Friday, April 20, 2012

Schrödinger has never met Newton

A Western European writer – let's call her Gertruda – believes that
Schrödinger meets Newton.

But is the story about the two physicists' encounter true?

Yes, these were just jokes. I don't think that Gertruda misunderstands the history in this way. Instead, what she completely misunderstands is the physics, especially quantum physics. She is in a good company. Aside from the authors of some nonsensical papers she mentions, e.g. van Meter, Giulini and Großardt, Harrison and Moroz with Tod, Diósi, and Carlip with Salzman, similar basic misconceptions about elementary quantum mechanics have been promoted by Penrose and Hameroff.

Hameroff is a physician who, along with Penrose, prescribed supernatural abilities to the gravitational field. It's responsible for the gravitationally induced "collapse of the wave function" which also gives us consciousness and may be even blamed for Penrose's (not to mention Hameroff's) complete inability to understand rudimentary quantum mechanics, among many other wonderful things; I am sure that many of you have read the Penrose-Hameroff crackpottery and a large percentage of those readers even fail to see why it is a crackpottery, a problem I will try to fix (and judging by the 85-year-long experience, I will fail).

It's really Penrose who should be blamed for the concept known as the

Schrödinger-Newton equations

So what are the equations? Gertruda reproduces them completely mindlessly and uncritically. They're supposed to be the symbiosis of quantum mechanics combined with the Newtonian limit of general relativity. They say:\[

i\hbar \pfrac {}{t} \Psi(t,\vec x) &= \zav{-\frac{\hbar^2}{2m} \Delta + m\Phi(t,\vec x)} \Psi(t,\vec x) \\
\Delta \Phi(t,\vec x) &= 4\pi G m \abs{\Psi(t,\vec x)}^2

\] Don't get misled by the beautiful form they take in \(\rm\LaTeX\) implemented by MathJax; superficial beauty of the letters doesn't guarantee the validity. Gertruda and others immediately talk about mechanically inserting numbers into these equations, and so on, but they never ask a basic question: Are these equations actually right? Can we prove that they are wrong? And if they are right, can they be responsible for anything important that shapes our observations?

Of course that the second one is completely wrong; it fundamentally misunderstands the basic concepts in physics. And even if you forgot the reasons why the second equation is completely wrong, they couldn't be responsible for anything important we observe – e.g. for well-defined perceptions after we measure something – because of the immense weakness of gravity (and because of other reasons).

Analyzing the equations one by one

So let us look at the equations, what they say, and whether they are the right equations describing the particular physical problems. We begin with the first one,\[

i\hbar \pfrac {}{t} \Psi(t,\vec x) = \zav{-\frac{\hbar^2}{2m} \Delta + m\Phi(t,\vec x)} \Psi(t,\vec x)

\] Is it right? Yes, it is a conventional time-dependent Schrödinger equation for a single particle that includes the gravitational potential. When the gravitational potential matters, it's important to include it in the Hamiltonian as well. The gravitational potential energy is of course as good a part of the energy (the Hamiltonian) as the kinetic energy, given by the spatial Laplacian term, and it should be included in the equations.

In reality, we may of course neglect the gravitational potential in practice. When we study the motion of a few elementary particles, their mutual gravitational attraction is negligible. For two electrons, the gravitational force is more than \(10^{40}\) times weaker than the electrostatic force. Clearly, we can't measure the transitions in a Hydrogen atom with the relative precision of \(10^{-40}\). The "gravitational Bohr radius" of an atom that is only held gravitationally would be comparably large to the visible Universe because the particles are very weakly bound, indeed. Of course, it makes no practical sense to talk about energy eigenstates that occupy similarly huge regions because well before the first revolution (a time scale), something will hit the particles so that they will never be in the hypothetical "weakly bound state" for a whole period.

