The influence of the constraints Friedmann type
$\phi \left(x,{x}^{\prime}\right)=\frac{{{x}^{\prime}}^{2}}{{a}_{0}^{2}}-\frac{{x}^{2}}{{b}_{0}^{2}}-k=0,$
*k* = 0, ± 1, *a*_{0}, *b*_{0} – constants on the dynamics of a free particle is investigated. In a phase space {*x*, *x*′}, these constraints
$\phi \left(x,{x}^{\prime}\right)=\frac{{{x}^{\prime}}^{2}}{{a}_{0}^{2}}-\frac{{x}^{2}}{{b}_{0}^{2}}-k=0,$
while *k* = ± 1 are canonical equations of hyperbola or equations of the second-orderhyperbolic cylinder in a phase space {*x*, *x*′, *z*}.

For the investigation the variational calculus is applied. From the action extremum condition with constraints *δS* = 0 Lagrange equation follows
$$\begin{array}{}{\displaystyle \frac{d}{dt}\frac{\mathrm{\partial}L}{\mathrm{\partial}{x}^{\prime}}-\frac{\mathrm{\partial}L}{\mathrm{\partial}x}=0,}\\ {\displaystyle \frac{\mathrm{\partial}L}{\mathrm{\partial}\lambda}=\frac{{{x}^{\prime}}^{2}}{{a}_{0}^{2}}-\frac{{x}^{2}}{{b}_{0}^{2}}-k=0,}\end{array}$$(1)

where Lagrange function is *L* (*x*, *x*’) =
$\frac{m{{x}^{\prime}}^{2}}{2}+\lambda \phi \left(x,{x}^{\prime}\right).$
Let us consider the dynamics of a free particle with hyperbolic constraint *φ* (*x*, *x*′) =
$\frac{{{x}^{\prime}}^{2}}{{a}_{0}^{2}}-\frac{{x}^{2}}{{b}_{0}^{2}}-k=0.$ In such a case, Lagrange equation takes the form
$$\begin{array}{}{\displaystyle \left(m+\frac{2\lambda}{{a}_{0}^{2}}\right){x}^{\u2033}=\frac{2\lambda}{{b}_{0}^{2}}x,}\\ {\displaystyle \left(\frac{{{x}^{\prime}}^{2}}{{a}_{0}^{2}}-\frac{{x}^{2}}{{b}_{0}^{2}}-k\right)=0.}\end{array}$$

The solution of the system of equations (1) with hyperbolic constraints is infinite and takes the form
$$\begin{array}{}{\displaystyle x=\left\{\begin{array}{l}{\stackrel{~}{x}}_{0}\mathrm{sh}\left({H}_{0}t\right),\phantom{\rule{thinmathspace}{0ex}}k=1\\ {\stackrel{~}{x}}_{0}\mathrm{ch}\left({H}_{0}t\right),\phantom{\rule{thinmathspace}{0ex}}k=-1\\ {\stackrel{~}{x}}_{0}\mathrm{exp}\left({H}_{0}t\right),\phantom{\rule{thinmathspace}{0ex}}k=0\end{array}\right.}\end{array}$$

Where
${a}_{0}={\stackrel{~}{x}}_{0}{H}_{0},\phantom{\rule{thinmathspace}{0ex}}{b}_{0}={\stackrel{~}{x}}_{0},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\lambda =-\frac{m{a}_{0}^{2}}{4},\phantom{\rule{thinmathspace}{0ex}}{\stackrel{~}{x}}_{0},\phantom{\rule{thinmathspace}{0ex}}{H}_{0}-$ integration constants. It should be noted that the system has a trivial solution describing the quiescent state. We can assume that for the constraint *φ*_{i} (*x*, *x*′) =
$\frac{{{x}^{\prime}}^{2}}{{a}_{0}^{2}}-1=0,$ the system of equations (1) has a Newton’s inertial solution of *x* = *v*_{0}*t*.

