p-adic Heisenberg-Robertson-Schrodinger and p-adic 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 the 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

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

In 1929, Robertson [1] derived the following mathematical form of the uncertainty principle of Heisenberg derived in 1927 [2]. Recall that for two 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 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

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

(1)

In 1930, Schrodinger made the following improvement of Inequality (1).

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

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

Surprisingly, in 2014, Maccone and Pati derived the following uncertainty principle which works for any unit vector which is orthogonal to given unit vector [7].

Theorem 3. 

[7] (Maccone-Pati 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 $h\in \mathcal{D}\left(A\right)\cap \mathcal{D}\left(B\right)$ with $\parallel h\parallel =1$, we have

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

As the study of p-adic Hilbert spaces is equally important as the study of Hilbert spaces, we naturally ask the following question.

Question 1.4. What are p-adic versions of Theorems 2 and 3?

In this paper, we answer Question 1.4.

2. p-Adic Heisenberg-Robertson-Schrodinger Uncertainty Principle and p-Adic Maccone-Pati Uncertainty Principle

We are going to consider the following notion of p-adic Hilbert space which is introduced by Kalisch [8] in 1947.

Definition 1. 

[8] Let $\mathbb{K}$ be a non-Archimedean valued field (with valuation $|·|$) and $\mathcal{X}$ be a non-Archimedean Banach space (with norm $\parallel ·\parallel$) over $\mathbb{K}$. We say that $\mathcal{X}$ is a p-adic Hilbert space if there is a map (called as p-adic inner product) $\langle ·,·\rangle :\mathcal{X}\times \mathcal{X}\to \mathbb{K}$ satisfying following.

(i) 

If $x\in \mathcal{X}$ is such that $\langle x,y\rangle =0$ for all $y\in \mathcal{X}$, then $x=0$.

(ii) 

$\langle x,y\rangle =\langle y,x\rangle$ for all $x,y\in \mathcal{X}$.

(iii) 

$\langle x,\alpha y+z\rangle =\alpha \langle x,y\rangle +\langle x,z\rangle$ for all $\alpha \in \mathbb{K}$, for all $x,y,z\in \mathcal{X}$.

(iv) 

$|\langle x,y\rangle |\le \parallel x\parallel \parallel y\parallel$ for all $x,y\in \mathcal{X}$.

Following are standard examples.

Example 1. 

Let $d\in \mathbb{N}$ and $\mathbb{K}$ be a non-Archimedean valued field. Then ${\mathbb{K}}^d$ is a p-adic Hilbert space w.r.t. norm

$$\begin{array}{c}\hfill \parallel {\left(x_j\right)}_{j=1}^d\parallel :=\underset{1\le j\le d}{max}\left|x_j\right|,\forall {\left(x_j\right)}_{j=1}^d\in {\mathbb{K}}^d\end{array}$$

and p-adic inner product

$$\begin{array}{c}\hfill \langle {\left(x_j\right)}_{j=1}^d,{\left(y_j\right)}_{j=1}^d\rangle :=\sum _{j=1}^d{x}_j{y}_j,\forall {\left(x_j\right)}_{j=1}^d,{\left(y_j\right)}_{j=1}^d\in {\mathbb{K}}^d.\end{array}$$

Example 2. 

Let $\mathbb{K}$ be a non-Archimedean valued field. Define

$$\begin{array}{c}\hfill c_0(\mathbb{N},\mathbb{K}):=\{{\left(x_n\right)}_{n=1}^{\infty }:x_n\in \mathbb{K},\forall n\in \mathbb{N},\underset{n\to \infty }{lim}{x}_n=0\}.\end{array}$$

