,B]x,x\rangle +\langle (A+iB)x,y\rangle \langle y,(A+iB)x\rangle -i[\langle Ax,x\rangle ,\langle Bx,x\rangle ]\parallel ,\hfill \\ \hfill & \forall y\in \mathcal{E}\mathrm{satisfying}\parallel y\parallel \le 1\mathrm{and}\langle x,y\rangle =0.\hfill \end{array}$$

(ix)

$$\begin{array}{cc}\hfill d_x{\left(A\right)}^2+d_x{\left(B\right)}^2& \ge i\langle [A,B]x,x\rangle +\langle (A-iB)x,y\rangle \langle y,(A-iB)x\rangle +i[\langle Ax,x\rangle ,\langle Bx,x\rangle ],\hfill \\ \hfill & \forall y\in \mathcal{E}\mathrm{satisfying}\parallel y\parallel \le 1\mathrm{and}\langle x,y\rangle =0.\hfill \end{array}$$

(x)

$$\begin{array}{cc}\hfill {\Delta }_x{\left(A\right)}^2+{\Delta }_x{\left(B\right)}^2& \ge \parallel i\langle [A,B]x,x\rangle +\langle (A-iB)x,y\rangle \langle y,(A-iB)x\rangle +i[\langle Ax,x\rangle ,\langle Bx,x\rangle ]\parallel ,\hfill \\ \hfill & \forall y\in \mathcal{E}\mathrm{satisfying}\parallel y\parallel \le 1\mathrm{and}\langle x,y\rangle =0.\hfill \end{array}$$

(xi)

$$\begin{array}{cc}\hfill d_x{\left(A\right)}^2+d_x{\left(B\right)}^2& \ge {\displaystyle \frac{\langle (A+iB)x,y\rangle \langle y,(A+iB)x\rangle +\langle (A-iB)x,y\rangle \langle y,(A-iB)x\rangle }{2}},\hfill \\ \hfill & \forall y\in \mathcal{E}\mathrm{satisfying}\parallel y\parallel \le 1\mathrm{and}\langle x,y\rangle =0.\hfill \end{array}$$

(xii)

$$\begin{array}{cc}\hfill {\Delta }_x{\left(A\right)}^2+{\Delta }_x{\left(B\right)}^2& \ge {\displaystyle \frac{\parallel \langle (A+iB)x,y\rangle \langle y,(A+iB)x\rangle +\langle (A-iB)x,y\rangle \langle y,(A-iB)x\rangle \parallel }{2}},\hfill \\ \hfill & \forall y\in \mathcal{E}\mathrm{satisfying}\parallel y\parallel \le 1\mathrm{and}\langle x,y\rangle =0.\hfill \end{array}$$

Proof. 

Let $x\in \mathcal{D}\left(AB\right)\cap \mathcal{D}\left(BA\right)$ with $\langle x,x\rangle =1$. Let $y\in \mathcal{E}$ be such that $\parallel y\parallel \le 1$ and $\langle x,y\rangle =0.$

(1)

$$\begin{array}{cc}\hfill & \langle (A+B)x,y\rangle \langle y,(A+B)x\rangle \hfill \\ \hfill & =\langle Ax-\langle Ax,x\rangle x+Bx-\langle Bx,x\rangle x,y\rangle \langle y,Ax-\langle Ax,x\rangle x+Bx-\langle Bx,x\rangle x\rangle \hfill \\ \hfill & \le \parallel \langle y,y\rangle \parallel \langle (Ax-\langle Ax,x\rangle x)+(Bx-\langle Bx,x\rangle x),(Ax-\langle Ax,x\rangle x)+(Bx-\langle Bx,x\rangle x)\rangle \hfill \\ \hfill & \le \langle Ax-\langle Ax,x\rangle x,Ax-\langle Ax,x\rangle x\rangle +\langle Bx-\langle Bx,x\rangle x,Bx-\langle Bx,x\rangle x\rangle \hfill \\ \hfill & +\langle Bx-\langle Bx,x\rangle x,Ax-\langle Ax,x\rangle x\rangle +\langle Ax-\langle Ax,x\rangle x,Bx-\langle Bx,x\rangle x\rangle \hfill \\ \hfill & =d_x{\left(A\right)}^2+d_x{\left(B\right)}^2+\langle Bx,Ax\rangle -\langle Bx,x\rangle \langle Ax,x\rangle +\langle Ax,Bx\rangle -\langle Ax,x\rangle \langle Bx,x\rangle \hfill \\ \hfill & =d_x{\left(A\right)}^2+d_x{\left(B\right)}^2+\langle \{A,B\}x,x\rangle -\{\langle Ax,x\rangle ,\langle Bx,x\rangle \}.\hfill \end{array}$$

Therefore

$$\begin{array}{c}\hfill d_x{\left(A\right)}^2+d_x{\left(B\right)}^2\ge -\langle \{A,B\}x,x\rangle +\langle (A+B)x,y\rangle \langle y,(A+B)x\rangle +\{\langle Ax,x\rangle ,\langle Bx,x\rangle \}.\end{array}$$

(ii)

Take norm on (i).

(iii)

Similar to (i).

(iv)

Take norm on (iii).