It should be also noted that if the variable *x* is equal to a scale factor *a* and the constant of integration which has a frequency dimension *H*_{0} (*ħ* = *c* = 1) is equal to the Hubble value
${H}_{0}=\sqrt{\frac{8\pi G\left|{U}_{0}\right|}{3}},$ then equations (1) are similar to Friedmann equations for a de Sitter universe filled with vacuum-like environment with the equation of state *ε* + *p* = 0, *p* = − *U*_{0} and with the metric *dS*^{2} = *N*^{2} (*dx*^{0})^{2} −
${a}^{2}\left[\frac{d{r}^{2}}{1-k{r}^{2}}+{r}^{2}\left(d{\vartheta}^{2}+{\mathrm{sin}}^{2}\left(\vartheta \right)d{\phi}^{2}\right)\right]$
and with time *dt* = *Ndx*^{0}, where *U*_{0} > 0– density of the potential energy of the scalar field:
$$\begin{array}{}{\displaystyle \frac{{x}^{\u2033}}{x}=\frac{8\pi G}{3}{U}_{0},}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}{\left(\frac{{x}^{\prime}}{x}\right)}^{2}=\frac{8\pi G}{3}{U}_{0}\pm \frac{{v}_{0}^{2}}{{x}^{2}},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{v}_{0}={\stackrel{~}{x}}_{0}{H}_{0}.}\end{array}$$

While *v*_{0} = *c*, these equations are equal to Friedmann equations, where *c* is the speed of light. In the general theory of relativity, the case *k* = 0 corresponds to the flat Universe, *k* = ± 1 corresponds to the closed and open Universe. Inertial constraint
${\phi}_{i}\left(x,{x}^{\prime}\right)=\frac{{{x}^{\prime}}^{2}}{{a}_{0}^{2}}-1=0$ in the theory of relativity corresponds to the quintessence with the equation of state
$p=-\phantom{\rule{thinmathspace}{0ex}}\frac{\epsilon}{3}.$

Let us consider the dynamics of a free particle with elliptic constraint *φ* (*x*, *x*′) =
$\phi \left(x,{x}^{\prime}\right)=\frac{{{x}^{\prime}}^{2}}{{a}_{0}^{2}}+\frac{{x}^{2}}{{b}_{0}^{2}}-1=0,$
which is similar to Friedmann energy equation with the density of the potential energy ε = *U*_{0} < 0. In such a case, Lagrange equation takes the form
$$\left(m+\frac{2\lambda}{{a}_{0}^{2}}\right){x}^{\u2033}=\frac{2\lambda}{{b}_{0}^{2}}x,$$(2)

$$\left(\frac{{{x}^{\prime}}^{2}}{{a}_{0}^{2}}+\frac{{x}^{2}}{{b}_{0}^{2}}-1\right)=0.$$(3)

The solution of the system (2, 3) is finite *x* =
${\stackrel{~}{x}}_{0}$ sin (*H*_{0}*t*),
$\lambda =-\frac{m{a}_{0}^{2}}{4},\phantom{\rule{thinmathspace}{0ex}}{a}_{0}={\stackrel{~}{x}}_{0}{H}_{0},\phantom{\rule{thinmathspace}{0ex}}{b}_{0}={\stackrel{~}{x}}_{0}.$
From the equations (2, 3), it follows
$\frac{d}{dt}\left[\frac{m{{x}^{\prime}}^{2}}{2}+\frac{m{H}_{0}^{2}{x}^{2}}{2}\right]=0$
so that
$\left[\frac{m{{x}^{\prime}}^{2}}{2}+\frac{m{H}_{0}^{2}{x}^{2}}{2}\right]=\frac{m{H}_{0}^{2}{\stackrel{~}{x}}_{0}^{2}}{2}=E=const$

For the elliptical constraint, the Hamiltonian corresponds to the constraints and
is equal to
$$H=\pi {x}^{\prime}-L=-\lambda \left[\frac{{{x}^{\prime}}^{2}}{{a}_{0}^{2}}+\frac{{x}^{2}}{{b}_{0}^{2}}-1\right]=\frac{1}{2}\left[h-E\right]=0$$(4)

where
$h=\frac{{\pi}^{2}}{2m}+\frac{m{H}_{0}^{2}}{2}{x}^{2},\pi =\frac{\mathrm{\partial}L}{\mathrm{\partial}{x}^{\prime}}=2\left(m+\frac{2\lambda}{{a}_{0}^{2}}\right){x}^{\prime}=m{x}^{\prime}-$ generalized momentum of the particle.

