Gravity and the Superposition Principle

Abstract

The relation between gravity and quantum mechanics is investigated in this work. The link is given by the wave packet expansion process, rooted from the Uncertainty Principle. The basic idea is to express the de Broglie wavelength used by Schrodinger for a massive particle in terms of the associated Compton wavelength which is replaced by the Michell-Laplace radius $Gm/c^{2}$ of the spherical object of mass $m\ge m_{P}$, where $m_{P}$ is the Planck mass. The wave packet spreading is studying in spherical coordinates, having the width $\sigma (t)$, expressed in terms of G and c instead of $\hbar$. Therefore, for masses larger than the Planck mass, a faster dispersion rate of $\sigma (t)$ is obtained, compared to the standard case. The dispersion of the wave packet is observed only by a free falling observer and the process breaks down once the observer hits the surface of the object. Different freely falling observers notice different rates of expansion of the wave packet and the source of gravity is in a quantum superposition. We further confront the Mita formula for the mean energy of the wave packet with the de Broglie-Bohm quantum potential energy when the Schrodinger equation is expressed in the Madelung form.

Appendices

Appendix A

Let us compare now the time intervals after which the width $\sigma (t)$ is twice the initial value $\sigma _{0}$. Inserting $\sigma (t) = 2\sigma _{0}$ in (3.2), we get

$$\begin{aligned} t_{q} = \frac{\sqrt{3}~m\sigma _{0}^{2}}{\hbar }. \end{aligned}$$

(5.1)

But from (3.3) one obtains

$$\begin{aligned} t_{g} = \frac{\sqrt{3}~c\sigma _{0}^{2}}{Gm}. \end{aligned}$$

(5.2)

With a mass, say, $m = 10^{6}$ grams and $\sigma _{0} = 10^{2}$ cm, we have $t_{q} \approx 10^{37}$s (much more than the age of the Universe) and $t_{g} \approx 10^{16}$s, less than the age of the Universe. When $m = m_{P}$ we obtain, of course, $t_{q} = t_{g} \approx 10^{26}$s.

We wish now to comment on the mean value of the energy of the wave packet. It was calculated by Mita [14]. His expression of the mean energy in one spatial dimension is $<p^{2}/2m> = mv_{0}^{2}/2 + \hbar ^{2}/2m\sigma _{0}^{2}$. We have in our situation $v_{0} = 0$ (no linear motion) and, for $m>m_{P}$

$$\begin{aligned} <\frac{p^{2}}{2m}> = \frac{mc^{2}}{2}\left( \frac{r_{g}}{\sigma _{0}}\right) ^{2} \end{aligned}$$

(5.3)

Mita [14] designated its energy expectation value as ”dispersion oscillations” of the particle, or energy of localization, without giving a precise nature of those oscillations. Our expression (5.3) for the mean energy could be interpreted as ”spreading energy” which becomes half of the rest energy when $R_{0} = \sigma _{0} = r_{g}$.

Appendix B

It is instructive to find out another type of energy, the so called de Broglie-Bohm quantum potential energy Q [15, 16]. Holland [15] suggested that Q may be regarded as the kinetic energy of additional “concealed” degrees of freedom. It is obtained from the Madelung form of the Schrodinger equation, when the wave function is written as $\Psi = \sqrt{\rho }~exp(iS/\hbar )$, where $\rho$ is the amplitude squared and S is the phase of the wave function. One equation is the continuity equation and the other is equivalent with the Hamilton-Jacobi equation, but with an extra term given by

$$\begin{aligned} Q = -\frac{\hbar ^{2}}{2m}\frac{\nabla ^{2}\sqrt{\rho }}{\sqrt{\rho }}. \end{aligned}$$

(6.1)

It is the only term from the two equations containing $\hbar$. According to Esposito [17] (see also [15]), Q is a kinetical energy for an internal motion of the object, the external motion being interpreted as the motion of the CM. Using $\rho$ from (3.1) it can be shown that, when $m>m_{P}$, Q is given by

$$\begin{aligned} Q(r,t) = \frac{mc^{2}}{2}\left( \frac{r_{g}}{\sigma (t)}\right) ^{2}\left[ 3 - \frac{r^{2}}{\sigma ^{2}(t)}\right] . \end{aligned}$$

(6.2)

When Q(r, t) is calculated in terms of $\hbar$, one obtains the same expression as that provided by Rahmani and Golshani [18]. The number “3” within the square parantheses comes from the three spatial dimensions (spherical coordinates). With one spatial dimension x we have “1” instead of “3” and x instead of r. It is clear that the Mita energy is just the quantum potential energy Q at $r = 0,~t = 0$, when Q equals $Q_{max} = mc^{2}r_{g}^{2}/2\sigma _{0}^{2}$ (in one spatial dimension).

We persuade ourselves of that by computing the expectation value of $Q_{0}$ at the initial time $t = 0$ or for a stationary state when the probability density of a quantum system does not depend on time. We have

$$\begin{aligned}<Q_{0}> \equiv <Q(r,0)> = \int {\Psi ^{*}(r,0)Q_{0}\Psi (r,0)dV}, \end{aligned}$$

(6.3)

where dV is the volume element in spherical coordinates. From (6.3) one obtains

$$\begin{aligned} <Q_{0}> = \frac{1}{\pi \sqrt{\pi }\sigma _{0}^{3}}~\frac{mc^{2}}{2}~\frac{r_{g}^{2}}{\sigma _{0}^{2}} \int _{0}^{\infty }\left( 3 - \frac{r^{2}}{\sigma _{0}^{2}}\right) e^{-\frac{2r^{2}}{\sigma _{0}^{2}}}~ 4\pi r^{2}dr \end{aligned}$$