But even if you consider the gravity between a microscopic particle (which must be there for our equation to be relevant) such as a proton and the whole Earth, it's pretty much negligible. For example, the protons are running around the LHC collider and the Earth's gravitational pull is dragging them down, with the usual acceleration of \(g=9.8\,\,{\rm m}/{\rm s}^2\). However, there are so many forces that accelerate the protons much more strongly in various directions that the gravitational pull exerted by the Earth can't be measured. But yes, it's true that the LHC magnets and electric fields are also preventing the protons from "falling down". The protons circulate for minutes if not hours and as skydivers know, one may fall pretty far down during such a time.

An exceptional experiment in which the Earth's gravity has a detectable impact on the quantum behavior of particles are the neutron interference experiments, those that may be used to prove that gravity cannot be an entropic force. To describe similar experiments, one really has to study the neutron's Schrödinger equation together with the kinetic term and the gravitational potential created by the Earth. Needless to say, much of the behavior is obvious. If you shoot neutrons through a pair of slits, of course that they will accelerate towards the Earth much like everything else so the interference pattern may be found again; it's just shifted down by the expected distance.

People have also studied neutrons that are jumping on a trampoline. There is an infinite potential energy beneath the trampoline which shoots the neutrons up. And there's also the Earth's gravity that attracts them down. Moreover, neutrons are described by quantum mechanics which makes their energy eigenstates quantized. It's an interesting experiment that makes one sure that quantum mechanics does apply in all situations, even if the Earth's gravity plays a role as well, and that's where the Schrödinger equation with the gravitational potential may be verified.

I want to say that while the one-particle Schrödinger equation written above is the right description for situations similar to the neutron interference experiments, it already betrays some misconceptions by the "Schrödinger meets Newton" folks. The fact that they write a one-particle equation is suspicious. The corresponding right description of many particles wouldn't contain wave functions that depend on the spacetime, \(\Psi(t,\vec x)\). Instead, the multi-particle wave function has to depend on positions of all the particles, e.g. \(\Psi(t,\vec x_1,\vec x_2)\). However, the Schrödinger equation above already suggests that the "Schrödinger meets Newton" folks want to treat the wave function as an object analogous to the gravitational potential, a classical field. This totally invalid interpretation of the objects becomes lethal in the second equation.

Confusing observables with their expectation values, mixing up probability waves with classical fields

The actual problem with the Schrödinger-Newton system of equations is the second equation, Poisson's equation for the gravitational potential,\[

\Delta \Phi(t,\vec x) = 4\pi G m \abs{\Psi(t,\vec x)}^2

\] Is this equation right under some circumstances? No, it is never right. It is a completely nonsensical equation which is nonlinear in the wave function \(\Psi\) – a fatal inconsistency – and which mixes apples with oranges. I will spend some time with explaining these points.

First, let me start with the full quantum gravity. Quantum gravity contains some complicated enough quantum observables that may only be described by the full-fledged string/M-theory but in the low-energy approximation of an "effective field theory", it contains quantum fields including the metric tensor \(\hat g_{\mu\nu}\). I added a hat to emphasize that each component of the tensor field at each point is a linear operator (well, operator distribution) acting on the Hilbert space.

I have already discussed the one-particle Schrödinger equation that dictates how the gravitational field influences the particles, at least in the non-relativistic, low-energy approximation. But we also want to know how the particles influence the gravitational field. That's given by Einstein's equations,\[

\hat{R}_{\mu\nu} - \frac{1}{2} \hat{R} \hat{g}_{\mu\nu} = 8\pi G \,\hat{T}_{\mu\nu}

\] In the quantum version, Einstein's equations become a form of the Heisenberg equations in the Heisenberg picture (Schrödinger's picture looks very complicated for gravity or other field theories) and these equations simply add hats above the metric tensor, Ricci tensor, Ricci scalar, as well as the stress-energy tensor. All these objects have to be operators. For example, the stress-energy tensor is constructed out of other operators, including the operators for the intensity of electromagnetic and other fields and/or positions of particles, so it must be an operator. If an equation relates it to something else, this something else has to be an operator as well.

