Time is not considered as a dimension in $3D$ Quantum Mechanics [1]. That is why gravity is excluded from the quantum spacetime. In this article, we introduce a $(3+1)D$ quantum formalism, which is developed neither in a Minkowski spacetime, nor in a purely Hilbert space and not its metric is Lorentzian. To include gravity in the quantum spacetime, this quantum mechanics yields the (renormalizable) quantum Einstein field equation in $(3+1)D$ quantum spacetime by introducing a purely quantum line element. Analysis of its quantum state spaces shows us that every observable $(3+1)D$ quantum spacetime contains internally hidden opposite spin quantum state. Application of Dimension Theory in this $(3+1)D$ quantum formalism yields that this hidden quantum state immediately acquires more extra hidden dimensions including string. This string, which is found to be fundamentally eleven-dimensional, is inevitably natural and universal but hidden inside every $(3+1)D$ observable quantum spacetime. This finding can make an effective impact on the field of studying and developing String/M-theory in a deeper level.

Let the line element $\mathrm{d}{s}^2=g_{\mu \nu }\mathrm{d}{x}^{\mu }\mathrm{d}{x}^{\nu }\equiv {\left(\frac{\mathrm{d}t}{m}\right)}^2{g}_{\mu \nu }{P}^{\mu }{P}^{\nu }$ for the ‘four-momentum’ $P^{\mu }$, then, for $\mu =0,1,2,3$, $i=1,2,3$ and $g_{\mu \nu }=\mathrm{diag}[1,-1,-1,-1]$, where, $m^2{\left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)}^2=m^2\left(1-\frac{{\left({\mathsf{v}}^i\right)}^2}{c^2}\right)c^2=m_0^2{c}^2$ for the rest mass $m_0$, whereas ${\mathsf{v}}^{\mu }$ is the ‘four-velocity’, then, the rearrangement of (1) as, $mE=p^i{p}^j+g_{\mu \nu }{P}^{\mu }{P}^{\nu }$, may yield the representation of a wave field $\psi (\overrightarrow{r},t)$ by superposition of a free particle (de Broglie wave) as, Thus, from (2), we can get the (total) energy operator $\widehat{E}\to i\hslash {\partial }_t$ (it is analogous with but not exactly the same as the classical Quantum Mechanics, since it is now related to $(3+1)D$ instead of $3D$ spacetime due to the presence of $g_{\mu \nu }$ in its wave field), and the three momentum operator $\widehat{\mathit{p}}\to -i\hslash {\overrightarrow{\nabla }}_i$, whereas the ‘Four-momentum’ operator, and the mass operator, for $c\left(\frac{\mathrm{d}t}{\mathrm{d}s}\right)=1/\sqrt{1-\frac{{\mathsf{v}}^2}{c^2}}=\gamma$. Let us prescribe quantum-to-classical metric tensor (i.e., $g_{\mu \nu }$) for the ‘Four-momentum’ operator ${\widehat{\mathcal{P}}}^{\mu }\to i\hslash {\overrightarrow{\nabla }}_{\mu }$ as, Here, the ‘quantum metric tensor’ ${\mathsf{g}}_{\mu \nu }$ is symmetric, i.e., ${\mathsf{g}}_{\mu \nu }={\mathsf{g}}_{\nu \mu }$, and $\det \left({\mathsf{g}}_{\mu \nu }\right)\ne 0$. Components of its inverse matrix ${\mathsf{g}}^{-1}$ are themselves the components of matrix $\mathsf{g}$, namely, ${\mathsf{g}}_{\mu \nu }{\mathsf{g}}^{\mu \gamma }={\mathsf{g}}^{\gamma \mu }{\mathsf{g}}_{\mu \nu }={\delta }_{\nu }^{\gamma }$, where, ${\delta }_{\nu }^{\gamma }$ is the Kronecker delta.

Then, $i\hslash (\partial /\partial s)$ in (4) may express as, But, (6) immediately tells us that (1) is possible to write as, The quantum line element $\mathrm{d}{\mathcal{S}}_{\mathcal{P}}^2$ has neither a Minkowski spacetime, nor a purely Hilbert space and not its metric is Lorentzian, since ${\mathsf{g}}_{\mu \nu }$ is satisfying (5).

