# Can we apply Schrödinger equation in Newton gravitational potential and derive the deterministic Newton's gravitation as a special case of it

We know the solutions for wave functions of a an atom of hydrogen, and the energy values as given by spectral analysis of radiation emitted by hydrogen confirms the possible energy states as predicted by the Schrödinger equation.

My question: in the above case the potential used is Coulomb potential which is mathematically the same as gravitational potential. Hence can it be created a similar hydrogen equation but with point masses (instead of point charges) in Newton's gravitational field (instead of charged particles like electrons in Coulomb field)?

And in this equation, making some appropriate parameter going to 0 or ∞, can we show that the deterministic Newton's law of gravitation emerges out of it?

Last but not the least, if not for the classical gravitation, can we do this for relativistic gravitation (GR)?

I guess this is the main theme of the research in quantum gravity. Please correct me if I am wrong.

• – Michael Jul 3 '13 at 9:27
• It works in the inverse way. You take a classical potential V(r), where r is a coordinate, and then this become an operator V(R) in the Schrodinger equation (R being an operator). – Trimok Jul 3 '13 at 9:30
• @Ben Crowel : Thanks for this clarification. This is what I had in mind. "The classical limit doesn't mean recovering Newton's law of gravity, which is what you put in the calculation. The classical limit is the one in which the particle follows Newton's second law." – Rajesh D Jul 3 '13 at 16:59
• Also, photons are not emergent from the standard treatment of the Hydrogen atom. Famously, they have to be put in as an approximation in order to derive the Lamb shift of the hydrogen atom. – Jerry Schirmer Jul 9 '13 at 1:14

I would like to add some details to @Ben Crowell's answer in regard to bouncing neutrons. I think this great experiment deserves more attention.

Schrodinger equation does work when the origin of the potential energy is gravitational. At least, in the Newtonian limit.

This is a real experiment that has been performed and proves that quantum mechanics works in a gravitational field, that is, the Newton gravitational potential (together with some boundary conditions) quantizes neutron's energy.

A beam of cold neutrons (with velocities $\sim 10$ m/s) go into a cavity, with a neutron mirror at the bottom and a neutron absorber at the top. The beam of neutrons flies with constant horizontal velocity component through the cavity. All the neutrons that reach the upper surface are absorbed and disappear from the experiment. Those that reach the lower surface are reflected elastically. The detector counts the transmission rate, that is, the total number of neutrons that reach the detector per unit time.

Then, one can observe that the vertical part of the motion and the energy of the neutrons is quantized due to the gravitational field of the Earth

(From Nature (Volume 415 page 299) copyright 2002 Macmillan Publishers Ltd)

The solid line is the classical expectation, which does not fit the experimental result (except for high enough heights). It is easy to see that the classical prediction is $N\sim H^{3/2}$ ($N$ is the number of neutrons that reach the detector and $H$ is the height of the cavity):

The rate of neutrons that come into the cavity at height $y$ (OY is the vertical axis, while OX is the horizontal one) and reach the detector is proportional to the range of allowed vertical velocities, that is, those that don't touch the absorber (we are assuming that the cavity's length is sufficiently long, so that all the neutrons that may be absorbed are actually absorbed).

$$dN/dy\propto \delta v_y\,$$

This range ($\delta v_y$) is given by energy conservation (the minimum — potential plus kinetic (note that the horizontal velocity is constant) — energy at the absorber is $mgH+{m\over 2}\,v_x^2$):

$$mgH+{m\over 2}\,v_x^2\,>\,mgy+{m\over 2}\,(v_x^2+v_y^2)\implies -\sqrt{2g(H-y)}<v_y(y)<\sqrt{2g(H-y)}\\ \implies \delta v_y(y)=2\sqrt{2g(H-y)}$$

Therefore the classical result is: $$N\propto\int_0^H \delta v_y(y)\,dy\propto \int_0^H 2\sqrt{2g(H-y)} \,dy\propto H^{3/2}\;.$$