It should be noted that for the hyperbolic constraints the value of the path curvature inverse of the radius of the circle of curvature in the phase space is nonzero
$${k}_{0}=\frac{{x}^{\prime}{y}^{\u2033}-{y}^{\prime}{x}^{\u2033}}{{\left[{\left({x}^{\prime}\right)}^{2}+{\left({y}^{\prime}\right)}^{2}\right]}^{3/2}}=\frac{\mp 1}{{\stackrel{~}{x}}_{0}{\left[1+2{\mathrm{sh}}^{2}\left({H}_{0}t\right)\right]}^{\frac{3}{2}}}\ne 0,$$

where
${x}^{\prime}=\frac{dx}{dt},{y}^{\prime}=\frac{dy}{dt},\phantom{\rule{thinmathspace}{0ex}}y=\frac{dx}{d\tau},\phantom{\rule{thinmathspace}{0ex}}\tau ={H}_{0}t.$
For the elliptic constraint
${k}_{0}=-\frac{1}{{\stackrel{~}{x}}_{0}}.$
For the inertial constraint *φ*_{i} (*x*, *x*′) =
$\frac{{{x}^{\prime}}^{2}}{{a}_{0}^{2}}-1=0,$
a path curvature *k*_{0} = 0. This means that the potential term in the Hamiltonian (4) is determined by the path curvature in the phase space. Obviously, a curved line in a flat phase space can be interpreted as a geodesic line in the corresponding curved phase space. At the same time it means that in such a case the constraints can modify Newton’s first law.

Since equality (4) *H* = 0 is equivalent to equality *h* = *E*, then from the expression (4), the unsteady-state equation follows
$$i\hslash \frac{\mathrm{\partial}}{\mathrm{\partial}t}\mathit{\Psi}\left(x,t\right)=\hat{h}\mathit{\Psi}\left(x,t\right),$$(5)

where
$\hat{h}=\left[\frac{{\hat{\pi}}^{2}}{2m}+\frac{m{H}_{0}^{2}}{2}{x}^{2}\right],\phantom{\rule{thinmathspace}{0ex}}\mathit{\Psi}\left(x,t\right)={e}^{-iEt}\mathit{\Psi}\left(x\right),$ so that owing to the energy constant
$\frac{m{H}_{0}^{2}{x}_{0}^{2}}{2}\phantom{\rule{1em}{0ex}}=\phantom{\rule{1em}{0ex}}E\phantom{\rule{1em}{0ex}}=$ *const* time does not vanish in quantum mechanics with constraints
$\phi \left(x,{x}^{\prime}\right)=\frac{{{x}^{\prime}}^{2}}{{a}_{0}^{2}}-\frac{{x}^{2}}{{b}_{0}^{2}}-k=0,$ while *k* = ± 1 and
$\phi \left(x,{x}^{\prime}\right)=\frac{{{x}^{\prime}}^{2}}{{a}_{0}^{2}}-\frac{{x}^{2}}{{b}_{0}^{2}}-1=0.$ Time vanishes only for the constraint
$\phi \left(x,{x}^{\prime}\right)=\frac{{{x}^{\prime}}^{2}}{{a}_{0}^{2}}-\frac{{x}^{2}}{{b}_{0}^{2}}-k=0$ if *k* = 0, since in this case instead of the equation (5) occurs equation *ĥ* Ψ (*x*,*t*) = 0,
$\hat{h}=\left[\frac{{\hat{\pi}}^{2}}{2m}-\frac{m{H}_{0}^{2}}{2}{x}^{2}\right].$

The solution of equation (5) and the condition for quantization have a known form
$$\begin{array}{}{\displaystyle {\mathit{\Psi}}_{n}\left(\xi \right)={N}_{0}\mathrm{exp}\left(-\frac{{\xi}^{2}}{2}\right){H}_{n}\left(\xi \right),}\\ {\displaystyle {H}_{n}\left(\xi \right)={\left(-1\right)}^{n}{e}^{{\xi}^{2}}\frac{{d}^{n}}{d{\xi}^{n}}\left[{e}^{-{\xi}^{2}}\right],}\\ {\displaystyle {E}_{n}=\hslash {H}_{0}\left(n+\frac{1}{2}\right),}\end{array}$$

where *N*_{0} – normalization constant, *H*_{n} (*ξ*) – Hermitian polynomial,
$\xi =\frac{x}{{\alpha}_{0}},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\alpha}_{0}=\sqrt{\frac{\hslash}{m{H}_{0}}}.$