(v)

$$\begin{array}{cc}\hfill & \langle (A+B)x,y\rangle \langle y,(A+B)x\rangle +\langle (A-B)x,y\rangle \langle y,(A-B)x\rangle =2\langle Ax,y\rangle \langle y,Ax\rangle +2\langle Bx,y\rangle \langle y,Bx\rangle \hfill \\ \hfill & =2\langle Ax-\langle Ax,x\rangle x,y\rangle \langle y,Ax-\langle Ax,x\rangle x\rangle +2\langle Bx-\langle Bx,x\rangle x,y\rangle \langle y,Bx-\langle Bx,x\rangle x\rangle \hfill \\ \hfill & \le 2\parallel \langle y,y\rangle \parallel \langle Ax-\langle Ax,x\rangle x,Ax-\langle Ax,x\rangle x\rangle +2\parallel \langle y,y\rangle \parallel \langle Bx-\langle Bx,x\rangle x,Bx-\langle Bx,x\rangle x\rangle \hfill \\ \hfill & \le 2d_x{\left(A\right)}^2+2d_x{\left(B\right)}^2.\hfill \end{array}$$

(vi)

Take norm on (v).

(vii)

$$\begin{array}{cc}\hfill & \langle (A+iB)x,y\rangle \langle y,(A+iB)x\rangle =\langle Ax+iBx,y\rangle \langle y,Ax+iBx\rangle \hfill \\ \hfill & =\langle Ax-\langle Ax,x\rangle x+i(Bx-\langle Bx,x\rangle x),y\rangle \langle y,Ax-\langle Ax,x\rangle x+i(Bx-\langle Bx,x\rangle x)\rangle \hfill \\ \hfill & \le \parallel \langle y,y\rangle \parallel \langle (Ax-\langle Ax,x\rangle x)+i(Bx-\langle Bx,x\rangle x),(Ax-\langle Ax,x\rangle x)+i(Bx-\langle Bx,x\rangle x)\rangle \hfill \\ \hfill & \le \langle Ax-\langle Ax,x\rangle x,Ax-\langle Ax,x\rangle x\rangle +\langle Bx-\langle Bx,x\rangle x,Bx-\langle Bx,x\rangle x\rangle \hfill \\ \hfill & +i\langle Bx-\langle Bx,x\rangle x,Ax-\langle Ax,x\rangle x\rangle -i\langle Ax-\langle Ax,x\rangle x,Bx-\langle Bx,x\rangle x\rangle \hfill \\ \hfill & =d_x{\left(A\right)}^2+d_x{\left(B\right)}^2+i(\langle Bx,Ax\rangle -\langle Bx,x\rangle \langle Ax,x\rangle )-i(\langle Ax,Bx\rangle -\langle Ax,x\rangle \langle Bx,x\rangle )\hfill \\ \hfill & =d_x{\left(A\right)}^2+d_x{\left(B\right)}^2+i\langle [A,B]x,x\rangle +i[\langle Ax,x\rangle ,\langle Bx,x\rangle ].\hfill \end{array}$$

Therefore$\begin{array}{c}\hfill d_x{\left(A\right)}^2+d_x{\left(B\right)}^2\ge -i\langle [A,B]x,x\rangle +\langle (A+iB)x,y\rangle \langle y,(A+iB)x\rangle -i[\langle Ax,x\rangle ,\langle Bx,x\rangle ].\end{array}$

(viii)

Take norm on (vii).

(ix)

Similar to (vii).

(x)

Take norm on (ix).

(xi)

Add (vii) and (ix).

(xii)

Add (viii) and (ix), then take norm.

□

Corollary 1.

Let $\mathcal{E}$ be a Hilbert C*-module over a unital commutative C*-algebra $\mathcal{A}$. Let $A:\mathcal{D}\left(A\right)\to \mathcal{E}$ and $B:\mathcal{D}\left(B\right)\to \mathcal{E}$ be self-adjoint morphisms. Then for all $x\in \mathcal{D}\left(AB\right)\cap \mathcal{D}\left(BA\right)$ with $\langle x,x\rangle =1$, we have

(1)

If $\mathcal{A}$ is commutative, then

$$\begin{array}{cc}\hfill d_x{\left(A\right)}^2+d_x{\left(B\right)}^2& \ge -i\langle [A,B]x,x\rangle +\langle (A+iB)x,y\rangle \langle y,(A+iB)x\rangle ,\hfill \\ \hfill & \forall y\in \mathcal{E}\mathrm{satisfying}\parallel y\parallel \le 1\mathrm{and}\langle x,y\rangle =0.\hfill \end{array}$$

(2)

If $\mathcal{A}$ is commutative, then

$$\begin{array}{cc}\hfill {\Delta }_x{\left(A\right)}^2+{\Delta }_x{\left(B\right)}^2& \ge \parallel -i\langle [A,B]x,x\rangle +\langle (A+iB)x,y\rangle \langle y,(A+iB)x\rangle \parallel ,\hfill \\ \hfill & \forall y\in \mathcal{E}\mathrm{satisfying}\parallel y\parallel \le 1\mathrm{and}\langle x,y\rangle =0.\hfill \end{array}$$

We also have following uncertainty principles.

Theorem 6.(Noncommutative Maccone-Pati Uncertainty Principles) Let $\mathcal{E}$ be a Hilbert C*-module over a unital C*-algebra $\mathcal{A}$. Let $A:\mathcal{D}\left(A\right)\to \mathcal{E}$ and $B:\mathcal{D}\left(B\right)\to \mathcal{E}$ be self-adjoint morphisms. Then for all $x\in \mathcal{D}\left(AB\right)\cap \mathcal{D}\left(BA\right)$ with $\langle x,x\rangle =1$, we have

(i)

$$\begin{array}{cc}\hfill {\Delta }_x{\left(A\right)}^2+{\Delta }_x{\left(B\right)}^2& \ge \langle y,(A+B)x\rangle \langle (A+B)x,y\rangle -\parallel \langle \{A,B\}x,x\rangle -\{\langle Ax,x\rangle ,\langle Bx,x\rangle \}\parallel ,\hfill \\ \hfill & \forall y\in \mathcal{E}\mathrm{satisfying}\parallel y\parallel \le 1\mathrm{and}\langle x,y\rangle =0.\hfill \end{array}$$