Note that, ${\left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)}^2=c^2-{\left({\mathsf{v}}^i\right)}^2\sim c^2$ since $c\gg {\mathsf{v}}^i$. Hence, for $c\gg {\mathsf{v}}^i$ and $c=1$, (4) becomes as, $\widehat{m}\to -i\hslash {\partial }_t$. For constant velocity ${\mathsf{v}}^i\ll c$ and $c=1$ in the rhs term of (4) yields an uncertainty principle describing the intrinsic indeterminacy with which $\widehat{m}$ and s can be determined by, The mass-energy relation in (1), i.e., $E=m{c}^2$, yields its quantum definition for the mass operator $\widehat{m}$ of (4) along with (3) and (6) as, The rhs term of this equation becomes very surprising due to $\gamma$, because it is not clear to whom $\gamma$ acts upon.

Relating the first line of (7) to its last line, the quantum line element yields, or, simply discarding $\left(1-{\mathsf{g}}^{00}\right)=\left(1+{\mathsf{g}}^{ij}\right)=0$, we can get, Then, (9) may give us the Schrödinger equation in $(3+1)D$ spacetime for (8) as, Again, applying the representation of wave field $\psi (\overrightarrow{r},t)$ of (2) into $mE=p^i{p}^j+g_{\mu \nu }{P}^{\mu }{P}^{\nu }$ from (1) and replacing $g_{\mu \nu }$ with ${\mathsf{g}}^{\mu \nu }$ for (5), we can get, Thus, (11) should be used as a general alternation of (10).

Now, let us check $(3+1)D$ Quantum Mechanics from the perspective of General theory of Relativity [2]. Let us consider a space $\mathcal{V}$. Let $\left({(i\hslash )}^{-1}{x}^0,\dots ,{(i\hslash )}^{-1}{x}^{n-1}\right)$ is a coordinate system of a point $p\in \mathcal{V}$. Let a line element ($\mathrm{d}\mathfrak{s}\ne \mathrm{d}s$) is, hence, this yields (7) as, where, ${\partial }^2/{(\partial \mathfrak{s})}^2\psi (\overrightarrow{r},t)={(i\hslash )}^2{\partial }^2/{(\partial s)}^2\psi (\overrightarrow{r},t)+\Pi \psi (\overrightarrow{r},t)$, for (6) and for some value of $\Pi$ (see (18) below for more details).

Let $(M^n,\mathsf{g})$ is a smooth, D-dimensional manifold, where $M^n$ is an n-dimensional differentiable manifold and $\mathsf{g}$ is a metric, which is either as a positive-definite section of the bundle of symmetric (covariant) 2-tensors $T^{*}M{\otimes }_S{T}^{*}M$ or as positive-definite bilinear maps, $\mathsf{g}\left({(i\hslash )}^{-1}x\right):T_{\left({(i\hslash )}^{-1}x\right)}M\times T_{\left({(i\hslash )}^{-1}x\right)}M\to \mathcal{V}$ 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 ${\left\{{(i\hslash )}^{-1}{x}^i\right\}}_{i=1}^n$ be local coordinates in a neighborhood U of some point of M. In U the vector fields ${\left\{{\left(i\hslash {\overrightarrow{\nabla }}_i\right)}^{-1}\right\}}_{i=1}^n$ form a local basis for $TM$ and the 1-forms ${\left\{i\hslash {\overrightarrow{\nabla }}_i\right\}}_{i=1}^n$ form a dual basis for $T^{*}M$, that is, $i\hslash {\overrightarrow{\nabla }}_j{\left(i\hslash {\overrightarrow{\nabla }}_i\right)}^{-1}={\delta }_j^i$. The metric may then be written in local coordinates as $\mathsf{g}={\mathsf{g}}_{ij}\left(i\hslash {\overrightarrow{\nabla }}_i\right)\otimes \left(i\hslash {\overrightarrow{\nabla }}_j\right)$. Let ${\nabla }^{\mathsf{g}}$ denote the Levi-Civita connection of the metric $\mathsf{g}$. The Christoffel symbols are the components of the Levi-Civita connection and are defined in U by ${\nabla }_{\left(i\hslash {\overrightarrow{\nabla }}_i\right)}\left(i\hslash {\overrightarrow{\nabla }}_j\right)\doteqdot {\Gamma }_{ij}^k\left(i\hslash {\overrightarrow{\nabla }}_k\right)$, and for $\left[\left(i\hslash {\overrightarrow{\nabla }}_i\right),\left(i\hslash {\overrightarrow{\nabla }}_j\right)\right]=0$, we see that they are given by, Let the curvature $(3,1)$-tensor $\mathsf{Rm}$ is defined by, Thus, the curvature tensor, $R_{ijk}^{\ell }\psi (\overrightarrow{r},t)=\left({\partial }_i{\Gamma }_{jk}^{\ell }-{\partial }_j{\Gamma }_{ik}^{\ell }+{\Gamma }_{jk}^p{\Gamma }_{ip}^{\ell }-{\Gamma }_{ik}^p{\Gamma }_{jp}^{\ell }\right)\psi (\overrightarrow{r},t)$, is purely Quantum Mechanical due to (12).

