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.

Now, according to the "Slow Roll Inflation" model, the total energy density will be:

$(1/2){\phi }^2=$ Kinetic Term of field $\phi$

$\mathrm{V}(\phi )=$ Potential Term of field $\phi$

Now, according to "Slow Roll Approximation"

$\phi \ll 1$ , but , therefore ${\phi }^2\approx 0$ may be considered. On the other hand, the potential energy of the field, that is $V(\phi )=(1/2){\mathrm{m}}^2{\phi }^2$ . Here $\mathfrak{m}$ is the mass of the field or the mass of the inflation boson. Thus simplifying equation (2):

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:

Another thing we get from "Slow Roll Approximation" is the pressure term $\mathrm{P}(\phi )$ , which is:

Since ${\phi }^2\approx 0$ for slow roll, what remains is:

Now substituting the value of $V(\phi )$ from equation (6) into equation (4):

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:

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):

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), $\mathrm{\Lambda }$ represents vacuum energy or the cosmological constant. And $\mathrm{G}$ is Newton's gravitational constant. Now substituting the value of $\mathrm{P}(\phi )$ from equation (7) into equation (9):

If we compare the scale factor at the beginning of inflation a(initial) and at the end a(final), we get:

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 , we can write:

Substituting the value of $\mathrm{H}$ from equation (12) into equation (10):

Thus in equation (13) we have $\phi$ on the left-hand side. According to the condition but , $\phi$ varies infinitesimally between 0 and 1. That means we may approximate $\phi =\mathrm{d}\phi /\mathrm{d}t$ as an infinitesimal variation. It is not exactly zero, but extremely close to zero. I call this the "Sensitive-$\phi$ 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 $\mathrm{\Lambda }\approx {10}^{-53}$ , but in Planck units $\mathrm{\Lambda }\approx {10}^{-123}\times l(p{)}^{-2}$ . Here $l(p)$ is the Planck length. In this equation $\mathrm{\Lambda }$ is multiplied by $H^{-2}\approx {10}^{-70}$ , therefore $\mathrm{\Lambda }={10}^{-53}\times {10}^{-70}$ or $\mathrm{\Lambda }\approx {10}^{-123}$ (Hubble Time Unit)! I denote this corrected $\mathrm{\Lambda }$ as $\mathrm{\Lambda }(\text{theo})$ . Therefore equation (13) becomes:

Now if Planck's constant is implemented into equation (14):

Thus the present form of equation (14) becomes:

Therefore:

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 $(3/2)\hslash$ . This reminds us of the quantum mechanical harmonic oscillator energy expression for the nth level:

Only for $n=1$ do we obtain $(3/2)\hslash$ :

In equation (17), we get the energy of the first excited state. $\omega =2\pi \nu$ represents the transition frequency. Now substituting $(3/2)\hslash$ from equation (17) into equation (16):

Or more clearly:

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 $[(1/2)\hslash \times \mathrm{\Lambda }(\mathrm{theo})-\mathrm{E}/2\pi \nu ]=\mathrm{E}(\mathrm{tot})$ , the total energy, which is actually a conserved quantity. Later we will see how energy is conserved. Then equation (18) becomes:

In equation (19), $\mathrm{\Delta }\phi =\phi (\mathrm{f})-\phi (\mathrm{i})$

If in $\mathrm{\Delta }t$ we put the upper time limit $t(f)$ and lower time limit $t(i)$ during inflation, then $\mathrm{\Delta }t=t(f)-t(i)$ . Where $t(f)={10}^{-33}$ seconds and $t(i)={10}^{-36}$ seconds. Therefore $\mathrm{\Delta }t\approx {10}^{-33}$ seconds. Thus:

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:

According to equation (20), and . Thus $\mathrm{C}\approx 0$ may be considered, meaning extremely close to zero. As we assumed a vanishing contribution for $\phi$ under the "Sensitive- $\phi$ " condition. The "Sensitive- $\phi$ " condition allows $\phi$ to stand simultaneously near zero and non-zero. For mathematical convenience $\phi$ can be expressed in multiple ways. And quantum mechanics inherently deals with microscopic or infinitesimal corrections, matching perfectly with the Sensitive- $\phi$ condition. Therefore:

Thus from equation (21), $\phi (\mathrm{f})$ is slightly larger than $\phi (\mathrm{i})$ . 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:

Now we will express H or the Hamiltonian in quantum mechanical terminology. For this we define $\hat{\mathrm{H}}$ , the Hamiltonian operator:

Here $\mathbf{E}$ is written in operator form $\hat{\mathbf{E}}$ :

Now if a state (particle state) is $|\mathrm{\Psi }⟩$ , acting $\hat{\mathrm{H}}$ on it gives: