Quantization of Dark Energy Particles and Sarkar Limit

# I. INTRODUCTION AND CALCULATION There is no end to scientists' curiosity about our universe. The greatest problem in the universe, especially in theoretical physics, is the problem associated with the cosmological constant. The value we obtain from theory does not match the value determined observationally. My equation has tried, in a particular way, to theoretically analyze this vacuum energy or cosmological constant and determine its nature. In the language of physics I call it the "Sarkar Limit of Inflationary Model". Here the nature of vacuum energy has been analyzed using the "Slow Roll Approximation" or model. Let us begin with the "Friedmann Equation". The "Friedmann Equation" is used to describe the dynamics of an expanding universe. $$ \mathrm {H} ^ {2} = (8 \pi \mathrm {G} / 3) \rho - (\mathrm {k c} ^ {2} / \mathrm {a} ^ {2}) + \Lambda / 3 \tag{1} $$ Now, according to the "Slow Roll Inflation" model, the total energy density will be: $$ \rho (\varphi) = (1 / 2) \varphi^ {2} + V (\varphi) \tag{2} $$ $(1 / 2)\varphi^{2} =$ Kinetic Term of field $\varphi$ $\mathrm{V}(\varphi) =$ Potential Term of field $\varphi$ Now, according to "Slow Roll Approximation" $\varphi \ll 1$ , but $\varphi > 0$ , therefore $\varphi^2 \approx 0$ may be considered. On the other hand, the potential energy of the field, that is $V(\varphi) = (1/2)\mathrm{m}^2\varphi^2$ . Here $\mathfrak{m}$ is the mass of the field or the mass of the inflation boson. Thus simplifying equation (2): $$ \rho (\varphi) = \mathrm {V} (\varphi) = (1 / 2) \mathrm {m} ^ {2} \varphi^ {2} \tag {3} $$ Now according to the Friedmann equation (that is, equation (1)), if we simplify slightly, we may set the curvature parameter $k = 0$ . Then equation (1) becomes: $$ \mathrm {H} ^ {2} = (8 \pi \mathrm {G} / 3) \rho (\varphi) + \Lambda / 3 $$ $$ 3 \mathrm {H} ^ {2} = 8 \pi \mathrm {G} \rho (\varphi) + \Lambda $$ $$ 3 \mathrm {H} ^ {2} = 4 \pi \mathrm {G} \varphi^ {2} \mathrm {m} ^ {2} + \Lambda $$ $$ \varphi^ {2} \mathrm {m} ^ {2} = (3 \mathrm {H} ^ {2} - \Lambda) / 4 \pi \mathrm {G} $$ $$ (1 / 2) \varphi^ {2} \mathrm {m} ^ {2} = (3 \mathrm {H} ^ {2} - \Lambda) / 8 \pi \mathrm {G} $$ $$ V (\varphi) = \left(3 H ^ {2} - \Lambda\right) / 8 \pi G \tag{4} $$ Another thing we get from "Slow Roll Approximation" is the pressure term $\mathrm{P}(\varphi)$ , which is: $$ \mathrm {P} (\varphi) = (1 / 2) \varphi^ {2} - \mathrm {V} (\varphi) \tag{5} $$ Since $\varphi^2\approx 0$ for slow roll, what remains is: $$ \mathrm {P} (\varphi) = - \mathrm {V} (\varphi) $$ $$ V (\varphi) = - P (\varphi) \tag{6} $$ Now substituting the value of $V(\varphi)$ from equation (6) into equation (4): $$ \mathrm {P} (\varphi) = \left(\Lambda - 3 \mathrm {H} ^ {2}\right) / 8 \pi \mathrm {G} \tag {7} $$ Now according to the mathematical definition of pressure, it is the force per unit area! The type of force depends on the pressure term. Thus: $$ \mathrm {P} (\varphi) = \mathrm {F} (\varphi) / \mathrm {A} \tag {8} $$ Here $A$ is the area. In cosmology we may think of area as the "Horizon Area". That is, $A = d(\mathrm{hor})^2$ . Again $d(\mathrm{hor}) \approx c^2\mathrm{H}^{-2}$ . Therefore from equation (8): $$ \mathrm {P} (\varphi) = \mathrm {F} (\varphi) / \mathrm {d} (\mathrm {hor}) ^ {2} = \mathrm {F} (\varphi) / \mathrm {c} ^ {2} \mathrm {H} ^ {- 2} \tag{9} $$ Here $\mathrm{H}$ is the Hubble Parameter, $\mathrm{c}$ is the speed of light. In the early moments of the universe, $\mathrm{H}$ was extremely large. Another important thing is that in equation (7), $\Lambda$ represents vacuum energy or the cosmological constant. And $\mathrm{G}$ is Newton's gravitational constant. Now substituting the value of $\mathrm{P}(\varphi)$ from equation (7) into equation (9): $$ \mathrm {F} (\varphi) / \mathrm {c} ^ {2} \mathrm {H} ^ {- 2} = \left(\Lambda - 3 \mathrm {H} ^ {2}\right) / 8 \pi \mathrm {G} $$ $$ \mathrm {F} (\varphi) = \left(\mathrm {c} ^ {2} / 8 \pi \mathrm {G}\right) \left[ \left(\Lambda / \mathrm {H} ^ {2}\right) - 3 \right] \tag{10} $$ If we compare the scale factor at the beginning of inflation a(initial) and at the end a(final), we get: $$ [ \mathrm {a (f) / a (i)} ] \approx 10^{43} $$ $$ a (f) = 10^{43} \times a (i) \tag{11} $$ Equation (11) is mathematically equivalent to $\exp(\mathrm{H}[t(\mathrm{f}) - t(\mathrm{i})])$ . Here $t(\mathrm{f})$ is the time inflation ends, and $t(\mathrm{i})$ is the time inflation begins. Since $t(\mathrm{f}) > t(\mathrm{i})$ , we can write: $$ \left[ \mathrm {a (f)} / \mathrm {a (i)} \right] = \exp (\mathrm {H [ t (f) - t (i) ]}) $$ $$ \mathrm {H} \Delta \mathrm {t} \times \ln (\mathrm {e}) = \ln [ \mathrm {a (f)} / \mathrm {a (i)} ] $$ $$ \mathrm {H} \times 10^{-33} \times \ln (2.7) = \ln (10^{43}) $$ $$ \mathrm {H} = 10^{33} \times [ \ln (10^{43}) / \ln (2.7) ] $$ $$ \mathrm {H} = 10^{35} \mathrm {km} / \sec / \mathrm {Mpc} \tag{12} $$ $$ \begin{array}{l} \text {Here} \Delta t = [ t (f) - t (i) ] \\ = 10^{-33} - 10^{-36} \\ \approx 10^{-33} \mathrm {seconds}. \\ \end{array} $$ Substituting the value of $\mathrm{H}$ from equation (12) into equation (10): $$ \mathrm {F} (\varphi) = \left(\mathrm {c} ^ {2} / 8 \pi \mathrm {G}\right) \left[ \left(\Lambda / \mathrm {H} ^ {2}\right) - 3 \right] $$ $$ \mathrm {m} (\mathrm {d} \varphi / \mathrm {d t}) = \left(\mathrm {c} ^ {2} / 8 \pi \mathrm {G}\right) \left[ \left(\Lambda / 10^{70}\right) - 3 \right] \tag{13} $$ $$ \mathrm {m} \varphi^ {\prime \prime} = (\mathrm {c} ^ {2} / 8 \pi \mathrm {G}) [ \Lambda \cdot 10^{-70} - 3 ] $$ Thus in equation (13) we have $\varphi$ on the left-hand side. According to the condition $\varphi < 1$ but $\varphi > 0$ , $\varphi$ varies infinitesimally between 0 and 1. That means we may approximate $\varphi = \mathrm{d}\varphi/\mathrm{d}t$ as an infinitesimal variation. It is not exactly zero, but extremely close to zero. I call this the "Sensitive-$\varphi$ Condition". On the left, $m$ denotes the mass of the inflaton boson, the field quanta of the inflation field. $m \approx 10^{12} \mathrm{GeV}$ . On the right side of equation (13), the theoretical value of the cosmological constant is $\Lambda \approx 10^{-53}$ , but in Planck units $\Lambda \approx 10^{-123} \times l(p)^{-2}$ . Here $l(p)$ is the Planck length. In this equation $\Lambda$ is multiplied by $H^{-2} \approx 10^{-70}$ , therefore $\Lambda = 10^{-53} \times 10^{-70}$ or $\Lambda \approx 10^{-123}$ (Hubble Time Unit)! I denote this corrected $\Lambda$ as $\Lambda(\text{theo})$ . Therefore equation (13) becomes: $$ \mathrm {m} \varphi^ {\prime \prime} = \left(\mathrm {c} ^ {2} / 8 \pi \mathrm {G}\right) \left[ \Lambda \cdot 10^{-70} - 3 \right] \tag{14} $$ $$ \mathrm {m} \varphi^ {\prime \prime} = (\mathrm {c} ^ {2} / 8 \pi \mathrm {G}) [ \Lambda (\text {theo}) - 3 ] $$ Now if Planck's constant is implemented into equation (14): $$ \hbar = \mathrm {h} / 2 \pi $$ $$ h = 2\pi \hbar \tag{15} $$ Thus the present form of equation (14) becomes: $$ \mathrm {m} \varphi^ {\prime \prime} = (\mathrm {c} ^ {2} / 4 \pi \mathrm {G}) \times (1 / 2 \pi) [ \Lambda (\mathrm {theo}) - 3 ] $$ $$ \mathrm {m} \varphi^ {\prime \prime} = \left(\mathrm {c} ^ {2} / 4 \pi \mathrm {G}\right) \times (\hbar / \mathrm {h}) \times [ \Lambda (\text{theo}) - 3 ] $$ $$ \mathrm {m} \varphi^ {\prime \prime} = (\mathrm {c} ^ {2} / 2 \mathrm {G h}) \times [ (1 / 2) \hbar \times \Lambda (\text {theo}) - (3 / 2) \hbar ] $$ $$ \mathrm {m} \varphi^ {\prime \prime} = (\mathrm {c} ^ {2} / 4 \pi \mathrm {G h}) \times [ (1 / 2) \hbar \times \Lambda (\mathrm{theo}) - (3 / 2) \hbar ] $$ Therefore: $$ \mathrm {m} \varphi^ {\prime \prime} = [ \mathrm {c} / 4 \pi \mathrm {M} (pl) ^ {2} ] \times [ (1 / 2) \hbar \times \Lambda (\text{theo}) - (3 / 2) \hbar ] \tag{16} $$ In equation (16) we beautifully obtain the Planck mass $\mathrm{M}(\mathrm{pl})$ from $\sqrt{(\mathrm{Gh} / \mathrm{c})}$ . You may verify this yourself. Now let us continue working with equation (16), especially with the final term $(3 / 2)\hbar$ . This reminds us of the quantum mechanical harmonic oscillator energy expression for the nth level: $$ \mathrm {E (Q M)} = (\mathrm {n + 1 / 2}) \hbar \omega $$ Only for $n = 1$ do we obtain $(3/2)\hbar$ : $$ \operatorname {E} (\mathrm {Q M}) = (1 + 1 / 2) \hbar \omega $$ $$ \mathrm {E} (\mathrm {Q M}) = (3 / 2) \mathrm {h} \omega $$ $$ (3 / 2) \mathrm {h} = \mathrm {E} (\mathrm {Q M}) / \omega = \mathrm {E} (\mathrm {Q M}) / 2 \pi v \tag{17} $$ In equation (17), we get the energy of the first excited state. $\omega = 2\pi \nu$ represents the transition frequency. Now substituting $(3 / 2)\hbar$ from equation (17) into equation (16): $$ \mathrm {m} \varphi^ {\prime \prime} = [ \mathrm {c} / 4 \pi \mathrm {M} (\mathrm {p l}) ^ {2} ] \times [ (1 / 2) \hbar \times \Lambda (\text{theo}) - (3 / 2) \hbar ] $$ $$ \mathrm {m} \varphi^ {\prime \prime} = [ \mathrm {c} / 4 \pi \mathrm {M} (\mathrm {p l}) ^ {2} ] \times [ (1 / 2) \hbar \times \Lambda (\text{theo}) - \mathrm {E} (\mathrm {Q M}) / 2 \pi \mathrm {v} ] $$ Or more clearly: $$ \varphi^ {\prime \prime} = [ c / 4 \pi \times 10^{12} ] \times [ 1 / M(pl) ^ {2} ] \times (1 / 2) [ \hbar \times \Lambda (\text {theo}) - E / \pi v ] \tag {18} $$ We represent $\mathrm{E}(\mathrm{QM})$ simply by $\mathrm{E}$ . Now $c / (4\pi \times 10^{12} \times \mathrm{M}(\mathrm{pl})^2) = \xi$ , which is a constant. And $[(1/2)\hbar \times \Lambda (\mathrm{theo}) - \mathrm{E} / 2\pi \nu] = \mathrm{E}(\mathrm{tot})$ , the total energy, which is actually a conserved quantity. Later we will see how energy is conserved. Then equation (18) becomes: $$ \varphi^ {\prime \prime} = \xi \times \mathrm {E} (\text{tot}) $$ $$ \left(\mathrm {d} \varphi / \mathrm {d t}\right) = \xi \times \mathrm {E} (\mathrm{tot}) $$ $$ \int (\mathrm {d} \varphi / \mathrm {d} t) \mathrm {d} t = \xi \times \mathrm {E} (\mathrm{tot}) \int \mathrm {d} t $$ $$ \varphi = \xi \times \mathrm {E} (\mathrm{tot}) \times \mathrm {t} + \mathrm {C} _ {1} $$ $$ \int d \varphi = \xi \times E(\mathrm{tot}) \times \int t d t + C _ {1} \int d t $$ $$ \Delta \varphi = \xi \times \mathrm {E} (\mathrm{tot}) \times (1 / 2) \Delta \mathrm {t} ^ {2} + \mathrm {C} _ {1} \Delta \mathrm {t} + \mathrm {C} _ {2} \tag{19} $$ In equation (19), $\Delta \varphi = \varphi (\mathrm{f}) - \varphi (\mathrm{i})$ If in $\Delta t$ we put the upper time limit $t(f)$ and lower time limit $t(i)$ during inflation, then $\Delta t = t(f) - t(i)$ . Where $t(f) = 10^{-33}$ seconds and $t(i) = 10^{-36}$ seconds. Therefore $\Delta t \approx 10^{-33}$ seconds. Thus: $$ \Delta \varphi = \xi \times \mathrm {E} (\text{tot}) \times (1 / 2) \Delta t ^ {2} + \mathrm {C} _ {1} \Delta t + \mathrm {C} _ {2} $$ $$ \varphi (f) - \varphi (i) = \xi \times E (\mathrm{tot}) \times (1 / 2) \times 10^{-66} + C _ {1} \times 10^{-33} + C _ {2} $$ $$ \Delta \varphi = \xi \times \mathrm {E} (\mathrm {tot}) \times (1 / 2) \times \mathrm {t} (\mathrm {pl}) ^ {2} \times \mathrm {M} (\mathrm {pl}) + (10^{-33} + 1) \mathrm {C} $$ In the above equation I set $\mathrm{C}_1 = \mathrm{C}_2 = \mathrm{C}$ for mathematical convenience. $\mathrm{t}(\mathrm{pl}) \approx 10^{-43}$ sec, and $\mathrm{M}(\mathrm{pl})$ is about $10^{20}$ GeV. Here $(10^{-33} + 1)\mathrm{C} \approx \mathrm{C}$ . Thus: $$ \Delta \varphi = (1 / 2) [ c / (4 \pi \times 10^{12}) ] E (\mathrm {t o t}) \{t (\mathrm {p l}) ^ {2} / M (\mathrm {p l}) \} + C $$ $$ \Delta \varphi = (1 / 2) \times \xi \times [ t (\mathrm {p l}) ^ {2} / \mathrm {M} (\mathrm {p l}) ] \times \mathrm {E} (\mathrm {tot}) + \mathrm {C} \tag {20} $$ According to equation (20), $\mathrm{C} < < 1$ and $\mathrm{C} > 0$ . Thus $\mathrm{C} \approx 0$ may be considered, meaning extremely close to zero. As we assumed a vanishing contribution for $\varphi$ under the "Sensitive- $\varphi$ " condition. The "Sensitive- $\varphi$ " condition allows $\varphi$ to stand simultaneously near zero and non-zero. For mathematical convenience $\varphi$ can be expressed in multiple ways. And quantum mechanics inherently deals with microscopic or infinitesimal corrections, matching perfectly with the Sensitive- $\varphi$ condition. Therefore: $$ \Delta \varphi = (1 / 2) \times \xi \times [ t(pl) ^ {2} / M(pl) ] \times E (\text{tot}) + C $$ $$ \varphi (f) - \varphi (i) = (1 / 2) \times \xi \times [ t(pl) ^ {2} / M(pl) ] \times E (\text{tot}) + C $$ $$ \varphi (\mathrm {f}) = \varphi (\mathrm {i}) + \left[ (1 / 2) \times \xi \times [ \mathrm {t} (\mathrm {pl}) ^ {2} / \mathrm {M} (\mathrm {pl}) ] \times \mathrm {E} (\mathrm {tot}) + \mathrm {C} \right] \tag {21} $$ Thus from equation (21), $\varphi (\mathrm{f})$ is slightly larger than $\varphi (\mathrm{i})$ . A minute deviation is noticeable. In quantum mechanics such minute variations cannot be ignored. In classical physics microscopic changes can often be neglected without affecting the system's macroscopic behavior. But quantum systems do not behave that way. From equation (21), E(tot) acts as the total energy term. We may call it the Hamiltonian: $\mathrm{E}(\mathrm{tot}) =$ Hamiltonian Of The System That is: $$ \mathrm {H} = \left[ (1 / 2) \hbar \times \Lambda (\text {theo}) - \mathrm {E} / 2 \pi \mathrm {v} \right] \tag{22} $$ Now we will express H or the Hamiltonian in quantum mechanical terminology. For this we define $\hat{\mathrm{H}}$ , the Hamiltonian operator: $$ \hat {\mathrm {H}} = \left[ (1 / 2) \hbar \times \Lambda (\text {theo}) - (\mathrm {i} \hbar \partial / \partial \mathrm {t}) \times (1 / 2 \pi \mathrm {v}) \right] \tag{23} $$ Here $\mathbf{E}$ is written in operator form $\hat{\mathbf{E}}$ : $$ \hat {\mathrm {E}} \rightarrow \mathrm {i} \hbar (\partial / \partial t) \tag{24} $$ Now if a state (particle state) is $|\Psi \rangle$ , acting $\hat{\mathrm{H}}$ on it gives: $$ \hat {\mathrm {H}} | \Psi \rangle = [ (1 / 2) \hbar \times \Lambda (\text {theo}) - (\mathrm {i} \hbar \partial / \partial t) \times (1 / 2 \pi v) ] | \Psi \rangle $$ $$ \hat {\mathrm {H}} | \Psi \rangle = (1 / 2) \mathrm {h} \times \Lambda (\text {theo}) | \Psi \rangle - (1 / 2 \pi \mathrm {v}) \times (\mathrm {i} \hbar \partial / \partial \mathrm {t}) | \Psi \rangle $$ $$ \hat {\mathrm {H}} | \Psi \rangle = 0 - (1 / 2 \pi \mathrm {v}) \times (\mathrm {i} \hbar \partial / \partial \mathrm {t}) | \Psi \rangle $$ $$ \hat {\mathrm {H}} | \Psi \rangle = - [ \mathrm {E} / (2 \pi \mathrm {v}) ] | \Psi \rangle \tag{25} $$ Here $-[\mathrm{E} / (2\pi \nu)]$ represents those states possessing negative energy eigenvalues. Another thing we obtain from quantum mechanics is the entity L, whose expectation value $\langle \mathbf{L}\rangle$ is defined: $$ \langle \mathrm {L} \rangle = \langle \Psi | \mathrm {L} | \Psi \rangle $$ $$ \left(\mathrm {d} / \mathrm {d t}\right) \langle \mathrm {L} \rangle = \left(\mathrm {d} / \mathrm {d t}\right)\langle \Psi | \mathrm{L} | \Psi \rangle $$ $$ \mathrm {d} \langle \mathrm {L} \rangle / \mathrm {d t} = \langle \Psi | \mathrm {L} | \Psi \rangle $$ $$ \mathrm {d} \langle \mathrm {L} \rangle / \mathrm {d t} = (\mathrm {i} \hbar / 2 \pi \mathrm {v}) | \langle \Psi | \mathrm {H L} | \Psi \rangle - \langle \Psi | \mathrm {L H} | \Psi \rangle | $$ Thus: $$ \mathrm {d} \langle \mathrm {L} \rangle / \mathrm {d t} = (\mathrm {i} \hbar / 2 \pi \mathrm {v}) | \langle \Psi | (\mathrm {H L} - \mathrm {L H}) | \Psi \rangle | $$ $$ \mathrm {d} \langle \mathrm {L} \rangle / \mathrm {d t} = (\mathrm {i} \hbar / 2 \pi \mathrm {v}) [ \mathrm {H}, \mathrm {L} ] \tag{26} $$ Now if $\mathrm{L} = \mathrm{H}$ , since both the Lagrangian $\mathrm{L}$ and Hamiltonian $\mathrm{H}$ are conserved quantities, we may say: $$ \mathrm {d} \langle \mathrm {H} \rangle / \mathrm {d t} = (\mathrm {i \hbar} / 2 \pi \mathrm {v}) [ \mathrm {H}, \mathrm {H} ] $$ $$ \mathrm {d} \langle \mathrm {H} \rangle / \mathrm {d t} = (\mathrm {i h} / 2 \pi \mathrm {v}) [ \mathrm {H H} - \mathrm {H H} ] $$ $$ \mathrm {d} \langle \mathrm {H} \rangle / \mathrm {d t} = 0 $$ $$ \langle \mathrm {H} \rangle = \text {Conserved} \tag{27} $$ Now let us see whether $\varphi$ and H exhibit a commutation relation: $$ \begin{array}{l} [ \boldsymbol {\varphi} ^ {\hat {\mathbf {\Gamma}}}, \hat {\mathbf {H}} ] \\ = (\varphi \mathrm {H} - \mathrm {H} \varphi) \\ = \varphi \left[ (1 / 2) \hbar \Lambda (\text {theo}) - (E / 2 \pi v) \right] - \left[ (1 / 2) \hbar \Lambda (\text {theo}) - (E / 2 \pi v) \right] \varphi \\ = (1 / 2 \pi \mathrm {v}) [ - \varphi \mathrm {i} \hbar (\partial / \partial \mathrm {t}) + \mathrm {i} \hbar (\partial / \partial \mathrm {t}) \varphi ] \\ \end{array} $$ Thus: $$ [ \hat {\boldsymbol {\varphi}}, \hat {\mathbf {H}} ] = (1 / 2 \pi \mathbf {v}) \mathrm {i} \hbar (\partial \boldsymbol {\varphi} / \partial t) $$ $$ \langle \dot {\varphi} \rangle = (2 \pi \mathrm {v} / \mathrm {i} \hbar) [ \varphi^ {\hat {\prime}}, \hat {\mathrm {H}} ] $$ $$ \langle \dot {\varphi} \rangle = (- \mathrm {i} 2 \pi \mathrm {v} / \hbar) [ \varphi^ {\hat {\prime}}, \hat {\mathrm {H}} ] \tag{28} $$ Now let us check if $\varphi$ and $V(\varphi)$ commute: $$ [ \varphi^ {\hat {\mathbf {\Gamma}}}, \mathrm {V} ^ {\hat {\mathbf {\Gamma}}} ] = (\varphi \mathrm {V} - \mathrm {V} \varphi) $$ $$ \mathrm {V} = (1 / 2) \mathrm {m} ^ {2} \varphi^ {2} $$ Thus: $$ [ \varphi^ {\hat {\mathbf {\Gamma}}}, \mathrm {V} ^ {\hat {\mathbf {\Gamma}}} ] = 1 / 2 (\mathrm {m} ^ {2} \varphi^ {3} - \mathrm {m} ^ {2} \varphi^ {3}) $$ $$ [ \varphi^ {\hat {\mathbf {\Gamma}}}, \mathrm {V} ^ {\hat {\mathbf {\Gamma}}} ] = 0 $$ $$ \langle 0 | [ \varphi^ {\hat {\prime}}, V ^ {\hat {\prime}} ] | 0 \rangle = 0 \tag{29} $$ Now returning to equation (20): $$ \Delta \varphi = (1 / 2) \times \xi \times [ t (\mathrm {p l}) ^ {2} / M (\mathrm {p l}) ] \times E (\mathrm{tot}) + C $$ If the "Sensitive-φ" condition is again implemented, we may say that: $(1/2) \times \xi \times [\mathrm{t}(\mathrm{pl})^2/\mathrm{M}(\mathrm{pl})] \times \mathrm{E}(\mathrm{tot})$ is much smaller than 1 but slightly greater than 0, essentially almost zero. The purpose of choosing such a limit is that it allows us to track how the mathematical terms change. In quantum systems such tiny variations cannot be ignored. Because minute variations give rise to large-scale effects. We stand right at the boundary of the slow roll approximation – the "Sensitive-φ" condition. Thus $\Delta \varphi \approx \mathrm{C}$ only when: $$ (1 / 2) \times \xi \times [ t(pl) ^ {2} / M(pl) ] \times E (\text {t o t}) \approx 0 \tag{30} $$ $$ \mathrm {E} (\text {t o t}) \approx 0 $$ $$ (1 / 2) \hbar \times \Lambda (\text {theo}) - (\mathrm {i} \hbar \partial / \partial \mathrm {t}) \times (1 / 2 \pi \mathrm {v}) \approx 0 $$ $$ (1 / 2) \mathrm {h} \times \Lambda (\text {theo}) - \mathrm {E} (\mathrm {Q . M}) \times (1 / 2 \pi \mathrm {v}) \approx 0 $$ $$ (1 / 