Think about Schrödinger's cat – or any other macroscopic physical system, for that matter. To make the thought experiment more spectacular, attach the whole Earth to the cat so if the cat dies, the whole Earth explodes and its gravitational field changes. It's clear that the values of microscopic quantities such as the decay stage of a radioactive nucleus may imprint themselves to the gravitational field around the Earth – something that may influence the Moon etc. (We may subjectively feel that we have already perceived one particular answer but a more perfect physicist has to evolve us into linear superpositions as well, in order to allow our wave function to interfere with itself and to negate the result of our perceptions. This more perfect and larger physicist will rightfully deny that in a precise calculation, it's possible to treat the wave function as a "collapsed one" at the moment right after we "feel an outcome".) Because the radioactive nucleus may be found in a linear superposition of dictinct states and because this state is imprinted onto the cat and the Earth, it's obvious that even the gravitational field around the (former?) Earth is generally found in a probabilistic linear superposition of different states. Consequently, the values of the metric tensors at various points have to be operators whose values may only be predicted probabilistically, much like the values of any observable in any quantum theory.

Let's now take the non-relativistic, weak-gravitational-field, low-energy limit of Einstein's equations written above. In this non-relativistic limit, \(\hat g_{00}\) is the only important component of the metric tensor (the gravitational redshift) and it gets translated to the gravitational potential \(\hat \Phi\) which is clearly an operator-valued field, too. We get\[

\Delta \hat\Phi(t,\vec x) = 4\pi G \hat\rho(t,\vec x).

\] It looks like Gertruda's version of Poisson's equation except that the gravitational potential on the left hand side has a hat; and the source \(\hat\rho\), i.e. the mass density, has replaced her \(m \abs{\Psi(t,\vec x)}^2\). Fine. There are some differences. But can I make special choices that will produce her equation out of the correct equation above? What is the mass density operator \(\hat\rho\) equal to in the case of the electron?

Well, it's easy to answer this question. The mass density coming from an electron blows up at the point where the electron is located; it's zero everywhere else. Clearly, the mass density is a three-dimensional delta-function:\[

\hat\rho(t,\vec x) = m \delta^{(3)}(\hat{\vec X} - \vec x)

\] Just to be sure, the arguments of the field operators such as \(\hat\rho\) – the arguments that the fields depend on – are ordinary coordinates \(\vec x\) which have no hats because they're not operators. In quantum field theories, whether they're relativistic or not, they're as independent variables as the time \(t\); after all, \((t,x,y,z)\) are mixed with each other by the relativistic Lorentz transformations which are manifest symmetries in relativistic quantum field theories. However, the equation above says that the mass density at the point \(\vec x\) blows up iff the eigenvalue of the electron's position \(\hat X\), an eigenvalue of an observable, is equal to this \(\vec x\). The equation above is an operator equation. And yes, it's possible to compute functions (including the delta-function) out of operator-valued arguments.

Semiclassical gravity isn't necessarily too self-consistent an approximation. It may resemble the equally named song by Savagery above.

Clearly, the operator \(\delta^{(3)}(\hat X - \vec x)\) is something different than Gertruda's \(\abs{\Psi(t,\vec x)}^2\) – which isn't an operator at all – so her equation isn't right. Can we obtain the squared wave function in some way? Well, you could try to take the expectation value of the last displayed equation:\[

\bra\Psi \Delta \hat\Phi(t,\vec x)\ket\Psi = 4\pi G m \abs{\Psi(t,\vec x)}^2

\] Indeed, if you compute the expectation value of the operator \(\delta^{(3)}(\hat X - \vec x)\) in the state \(\ket\Psi\), you will obtain \(\abs{\Psi(t,\vec x)}^2\). However, note that the equation above still differs from the Gertruda-Poisson equation: our right equation properly sandwiches the gravitational potential, which is an operator-valued field, in between the two copies of the wave functions.