(ii)

$$\begin{array}{cc}\hfill {\Delta }_x{\left(A\right)}^2+{\Delta }_x{\left(B\right)}^2& \ge \langle y,(A-B)x\rangle \langle (A-B)x,y\rangle -\parallel \langle \{A,B\}x,x\rangle -\{\langle Ax,x\rangle ,\langle Bx,x\rangle \}\parallel \hfill \\ \hfill & \forall y\in \mathcal{E}\mathrm{satisfying}\parallel y\parallel \le 1\mathrm{and}\langle x,y\rangle =0.\hfill \end{array}$$

(iii)

$$\begin{array}{cc}\hfill {\Delta }_x{\left(A\right)}^2+{\Delta }_x{\left(B\right)}^2& \ge {\displaystyle \frac{\parallel \langle y,(A+B)x\rangle \langle (A+B)x,y\rangle +\langle (A-B)x,y\rangle \langle y,(A-B)x\rangle \parallel }{2}},\hfill \\ \hfill & \forall y\in \mathcal{E}\mathrm{satisfying}\parallel y\parallel \le 1\mathrm{and}\langle x,y\rangle =0.\hfill \end{array}$$

(iv)

$$\begin{array}{cc}\hfill {\Delta }_x{\left(A\right)}^2+{\Delta }_x{\left(B\right)}^2& \ge \langle y,(A+iB)x\rangle \langle (A+iB)x,y\rangle -\parallel \langle [A,B]x,x\rangle +[\langle Ax,x\rangle ,\langle Bx,x\rangle ]\parallel ,\hfill \\ \hfill & \forall y\in \mathcal{E}\mathrm{satisfying}\parallel y\parallel \le 1\mathrm{and}\langle x,y\rangle =0.\hfill \end{array}$$

(v)

$$\begin{array}{cc}\hfill {\Delta }_x{\left(A\right)}^2+{\Delta }_x{\left(B\right)}^2& \ge \langle y,(A-iB)x\rangle \langle (A-iB)x,y\rangle -\parallel \langle [A,B]x,x\rangle +[\langle Ax,x\rangle ,\langle Bx,x\rangle ]\parallel ,\hfill \\ \hfill & \forall y\in \mathcal{E}\mathrm{satisfying}\parallel y\parallel \le 1\mathrm{and}\langle x,y\rangle =0.\hfill \end{array}$$

Proof. 

Let $x\in \mathcal{D}\left(AB\right)\cap \mathcal{D}\left(BA\right)$ with $\langle x,x\rangle =1$. Let $y\in \mathcal{E}$ be such that $\parallel y\parallel \le 1$ and $\langle x,y\rangle =0.$

(i)

$$\begin{array}{cc}\hfill & \langle y,(A+B)x\rangle \langle (A+B)x,y\rangle \hfill \\ \hfill & =\langle y,Ax-\langle Ax,x\rangle x+(Bx-\langle Bx,x\rangle x)\rangle \langle Ax-\langle Ax,x\rangle x+(Bx-\langle Bx,x\rangle x),y\rangle \hfill \\ \hfill & \le \parallel \langle (Ax-\langle Ax,x\rangle x)+(Bx-\langle Bx,x\rangle x),(Ax-\langle Ax,x\rangle x)+(Bx-\langle Bx,x\rangle x)\rangle \parallel \langle y,y\rangle \hfill \\ \hfill & \le \parallel \langle (Ax-\langle Ax,x\rangle x)+(Bx-\langle Bx,x\rangle x),(Ax-\langle Ax,x\rangle x)+(Bx-\langle Bx,x\rangle x)\rangle \parallel \parallel \langle y,y\rangle \parallel \hfill \\ \hfill & \le \parallel \langle Ax-\langle Ax,x\rangle x,Ax-\langle Ax,x\rangle x\rangle +\langle Bx-\langle Bx,x\rangle x,Bx-\langle Bx,x\rangle x\rangle \hfill \\ \hfill & +\langle Bx-\langle Bx,x\rangle x,Ax-\langle Ax,x\rangle x\rangle +\langle Ax-\langle Ax,x\rangle x,Bx-\langle Bx,x\rangle x\rangle \parallel \hfill \\ \hfill & =\parallel d_x{\left(A\right)}^2+d_x{\left(B\right)}^2+(\langle Bx,Ax\rangle -\langle Bx,x\rangle \langle Ax,x\rangle )+\langle Ax,Bx\rangle -\langle Ax,x\rangle \langle Bx,x\rangle \parallel \hfill \\ \hfill & \le \parallel d_x{\left(A\right)}^2+d_x{\left(B\right)}^2\parallel +\parallel \langle \{A,B\}x,x\rangle -\{\langle Ax,x\rangle ,\langle Bx,x\rangle \}\parallel .\hfill \end{array}$$

Therefore$\begin{array}{c}\hfill {\Delta }_x{\left(A\right)}^2+{\Delta }_x{\left(B\right)}^2\ge \parallel d_x{\left(A\right)}^2+d_x{\left(B\right)}^2\parallel \ge \langle y,(A+B)x\rangle \langle (A+B)x,y\rangle -\parallel \langle \{A,B\}x,x\rangle -\{\langle Ax,x\rangle ,\langle Bx,x\rangle \}\parallel .\end{array}$

(ii)

Similar to (i).