Let the tensor $\mathsf{Rc}$ is the trace of $\mathsf{Rm}$ curvature tensor: $\mathsf{Rc}(Y,Z)\psi (\overrightarrow{r},t)\doteqdot \mathrm{trace}\left(X\mapsto \mathsf{Rm}(X,Y)Z\right)\psi (\overrightarrow{r},t)$, defined by $R_{ij}\psi (\overrightarrow{r},t)\doteqdot \mathsf{Rc}\left(\left(i\hslash {\overrightarrow{\nabla }}_i\right),\left(i\hslash {\overrightarrow{\nabla }}_j\right)\right)\psi (\overrightarrow{r},t)$, and the scalar curvature $\mathsf{R}$ is the trace of $\mathsf{Rc}$ tensor: $\mathsf{R}\psi (\overrightarrow{r},t)\doteqdot {\sum }_{a=1}^n\mathsf{Rc}(e_a,e_a)\psi (\overrightarrow{r},t)$ where $e_a\in T_{\left({(i\hslash )}^{-1}x\right)}{M}^n$ is a unit vector spanning $L\subset T_{\left({(i\hslash )}^{-1}x\right)}{M}^n$. Then, the Einstein-like tensor $\mathsf{Rc}-\frac{1}{2}\mathsf{g}\mathsf{R}$ directly acts on a quantum space. Thus, Einstein-like field equation, $\left[\mathsf{Rc}-\frac{1}{2}\mathsf{g}\mathsf{R}\right]\psi (\overrightarrow{r},t)={\left(i\hslash \right)}^{2}8\pi G\mathsf{T}\psi (\overrightarrow{r},t)$, is now ‘purely’ Quantum Mechanical for (12). But the Ricci tensor $R_{ij}\psi (\overrightarrow{r},t)\doteqdot \mathsf{Rc}\left(\left(i\hslash {\overrightarrow{\nabla }}_i\right),\left(i\hslash {\overrightarrow{\nabla }}_j\right)\right)\psi (\overrightarrow{r},t)\doteqdot {\left(i\hslash \right)}^2\mathrm{Rc}\left(\frac{\partial }{\partial x^i},\frac{\partial }{\partial x^j}\right)\psi (\overrightarrow{r},t)$, thus, Einstein field equation (in quantum spacetime) should become as, where g is satisfying (5), and where $\left[\mathrm{Rc}-\frac{1}{2}g\mathrm{R}\right]$ is Einsteinian and not renormalizable, though, in the first line of (13), mass dimension of gravitational constant vanishes due to ${\hslash }^{2}G$ and if divergences are to be present, they could now be disposed of by the technique of renormalization (though, this will not play a role in our present discussion). Hence, (13) should be used as a renormalizable Einstein field equation in quantum spacetime for general purposes.