The observed quantum levels are given by the Schrödinger equation with a lineal potential energy $U(y)=mgy$ for $y$ lower than $H$ and $U(y)=\infty$ for $y$ higher than $H$, where $m$ is neutron's mass, $g\approx 9.8$ m/s$^2$ the gravitational field at the Earth surface, $y$ the vertical coordinate and $H$ the absorber's height. However, a good estimate is given by a semi-classical Bohr-Sommerfeld quantization (which is a semiclassical limit of the Schrödinger/Heisenberg quantization): $$S\equiv\int p(y)\,dy=n\, h\,,$$ with $p(y)$ the vertical component of neutron's momentum, $n$ a quantum number, and $h$ the Planck constant. Since $$S={4\over 3}\sqrt{2g}\,m\,H^{3/2}$$ and $E_n=mgH_n$, one obtains $$E_n^{\,3}\propto m\,g^2\,h^2\,n^2\,$$ i.e., neutron's gravitational energy is quantized. Order or magnitude $E_1\sim 10^{-12}$eV.

Note that this does not imply a quantization of the gravitational field. It is just a quantum particle in a classical, external, weak gravitational potential.

References: V. V. Nesvizhevsky et al., Nature 415 (2002) 297; Phys. Rev. D67 (2003) 102002.

Michael Brown on Jul 30 '13:

Agreed that GRANIT deserves more attention. They are upgrading (or have done - not sure what stage they are at the moment) the experiment so that it no longer operates in the "flow through" mode you describe. Instead they'll trap neutrons in a well and measure the energy levels directly using resonance spectroscopy. Remarkably the transitions are in the audio frequency range with energies of the order of pico-eV!

Googling around for GRANIT should turn up the latest details. –

• Fascinating.Thus nobody can dispute that gravity will quantize the energy of a neutron. It is not an elementary graviton though, ie the analogue of hydrogen. The whole earth is the nucleus. – anna v Jul 9 '13 at 4:03
• Agreed that GRANIT deserves more attention. They are upgrading (or have done - not sure what stage they are at the moment) the experiment so that it no longer operates in the "flow through" mode you describe. Instead they'll trap neutrons in a well and measure the energy levels directly using resonance spectroscopy. Remarkably the transitions are in the audio frequency range with energies of the order of pico-eV! Googling around for GRANIT should turn up the latest details. – Michael Jul 30 '13 at 12:40
• @MichaelBrown Thanks for the information! I didn't know it. – Diego Mazón Jul 30 '13 at 21:10

Hence can be create a similar hydrogen atom like setup but with point masses (instead of point charges) in Newton's gravitational field instead of charged particles like electrons in coloumb field?

Yes, in theory this can be done with any two electrically neutral particles. See Floratos 2010. In practice, the kinds of experiments people have done have been to demonstrate the quantized states of a neutron bouncing in a uniform field. (This is described briefly in Floratos.)

And in this set up, by in a limit of making some appropriate parameter going to zero/infty, can we show that the deterministic Newton's law of gravitation emerges out of it?

There is a classical limit, which can be realized, for example, with coherent superpositions of states having a very high principal quantum number $n$. The classical limit doesn't mean recovering Newton's law of gravity, which is what you put in the calculation. The classical limit is the one in which the particle follows Newton's second law.

Last but not the least, if not for the classical gravitation, can we do this for relativistic gravitation (GR)?

No. In GR, gravity is described by a field that propagates, not by a force law such as Newton's law of gravity.

Floratos, http://arxiv.org/abs/1008.0765

• Your answer does not address the question, whether a Hydrogen atom solution of the shroedinger equation could exist. It addresses a different aspect, the effect of gravity on the solutions. – anna v Jul 3 '13 at 13:51
• @annav: Sorry, but I don't see how you got that interpretation from reading my answer. – user4552 Jul 3 '13 at 14:23
• But a neutron is a composite object and it has spill over diagrams electromagnetic and strong which will affect how a neutron behaves at short scales. – anna v Jul 3 '13 at 14:31
• @anna "Hence can be create a similar hydrogen atom like setup but with point masses(instead of point charges)", This is puerly a theoretical setup i am talking about, not expecting to do it in real world. This is actually a thought experiment, and the point masses are just point massess, nothing more. Neverthless your answer was really useful in being informed about the real world situation. Just thought I'd clarify this to you. – Rajesh D Jul 3 '13 at 16:33

My question: in the above case the potential used is coloumb potential which is mathematically the same as gravitational potential. Hence can be create a similar hydrogen atom like setup but with point masses (instead of point charges) in Newton's gravitational field instead of charged particles like electrons in coloumb field?