Then $c_0(\mathbb{N},\mathbb{K})$ is a p-adic Hilbert space w.r.t. norm

$$\begin{array}{c}\hfill \parallel {\left(x_n\right)}_{n=1}^{\infty }\parallel :=\underset{n\in \mathbb{N}}{sup}\left|x_n\right|,\forall {\left(x_n\right)}_{n=1}^{\infty }\in c_0(\mathbb{N},\mathbb{K})\end{array}$$

and p-adic inner product

$$\begin{array}{c}\hfill \langle {\left(x_n\right)}_{n=1}^{\infty },{\left(y_n\right)}_{n=1}^{\infty }\rangle \sum _{n=1}^{\infty }{x}_n{y}_n,\forall {\left(x_n\right)}_{n=1}^{\infty },{\left(y_n\right)}_{n=1}^{\infty }\in c_0(\mathbb{N},\mathbb{K}).\end{array}$$

Let $\mathcal{X},\mathcal{Y}$ be p-adic Hilbert spaces and $T:\mathcal{X}\to \mathcal{Y}$ be a linear operator. We say that T is adjointable if there is a linear operator, denoted by $T^{*}:\mathcal{Y}\to \mathcal{X}$ such that $\langle Tx,y\rangle =\langle x,T^{*}y\rangle$, $\forall x\in \mathcal{X},\forall y\in \mathcal{X}$. Note that (i) in Definition 1 says that adjoint, if exists, is unique. An adjointable linear operator $T:\mathcal{X}\to \mathcal{X}$ is said to be self-adjoint if $T^{*}=T$.

• Let A be a possibly unbounded linear operator (need not be self-adjoint) defined on domain $\mathcal{D}\left(A\right)\subseteq \mathcal{X}$. For $x\in \mathcal{D}\left(A\right)$ with $\langle x,x\rangle =1$, define the uncertainty of A at the point x as

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

We now have the p-adic version of Theorem 2.

Theorem 4. (p-adic Heisenberg-Robertson-Schrodinger Uncertainty Principles) Let $\mathcal{X}$ be a p-adic Hilbert space. Let $A:\mathcal{D}\left(A\right)\subseteq \mathcal{X}\to \mathcal{X}$ and $B:\mathcal{D}\left(B\right)\subseteq \mathcal{X}\to \mathcal{X}$ be linear operators. 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) 

$$\begin{array}{c}\hfill {\mathsf{\Delta }}_x\left(A\right){\mathsf{\Delta }}_x\left(B\right)\ge |\langle Ax,Bx\rangle -\langle Ax,x\rangle \langle Bx,x\rangle |=|\langle Bx,Ax\rangle -\langle Ax,x\rangle \langle Bx,x\rangle |.\end{array}$$

In particular, if A and B are self-adjoint, then

$$\begin{array}{c}\hfill {\mathsf{\Delta }}_x\left(A\right){\mathsf{\Delta }}_x\left(B\right)\ge |\langle BAx,x\rangle -\langle Ax,x\rangle \langle Bx,x\rangle |=|\langle ABx,x\rangle -\langle Ax,x\rangle \langle Bx,x\rangle |.\end{array}$$

(ii) 

$$\begin{array}{c}\hfill max\{{\mathsf{\Delta }}_x\left(A\right),{\mathsf{\Delta }}_x\left(B\right)\}\ge {\displaystyle \frac{\sqrt{|{〈[A,B]x,x〉}^2+{\left(\langle \{A,B\}x,x\rangle -2\langle Ax,x\rangle \langle Bx,x\rangle \right)}^2|}}{\sqrt{\left|2\right|}}}.\end{array}$$

(iii) 

If A and B are adjointable, then

$$\begin{array}{c}\hfill max\{{\mathsf{\Delta }}_x\left(A\right),{\mathsf{\Delta }}_x\left(B\right)\}\ge \sqrt{\left|\langle (A^{*}A+B^{*}B)x,x\rangle -{\displaystyle \frac{{\langle (A+B)x,x\rangle }^2+{\langle (A-B)x,x\rangle }^2}{2}}\right|}.\end{array}$$