Both of the above perspectives of $(3+1)D$ Quantum Mechanics yield extra eleven dimensions from its quantum state spaces quite naturally. The wave field $\psi (\overrightarrow{r},t)$ in (2) must satisfy the eigenfunctions for a discrete Lorentz transformation as, for $i=(1,2,3)$, when ${\psi }^{†}$ is the complex conjugate of $\psi$. Then, using summation convention and (6), we can write the joint state formalism as, But, (15) may also intend to, for ${\partial }_{\mu }^2={\widehat{\partial }}_{\mu }^2=\left[\frac{{\partial }^2}{\partial t^2},\frac{{\partial }^2}{\partial x^i\partial x^j}\right]$ and ${\mathsf{g}}_{\mu \nu }={\widehat{\mathsf{g}}}_{\mu \nu }=\mathrm{diag}[1,-1,-1,-1]$, whereas $\widehat{\partial }$ implies either ${\partial }_t$ is in company with ${\psi }_i$ or ${\partial }_i$ is in company with ${\psi }_0$, exclusively, i.e., the ‘wrong’ distribution of indices for ∂ and $\psi$. Here, $\rho$ and $\Gamma$ have been chosen arbitrarily. Hence, the complex conjugate of (16) is, If we take the line element as, $\mathrm{d}{s}^{-2}\phi ={(\mathrm{d}{t}^2-\sum \mathrm{d}{x}^i\mathrm{d}{x}^j)}^{-1}\phi$, where $\phi$ is an operator, then we can say that, ${(\mathrm{d}{t}^2-\sum \mathrm{d}{x}^i\mathrm{d}{x}^j)}^{-1}\ne \mathrm{d}{t}^{-2}-\sum {\left(\mathrm{d}{x}^i\mathrm{d}{x}^j\right)}^{-1}$, though, we may assume without any objection that, ${(\mathrm{d}{t}^2-\sum \mathrm{d}{x}^i\mathrm{d}{x}^j)}^{-1}\doteqdot \frac{{\partial }^2}{\partial t^2}-\sum \frac{{\partial }^2}{\partial x^i\partial x^j}-\Pi$, for some value of $\Pi$. Thus, Then, comparing with (16), if ${\mathsf{g}}^{\mu \nu }{\partial }_{\mu }{\partial }_{\nu }\equiv {\mathsf{g}}^{\mu \nu }\left(\rho \right){\partial }_{\mu }{\partial }_{\nu }$, we can write that, This yields, $\frac{{\partial }^2}{\partial s^2}\Psi {|}_{s_0}-\Gamma \Psi {|}_{x_0^{\mu }}\doteqdot \mathrm{d}{s}^{-2}\phi +\Pi \phi$, if $\Gamma \to -\Pi$, which gives the equivalency of (18) and (16).

No more combinations are possible from (16) apart from $\rho$, $\Gamma$ and (15) itself. The arrangement of $\rho$ and $\Gamma$ implies that ${\mathsf{g}}_{\mu \nu }\left(s\right)=\frac{1}{2}\left({\mathsf{g}}_{\mu \nu }\left(\rho \right)+{\widehat{\mathsf{g}}}_{\mu \nu }(\Gamma )\right)$. Thus, (16) should be rewritten by considering (15) and additionally replacing ${\mathsf{g}}_{\mu \nu }\left(s\right)$ with ${\mathsf{g}}_{\mu \nu }\left(s\right)=\frac{1}{2}\left({\mathsf{g}}_{\mu \nu }\left(\rho \right)+{\widehat{\mathsf{g}}}_{\mu \nu }(\Gamma )\right)$ as follows, Note that, $\Pi$ exclusively has to depend upon spacetime. Comparing (18) with (16), let us say that, when $\Gamma \to -\Pi$, Since ${\mathsf{g}}_{\mu \nu }\left(s\right)\ne \frac{1}{2}\left({\mathsf{g}}_{\mu \nu }\left(\rho \right)-{\widehat{\mathsf{g}}}_{\mu \nu }(\Pi )\right)\equiv 0$ as long as ${\mathsf{g}}_{\mu \nu }={\widehat{\mathsf{g}}}_{\mu \nu }=\mathrm{diag}[1,-1,-1,-1]$, then, for ${\mathsf{g}}_{\mu \nu }\left(s\right)\ne 0$, the ‘−’-sign of (20) must have to switch its position in such a way that, Reduction of (19) or (21) is impossible since both of their lhs are only depended upon ∂, thus,