We could, using the gravitational constants, in a hypothetical world where these point particles will have no other interactions. The gravitational potential is very weak; with respect to the electromagnetic and the other forces :

Coupling Constants

Strong 1

Electromagnetic 1/137

Weak 10^-6

Gravity 10^-39

So even though mathematically one could describe this "atom" substituting the appropriate constants, the units are way out of our real world and this is the second reason such "atoms" are impossible. The first reason is that all particles in our world have other interactions than gravitational.

A calculation can be found here, taking the hydrogen atom as a prototype with a gravitational force.

According to pre-SSCP physics, one can calculate what the radius of the hydrogen atom would be if the atom is governed by the conventional gravitational interaction between the proton and the electron. This radius is referred to as the Gravitational Bohr Radius (R), and it can be determined by:

R = ħ2/Gm2M (1)

where ħ is Planck's constant divided by 2π, G is the gravitational coupling factor, m is the mass of the electron and M is the mass of the proton. The conventional calculation of R, using 6.67 x 10-8 cm3/g sec2 as the appropriate value for G, yields

R = 1.20 x 10^31 cm.

This radius is larger than the radius of the observable universe. It is clearly a ridiculously large value and is usually cited as iron-clad proof that Atomic Scale systems are primarily bound by electrostatic rather than gravitational interactions.

The author has a series of papers where he modifies G, but it is not a mainstream route.

(side speculation :It is intriguing to wonder if dark matter particles have only gravitational interactions whether over the enormous distances of the universe such bindings could happen of course the masses involved should be much larger than the proton's and electron's, from the calculation above).

And in this set up, by in a limit of making some appropriate parameter going to zero/infty, can we show that the deterministic Newton's law of gravitation emerges out of it?

But already you have inserted Newton's law in the equation and it is a tautology, same as Coulombs law coming out macroscopically. The link by @MichaelBrown on the correspondence principle applies.

Last but not the least, if not for the classical gravitation, can we do this for relativistic gravitation (GR)?

As General Relativity has not been as yet quantized in a rigorous way it is a moot question, but again the equivalence principle should apply.

I guess this is the main theme of the research in quantum gravity. Please correct me if I am wrong.

No. The main theme of research in quantum gravity is how to quantize classical General Relativity rigorously and connect it with the Standard Model of particle physics, which encapsulates all the particle physics data up to now.

• The first half of the answer is wrong, because for electrically neutral objects the only significant force (at long ranges) is gravity. – user4552 Jul 3 '13 at 13:28
• But that is exactly what I am saying: if you only have gravity then you can have a potential solution similar to the hydrogen atom. But if we suppose neutral composite objects they will have other forces entering short range and the Schroedinger solutions cover from 0 to infinity. Thus one should have completely neutral to the other three interactions point particles. – anna v Jul 3 '13 at 13:37
• @BenCrowell: first, what anna v said. Second, even if the system is electrically neutral, you still have other multipole moments to tangle with, and they will dominate gravity--spin-spin corrections to the hydrogen atom swamp gravitational corrections, to name one obvious case (not to mention the fact that your refrigerator magnet is ALSO electrically neutral). – Jerry Schirmer Jul 9 '13 at 1:18
• @drake I think you misunderstood Jerry . He just says that a gravitational atom in the presence of all the other known forces is not viable.Nobody disputes that gravity will affect a neutral object. It is fascinating that the neutron shows quantized levels under gravitational fields. It is not an elementary graviton though, ie the analogue of the hydrogen. The whole earth is the nucleus. I will add this lat to your answer and give you a +1 since I did not know of this experimental fact. – anna v Jul 9 '13 at 3:57
• @drake , yes, :), like grexit ( the threatened exit of greece from the eurozone) – anna v Jul 9 '13 at 8:00