For the hyperbolic constraints
$\phi \left(x,{x}^{\prime}\right)=\frac{{{x}^{\prime}}^{2}}{{a}_{0}^{2}}-\frac{{x}^{2}}{{b}_{0}^{2}}-k=$ 0 if *k* = ± 1 Schrodinger steady-state equation has a form
$$\left[\frac{{\hat{\pi}}^{2}}{2m}-\frac{m{H}_{0}^{2}}{2}{x}^{2}\right]\mathit{\Psi}\left(x\right)=\pm E\mathit{\Psi}\left(x\right),$$(6)

so that in this case the potential term of geometric origin determined by the hyperbolic constraint
$-\frac{m{H}_{0}^{2}}{2}{x}^{2}$ is the barrier while the energy spectrum is continuous.

In this case, quantum phenomena can occur as well. For example, a barrier equation (6) takes place in the theory of particle production from vacuum by intensive homogeneous electric field. Functions (7) essentially coincide with the solutions leading to Klein paradox. It implies that the current of the passing particles is higher than the current of the falling particles by the potential barrier with the height > 2*m*_{0}*c*^{2}. The exceedance of current, mentioned by Klein, is determined by the increase of the whole number of the particles as a result of pair production by the field of the barrier.

This process is investigated within the field theory [19]. In case of the potential barrier, equation (6) in dimensionless variables takes the form
$$\left[\frac{{d}^{2}}{d{\xi}^{2}}-\mu +{\xi}^{2}\right]\mathit{\Psi}\left(\xi \right)=0,\phantom{\rule{thinmathspace}{0ex}}\mu =\frac{2E}{{H}_{0}}$$(7)

The solutions of equation (7) are the function of parabolic cylinder
$${D}_{\nu}\left(z\right),\phantom{\rule{thinmathspace}{0ex}}{D}_{{\nu}^{\bullet}}\left({z}^{\bullet}\right),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\nu =-\frac{1}{2}\pm i\frac{\mu}{2},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}z=\pm \left(1\pm i\right)\xi .$$

For the equation
$$\left[\frac{{d}^{2}}{d{\xi}^{2}}+\mu +{\xi}^{2}\right]\mathit{\Psi}\left(\xi \right)=0$$

the solutions take the form
$${D}_{\nu}\left(z\right),\phantom{\rule{thinmathspace}{0ex}}{D}_{{\nu}^{\bullet}}\left({z}^{\bullet}\right),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\nu =-\frac{1}{2}\pm i\frac{\mu}{2},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}z=\pm \left(1\mp i\right)\xi .$$

For the function of the parabolic cylinder there is a relation
$$D}_{\nu}\left(z\right)={D}_{\nu}\left(-z\right){e}^{i\pi \nu}+\frac{\sqrt{2\pi}}{\mathit{\Gamma}\left(-\nu \right)}{D}_{-\nu -1}\left(-iz\right){e}^{i\frac{\pi \left(1+\nu \right)}{2}$$(8)

The scalar product of such solutions is the Wronskian determinant
$${D}_{\nu}\left[z\right]\phantom{\rule{thinmathspace}{0ex}}\frac{\overleftrightarrow{\mathrm{\partial}}}{\mathrm{\partial}z}{D}_{-\nu -1}\left[iz\right]\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=-i{e}^{-\phantom{\rule{thinmathspace}{0ex}}\frac{\pi \nu i}{2}},$$

so, that if
$\nu =-\frac{1}{2}+i\frac{\mu}{2},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}z=-\left(1-i\right)\xi $
$$D}_{-\frac{1}{2}-i\frac{\mu}{2}}\left[-\left(1+i\right)\xi \right]\phantom{\rule{thinmathspace}{0ex}}i\frac{\overleftrightarrow{\mathrm{\partial}}}{\mathrm{\partial}x}{D}_{-\frac{1}{2}+i\frac{\mu}{2}}\left[-\left(1-i\right)\xi \right]\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}={\alpha}_{0}^{-1}\sqrt{2}{e}^{\frac{\pi \mu}{4}$$(9)

