I am asking problem from the 'J.J. Sakurai and Jim J. Napolitano' Book Modern Quantum Mechanics 2nd edition. It reads as follows:

"...This experiment also shows that gravity is not purely geometric at the quantum level because effect depends on $(m/\hbar)^2$."

Is it because $m$ is comparable to $\hbar$ or this is the term that does appear in pair or anything else? Page 134,135.

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    $\begingroup$ Is it because m is comparable to hbar? Quantities with different dimensions can’t be compared. $\endgroup$
    – Ghoster
    Jul 7, 2023 at 17:27
  • 2
    $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Jul 7, 2023 at 18:19

3 Answers 3


In page 131, it is said :

Because the mass does not appear in the equation of a particle trajectory, gravity in classical mechanics is often said to be a purely geometric theory.

Basically what it has in mind, that if you want to calculate classically where particle will land on the screen in the gravitational field acting,- you employ classical mechanics trajectory equation : $$ s = s_{0}+ v_{0} t+{\tfrac {1}{2}}{g}t^{2} \tag 1$$

This equation does not depend on particle mass, i.e. all other conditions being equal,- particles with different mass will land on the same screen spot.

While gravity-induced quantum interference of neutron beams, shows that interference pattern or where exactly neutron will land on the screen depends on particle mass, i.e. phase difference between neutron beams

$$ \Delta \varPhi \propto - \frac {m_n^{~2}}{\hbar^2} \tag 2$$

is dependent on neutron mass. Particles having different mass will induce different interference patterns, and hence gravitational field acting on particles is not purely geometric theory.

As an additional note, term $\hbar$ (or just $h$) in any equation shows that some quantity is not continuous , but rather quantized. You will not see quanta of action in any classical mechanics or gravity theory. This is an additional argument of author mentioned experiment against gravity as a purely geometric theory.


Your text, in §2.7.2, 3rd ed, points out that the coupling to the gravitational potential in the Schrödinger equation, $$ \left (i{\hbar\over m} \partial_t + {\hbar^2\over 2m^2}\nabla^2 -\Phi_{grav}\right )\psi=0,\tag{2.331} $$ does not drop out as in classical physics equations of motion and hence trajectories; therefore it is the quantity $(m/\hbar)^2$ with dimensions $T^2L^{-4}$ that will have to appear in gravity-induced quantum interference experiment phases such as the important Colella, Overhauser, & Werner, (1975) "Observation of gravitationally induced quantum interference" Physical Review Letters 34(23), p1472 experiment.

That is, the neutron mass in this interference experiment does not drop out, as it would in classical physics, leading to the celebrated Einstein geometrical formulation, where all objects move the same regardless of mass, there.

The scale of the masses involved is just right to actually observe quantum interference (which vanishes in the classical limit, $\hbar/m\to 0$). This is not altogether surprising since, after all, $\hbar$ appears/presents at the microscopic level of "elementary" particles.

Of course, on similar dimensional grounds, the same type of functional integral combination of dimensionful quantities appears in the functional integral undergirding most attempts at quantizing gravity, a willful forfeiture of geometry: individual masses matter.

Edit in response to comment by @flippiefanus

Another answer links the entire 2nd edition of S&N, whose 3rd edition I use and reference. (2.331) of the 3rd edition above, is (2.7.11) there. "Pure geometric", in its language, is merely the dropping out of mass from particle trajectories in a gravitational field, so individual masses of the moving "test" particles don't matter. Through an admirably sly footnote, they avoid exegesis of the equivalence principle, that GR readers might relish dragging them into the rabbit hole of. See next.

The COW experiment, in footnote 6, points out the gravitational potential entering the Schrödinger equation can be eliminated if all the macroscopic equipment is in an equivalent accelerated frame, which I understand as a reassurance that quantum experiments do not unsubtly violate the equivalence principle.

But, as S&N remark, individual test particle masses do enter in the interference phase, so, in that sense (perhaps only), QM interference phases depend on the test particle's mass, here the neutron, which is their seat-of-the pants effective definition of non-geometrical. (It's almost as though the tower of Pisa paradigm-experiment involves interferometers for particles of different mass? )

  • $\begingroup$ The answer here is not self-contained enough to understand without the textbook, which I don't have access to. However, I could find the referenced paper, though it seems that their argument is flawed. They seem to derive a "gravity" induced phase change based on the equivalence principle, thus employing the geometric nature of gravity, but eventually conclude that gravity is not purely geometrical. That seems contradictory to me. $\endgroup$ Jul 9, 2023 at 11:08
  • $\begingroup$ Two more refs you might like are [Peters et al, 1999]( doi.org/10.1038/23655) and Safronova et al, 2018. $\endgroup$ Jul 12, 2023 at 14:29
  • $\begingroup$ Thanks for the references. Unfortunately both are behind paywalls. However, from their abstracts it seems that their relevance is that they are concerned with tests of the equivalence principle, yet neither seemed to report an observed violation. $\endgroup$ Jul 13, 2023 at 12:30
  • $\begingroup$ Right, the quantum result is completely consistent with the equivalence principle. S&N do not claim that principle is violated by QM, of course! $\endgroup$ Jul 13, 2023 at 13:40

It should be pointed out that depending on how you phrase the equivalence principle, the principle does not hold in classical mechanics either. In Hamiltonian mechnaics, the initial conditions of a system are characterized by position and momentum. If one does an experiment with the same initial positions and momenta with two particles of different masses, the experiments would follow different trajectories. It's only when you characterize the initial conditions as position and velocity, the same initial conditions would lead to the same trajectory for different masses in the presence of gravity, leading to the equivalence principle

In fact, this exact phenomenon is what's happening in quantum mechanics. When we initialise two particles of different masses with the same initial wavefunction, we have chosen to have the same initial momentum for both the particles, instead of the same initial velocity. This is reflected by the fact that the momentum space wavefunction, as well as the momentum expectation value, is the same for both the particles.

Why does gravity-induced quantum interference in quantum mechanics show that gravity is not purely geometric at the quantum level?

In conclusion, we cannot conclude in QM that gravity isn't purely geometrical anymore than we can conclude it in classical mechanics.


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