2) \mathrm {h} \times \Lambda (\text {theo}) \approx \mathrm {E} (\mathrm {Q}. \mathrm {M}) \times (1 / 2 \pi \mathrm {v}) $$ $$ \hbar \times \Lambda (\text {theo}) \approx \mathrm {E} (\mathrm {Q . M}) \times (1 / \pi v) $$ $$ \mathrm {E} (\mathrm {Q . M}) \approx \pi \mathrm {v h} \times \Lambda (\text {theo}) \tag {31} $$ Here $\pi$ is a number, $\nu$ or transition frequency is a number, $\hbar$ is a very small positive number, and $\Lambda(\mathrm{theo})$ is the positive cosmological constant, which itself is very small. A universe with a positive cosmological constant always expands. If there were a negative sign before $\Lambda(\mathrm{theo})$ , it would act not as a repulsive field but as an attractive one. In that case expansion would slow down or the universe would collapse back into a singularity. Dark energy field does not arise from ordinary matter particles. Instead, it determines energy states that possess negative energy eigenvalues, as seen from equation (25). That is why in the "Slow Roll Inflation" model we obtained $\mathrm{P}(\varphi) = -\mathrm{V}(\varphi)$ , $\rho(\varphi) = \mathrm{V}(\varphi)$ , or $\rho(\varphi) =$ $-\mathrm{P}(\varphi)$ , or $\mathrm{P}(\varphi) = -\rho (\varphi)$ , derived from $\mathrm{P} = \omega \rho$ where $\omega = -1$ for vacuum energy or cosmological constant. In the early universe or at the beginning of inflation, the universe starts from a non-zero cosmological constant. When inflation ends and reheating begins, through thermalization dark energy or cosmological constant transforms into ordinary matter and dark matter. Despite its very small value, the effective repulsive strength of dark energy is much greater than the attractive gravity of ordinary matter or dark matter. That is why the universe is expanding. # II. CONCLUSION In conclusion, we may say that even vacuum carries energy. Because it contributes to a type of particle which possesses negative energy eigenvalues. And my method has beautifully quantified the dark energy particle. According to equation (23): $$ \begin{array}{l} \hat {\mathrm {H}} = \left[ (1 / 2) \hbar \times \Lambda (\text{theo}) - (\mathrm {i} \hbar \partial / \partial t) \times (1 / 2 \pi v) \right] \tag{23} \\ \hat {\mathrm {H}} = \left[ (1 / 2) \hbar \times \Lambda (\text{theo}) + \{- (\mathrm {i} \hbar \partial / \partial t) \times (1 / 2 \pi v) \} \right] \\ \hat {\mathrm {H}} = (1 / 2) \hbar \times \Lambda (\text{theo}) + \mathrm{E} ^ {\prime} \times (1 / 2 \pi \mathrm{v}) \tag{32} \\ \end{array} $$ It may also be written this way. Notably, in $\hat{\mathrm{H}}$ we do not obtain any potential term. Because in the earliest epoch of the universe gravity was repulsive. Only in ordinary Newtonian or classical gravity does a potential term appear. However, in one sense $\Lambda$ (theo) behaves like a negative potential of the field. The entire matter depends on how mathematical terms may be related to each other. And the force, that's I got in the equation (8), I called it Scalar force! It's probably the Sakharov or the Sakharov definition of scalar force.

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