The asymptotics of a function
${D}_{-\frac{1}{2}+i\frac{\mu}{2}}\left[-\left(1-i\right)\xi \right]\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{for}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mu >>1$ have the form
$$\begin{array}{}{D}_{-\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{1}{2}+i\frac{\mu}{2}}\left[-\left(1-i\right)\xi \right]\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{1}{\sqrt[4]{2{\xi}^{2}}}\mathrm{exp}\left[\phi \left(\xi \right)+\frac{i{\xi}^{2}}{2}\right],\\ {D}_{-\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{1}{2}+i\frac{\mu}{2}}\left[-\left(1-i\right)\xi \right]\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{1}{\sqrt[4]{2{\xi}^{2}}}\mathrm{exp}\left[-3\phi \left(-\xi \right)+\frac{i{\xi}^{2}}{2}\right],\end{array}$$

where
$\phi \left(\xi \right)\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\frac{\pi}{8}\left(i+\mu \right)\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}\frac{i\mu}{4}\mathrm{ln}2\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}\frac{i\mu}{2}\mathrm{ln}\left(\xi \right).$

This means that the function
${D}_{-\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{1}{2}+i\frac{\mu}{2}}\left[-\left(1-i\right)\xi \right]$ corresponds to a positive spatial frequency solution for *ξ* → ± ∞, *i.e*. antiparticle. The complex conjugate solution corresponds to a negative spatial frequency, *i.e*. particle. This means that, although the field exists everywhere, it is possible to specify regions of space in which the states of particles and antiparticles can be determined. From the asymptotics it can be obtained that with μ ≥ 1 the length of the formation of the process *Δx* ≈ *μα*_{0}, and with *μ* < 1 this length equal to *Δx* ≈ *α*_{0} [19].

From the asymptotics of the functions of a parabolic cylinder it follows that ^{(±)} *Ψ* (*ξ*) =
${N}_{\nu}{D}_{-\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{1}{2}\mp \frac{i\mu}{2}}\left[-\left(1\pm i\right)\xi \right]\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left({}_{\left(\pm \right)}\mathit{\Psi}\left(\xi \right)={N}_{\nu}{D}_{-\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{1}{2}\pm \frac{i\mu}{2}}\left[\left(1\mp i\right)\xi \right]\right)$ are the solutions of equations (5, 7) thathavepositiveornegativefrequencies corresponding to the index (+) and (−), while *ξ* → − ∞ ( *ξ* → + ∞). The functions _{(±)}*Ψ* (*ξ*) satisfy normalization and orthogonality conditions
$$\begin{array}{}\left({}_{\left(\pm \right)}\mathit{\Psi},{}_{\left(\pm \right)}\mathit{\Psi}\right)\equiv {\phantom{\rule{thinmathspace}{0ex}}}_{\left(\pm \right)}{\mathit{\Psi}}^{\bullet}\left(\xi \right)\phantom{\rule{thinmathspace}{0ex}}i\frac{\overleftrightarrow{\mathrm{\partial}}}{\mathrm{\partial}x}\phantom{\rule{thinmathspace}{0ex}}{\phantom{\rule{thinmathspace}{0ex}}}_{\left(\pm \right)}\mathit{\Psi}\left(\xi \right)=\pm 1,\\ {}_{\left(\mp \right)}{\mathit{\Psi}}^{\bullet}\left(\xi \right)\phantom{\rule{thinmathspace}{0ex}}i\frac{\overleftrightarrow{\mathrm{\partial}}}{\mathrm{\partial}x}\phantom{\rule{thinmathspace}{0ex}}{\phantom{\rule{thinmathspace}{0ex}}}_{\left(\pm \right)}\mathit{\Psi}\left(\xi \right)\phantom{\rule{thinmathspace}{0ex}}=0.\end{array}$$

The similar conditions are also for the functions ^{(±)}*Ψ*(*ξ*).

