Let us postulate the existence of a single space-time M, on which all events occur, with no particular coordinates attached to it. Let us consider the differential operator $i\hslash {\overrightarrow{\nabla }}_{\mu }=\left(\left(i\hslash (\partial /\partial \left(ct\right)),-{\sum }_{i=1}^{3}i\hslash (\partial /\partial x^i)\right)\right)$ in $\mathcal{H}=L^2\left({\mathbb{R}}^{3,1}\right)$. For a diffeomorphism $\Phi :M\to L^2\left({\mathbb{R}}^{3,1}\right)$, let an event at a point $m\in M$, is occurred at $\Phi \left(m\right)$. Similarly, for another diffeomorphism $\Psi :M\to L^2\left({\mathbb{R}}^{3,1}\right)$, the same event is considered to be occurred at $\Psi \left(m\right)$. From inertial reference frames, the changing coordinates is then given by the map $F:\Psi \circ {\Phi }^{-1}:L^2\left({\mathbb{R}}^{3,1}\right)\to L^2\left({\mathbb{R}}^{3,1}\right)$. Let $Q:L^2\left({\mathbb{R}}^{3,1}\right)\to L^2\left(\mathbb{R}\right)$ be the (quantum) quadratic form defined by, so that, for the basic axiom of special relativity, F preserves Q, i.e., $Q\circ F=Q$.

Let us consider an infinite-dimensional complex vector space $\mathcal{V}\subset \mathcal{H}=L^2\left({\mathbb{R}}^{n-1,1}\right)$. Let $(M^n,\mathrm{q})$ is a smooth manifold, that is, $M^n$ is an n-dimensional differentiable manifold and $\mathrm{q}$ is a (quantum) metric tensor satisfying, when $g_{\mu \nu }=\mathrm{diag}[1,-1,\dots ,-1]$ is the n-dimensional Lorentz metric. This $\mathrm{q}$ is either a positive-definite section of the bundle of symmetric (covariant) 2-tensors $T^{*}M{\otimes }_S{T}^{*}M$ or a positive-definite bilinear maps, $\mathrm{q}\left({(i\hslash )}^{-1}x\right):T_{\left({(i\hslash )}^{-1}x\right)}M\times T_{\left({(i\hslash )}^{-1}x\right)}M\to \mathcal{V}$ for all ${(i\hslash )}^{-1}x\in M$. Here, the metric tensor ${\mathrm{q}}_{\mu \nu }$ is symmetric, i.e., ${\mathrm{q}}_{\mu \nu }={\mathrm{q}}_{\nu \mu }$, and $\det \left({\mathrm{q}}_{\mu \nu }\right)\ne 0$ . Components of its inverse matrix ${\mathrm{q}}^{-1}$ are themselves the components of matrix $\mathrm{q}$, namely, ${\mathrm{q}}_{\mu \nu }{\mathrm{q}}^{\mu \gamma }={\mathrm{q}}^{\gamma \mu }{\mathrm{q}}_{\mu \nu }={\delta }_{\nu }^{\gamma }$, where ${\delta }_{\nu }^{\gamma }$ is the Kronecker delta. The basic axiom of special relativity then ensures that F preserves $\mathrm{q}$.

If we consider Λ matrix as then a standard form of (Quantum) Lorentz-like transformation (QLT) is, Thus,