(iii)

$$\begin{array}{cc}\hfill & \langle y,(A+B)x\rangle \langle (A+B)x,y\rangle +\langle y,(A-B)x\rangle \langle (A-B)x,y\rangle =2\langle y,Ax\rangle \langle Ax,y\rangle +2\langle y,Bx\rangle \langle Bx,y\rangle \hfill \\ \hfill & =2\langle y,Ax-\langle Ax,x\rangle x\rangle \langle Ax-\langle Ax,x\rangle x,y\rangle +2\langle y,Bx-\langle Bx,x\rangle x\rangle \langle Bx-\langle Bx,x\rangle x,y\rangle \hfill \\ \hfill & \le 2\parallel \langle Ax-\langle Ax,x\rangle x,Ax-\langle Ax,x\rangle x\rangle \parallel \langle y,y\rangle +2\parallel \langle Bx-\langle Bx,x\rangle x,Bx-\langle Bx,x\rangle x\rangle \parallel \langle y,y\rangle \hfill \\ \hfill & \le 2{\Delta }_x{\left(A\right)}^2+2{\Delta }_x{\left(B\right)}^2.\hfill \end{array}$$

(iv)

$$\begin{array}{cc}\hfill & \langle y,(A+iB)x\rangle \langle (A+iB)x,y\rangle =\langle y,Ax+iBx\rangle \langle Ax+iBx,y\rangle \hfill \\ \hfill & =\langle y,Ax-\langle Ax,x\rangle x+i(Bx-\langle Bx,x\rangle x)\rangle \langle Ax-\langle Ax,x\rangle x+i(Bx-\langle Bx,x\rangle x),y\rangle \hfill \\ \hfill & \le \parallel \langle (Ax-\langle Ax,x\rangle x)+i(Bx-\langle Bx,x\rangle x),(Ax-\langle Ax,x\rangle x)+i(Bx-\langle Bx,x\rangle x)\rangle \parallel \langle y,y\rangle \hfill \\ \hfill & \le \parallel \langle Ax-\langle Ax,x\rangle x,Ax-\langle Ax,x\rangle x\rangle +\langle Bx-\langle Bx,x\rangle x,Bx-\langle Bx,x\rangle x\rangle \hfill \\ \hfill & +i\langle Bx-\langle Bx,x\rangle x,Ax-\langle Ax,x\rangle x\rangle -i\langle Ax-\langle Ax,x\rangle x,Bx-\langle Bx,x\rangle x\rangle \parallel \hfill \\ \hfill & =\parallel d_x{\left(A\right)}^2+d_x{\left(B\right)}^2+i(\langle Bx,Ax\rangle -\langle Bx,x\rangle \langle Ax,x\rangle )-i(\langle Ax,Bx\rangle -\langle Ax,x\rangle \langle Bx,x\rangle )\parallel \hfill \\ \hfill & =\parallel d_x{\left(A\right)}^2+d_x{\left(B\right)}^2\parallel +\parallel (\langle Bx,Ax\rangle -\langle Bx,x\rangle \langle Ax,x\rangle )-(\langle Ax,Bx\rangle -\langle Ax,x\rangle \langle Bx,x\rangle )\parallel \hfill \\ \hfill & =\parallel d_x{\left(A\right)}^2+d_x{\left(B\right)}^2\parallel +\parallel \langle [A,B]x,x\rangle +[\langle Ax,x\rangle ,\langle Bx,x\rangle ]\parallel .\hfill \end{array}$$

Therefore

$$\begin{array}{cc}\hfill & {\Delta }_x{\left(A\right)}^2+{\Delta }_x{\left(B\right)}^2\ge \parallel d_x{\left(A\right)}^2+d_x{\left(B\right)}^2\parallel \hfill \\ \hfill & \ge \langle y,(A+iB)x\rangle \langle (A+iB)x,y\rangle -\parallel \langle [A,B]x,x\rangle +[\langle Ax,x\rangle ,\langle Bx,x\rangle ]\parallel .\hfill \end{array}$$

(v)

Similar to (iv).

□

Theorem 7.

Let $\mathcal{E}$ be a Hilbert C*-module over a unital commutative C*-algebra $\mathcal{A}$. Let $A:\mathcal{D}\left(A\right)\to \mathcal{E}$ and $B:\mathcal{D}\left(B\right)\to \mathcal{E}$ be self-adjoint morphisms. Then for all $x\in \mathcal{D}\left(AB\right)\cap \mathcal{D}\left(BA\right)$ with $\langle x,x\rangle =1$, we have

(i)

If $d_x\left(A\right)$, $d_x\left(B\right)$ are invertible, then

$$\begin{array}{cc}\hfill d_x\left(A\right)d_x\left(B\right)& \ge {\displaystyle \frac{-i\langle [A,B]x,x\rangle }{2-〈\left(\frac{A}{d_x\left(A\right)}+\frac{iB}{d_x\left(B\right)}\right)x,y〉〈y,\left(\frac{A}{d_x\left(A\right)}+\frac{iB}{d_x\left(B\right)}\right)x〉}},\hfill \\ \hfill & \forall y\in \mathcal{E}\mathrm{satisfying}\parallel y\parallel \le 1\mathrm{and}\langle x,y\rangle =0.\hfill \end{array}$$

(ii)

$$\begin{array}{cc}\hfill {\Delta }_x\left(A\right){\Delta }_x\left(B\right)& \ge {\displaystyle \frac{-i\langle [A,B]x,x\rangle }{2-〈\left(\frac{A}{{\Delta }_x\left(A\right)}+\frac{iB}{{\Delta }_x\left(B\right)}\right)x,y〉〈y,\left(\frac{A}{{\Delta }_x\left(A\right)}+\frac{iB}{{\Delta }_x\left(B\right)}\right)x〉}},\hfill \\ \hfill & \forall y\in \mathcal{E}\mathrm{satisfying}\parallel y\parallel \le 1\mathrm{and}\langle x,y\rangle =0.\hfill \end{array}$$

(iii)

If $d_x\left(A\right)$, $d_x\left(B\right)$ are invertible, then

$$\begin{array}{cc}\hfill d_x\left(A\right)d_x\left(B\right)& \ge {\displaystyle \frac{i\langle [A,B]x,x\rangle }{2-〈\left(\frac{A}{d_x\left(A\right)}-\frac{iB}{d_x\left(B\right)}\right)x,y〉〈y,\left(\frac{A}{d_x\left(A\right)}-\frac{iB}{d_x\left(B\right)}\right)x〉}},\hfill \\ \hfill & \forall y\in \mathcal{E}\mathrm{satisfying}\parallel y\parallel \le 1\mathrm{and}\langle x,y\rangle =0.\hfill \end{array}$$

(iv)

$$\begin{array}{cc}\hfill {\Delta }_x\left(A\right){\Delta }_x\left(B\right)& \ge {\displaystyle \frac{i\langle [A,B]x,x\rangle }{2-〈\left(\frac{A}{{\Delta }_x\left(A\right)}-\frac{iB}{{\Delta }_x\left(B\right)}\right)x,y〉〈y,\left(\frac{A}{{\Delta }_x\left(A\right)}-\frac{iB}{{\Delta }_x\left(B\right)}\right)x〉}},\hfill \\ \hfill & \forall y\in \mathcal{E}\mathrm{satisfying}\parallel y\parallel \le 1\mathrm{and}\langle x,y\rangle =0.\hfill \end{array}$$

(v)

If $d_x\left(A\right)$, $d_x\left(B\right)$ are invertible, then

$$\begin{array}{cc}\hfill d_x\left(A\right)d_x\left(B\right)& \ge {\displaystyle \frac{2\langle Ax,x\rangle \langle Bx,x\rangle -\langle \{A,B\}x,x\rangle }{2-〈\left(\frac{A}{d_x\left(A\right)}+\frac{B}{d_x\left(B\right)}\right)x,y〉〈y,\left(\frac{A}{d_x\left(A\right)}+\frac{B}{d_x\left(B\right)}\right)x〉}},\hfill \\ \hfill & \forall y\in \mathcal{E}\mathrm{satisfying}\parallel y\parallel \le 1\mathrm{and}\langle x,y\rangle =0.\hfill \end{array}$$

(vi)

$$\begin{array}{cc}\hfill {\Delta }_x\left(A\right){\Delta }_x\left(B\right)& \ge {\displaystyle \frac{2\langle Ax,x\rangle \langle Bx,x\rangle -\langle \{A,B\}x,x\rangle }{2-〈\left(\frac{A}{{\Delta }_x\left(A\right)}+\frac{B}{{\Delta }_x\left(B\right)}\right)x,y〉〈y,\left(\frac{A}{{\Delta }_x\left(A\right)}+\frac{B}{{\Delta }_x\left(B\right)}\right)x〉}},\hfill \\ \hfill & \forall y\in \mathcal{E}\mathrm{satisfying}\parallel y\parallel \le 1\mathrm{and}\langle x,y\rangle =0.\hfill \end{array}$$

(vii)

If $d_x\left(A\right)$, $d_x\left(B\right)$ are invertible, then

$$\begin{array}{cc}\hfill d_x\left(A\right)d_x\left(B\right)& \ge {\displaystyle \frac{-2\langle Ax,x\rangle \langle Bx,x\rangle +\langle \{A,B\}x,x\rangle }{2-〈\left(\frac{A}{d_x\left(A\right)}-\frac{B}{d_x\left(B\right)}\right)x,y〉〈y,\left(\frac{A}{d_x\left(A\right)}-\frac{B}{d_x\left(B\right)}\right)x〉}},\hfill \\ \hfill & \forall y\in \mathcal{E}\mathrm{satisfying}\parallel y\parallel \le 1\mathrm{and}\langle x,y\rangle =0.\hfill \end{array}$$

(viii)

$$\begin{array}{cc}\hfill {\Delta }_x\left(A\right){\Delta }_x\left(B\right)& \ge {\displaystyle \frac{-2\langle Ax,x\rangle \langle Bx,x\rangle +\langle \{A,B\}x,x\rangle }{2-〈\left(\frac{A}{{\Delta }_x\left(A\right)}-\frac{B}{{\Delta }_x\left(B\right)}\right)x,y〉〈y,\left(\frac{A}{{\Delta }_x\left(A\right)}-\frac{B}{{\Delta }_x\left(B\right)}\right)x〉}},\hfill \\ \hfill & \forall y\in \mathcal{E}\mathrm{satisfying}\parallel y\parallel \le 1\mathrm{and}\langle x,y\rangle =0.\hfill \end{array}$$

Proof. 

Let $x\in \mathcal{D}\left(AB\right)\cap \mathcal{D}\left(BA\right)$ with $\langle x,x\rangle =1$. Let $y\in \mathcal{E}$ be such that $\parallel y\parallel \le 1$ and $\langle x,y\rangle =0.$

(i)