In particular, if A and B are self-adjoint, then

$$\begin{array}{cc}\hfill max\{{\mathsf{\Delta }}_x\left(A\right),{\mathsf{\Delta }}_x\left(B\right)\}& \ge \sqrt{\left|\langle (A^2+B^2)x,x\rangle -{\displaystyle \frac{{\langle (A+B)x,x\rangle }^2+{\langle (A-B)x,x\rangle }^2}{2}}\right|}\hfill \\ \hfill & =\sqrt{\left|{\displaystyle \frac{\langle {(A+B)}^{2}x,x\rangle +\langle {(A-B)}^{2}x,x\rangle -{\langle (A+B)x,x\rangle }^2-{\langle (A-B)x,x\rangle }^2}{2}}\right|}.\hfill \end{array}$$

(iv) 

If A and B are adjointable, then

$$\begin{array}{c}\hfill max\{{\mathsf{\Delta }}_x\left(A\right),{\mathsf{\Delta }}_x\left(B\right)\}\ge \sqrt{|\langle (A^{*}A-B^{*}B)x,x\rangle -\langle (A+B)x,x\rangle \langle (A-B)x,x\rangle |}.\end{array}$$

In particular, if A and B are self-adjoint, then

$$\begin{array}{c}\hfill max\{{\mathsf{\Delta }}_x\left(A\right),{\mathsf{\Delta }}_x\left(B\right)\}\ge \sqrt{|\langle (A^2-B^2)x,x\rangle -\langle (A+B)x,x\rangle \langle (A-B)x,x\rangle |}.\end{array}$$

(v) 

$$\begin{array}{c}\hfill max\{{\mathsf{\Delta }}_x\left(A\right),{\mathsf{\Delta }}_x\left(B\right)\}\ge \sqrt{{|\langle (A+B)x,(A+B)x\rangle -\langle (A+B)x,x\rangle }^2|}.\end{array}$$

(vi) 

$$\begin{array}{c}\hfill max\{{\mathsf{\Delta }}_x\left(A\right),{\mathsf{\Delta }}_x\left(B\right)\}\ge \sqrt{{|\langle (A-B)x,(A-B)x\rangle -\langle (A-B)x,x\rangle }^2|}.\end{array}$$

Proof. 

Let $x\in \mathcal{D}\left(AB\right)\cap \mathcal{D}\left(BA\right)$ be such that $\langle x,x\rangle =1$.

(i)

By using the definition of p-adic inner product,

$$\begin{array}{cc}\hfill {\mathsf{\Delta }}_x\left(A\right){\mathsf{\Delta }}_x\left(B\right)& =\parallel Ax-\langle Ax,x\rangle x\parallel \parallel Bx-\langle Bx,x\rangle x\parallel \hfill \\ \hfill & \ge |\langle Ax-\langle Ax,x\rangle x,Bx-\langle Bx,x\rangle x\rangle |\hfill \\ \hfill & =|\langle Ax,Bx\rangle -\langle Ax,x\rangle \langle Bx,x\rangle |.\hfill \end{array}$$

(ii)

By making a direct expansion and simplification, we see that

$$\begin{array}{cc}\hfill & {〈[A,B]x,x〉}^2+{\left(\langle \{A,B\}x,x\rangle -2\langle Ax,x\rangle \langle Bx,x\rangle \right)}^2\hfill \\ \hfill & ={\left(\langle ABx,x\rangle -\langle BAx,x\rangle \right)}^2+{〈\{A,B\}x,x〉}^2+4{\langle Ax,x\rangle }^2{\langle Bx,x\rangle }^2-4\langle \{A,B\}x,x\rangle \langle Ax,x\rangle \langle Bx,x\rangle \hfill \\ \hfill & ={\left(\langle ABx,x\rangle -\langle BAx,x\rangle \right)}^2+{\left(\langle ABx,x\rangle +\langle BAx,x\rangle \right)}^2+4{\langle Ax,x\rangle }^2{\langle Bx,x\rangle }^2\hfill \\ \hfill & -4\langle ABx,x\rangle \langle Ax,x\rangle \langle Bx,x\rangle -4\langle BAx,x\rangle \langle Ax,x\rangle \langle Bx,x\rangle \hfill \\ \hfill & =2{\langle ABx,x\rangle }^2+2{\langle BAx,x\rangle }^2+4{\langle Ax,x\rangle }^2{\langle Bx,x\rangle }^2-4\langle ABx,x\rangle \langle Ax,x\rangle \langle Bx,x\rangle -4\langle BAx,x\rangle \langle Ax,x\rangle \langle Bx,x\rangle \hfill \\ \hfill & =2{(\langle ABx,x\rangle -\langle Ax,x\rangle \langle Bx,x\rangle )}^2+2{(\langle BAx,x\rangle -\langle Ax,x\rangle \langle Bx,x\rangle )}^2.\hfill \end{array}$$