There are the following relations for these solutions
$${}_{\left(+\right)}\mathit{\Psi}={{c}_{1}}^{\left(+\right)}\mathit{\Psi}+{{c}_{2}}^{\left(-\right)}\mathit{\Psi},$$(10)

$$}_{\left(-\right)}\mathit{\Psi}={{c}_{1}^{\prime}}^{\left(+\right)}\mathit{\Psi}+{{c}_{2}^{\prime}}^{\left(-\right)}\mathit{\Psi$$(11)

From (10, 11) it follows
$$\begin{array}{}{}^{\left(+\right)}\mathit{\Psi}={\mathit{\Delta}}^{-1}\left({{c}^{\prime}}_{2}{\phantom{\rule{thinmathspace}{0ex}}}_{\left(+\right)}\mathit{\Psi}-{c}_{2}{\phantom{\rule{thinmathspace}{0ex}}}_{\left(-\right)}\mathit{\Psi}\right),\\ {}^{\left(-\right)}\mathit{\Psi}={\mathit{\Delta}}^{-1}\left(-{{c}^{\prime}}_{1}{\phantom{\rule{thinmathspace}{0ex}}}_{\left(+\right)}\mathit{\Psi}+{c}_{1}{\phantom{\rule{thinmathspace}{0ex}}}_{\left(-\right)}\mathit{\Psi}\right),\\ \mathit{\Delta}={c}_{1}{{c}^{\prime}}_{2}-{{c}^{\prime}}_{1}{c}_{2},\end{array}$$(12)

where
$$\begin{array}{}{}^{\left(-\right)}\mathit{\Psi}\left(\xi \right)=N{D}_{-\frac{1}{2}+i\frac{\mu}{2}}\left[-\left(1-i\right)\xi \right],\\ {}^{\left(+\right)}\mathit{\Psi}\left(\xi \right)=N{D}_{-\frac{1}{2}-i\frac{\mu}{2}}\left[-\left(1+i\right)\xi \right],\\ {}_{\left(+\right)}\mathit{\Psi}\left(\xi \right)=N{D}_{-\frac{1}{2}+i\frac{\mu}{2}}\left[\left(1-i\right)\xi \right],\\ {}_{\left(-\right)}\mathit{\Psi}\left(\xi \right)=N{D}_{-\phantom{\rule{thinmathspace}{0ex}}\frac{1}{2}-i\frac{\mu}{2}}\left[\left(1+i\right)\xi \right].\end{array}$$

From normalization conditions
$\left({}^{\left(\pm \right)}\mathit{\Psi}{,}^{\left(\pm \right)}\mathit{\Psi}\right)\phantom{\rule{1em}{0ex}}\equiv \phantom{\rule{1em}{0ex}}{\phantom{\rule{thinmathspace}{0ex}}}^{\left(\pm \right)}{\mathit{\Psi}}^{\bullet}\left(\xi \right)\phantom{\rule{thinmathspace}{0ex}}i\frac{\overleftrightarrow{\mathrm{\partial}}}{\mathrm{\partial}x}\phantom{\rule{thinmathspace}{0ex}}{\phantom{\rule{thinmathspace}{0ex}}}^{\left(\pm \right)}\mathit{\Psi}\left(\xi \right)=\pm 1$
we can obtain
$\left({}_{\left(+\right)}\mathit{\Psi}{,}^{\left(+\right)}\mathit{\Psi}\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}{c}_{1}^{\bullet}\phantom{\rule{thinmathspace}{0ex}}=\frac{{{c}^{\prime}}_{2}}{\mathit{\Delta}},\phantom{\rule{thinmathspace}{0ex}}\left({}_{\left(+\right)}\mathit{\Psi}{,}^{\left(-\right)}\mathit{\Psi}\right)\phantom{\rule{thinmathspace}{0ex}}=-{c}_{2}^{\bullet}\phantom{\rule{thinmathspace}{0ex}}=-\frac{{{c}^{\prime}}_{1}}{\mathit{\Delta}},{\left|{c}_{1}\right|}^{2}-{\left|{c}_{2}\right|}^{2}=1,$
so that
$$\begin{array}{}{}_{\left(+\right)}\mathit{\Psi}={{c}_{1}}^{\left(+\right)}\mathit{\Psi}+{{c}_{2}}^{\left(-\right)}\mathit{\Psi},\\ {}_{\left(-\right)}\mathit{\Psi}={c}_{2}^{\bullet \left(+\right)}\mathit{\Psi}+{c}_{1}^{\bullet \left(-\right)}\mathit{\Psi},\\ {}^{\left(+\right)}\mathit{\Psi}=\left({c}_{1}^{\bullet}{\phantom{\rule{thinmathspace}{0ex}}}_{\left(+\right)}\mathit{\Psi}-{c}_{2}{\phantom{\rule{thinmathspace}{0ex}}}_{\left(-\right)}\mathit{\Psi}\right),{\phantom{\rule{thinmathspace}{0ex}}}^{\left(-\right)}\mathit{\Psi}=-{c}_{2\phantom{\rule{thinmathspace}{0ex}}}^{\bullet}{\phantom{\rule{thinmathspace}{0ex}}}_{\left(+\right)}\mathit{\Psi}+{c}_{1}{\phantom{\rule{thinmathspace}{0ex}}}_{\left(-\right)}\mathit{\Psi}.\end{array}$$