Since $\mathcal{A}$ is commutative and $d_x\left(A\right)$, $d_x\left(B\right)$ are invertible,

$$\begin{array}{cc}\hfill & 〈\left({\displaystyle \frac{A}{d_x\left(A\right)}}+{\displaystyle \frac{iB}{d_x\left(B\right)}}\right)x,y〉〈y,\left({\displaystyle \frac{A}{d_x\left(A\right)}}+{\displaystyle \frac{iB}{d_x\left(B\right)}}\right)x〉\hfill \\ \hfill & =〈{\displaystyle \frac{Ax-\langle Ax,x\rangle x}{d_x\left(A\right)}}+{\displaystyle \frac{i(Bx-\langle Bx,x\rangle x)}{d_x\left(B\right)}},y〉〈y,{\displaystyle \frac{Ax-\langle Ax,x\rangle x}{d_x\left(A\right)}}+{\displaystyle \frac{i(Bx-\langle Bx,x\rangle x)}{d_x\left(B\right)}}〉\hfill \\ \hfill & \le \parallel \langle y,y\rangle \parallel 〈{\displaystyle \frac{Ax-\langle Ax,x\rangle x}{d_x\left(A\right)}}+{\displaystyle \frac{i(Bx-\langle Bx,x\rangle x)}{d_x\left(B\right)}},{\displaystyle \frac{Ax-\langle Ax,x\rangle x}{d_x\left(A\right)}}+{\displaystyle \frac{i(Bx-\langle Bx,x\rangle x)}{d_x\left(B\right)}}〉\hfill \\ \hfill & \le 〈{\displaystyle \frac{Ax-\langle Ax,x\rangle x}{d_x\left(A\right)}},{\displaystyle \frac{Ax-\langle Ax,x\rangle x}{d_x\left(A\right)}}〉+〈{\displaystyle \frac{Bx-\langle Bx,x\rangle x}{d_x\left(B\right)}},{\displaystyle \frac{Bx-\langle Bx,x\rangle x}{d_x\left(B\right)}}〉\hfill \\ \hfill & +i〈{\displaystyle \frac{Bx-\langle Bx,x\rangle x}{d_x\left(B\right)}},{\displaystyle \frac{Ax-\langle Ax,x\rangle x}{d_x\left(A\right)}}〉-i〈{\displaystyle \frac{Ax-\langle Ax,x\rangle x}{d_x\left(A\right)}},{\displaystyle \frac{Bx-\langle Bx,x\rangle x}{d_x\left(B\right)}}〉\hfill \\ \hfill & =1+1+{\displaystyle \frac{i\langle Bx,Ax\rangle -i\langle Bx,x\rangle \langle Ax,x\rangle }{d_x\left(A\right)d_x\left(B\right)}}+{\displaystyle \frac{-i\langle Ax,Bx\rangle +i\langle Ax,x\rangle \langle Bx,x\rangle }{d_x\left(A\right)d_x\left(B\right)}}\hfill \\ \hfill & =2+{\displaystyle \frac{i\langle [A,B]x,x\rangle }{d_x\left(A\right)d_x\left(B\right)}}.\hfill \end{array}$$

Therefore

$$\begin{array}{cc}\hfill & {\displaystyle \frac{-i\langle [A,B]x,x\rangle }{d_x\left(A\right)d_x\left(B\right)}}\le 2-〈\left({\displaystyle \frac{A}{d_x\left(A\right)}}+{\displaystyle \frac{iB}{d_x\left(B\right)}}\right)x,y〉〈y,\left({\displaystyle \frac{A}{d_x\left(A\right)}}+{\displaystyle \frac{iB}{d_x\left(B\right)}}\right)x〉\hfill \\ \hfill & \Rightarrow {\displaystyle \frac{-i\langle [A,B]x,x\rangle }{2-〈\left(\frac{A}{d_x\left(A\right)}+\frac{iB}{d_x\left(B\right)}\right)x,y〉〈y,\left(\frac{A}{d_x\left(A\right)}+\frac{iB}{d_x\left(B\right)}\right)x〉}}\le d_x\left(A\right)d_x\left(B\right).\hfill \end{array}$$

(ii)

Since $\mathcal{A}$ is commutative,

$$\begin{array}{cc}\hfill & 〈\left({\displaystyle \frac{A}{{\Delta }_x\left(A\right)}}+{\displaystyle \frac{iB}{{\Delta }_x\left(B\right)}}\right)x,y〉〈y,\left({\displaystyle \frac{A}{{\Delta }_x\left(A\right)}}+{\displaystyle \frac{iB}{{\Delta }_x\left(B\right)}}\right)x〉\hfill \\ \hfill & =〈{\displaystyle \frac{Ax-\langle Ax,x\rangle x}{{\Delta }_x\left(A\right)}}+{\displaystyle \frac{i(Bx-\langle Bx,x\rangle x)}{{\Delta }_x\left(B\right)}},y〉〈y,{\displaystyle \frac{Ax-\langle Ax,x\rangle x}{{\Delta }_x\left(A\right)}}+{\displaystyle \frac{i(Bx-\langle Bx,x\rangle x)}{{\Delta }_x\left(B\right)}}〉\hfill \\ \hfill & \le \parallel \langle y,y\rangle \parallel 〈{\displaystyle \frac{Ax-\langle Ax,x\rangle x}{{\Delta }_x\left(A\right)}}+{\displaystyle \frac{i(Bx-\langle Bx,x\rangle x)}{{\Delta }_x\left(B\right)}},{\displaystyle \frac{Ax-\langle Ax,x\rangle x}{{\Delta }_x\left(A\right)}}+{\displaystyle \frac{i(Bx-\langle Bx,x\rangle x)}{{\Delta }_x\left(B\right)}}〉\hfill \\ \hfill & \le 〈{\displaystyle \frac{Ax-\langle Ax,x\rangle x}{{\Delta }_x\left(A\right)}},{\displaystyle \frac{Ax-\langle Ax,x\rangle x}{{\Delta }_x\left(A\right)}}〉+〈{\displaystyle \frac{Bx-\langle Bx,x\rangle x}{{\Delta }_x\left(B\right)}},{\displaystyle \frac{Bx-\langle Bx,x\rangle x}{{\Delta }_x\left(B\right)}}〉\hfill \\ \hfill & +i〈{\displaystyle \frac{Bx-\langle Bx,x\rangle x}{{\Delta }_x\left(B\right)}},{\displaystyle \frac{Ax-\langle Ax,x\rangle x}{{\Delta }_x\left(A\right)}}〉-i〈{\displaystyle \frac{Ax-\langle Ax,x\rangle x}{{\Delta }_x\left(A\right)}},{\displaystyle \frac{Bx-\langle Bx,x\rangle x}{{\Delta }_x\left(B\right)}}〉\hfill \\ \hfill & =1+1+{\displaystyle \frac{i\langle Bx,Ax\rangle -i\langle Bx,x\rangle \langle Ax,x\rangle }{{\Delta }_x\left(A\right){\Delta }_x\left(B\right)}}+{\displaystyle \frac{-i\langle Ax,Bx\rangle +i\langle Ax,x\rangle \langle Bx,x\rangle }{{\Delta }_x\left(A\right){\Delta }_x\left(B\right)}}\hfill \\ \hfill & =2+{\displaystyle \frac{i\langle [A,B]x,x\rangle }{{\Delta }_x\left(A\right){\Delta }_x\left(B\right)}}.\hfill \end{array}$$