But, the rhs of (21) gives us, where, the swap operator $F|\Psi \rangle =exp\left[i\theta \right]$ for some phase $exp\left[i\theta \right]$, whereas V is a vector space. Then the corresponding eigenspaces are called the symmetric and antisymmetric subspaces and are denoted by the state spaces ${\mathrm{Sym}}^{2}V$ and ${\mathrm{Anti}}^{2}V$, respectively. Note that, we have not intended here that $\rho$ and $\Pi$ individually are two distinguishable particles for the state space ${\mathrm{Sym}}^{2}V$ or ${\mathrm{Anti}}^{2}V$; the above equations are just the generalization forms of their kinds, because $\rho$ and $\Pi$ do not have distinguished (opposite) spins until otherwise they are dissociated as free particles at very high energy scale; so, the observable spin is always the spin of $\rho$, since $\Pi$ is an internally hidden property of the overall system and the observable spacetime is always ∂-dependent. Thus, (21) tells us that, if we allow $\Pi$ to be dissociated as a free particle at very high energy, the internally hidden spacetime of $\Pi$ then must be transformed into a fermionic particle, whereas, the overall ∂-dependent system remains bosonic, since, the observable spacetime is always ∂-dependent.

Similarly, (17) yields, which tells us that the internally hidden spacetime of $\Pi$ must be now bosonic, whereas, the overall system is fermionic, since, the observable spacetime is always ∂-dependent. So, whatever (21) and (22) want to tell us is that the ∂-dependent overall system has Supersymmetry and since the spacetimes of $\rho$ and its supersymmetric partner $\Pi$ are not easily dissociative even upto a very high energy scale; thus, $\Pi$ must require extremely high energy to dissociate itself from the overall system as a free particle. Instead of being a free supersymmetric partner, $\Pi$ actually works quite differently inside of the observable spacetime $\rho$, though, at the same time, $\Pi$ is still satisfying the properties of Supersymmetry. We will show you $\Pi$’s actual purpose very soon in the below. But before that, Supersymmetry needs extra dimensions and we have to discuss it now.

The internally hidden spacetime of $\Pi$ in (21) and (22) also provides us some additional geometries for its ${\widehat{\mathsf{g}}}^{\mu \nu }(\Pi ){\widehat{\partial }}_{\mu }{\widehat{\partial }}_{\nu }\left[\pm \Psi \right]$ structures. Suppose, for ${\widehat{\mathsf{g}}}^{\mu \nu }(\Gamma ){\widehat{\partial }}_{\mu }{\widehat{\partial }}_{\nu }\left[\Psi \right]$, we have, $\frac{1}{\sqrt{2}}\left\{\frac{{\partial }^2}{\partial t^2}\left[-{\psi }_i\right]-\frac{{\partial }^2}{\partial x^i\partial x^j}\left[{\psi }_0\right]\right\}$, where, both spacetimes have the ‘wrong’ distribution of indices for ∂ and $\psi$ within the curvey brackets. These ‘wrong’ distributions must have a noticeable effect on the acceptable spacetime, i.e., its temporal part must influence over the spatially depended ${\psi }_i$, or its spatial part must influence over the temporally depended ${\psi }_0$, or vice versa. In other words, the acceptable spacetime may not be four-dimensional in this case. Let us check it.

Let, $\left({\widehat{\partial }}_{[0,i]}^2\otimes {\Psi }_{[i,0]}\right)\equiv \left({\widehat{\partial }}_{\left[0\right]}^2{\Psi }_{\left[i\right]}\mp {\widehat{\partial }}_{\left[i\right]}^2{\Psi }_{\left[0\right]}\right)$, when ${\widehat{\partial }}_{\left[0\right]}^2{\Psi }_{\left[i\right]}=\frac{{\partial }^2}{\partial t^2}\left[\mp {\psi }_i\right]$ and ${\widehat{\partial }}_{\left[i\right]}^2{\Psi }_{\left[0\right]}=\frac{{\partial }^2}{\partial x^i\partial x^j}[\pm {\psi }_0]$. Suppose, for $\frac{1}{\sqrt{2}}\left\{\frac{{\partial }^2}{\partial t^2}\left[-{\psi }_i\right]-\frac{{\partial }^2}{\partial x^i\partial x^j}\left[{\psi }_0\right]\right\}$, we should consider a dimension function, Let the space $\left({\widehat{\partial }}_{[0,i]}^2\otimes {\Psi }_{[i,0]}\right)$ satisfy a normal $T_1$-space. Let $\mathcal{U}$ be a collection in an initially $(3+1)D$ topological spacetime $\left({\widehat{\partial }}_{[0,i]}^2\otimes {\Psi }_{[i,0]}\right)$, i.e., $\dim \left({\widehat{\partial }}_{[0,i]}^2\otimes {\Psi }_{[i,0]}\right)=(3+1)$, which is actually hidden inside an observable $(3+1)D$ spacetime, $\left({\partial }_{[0,i]}^2\otimes {\Psi }_{[0,i]}\right)$ (be careful about the subscript indices $[0,i]$ here – do not confuse observable spacetime with the hidden spacetime, which always carries subscript indices $[0,i]$ and $[i,0]$ both at the same time).