The relations similar to (10 - 12) can be obtained for the creation operator
${\hat{a}}^{+},{\hat{\alpha}}^{+}\left({\hat{\beta}}^{+}\right)$
and annihilation operator
$\hat{a},\hat{\alpha}\left(\hat{\beta}\right)$ of particles (antiparticles). It is true that from (10 - 12) and from the relation
$\mathit{\Psi}=\hat{a}{\phantom{\rule{thinmathspace}{0ex}}}_{\left(+\right)}\mathit{\Psi}+{b}^{+}{\phantom{\rule{thinmathspace}{0ex}}}_{\left(-\right)}\mathit{\Psi}=\hat{\alpha}\phantom{\rule{thinmathspace}{0ex}}{\phantom{\rule{thinmathspace}{0ex}}}^{\left(+\right)}\mathit{\Psi}+{\beta}^{+}{\phantom{\rule{thinmathspace}{0ex}}}^{\left(+\right)}\mathit{\Psi}$ there can be obtained the relations
$$\begin{array}{}\alpha ={c}_{1}\hat{a}+{c}_{2}^{\bullet}\hat{b}{}^{+},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\hat{\beta}{}^{+}={c}_{2}\hat{a}+{c}_{1}^{\bullet}\hat{b}{}^{+},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\hat{a}={c}_{1}^{\bullet}\hat{\alpha}-{c}_{2}^{\bullet}\hat{\beta}{}^{+},\\ \hat{b}{}^{+}=-{c}_{2}\hat{\alpha}+{c}_{1}\hat{\beta}{}^{+}.\end{array}$$(13)

From the normalization conditions with account of (9), it follows that the normalization factor is *N*_{ν} =
$\frac{\sqrt{{\alpha}_{0}}{e}^{-\phantom{\rule{thinmathspace}{0ex}}\frac{\pi i\left(\nu +\frac{1}{2}\right)}{4}}}{\sqrt[4]{2}}=\frac{{e}^{-\phantom{\rule{thinmathspace}{0ex}}\frac{\pi \mu}{8}}\sqrt{{\alpha}_{0}}}{\sqrt[4]{2}}.$ It is easy to demonstrate that the orthogonality condition is met
${}^{\left(\mp \right)}{\mathit{\Psi}}^{\bullet}\left(\xi \right)\phantom{\rule{thinmathspace}{0ex}}i\frac{\overleftrightarrow{\mathrm{\partial}}}{\mathrm{\partial}x}\phantom{\rule{thinmathspace}{0ex}}{\phantom{\rule{thinmathspace}{0ex}}}^{\left(\pm \right)}\mathit{\Psi}\left(\xi \right)=0.$
In case of
$\nu =-\frac{1}{2}+i\frac{\mu}{2},z=-\left(1-i\right)\xi $ the lowering (*a*) and raising (*b*^{+}) operators can be expressed in a differential form
$$\begin{array}{}a=\frac{{e}^{-i\frac{\pi}{4}}}{2}\left(z+2\frac{d}{dz}\right)=\frac{i}{\sqrt{2}}\left[\xi +i\frac{d}{d\xi}\right],\\ {b}^{+}=\frac{{e}^{i\frac{\pi}{4}}}{2}\left(z-2\frac{d}{dz}\right)=\frac{-1}{\sqrt{2}}\left[\xi -i\frac{d}{d\xi}\right].\end{array}$$

