# Does quantum mechanics violate the equivalence principle?

I have a question about equivalence principle in quantum mechanics.

Consider a Schroedinger equation under gravitional field $$\left[ - \frac{1}{2m_I} \nabla^2 + m_g \Phi_{\mathrm{grav}} \right]\psi = i \partial_t \psi \tag{1}$$

where $m_I$ and $m_g$ are the inertia and gravitational masses, respectively. $\hbar=1$ unit is adopted.

To the contrary as the classical mechanics $$m_I \frac{ d^2 x}{ dt^2} = m_g g \tag{2}$$ we can choose a transformation $x'=x-\frac{1}{2} g t^2$ to "switch off" the gravity. But it seems the transformation will not switch off gravity in quantum mechanics, Eq. (1). Does it mean quantum mechanics break the equivalence principle? (I can think about relativistic Hamiltonian, but it will not resolve the problem, as far as I can see)

• There is a nice discussion of this issue in section XIV of arXiv:1205.3365. Summary: "This concludes the section on the weak equivalence principle in quantum mechanics. The result is mixed. We have shown that, very often, it is satisfied in the frame-work of quantum mechanics. But, on the other hand, it seems sometimes to be modified by quantum effects, ... . Maybe this indicates that its application to the more complicated case of quantum vacuum fluctuations of a field is rather suspicious?" – Michael Aug 1 '13 at 14:47
• Eq. (871) seems to do the job like "free falling"...although i am not sure about the meaning in (878) yet.. – user26143 Aug 1 '13 at 20:15
• @user26143 : There were errors in the answer, so I delete it. – Trimok Aug 1 '13 at 20:57

The Weak Equivalence Principle, or WEP for short, states that under identical initial conditions, the motion of particles of different masses in a given gravitational field is identical. Or in other words, there are no physical effects that depend on the mass of a point particle in an external gravitational field. This is just the equivalence between the inertial and gravitational mass.

Below I'll present some results that are on either sides of the camp (that WEP is or isn't violated). From what I read, the consensus on this matter is that the equivalence principle in the general sense in violated. However, there are special cases in which the principle holds true for quantum mechanics (one such case in given below).

Let us take the Schrodinger equation for a particle of inertial mass $m_i$ and gravitational mass $m_g$, that is falling toward mass $M$.

$$i\hbar\partial_{t}\psi=-\frac{\hbar^2}{2m_{i}}\nabla^{2}\psi - G\frac{m_{g}M}{r}$$

It is obviously that even for $m_{i}=m_{g}$ the mass does not cancel out of the equations of motions. This fact is even more apparent for $m_{i}=m_{g}=m$ in a uniform gravitational field in the $x$ direction, of acceleration $g$

$$i\hbar\partial_{t}\psi=-\frac{\hbar^2}{2m}\partial_{x}^{2}\psi +mgx\psi$$

whose solution will depend parametrically on $\hbar/m$. At this point we could say that the wave functions, the propagators and probability density distributions violate the WEP. Also, Rabinowitz in the 90's examined the possibility of gravitationally bound atoms, and he found that the mass, $m$, remains in the quantized equations of motions (although the mass cancels out in the classical equations of motions). We would expect $m$ to cancel out when averaging over states with large quantum numbers, but that puts them effectively in the classical continuum.

Consider a variant of the simple Galileo experiment, where particles of different mass are projected vertically in a uniform gravitational ﬁeld with a given initial velocity $v$. Classically, it is predicted that the particles will return a time $2v/g$ later, having risen to a height $x_{max}=v^{2}/ 2g$. But quantum particles are able to tunnel into the classically forbidden region above xmax. Moreover, the tunnelling depth depends on the mass. One might therefore expect a small, but highly signiﬁcant mass-dependant ‘quantum delay’ in the return time. Such a delay would represent a violation of the equivalence principle.

At the end of section $3$ he proves that the expectation value for the turn-around time of a quantum particle is identical, when the measurement is performed far from the classical turning point. In this sense, the WEP holds for a quantum particle.

This result suggest that a uniform gravitational potential—which applies locally to any non-singular gravitational ﬁeld—has a special property in relation to quantum mechanics, namely that the expectation time for the propagation of a quantum particle in this background is identical to the classical propagation time. This may be taken as an extension of the principle of equivalence into the quantum regime (for a broader discussion of what is entailed by a ‘quantum equivalence principle’). This special property seems to depend on the form of the potential; it does not apply in the case of a sharp potential step, or an exponential potential.

Finally I would like to point out this article, where the authors compute low corrections to the cross-section for the scattering of different quantum particles by an external gravitational field (taken as an external field in linearized gravity). They show that to first order, the cross-sections are spin-dependent. In the second order, they are dependent on energy as well. So, they prove that the equivalence principle is violated in both cases.