See [3,4,5,6,7] for the required background of Dimension Theory to construct a mapping f of the spacetime $\left({\widehat{\partial }}_{[0,i]}^2\otimes {\Psi }_{[i,0]}\right)$ into a spacetime S, which is a closed (open) mapping, if the image of every closed (open) set of $\left({\widehat{\partial }}_{[0,i]}^2\otimes {\Psi }_{[i,0]}\right)$ is closed (open) in S. Then the continuous mappings which lower dimensions of the spacetime $\left({\widehat{\partial }}_{[0,i]}^2\otimes {\Psi }_{[i,0]}\right)$ should be defined as follows,

Note that, here and hereafter, the sign ‘≤’ always intends to give meaning to Dimension Theory [3], rather than its traditional algebraic meaning.

Since, the temporal axis is unaltered in Lorentz transformation, as we have already seen it in (14), we can express the maximal continuous mapping of the $(3+1)D$ spacetime $\left({\widehat{\partial }}_{[0,i]}^2\otimes {\Psi }_{[i,0]}\right)$ onto the spacetime S of (26) as, since i should not be zero in (25), then the spacetime S definitely intends the basic structure of a 2-dimensional worldsheet $X^{\mu }(\tau ,\sigma )$ with the joint states, $\frac{1}{\sqrt{2}}\left\{\frac{{\partial }^2}{\partial {\tau }^2}\left[-{\psi }_{\sigma }\right]-\frac{{\partial }^2}{\partial {\sigma }^2}\left[{\psi }_{\tau }\right]\right\}$ for the spacetime ${\mathfrak{X}}^{\mu }$, where S is a $(1+1)D$ spacetime, i.e., string. Obviously, a string can sweep out the 2-dimensional worldsheet $X^{\mu }(\tau ,\sigma )$ for the spacetime ${\mathfrak{X}}^{\mu }$. But for the space K, we need to discuss it in more details, what we are going to do below in Theorem 2.

Before that, since the spacetime of $\Pi$ is hidden inside the overall system of (21), i.e., in other words, inside the observable spacetime of $\rho$, then the increment of spacetime ${\mathfrak{X}}^{\mu }$ should not be observable by any means, i.e., the extra dimensions of ${\mathfrak{X}}^{\mu }$ remain hidden forever inside the observable spacetime of $\rho$. As these internally hidden extra dimensions inside the observable spacetime of $\rho$ are considered as the representation of the spacetime S and the space K, thus, we can conclude,

But beyond $\dim K\le k$, the space K can raise more extra hidden dimensions by a closed continuous mapping by adopting the following,

Then, we can say for the overall spacetime ${\mathfrak{X}}_{\mathrm{ov}}^{\mu }$ that, hence, for the value of K, Note here that stringy spacetime S vanishes in the overall spacetime ${\mathfrak{X}}_{\mathrm{ov}}^{\mu }$ of (27) for the space K, leaving behind the forever hidden extra dimensions m in ${\mathfrak{X}}_{\mathrm{ov}}^{\mu }$. Thus, in other words, strings are experimentally unobservable forever, whereas, their actions should be mandatory in the purpose of overall spacetime ${\mathfrak{X}}_{\mathrm{ov}}^{\mu }$ (see end of this section for more details). Also notice that Supersymmetry (now having extra dimensions m for ${\mathfrak{X}}^{\mu }$ due to (28)) remains unchanged in ${\mathfrak{X}}_{\mathrm{ov}}^{\mu }$ of (27). Thus, with these extra dimensions, the above scenario is now perfect for Supersymmetry and String Theory without any further objections and/or adjustments.