Therefore

$$\begin{array}{cc}\hfill & \left|{〈[A,B]x,x〉}^2+{\left(\langle \{A,B\}x,x\rangle -2\langle Ax,x\rangle \langle Bx,x\rangle \right)}^2\right|\hfill \\ \hfill & =\left|2\right|\left|{(\langle ABx,x\rangle -\langle Ax,x\rangle \langle Bx,x\rangle )}^2+{(\langle BAx,x\rangle -\langle Ax,x\rangle \langle Bx,x\rangle )}^2\right|\hfill \\ \hfill & \le \left|2\right|max\left\{\left|{(\langle ABx,x\rangle -\langle Ax,x\rangle \langle Bx,x\rangle )}^2\right|,\left|{(\langle BAx,x\rangle -\langle Ax,x\rangle \langle Bx,x\rangle )}^2\right|\right\}\hfill \\ \hfill & =\left|2\right|max\left\{{\left|\langle ABx,x\rangle -\langle Ax,x\rangle \langle Bx,x\rangle \right|}^2,{\left|\langle BAx,x\rangle -\langle Ax,x\rangle \langle Bx,x\rangle \right|}^2\right\}\hfill \\ \hfill & \le \left|2\right|max\left\{max\{{\mathsf{\Delta }}_x{\left(A\right)}^2,{\mathsf{\Delta }}_x{\left(B\right)}^2\},max\{{\mathsf{\Delta }}_x{\left(B\right)}^2,{\mathsf{\Delta }}_x{\left(A\right)}^2\}\right\}=\left|2\right|max\{{\mathsf{\Delta }}_x{\left(A\right)}^2,{\mathsf{\Delta }}_x{\left(B\right)}^2\}.\hfill \end{array}$$

(iii)

By using the non-Archimedean triangle inequality and the definition of p-adic inner product,

$$\begin{array}{cc}\hfill max\{{\mathsf{\Delta }}_x\left(A\right),{\mathsf{\Delta }}_x\left(B\right)\}& =max\{\parallel Ax-\langle Ax,x\rangle x\parallel ,\parallel Bx-\langle Bx,x\rangle x\parallel \}\hfill \\ \hfill & \ge max\{\sqrt{|\langle Ax-\langle Ax,x\rangle x,Ax-\langle Ax,x\rangle x\rangle |},\sqrt{|\langle Bx-\langle Bx,x\rangle x,Bx-\langle Bx,x\rangle x\rangle |}\}\hfill \\ \hfill & =max\{\sqrt{{|\langle Ax,Ax\rangle -\langle Ax,x\rangle }^2|},\sqrt{{|\langle Bx,Bx\rangle -\langle Bx,x\rangle }^2|}\}\hfill \\ \hfill & =\sqrt{max\left\{\right|\langle Ax,Ax\rangle -{\langle Ax,x\rangle }^2|,|\langle Bx,Bx\rangle -{\langle Bx,x\rangle }^2\left|\right\}}\hfill \\ \hfill & \ge \sqrt{{|\langle Ax,Ax\rangle -\langle Ax,x\rangle }^2+\langle Bx,Bx\rangle -{\langle Bx,x\rangle }^2|}\hfill \\ \hfill & =\sqrt{|\langle A^{*}Ax,x\rangle +\langle B^{*}Bx,x\rangle -({\langle Ax,x\rangle }^2+{\langle Bx,x\rangle }^2)|}\hfill \\ \hfill & =\sqrt{\left|\langle (A^{*}A+B^{*}B)x,x\rangle -{\displaystyle \frac{{\langle (A+B)x,x\rangle }^2+{\langle (A-B)x,x\rangle }^2}{2}}\right|}.\hfill \end{array}$$