Can't you just introduce a new symbol \(\Delta\Phi\), one without any hats, for the expectation value entering the left hand side of the last equation? You may but it's just an expectation value, a number that depends on the state. The proper Schrödinger equation with the gravitational potential that we started with contains the operator \(\hat\Phi(t,\vec x)\) that is manifestly independent of the wave function (either because it is an external classical field – if we want to treat it as a deterministically evolving background field – or because it is a particular operator acting on the Hilbert space). So they're different things. At any rate, the original pair of equations is wrong.

Nonlinearity in the wave function is lethal

Those deluded people are obsessed with expectation values because they don't want to accept quantum mechanics. The expectation value of an operator "looks like" a classical quantity and classical quantities are the only physical quantities they have really accepted – and 19th century classical physics is the newest framework for physics that they have swallowed – so they try to deform and distort everything so that it resembles classical physics. An arbitrarily silly caricature of the reality is always preferred by them over the right equations as long as it looks more classical.

But Nature obeys quantum mechanics. The observables we can see – all of them – are indeed linear operators acting on the Hilbert space. If something may be measured and seen to be equal to something or something else (this includes Yes/No questions we may answer by an experiment), then "something" is always associated with a linear operator on the Hilbert space (Yes/No questions are associated with Hermitian projection operators). If you are using a set of concepts that violate this universal postulate, then you contradict basic rules of quantum mechanics and what you say is just demonstrably wrong. This basic rule doesn't depend on any dynamical details of your would-be quantum theory and it admits no loopholes.

Two pieces of the wave function don't attract each other at all

You could say that one may talk about the expectation values in some contexts because they may give a fair approximation to quantum mechanics. The behavior of some systems may be close to the classical one, anyway, so why wouldn't we talk about the expectation values only?

However, this approximation is only meaningful if the variations of the physical observables (encoded in the spread of the wave function) are much smaller than their characteristic values such as the (mean) distances between the particles which we want to treat as classical numbers, e.g.\[

\abs{\Delta \vec x} \ll O(\abs{\vec x_1-\vec x_2})

\] However, the very motivation that makes those confused people study the Schrödinger-Newton system of equations is that this condition isn't satisfied at all. What they typically want to achieve is to "collapse" the wave function packets. They're composed of several distant enough pieces, otherwise they wouldn't feel the need to collapse them. In their system of equations, two distant portions of the wave function attract each other in the same way as two celestial bodies do – because \(m \abs{\Psi}^2\) enters as the classical mass density to Poisson's equation for the gravitational potential. They write many papers studying whether this self-attraction of "parts of the electron" or another object may be enough to "keep the wave function compact enough".

Of course, it is not enough. The gravitational force is extremely weak and cannot play such an essential role in the experiments with elementary particles. In
Ghirardi-Rimini-Weber: collapsed pseudoscience,
I have described somewhat more sophisticated "collapse theories" that are trying to achieve a similar outcome: to misinterpret the wave function as a "classical object" and to prevent it from spreading. Of course, these theories cannot work, either. To keep these wave functions compact enough, they have to introduce kicks that are so large that we are sure that they don't exist. You simply cannot find any classical model that agrees with observations in which the wave function is a classical object – simply because the wave function isn't a classical object and this fact is really an experimentally proven one as you know if you think a little bit.

But what the people studying the Schrödinger-Newton system of equations do is even much more stupid than what the GRW folks attempted. It is internally inconsistent already at the mathematical level. You don't have to think about some sophisticated experiments to verify whether these equations are viable. They can be safely ruled out by pure thought because they predict things that are manifestly wrong.

I have already said that the Gertruda-Poisson equation for the gravitational potential treats the squared wave function as if it were a mass density. If your wave function is composed of two major pieces in two regions, they will behave as two clouds of interplanetary gas and these two clouds will attract because each of them influences the gravitational potential that influences the motion of the other cloud, too.