Therefore

$$\begin{array}{cc}\hfill & {\displaystyle \frac{-i\langle [A,B]x,x\rangle }{{\Delta }_x\left(A\right){\Delta }_x\left(B\right)}}\le 2-〈\left({\displaystyle \frac{A}{{\Delta }_x\left(A\right)}}+{\displaystyle \frac{iB}{{\Delta }_x\left(B\right)}}\right)x,y〉〈y,\left({\displaystyle \frac{A}{{\Delta }_x\left(A\right)}}+{\displaystyle \frac{iB}{{\Delta }_x\left(B\right)}}\right)x〉\hfill \\ \hfill & \Rightarrow {\displaystyle \frac{-i\langle [A,B]x,x\rangle }{2-〈\left(\frac{A}{{\Delta }_x\left(A\right)}+\frac{iB}{{\Delta }_x\left(B\right)}\right)x,y〉〈y,\left(\frac{A}{{\Delta }_x\left(A\right)}+\frac{iB}{{\Delta }_x\left(B\right)}\right)x〉}}\le {\Delta }_x\left(A\right){\Delta }_x\left(B\right).\hfill \end{array}$$

(iii)

Similar to (i).

(iv)

Similar to (ii).

(v)

Since $\mathcal{A}$ is commutative and $d_x\left(A\right)$, $d_x\left(B\right)$ are invertible,

$$\begin{array}{cc}\hfill & 〈\left({\displaystyle \frac{A}{d_x\left(A\right)}}+{\displaystyle \frac{B}{d_x\left(B\right)}}\right)x,y〉〈y,\left({\displaystyle \frac{A}{d_x\left(A\right)}}+{\displaystyle \frac{B}{d_x\left(B\right)}}\right)x〉\hfill \\ \hfill & =〈{\displaystyle \frac{Ax-\langle Ax,x\rangle x}{d_x\left(A\right)}}+{\displaystyle \frac{Bx-\langle Bx,x\rangle x}{d_x\left(B\right)}},y〉〈y,{\displaystyle \frac{Ax-\langle Ax,x\rangle x}{d_x\left(A\right)}}+{\displaystyle \frac{Bx-\langle Bx,x\rangle x}{d_x\left(B\right)}}〉\hfill \\ \hfill & \le \parallel \langle y,y\rangle \parallel 〈{\displaystyle \frac{Ax-\langle Ax,x\rangle x}{d_x\left(A\right)}}+{\displaystyle \frac{Bx-\langle Bx,x\rangle x}{d_x\left(B\right)}},{\displaystyle \frac{Ax-\langle Ax,x\rangle x}{d_x\left(A\right)}}+{\displaystyle \frac{Bx-\langle Bx,x\rangle x}{d_x\left(B\right)}}〉\hfill \\ \hfill & \le 〈{\displaystyle \frac{Ax-\langle Ax,x\rangle x}{d_x\left(A\right)}},{\displaystyle \frac{Ax-\langle Ax,x\rangle x}{d_x\left(A\right)}}〉+〈{\displaystyle \frac{Bx-\langle Bx,x\rangle x}{d_x\left(B\right)}},{\displaystyle \frac{Bx-\langle Bx,x\rangle x}{d_x\left(B\right)}}〉\hfill \\ \hfill & +〈{\displaystyle \frac{Bx-\langle Bx,x\rangle x}{d_x\left(B\right)}},{\displaystyle \frac{Ax-\langle Ax,x\rangle x}{d_x\left(A\right)}}〉+〈{\displaystyle \frac{Ax-\langle Ax,x\rangle x}{d_x\left(A\right)}},{\displaystyle \frac{Bx-\langle Bx,x\rangle x}{d_x\left(B\right)}}〉\hfill \\ \hfill & =1+1+{\displaystyle \frac{\langle Bx,Ax\rangle -\langle Bx,x\rangle \langle Ax,x\rangle }{d_x\left(A\right)d_x\left(B\right)}}+{\displaystyle \frac{\langle Ax,Bx\rangle -\langle Ax,x\rangle \langle Bx,x\rangle }{d_x\left(A\right)d_x\left(B\right)}}\hfill \\ \hfill & =2+{\displaystyle \frac{-2\langle Ax,x\rangle \langle Bx,x\rangle +\langle \{A,B\}x,x\rangle }{d_x\left(A\right)d_x\left(B\right)}}.\hfill \end{array}$$