(iv)

Using initial calculations in (iii),

$$\begin{array}{cc}\hfill max\{{\mathsf{\Delta }}_x\left(A\right),{\mathsf{\Delta }}_x\left(B\right)\}& \ge \sqrt{max\left\{\right|\langle Ax,Ax\rangle -{\langle Ax,x\rangle }^2|,|\langle Bx,Bx\rangle -{\langle Bx,x\rangle }^2\left|\right\}}\hfill \\ \hfill & \ge \sqrt{{|\langle Ax,Ax\rangle -\langle Ax,x\rangle }^2-\langle Bx,Bx\rangle +{\langle Bx,x\rangle }^2|}\hfill \\ \hfill & =\sqrt{|\langle A^{*}Ax,x\rangle -\langle B^{*}Bx,x\rangle -({\langle Ax,x\rangle }^2-{\langle Bx,x\rangle }^2)|}\hfill \\ \hfill & =\sqrt{|\langle (A^{*}A-B^{*}B)x,x\rangle -\langle (A+B)x,x\rangle \langle (A-B)x,x\rangle |}.\hfill \end{array}$$

(v)

Using ultrametric inequality first and then using p-adic inner product we get

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

(vi)

Using initial calculations in (v),

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

□

Note that for self-adjoint operators A and B, we have

$$\begin{array}{c}\hfill \langle [A,B]x,x\rangle =\langle ABx,x\rangle -\langle BAx,x\rangle =\langle Bx,Ax\rangle -\langle Ax,Bx\rangle =\langle Bx,Ax\rangle -\langle Bx,Ax\rangle =0\end{array}$$

and

$$\begin{array}{cc}\hfill \langle \{A,B\}x,x\rangle & =\langle ABx,x\rangle +\langle BAx,x\rangle =\langle Bx,Ax\rangle +\langle Ax,Bx\rangle =\langle Bx,Ax\rangle +\langle Bx,Ax\rangle \hfill \\ \hfill & =2\langle Bx,Ax\rangle =2\langle ABx,x\rangle .\hfill \end{array}$$

We next derive p-adic version of Theorem 3.

Theorem 5. (p-adic Maccone-Pati Uncertainty Principle) Let $\mathcal{X}$ be a p-adic Hilbert space. Let $A:\mathcal{D}\left(A\right)\subseteq \mathcal{X}\to \mathcal{X}$ and $B:\mathcal{D}\left(B\right)\subseteq \mathcal{X}\to \mathcal{X}$ be linear operators. Then for all $x\in \mathcal{D}\left(A\right)\cap \mathcal{D}\left(B\right)$ with $\langle x,x\rangle =1$, we have

$$\begin{array}{c}\hfill max\{{\mathsf{\Delta }}_x\left(A\right),{\mathsf{\Delta }}_x\left(B\right)\}\ge |\langle (A+B)x,y\rangle |,\forall y\in \mathcal{X}\mathit{satisfying}\parallel y\parallel \le 1,\langle x,y\rangle =0\end{array}$$

and

$$\begin{array}{c}\hfill max\{{\mathsf{\Delta }}_x\left(A\right),{\mathsf{\Delta }}_x\left(B\right)\}\ge |\langle (A-B)x,y\rangle |,\forall y\in \mathcal{X}\mathit{satisfying}\parallel y\parallel \le 1,\langle x,y\rangle =0.\end{array}$$

Proof. 

Let $x\in \mathcal{D}\left(A\right)\cap \mathcal{D}\left(B\right)$ be such that $\langle x,x\rangle =1$. Let $y\in \mathcal{X}$ satisfies $\parallel y\parallel \le 1$ and $\langle x,y\rangle =0.$ Then

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

□