Let $(M^n,\mathrm{q})$ is a smooth manifold, that is, $M^n$ is an n-dimensional differentiable manifold and $\mathrm{q}$ is the quantum metric tensor satisfying (2), which is either a positive-definite section of the bundle of symmetric (covariant) 2-tensors $T^{*}M{\otimes }_S{T}^{*}M$ or a positive-definite bilinear maps, $\mathrm{q}\left({(i\hslash )}^{-1}x\right):T_{\left({(i\hslash )}^{-1}x\right)}M\times T_{\left({(i\hslash )}^{-1}x\right)}M\to \mathcal{V}$ for all ${(i\hslash )}^{-1}x\in M$. Here, $T^{*}M{\otimes }_S{T}^{*}M$ is the subspace of $T^{*}M\otimes T^{*}M$ generated by elements of the form $X\otimes Y+Y\otimes X$. Let ${\left\{{(i\hslash )}^{-1}{x}^{\mu }\right\}}_{\mu =1}^n$ be local coordinates in a neighborhood U of some point of M. In U the vector fields ${\left\{i\hslash {\overrightarrow{\nabla }}_{\mu }\right\}}_{\mu =1}^n$ form a local basis for $TM$ and the 1-forms ${\left\{{(i\hslash )}^{-1}\mathrm{d}{x}^{\mu }\right\}}_{\mu =1}^n$ form a dual basis for $T^{*}M$, that is, ${(i\hslash )}^{-1}\mathrm{d}{x}^{\mu }·(i\hslash ){\overrightarrow{\nabla }}_{\nu }={\delta }_{\nu }^{\mu }$. The metric may then be written in local coordinates as $\mathrm{q}={(i\hslash )}^{-2}{\mathrm{q}}_{\mu \nu }\mathrm{d}{x}^{\mu }\otimes \mathrm{d}{x}^{\nu }$, where ${\mathrm{q}}_{\mu \nu }\doteqdot \mathrm{q}\left(i\hslash {\overrightarrow{\nabla }}_{\mu },i\hslash {\overrightarrow{\nabla }}_{\nu }\right)$. Let ${\nabla }^{\mathrm{q}}$ denote the Levi-Civita connection of the metric $\mathrm{q}$. The Christoffel symbol is the components of the Levi-Civita connection and is defined in U by ${\nabla }_{\left(i\hslash {\overrightarrow{\nabla }}_{\mu }\right)}\left(i\hslash {\overrightarrow{\nabla }}_{\nu }\right)\doteqdot {\Gamma }_{\mu \nu }^k\left(i\hslash {\overrightarrow{\nabla }}_k\right)$, and for $\left[\left(i\hslash {\overrightarrow{\nabla }}_{\mu }\right),\left(i\hslash {\overrightarrow{\nabla }}_{\nu }\right)\right]=0$, we see that they are given by, Let the curvature $(3,1)$-tensor $\mathsf{Rm}$ is defined by, Let the tensor $\mathsf{Rc}$ is the trace of $\mathsf{Rm}$ curvature tensor, Its components are defined by ${\mathcal{R}}_{\mu \nu }|\psi \rangle \doteqdot \mathsf{Rc}\left(\left(i\hslash {\overrightarrow{\nabla }}_{\mu }\right),\left(i\hslash {\overrightarrow{\nabla }}_{\nu }\right)\right)|\psi \rangle$, thus, the Ricci tensor, is purely Quantum Mechanical due to (13). Again, the scalar curvature $\mathsf{R}$ is the trace of $\mathsf{Rc}$ tensor: $\mathsf{R}|\psi \rangle \doteqdot {\sum }_{a=1}^n\mathsf{Rc}(e_a,e_a)|\psi \rangle$ where $e_a\in T_{\left({(i\hslash )}^{-1}x\right)}{M}^n$ is a unit vector spanning of a line $L\subset T_{\left({(i\hslash )}^{-1}x\right)}{M}^n$. Then, the non-vacuum quantum Einstein tensor $\mathsf{Rc}-{\displaystyle \frac{1}{2}}\mathrm{q}\mathsf{R}$ directly acts on a quantum spacetime. Thus, the non-vacuum quantum Einstein field equation, $\left[\mathsf{Rc}-{\displaystyle \frac{1}{2}}\mathrm{q}\mathsf{R}\right]|\psi \rangle ={\left(i\hslash \right)}^2(8\pi G/c^4)T|\psi \rangle$, is now ‘purely’ Quantum Mechanical for (13), where the rhs${\left(i\hslash \right)}^2$ comes from the energy-momentum tensor, for example, ${\mathcal{T}}_{\mu \nu }={\left(i\hslash \right)}^2{\partial }_{\mu }\varphi {\partial }_{\nu }\varphi -{\displaystyle \frac{1}{2}}{\left(i\hslash \right)}^2{g}_{\mu \nu }{\left(\partial \varphi \right)}^2-{\left(i\hslash \right)}^2\left(a/{\ell }^2\right){\varphi }^2{g}_{\mu \nu }={\left(i\hslash \right)}^2{T}_{\mu \nu }$, where ℓ is related to the cosmological constant $\Lambda$ as $\Lambda =-3/{\ell }^2$ and a is the (squared) Klein-Gordon mass [5]. 3 Again, the Ricci tensor is ${\mathcal{R}}_{\mu \nu }|\psi \rangle \doteqdot \mathsf{Rc}\left(\left(i\hslash {\overrightarrow{\nabla }}_{\mu }\right),\left(i\hslash {\overrightarrow{\nabla }}_{\nu }\right)\right)|\psi \rangle$$\doteqdot {\left(i\hslash \right)}^2\mathrm{Rc}\left({\partial }_{\mu },{\partial }_{\nu }\right)|\psi \rangle$, thus, the non-vacuum (Quantum) Einstein field equation in $(3+1)D$ quantum spacetime would become as, where the Lorentz metric tensor $g_{\mu \nu }$ is satisfying (2) and the ‘naive’ cosmological constant (in SI unit) ${\Lambda }^{\prime }=1.112121717\times {10}^{-150}{\mathrm{J}}^2{\mathrm{s}}^2{G}^{-1}$, which yields the cosmological constant $\Lambda =-{\left(i\hslash \right)}^{-2}{\Lambda }^{\prime }$. □

For $\left[{(i\hslash )}^2{\overrightarrow{\nabla }}_0^2-{(i\hslash )}^2{\overrightarrow{\nabla }}_i{\overrightarrow{\nabla }}_j\right]|\psi \rangle =({\partial }^2/\partial {\widehat{\mathfrak{s}}}^2)|\psi \rangle$ of (3), there exists a first order quantum equation in $(3+1)D$ as, while V is the external potential energy.

Let $\left[{(i\hslash )}^2{\overrightarrow{\nabla }}_0^2-{(i\hslash )}^2{\overrightarrow{\nabla }}_i{\overrightarrow{\nabla }}_j\right]|\psi \rangle ={(i\hslash )}^2{\gamma }_Q^2{\overrightarrow{\nabla }}_0^2|\psi \rangle$ for (10). Then the dynamics of the quantum state is possible to be defined as, while ${\overrightarrow{\nabla }}_0^{-1}{\overrightarrow{\nabla }}_i={\left({\mathsf{v}}^i\right)}^{-1}c$. Therefore, by introducing external potential energy V and the Hamiltonian (or total energy) operator $\mathcal{H}$, we have a first order quantum equation in $(3+1)D$ as, where ${\Phi }_i={\left[(1-{\gamma }_Q^2){\mathsf{v}}^i\right]}^{-1}{c}^2$. □

If ${\displaystyle \frac{1}{2{\mathbf{I}}_4}}\left\{{\gamma }^{\mu },{\gamma }^{\nu }\right\}\equiv {\mathrm{q}}_{\mu \nu }=g^{\mu \nu }$, then it must satisfy,

In (19), the anticommutator relation ${\displaystyle \frac{1}{2{\mathbf{I}}_4}}\left\{{\gamma }^{\mu },{\gamma }^{\nu }\right\}$ must have to satisfy that, for $\left\{{\gamma }^{\beta },{\gamma }^{\alpha }\right\}=-\left\{{\gamma }^{\alpha },{\gamma }^{\beta }\right\}$ and $(i\hslash {\overrightarrow{\nabla }}_{\beta })(i\hslash {\overrightarrow{\nabla }}_{\alpha })=-(i\hslash {\overrightarrow{\nabla }}_{\alpha })(i\hslash {\overrightarrow{\nabla }}_{\beta })$, etc., we have, inputting ${\displaystyle \frac{1}{2{\mathbf{I}}_4}}\left\{{\gamma }^{\alpha },{\gamma }^{\beta }\right\}\equiv {\mathrm{q}}_{\alpha \beta }=g^{\alpha \beta }$, etc., Hence, this gives the lemma. □