Then
$a{a}^{+}-{a}^{+}a=\phantom{\rule{thinmathspace}{0ex}}b{b}^{+}-{b}^{+}b=i,a{b}^{+}+{b}^{+}a=2\nu +1=$ *i**μ* so that the equation
$\left[\frac{{d}^{2}}{d{\xi}^{2}}+\mu +{\xi}^{2}\right]{\mathit{\Psi}}_{\nu}\left(\xi \right)=0$ can be expressed as
$\hat{H}{\mathit{\Psi}}_{\nu}\left(\xi \right)=\left(2\nu +1\right){\mathit{\Psi}}_{\nu}\left(\xi \right),$ where the operator is
$\hat{H}=a{b}^{+}+{b}^{+}a.$

The average number of scalar particle pairs generated from the vacuum state
$\left|0\right.\u3009={N}_{0}{e}^{-\frac{{z}^{2}}{4}}$ is determined by the relation
$\left.\u30080\right|\hat{\alpha}{}^{+}\hat{\alpha}\left|0\right.\u3009,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{where}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\hat{a}\left|0\right.\u3009\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}0,\phantom{\rule{thinmathspace}{0ex}}\left.\u30080\right|\hat{a}{}^{+}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}0.$ If according to the relations (13),
$\hat{\alpha}={c}_{1}\hat{a}+{c}_{2}^{\bullet}\hat{b}{}^{+},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{then}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\hat{\alpha}\left|0\right.\u3009={c}_{2}^{\bullet}{\hat{b}}^{+}\left|0\right.\u3009,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left.\u30080\right|{\hat{\alpha}}^{+}={c}_{2}\left.\u30080\right|\phantom{\rule{thinmathspace}{0ex}}\hat{b}.$
In this case the average number of scalar particle pairs is
$$\left|\left.\u30080\right|{\hat{\alpha}}^{+}\hat{\alpha}\left|0\right.\u3009\right|={\left|{c}_{2}\right|}^{2}\left|\left.\u30080\right|\hat{b}{\hat{b}}^{+}\left|0\right.\u3009\right|={\left|{c}_{2}\right|}^{2},$$

where
$\left.\u30080\right|b{\hat{b}}^{+}\left|0\right.\u3009=i.$ From relation (8, 10), it follows that
${c}_{1}=\frac{\sqrt{2\pi}{e}^{-\frac{\pi}{4}\left(\mu +i\right)}}{\mathit{\Gamma}\left(\frac{1}{2}+i\frac{\mu}{2}\right)},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{c}_{2}={e}^{-\frac{\pi}{2}\left(\mu +i\right)},$ so that the average number of scalar particle pairs produced by the barrier is
$$\begin{array}{}{\displaystyle \overline{N}=\underset{{m}_{0}}{\overset{\mathrm{\infty}}{\int}}{\left|{c}_{2}\right|}^{2}\frac{TdE}{2\pi}=\frac{T{H}_{0}}{4{\pi}^{2}}{e}^{-\phantom{\rule{thinmathspace}{0ex}}\frac{2\pi {m}_{0}}{{H}_{0}}}.}\end{array}$$

Therefore, the average number of pairs generated per unit time
$P=\frac{\overline{N}}{T}=\frac{{H}_{0}}{4{\pi}^{2}}{e}^{-\phantom{\rule{thinmathspace}{0ex}}\frac{2\pi {m}_{0}}{{H}_{0}}}$ where the function of a homogeneous electric field is performed by the value determined
by the Hubble constant
$\stackrel{~}{E}=\frac{{m}_{0}{H}_{0}}{2e}=\frac{{m}_{0}}{2e}\sqrt{\frac{8\pi G{U}_{0}}{3}}.$ As this takes place, in the corresponding notations the probability takes the standard form [19]
$P=\frac{{H}_{0}}{4{\pi}^{2}}{e}^{-\phantom{\rule{thinmathspace}{0ex}}\frac{2\pi {m}_{0}}{{H}_{0}}}=\frac{\chi}{2{\pi}^{2}{\tau}_{0}}{e}^{-\phantom{\rule{thinmathspace}{0ex}}\frac{\pi}{\chi}},\chi =\frac{\stackrel{~}{E}}{{\stackrel{~}{E}}_{0}},\phantom{\rule{thinmathspace}{0ex}}{\stackrel{~}{E}}_{0}=\frac{{m}_{0}^{2}{c}^{3}}{e\hslash},\phantom{\rule{thinmathspace}{0ex}}{\tau}_{0}=\frac{\hslash}{{m}_{0}{c}^{2}}.$

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