Therefore

$$\begin{array}{cc}\hfill & -{\displaystyle \frac{-2\langle Ax,x\rangle \langle Bx,x\rangle +\langle \{A,B\}x,x\rangle }{d_x\left(A\right)d_x\left(B\right)}}\le 2-〈\left({\displaystyle \frac{A}{d_x\left(A\right)}}+{\displaystyle \frac{B}{d_x\left(B\right)}}\right)x,y〉〈y,\left({\displaystyle \frac{A}{d_x\left(A\right)}}+{\displaystyle \frac{B}{d_x\left(B\right)}}\right)x〉\hfill \\ \hfill & \Rightarrow {\displaystyle \frac{2\langle Ax,x\rangle \langle Bx,x\rangle -\langle \{A,B\}x,x\rangle }{2-〈\left(\frac{A}{d_x\left(A\right)}+\frac{B}{d_x\left(B\right)}\right)x,y〉〈y,\left(\frac{A}{d_x\left(A\right)}+\frac{B}{d_x\left(B\right)}\right)x〉}}\le d_x\left(A\right)d_x\left(B\right).\hfill \end{array}$$

(vi)

Since $\mathcal{A}$ is commutative,

$$\begin{array}{cc}\hfill & 〈\left({\displaystyle \frac{A}{{\Delta }_x\left(A\right)}}+{\displaystyle \frac{B}{{\Delta }_x\left(B\right)}}\right)x,y〉〈y,\left({\displaystyle \frac{A}{{\Delta }_x\left(A\right)}}+{\displaystyle \frac{B}{{\Delta }_x\left(B\right)}}\right)x〉\hfill \\ \hfill & =〈{\displaystyle \frac{Ax-\langle Ax,x\rangle x}{{\Delta }_x\left(A\right)}}+{\displaystyle \frac{Bx-\langle Bx,x\rangle x}{{\Delta }_x\left(B\right)}},y〉〈y,{\displaystyle \frac{Ax-\langle Ax,x\rangle x}{{\Delta }_x\left(A\right)}}+{\displaystyle \frac{Bx-\langle Bx,x\rangle x}{{\Delta }_x\left(B\right)}}〉\hfill \\ \hfill & \le \parallel \langle y,y\rangle \parallel 〈{\displaystyle \frac{Ax-\langle Ax,x\rangle x}{{\Delta }_x\left(A\right)}}+{\displaystyle \frac{Bx-\langle Bx,x\rangle x}{{\Delta }_x\left(B\right)}},{\displaystyle \frac{Ax-\langle Ax,x\rangle x}{{\Delta }_x\left(A\right)}}+{\displaystyle \frac{Bx-\langle Bx,x\rangle x}{{\Delta }_x\left(B\right)}}〉\hfill \\ \hfill & \le 〈{\displaystyle \frac{Ax-\langle Ax,x\rangle x}{{\Delta }_x\left(A\right)}},{\displaystyle \frac{Ax-\langle Ax,x\rangle x}{{\Delta }_x\left(A\right)}}〉+〈{\displaystyle \frac{Bx-\langle Bx,x\rangle x}{{\Delta }_x\left(B\right)}},{\displaystyle \frac{Bx-\langle Bx,x\rangle x}{{\Delta }_x\left(B\right)}}〉\hfill \\ \hfill & +〈{\displaystyle \frac{Bx-\langle Bx,x\rangle x}{{\Delta }_x\left(B\right)}},{\displaystyle \frac{Ax-\langle Ax,x\rangle x}{{\Delta }_x\left(A\right)}}〉+〈{\displaystyle \frac{Ax-\langle Ax,x\rangle x}{{\Delta }_x\left(A\right)}},{\displaystyle \frac{Bx-\langle Bx,x\rangle x}{{\Delta }_x\left(B\right)}}〉\hfill \\ \hfill & =1+1+{\displaystyle \frac{\langle Bx,Ax\rangle -\langle Bx,x\rangle \langle Ax,x\rangle }{{\Delta }_x\left(A\right){\Delta }_x\left(B\right)}}+{\displaystyle \frac{\langle Ax,Bx\rangle -\langle Ax,x\rangle \langle Bx,x\rangle }{{\Delta }_x\left(A\right){\Delta }_x\left(B\right)}}\hfill \\ \hfill & =2+{\displaystyle \frac{2\langle Ax,x\rangle \langle Bx,x\rangle +\langle \{A,B\}x,x\rangle }{{\Delta }_x\left(A\right){\Delta }_x\left(B\right)}}.\hfill \end{array}$$

Therefore

$$\begin{array}{cc}\hfill & -{\displaystyle \frac{-2\langle Ax,x\rangle \langle Bx,x\rangle +\langle \{A,B\}x,x\rangle }{{\Delta }_x\left(A\right){\Delta }_x\left(B\right)}}\le 2-〈\left({\displaystyle \frac{A}{{\Delta }_x\left(A\right)}}+{\displaystyle \frac{B}{{\Delta }_x\left(B\right)}}\right)x,y〉〈y,\left({\displaystyle \frac{A}{{\Delta }_x\left(A\right)}}+{\displaystyle \frac{B}{{\Delta }_x\left(B\right)}}\right)x〉\hfill \\ \hfill & \Rightarrow {\displaystyle \frac{2\langle Ax,x\rangle \langle Bx,x\rangle -\langle \{A,B\}x,x\rangle }{2-〈\left(\frac{A}{{\Delta }_x\left(A\right)}+\frac{B}{{\Delta }_x\left(B\right)}\right)x,y〉〈y,\left(\frac{A}{{\Delta }_x\left(A\right)}+\frac{B}{{\Delta }_x\left(B\right)}\right)x〉}}\le {\Delta }_x\left(A\right){\Delta }_x\left(B\right).\hfill \end{array}$$

(vii)

Similar to (v).

(viii)

Similar to (vi).

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