(6.4)

With the help of the well-known relations

$$\begin{aligned} \int _{0}^{\infty }y^{2}e^{-\frac{y^{2}}{b^{2}}} dy = \frac{\sqrt{\pi }}{4}b^{3},~~~~~~\int _{0}^{\infty }y^{4}e^{-\frac{y^{2}}{b^{2}}} dy = \frac{3\sqrt{\pi }}{8}b^{5}, \end{aligned}$$

(6.5)

we get from (6.4)

$$\begin{aligned} <Q_{0}> = \frac{9}{32\sqrt{2}} ~\frac{r_{g}^{2}}{\sigma _{0}^{2}}~ mc^{2},~~~~m\ge m_{P}, \end{aligned}$$

(6.6)

which, for $\sigma _{0} = r_{g}$ it will be of the order of $mc^{2}$. For $m< m_{P}, <Q_{0}>$ acquires the form

$$\begin{aligned} <Q_{0}> = \frac{9}{16\sqrt{2}} \frac{\hbar ^{2}}{2m\sigma _{0}^{2}} , \end{aligned}$$

(6.7)

which has the same form as the Mita “localization energy” $E_{L}$. In our situation, $<Q_{0}>$ is not a kinetic energy but a potential one, that will become kinetic from the point of view of a free falling observer. Let us exhibit an estimation of the mean value of $Q_{0}$ from (6.6). Take for m the mass of the Earth, $m\approx 6\cdot 10^{27}$ grams, $r_{g} = 2.5 ~cm$, and $\sigma _{0} = 6.37 \cdot 10^{8}$ cm. With these values, one obtains $<Q_{0}> \approx 10^{31}$ ergs, a reasonable value. If one uses the formula (6.7) for the same mass, we get a very tiny value, completely negligible.

From Q(r, t) the expression of the quantum force appears as

$$\begin{aligned} {\textbf {F}}_{Q} = -\nabla Q = -\frac{\partial Q(r,t)}{\partial r}{} {\textbf {e}}_{r} = \frac{mc^{2}r_{g}^{2}}{\sigma ^{4}(t)}{} {\textbf {r}}. \end{aligned}$$

(6.8)

Note that the quantum force (which is repulsive, being positive) is proportional to ${\textbf {r}}$, as the expansion force in the case of the de Sitter Universe in static coordinates.

Appendix C

As far as the Uncertainty Principle is concerned, it is worth noting that the appearance of the Planck constant on the r.h.s. of the Heisenberg uncertainty relation originates from the commutator of two operators, and there from experiments in Microphysics. Therefore, macroscopically we may replace $\hbar$ by its macroscopic counterpart, namely $Gm^{2}/c$. It leads us to

$$\begin{aligned} \Delta r \Delta p = \frac{\sigma _{0}}{\sqrt{2}}\sqrt{1 + \frac{a^{2}t^{2}}{c^{2}}}\frac{\hbar }{\sqrt{2}\sigma _{0}}\ge \frac{\hbar }{2} ~~\rightarrow ~~ \frac{Gm^{2}}{2c} = \frac{1}{2}r_{g} mc, \end{aligned}$$

(7.1)

with $r_{g} \approx (\Delta r)_{min}$. We notice that the momentum mc in (7.1) plays the role of the constant $\Delta p \approx \hbar /\sigma _{0}$ and we get the maximal value $\Delta p = mc$ when $\sigma _{0}$ equals the Compton wavelength of the particle or its radius $r_{g}$ when $m\ge m_{P}$.

If we keep track of the Generalized Uncertainty Principle (GUP) [19] (see also [20, 21])

$$\begin{aligned} \Delta x \ge \frac{\hbar }{\Delta p} + l_{P}^{2}\frac{\Delta p}{\hbar }, \end{aligned}$$

(7.2)

where $l_{P} = 10^{-33}$cm is the Planck length, one observes that the 2nd term of the r.h.s. of (7.2) does not depend on $\hbar$, so it appears as

$$\begin{aligned} \Delta x \ge \frac{\hbar }{\Delta p} + \frac{G}{c^{3}}{\Delta p}, \end{aligned}$$

(7.3)

If we look for $(\Delta x)_{min}$, it is obtained when $\Delta p = mc$, i.e., the value from (7.1). Taking into consideration that, in our view, for $m\ge m_{P} = 10^{-5}$grams, $\hbar \rightarrow Gm^{2}/c$, (7.3) yields

$$\begin{aligned} (\Delta x)_{min}= \frac{2Gm}{c^{2}}, \end{aligned}$$

(7.4)

due to the contribution from both terms. Equation (7.3) may be also written as

$$\begin{aligned} \Delta x \Delta p \ge \hbar \left[ 1 + \left( \frac{l_{P} \Delta p}{\hbar }\right) ^{2}\right] . \end{aligned}$$

(7.5)

We distinguish here two situations:

• i) if $\Delta p<< \hbar /l_{P} = m_{P}c$,    $\Delta x \Delta p \ge \hbar$ .

• ii) if $\Delta p>> \hbar /l_{P} = m_{P}c$,    $\Delta x \Delta p \ge (G/c^{3})(\Delta p)^{2}$, or    $\Delta x \ge (G/c^{3})\Delta p$,

that is always valid if $\Delta x \ge (G/c^{3})(\Delta p)_{max} = Gm/c^{2}$ (the factor of 2 is missing because we started from Carlip’s [19] (7.2)). As we anticipated before, the Planck mass (or the Planck momentum) decides which term is more important in the r.h.s. of (7.3). The gravitational term (depending on G) dominates if the mass of the object is bigger than the Planck mass.