• +1 Thanks for a well written, readily grasped review. I'm still a bit weirded out by all this: I really had no idea of this yet-another-subtlety in bringing GR and QM together. – Selene Routley Dec 3 '13 at 9:50
• How are these violations of the equivalence principle consistent with Weinberg's 1960's general proofs of the equivalence for spin-2 fields? (see: journals.aps.org/pr/abstract/10.1103/PhysRev.135.B1049). – user1247 May 1 '14 at 14:41

The problem only arises when considering energy eigenstates, which are completely delocalized. One of the tenets of the equivalence principle is that the equivalence between gravitational systems and accelerated frames is only true locally. All that the equivalence principle states is that there is a neighbourhood in any point small enough such that the physical systems are equivalent. But an energy eigenstate will be dependent on the physics outside any such neighbourhood. Because of that, eigenstates being highly delocalized states, do not need to satisfy the equivalence principle in any way

A more detailed explanation would be as follows: The transformation induces a Hamiltonian on the accelerated frame. The equivalence principle says that if the locality is small enough, you can't do a measurement inside the locality that will tell me if I'm in a gravitational potential or in an accelerated frame. But, by definition you can't make perfect momentum or energy measurements inside such locality because your locality has an extent in space, so the position can't be outside of that extent. Hence no possible measurement inside the locality can tell apart extended eigenstates of energy or momentum. If the packets are narrower than the width of the locality, the equivalence principle should hold. If the packets are widers than the locality, the equivalence principle should NOT hold

• The time-evolution, Eq. (1), is valid for any state, if i am not mistaken... – user26143 Dec 8 '13 at 18:06
• Indeed the equation is valid. – lurscher Dec 9 '13 at 6:59
• How are these violations of the equivalence principle consistent with Weinberg's 1960's general proofs of the equivalence for spin-2 fields? (see: journals.aps.org/pr/abstract/10.1103/PhysRev.135.B1049). Doesn't his argument apply specifically to momentum eigenstates? – user1247 May 1 '14 at 14:42

I'll take weak equivalence principle in the formulation as given on Wikipedia page:

The local effects of motion in a curved space (gravitation) are indistinguishable from those of an accelerated observer in flat space, without exception.

Consider a wave function $\Psi(r,t)$ and suppose that the potential energy is constant. Now let's switch$^\dagger$ to a reference frame, which moves with respect to original one with velocity $At$ with $A$ constant:

$$\Psi'(r,t)=\Psi\left(r-\frac{At^2}2,t\right)\exp\left[\frac {im}\hbar\left(Atr-\frac {A^2t^3}6\right)\right].$$

$\Psi'(r,t)$ is the wavefunction in accelerated (with acceleration $A$) frame.

$$i\hbar \partial_t\Psi(r,t)=-\frac{\hbar^2}{2m}\partial_{rr}\Psi(r,t),$$

we can get the effective potential energy for the $\Psi'(r,t)$ wave function:

$$U_\text{eff}(r)=\frac{i\hbar \partial_t\Psi'(r,t)+\frac{\hbar^2}{2m}\partial_{rr}\Psi'(r,t)}{\Psi'(r,t)}=-mAr.$$

But this is nothing than potential energy in uniform gravitational field:

$$U_\text{grav}(r)=mgh,$$

where we use $g=-A$ is free fall acceleration and $h=r$ is height.

What do we get from this? Indeed, it appears that motion in uniformly accelerated frame is indistinguishable from motion in gravitational potential, i.e. weak equivalence principle is satisfied, if we take the formulation I've cited above.

What do then papers like e.g. this talk about? They say about "strong quantum violation of weak equivalence principle"! The answer, as it seems to me, is that they confuse weak equivalence principle with mass-dependent effects. See, most of discussion is about dependence of some wave packet properties on particle mass. But this doesn't have anything to do with weak equivalence principle: we have mass-dependent wave packet broadening even without any gravitation — even in free space!

Maybe there's some inequivalent formulation of weak equivalence principle, which speaks about mass-dependent effects in cases where classical mechanics doesn't have them, but then it should be unrelated to gravity and general relativity theory at all.

$^\dagger$ The switch is similar to the one described in e.g. Landau, Lifshitz "Quantum mechanics. Non-relativistic theory" — in a problem after $\S$17, but taking time-dependent velocity into account (i.e. not forgetting to integrate $\frac12mV^2$ with respect to time instead of just multiplying by $t$).

• I can't seem to reproduce the factor of $1/6$ for $A^2 t^3$ ... Can you please explain how you got that? – drewdles Aug 8 '16 at 8:28
• @AnantSaxena please see the footnote. To get the correct result, you don't simply multiply by $t$, you integrate with $\mathrm dt$, since $V\ne\operatorname{const}$. – Ruslan Aug 8 '16 at 9:24

If Quantum Mechanics violates the equivalence principle then this would mean that Einstein's theory is wrong. But we know from experiments (Gravity Probe B...etc) that Einstein's theory is not wrong. Thus Quantum Mechanics doesn't violate Einstein's theory so it doesn't violate the equivalence principle.

• This is totally unhelpful until you answer "how wrong does QM say the equivalence principle should be in the actual experiments where it has been checked?" Everything is known only to finite precision, including the equivalence principle. You can't just assume the answer a priori as you have done. You must actually do a calculation. – Michael Oct 18 '13 at 16:21
• I should also mention that experiments with ultra-cold neutrons have confirmed the non-trivial mass dependence of gravitational energy levels predicted by quantum mechanics. (The energy is of the order $mg\ell$ where $\ell$ is the de Broglie wavelength $(\hbar^2/m^2 g)^{1/3}$. So the total mass dependence of the energy is $m^{1/3}$.) So the interface of QM and the equivalence principle is indeed a subtle matter. By the way I'm not the downvoter. – Michael Oct 18 '13 at 16:47
• @MichaelBrown Wow! WOW!! There's more than enough material in your comments for a highly interesting answer to the uninitiated like me: I had no idea of this yet-another-subtlety in bringing GR and QM together. – Selene Routley Dec 3 '13 at 3:37