However, this attraction between two "pieces" of a wave function definitely doesn't exist, in a sharp contrast with the immensely dumb opinion held by pretty much every "alternative" kibitzer about quantum mechanics i.e. everyone who has ever offered any musings that something is fundamentally wrong with the proper Copenhagen quantum mechanics. There would only be an attraction if the matter (electron) existed at both places because the attraction is proportional to \(M_1 M_2\). However, one may easily show that the counterpart of \(M_1M_2\) is zero: the matter is never at both places at the same time.

Imagine that the wave function has the form\[

\ket\psi = 0.6\ket \phi+ 0.8 i \ket \chi

\] where the states \(\ket\phi\) and \(\ket\chi\) are supported by very distant regions. As you know, this state vector implies that the particle has 36% odds to be in the "phi" region and 64% odds to be in the "chi" region. I chose probabilities that are nicely rational, exploiting the famous 3-4-5 Pythagorean triangle, but there's another reason why I didn't pick the odds to be 50% and 50%: there is absolutely nothing special about wave functions that predict exactly the same odds for two different outcomes. The number 50 is just a random number in between \(0\) and \(100\) and it only becomes special if there is an exact symmetry between \(p\) and \((1-p)\) which is usually not the case. Much of the self-delusion by the "many worlds" proponents is based on the misconception that predictions with equal odds for various outcomes are special or "canonical". They're not.

Fine. So if we have the wave function \(\ket\psi\) above, do the two parts of the wave function attract each other? The answer is a resounding No. The basic fact about quantum mechanics that all these Schrödinger-Newton and many-worlds and other pseudoscientists misunderstand is the following point. The wave function above doesn't mean that
there is 36% of an object here AND 64% of an object there. (WRONG.)
Note that there is "AND" in the sentence above, indicating the existence of two objects. Instead, the right interpretation is that
the particle is here (36% odds) OR there (64% odds). (RIGHT.)
The correct word is "OR", not "AND"! However, unlike in classical physics, you're not allowed to assume that one of the possibilities is "objectively true" in the classical sense even if the position isn't measured. On the other hand, even in quantum mechanics, it's still possible to strictly prove that the particle isn't found at both places simultaneously; the state vector is an eigenstate of the "both places" projection operator (product of two projection operators) with the eigenvalue zero. (The same comments apply to two slits in a double-slit experiment.)

The mutually orthogonal terms contributing to the wave function or density matrix aren't multiple objects that simultaneously exist, as the word "AND" would indicate. You would need (tensor) products of Hilbert spaces and/or wave functions, not sums, to describe multiple objects! Instead, they are mutually excluding alternatives for what may exist, alternative properties that one physical system (e.g. one electron) may have. And mutually excluding alternatives simply cannot interact with each other, gravitationally or otherwise.

Imagine you throw dice. The result may be "1" or "2" or "3" or "4" or "5" or "6". But you know that only one answer is right. There can't be any interaction that would say that because both "1" and "6" may occur, they attract each other which is why you probably get "3" or "4" in the middle. It's nonsense because "1" and "6" are never objects that simultaneously exist. If they don't simultaneously exist, they can't attract each other, whatever the rules are. They can't interact with one another at all! While the expectation value of the electron's position may be "somewhere in between" the regions "phi" and "chi", we may use the wave function to prove with absolute certainty that the electron isn't in between.

The proponents of the "many-worlds interpretation" often commit the same trivial mistake. They are imagining that two copies of you co-exist at the same moment – in some larger "multiverse". That's why they often talk about one copy's thinking how the other copy is feeling in another part of a multiverse. But the other copy can't be feeling anything at all because it doesn't exist if you do! You and your copy are mutually excluding. If you wanted to describe two people, you would need a larger Hilbert space (a tensor product of two copies of the space for one person) and if you produced two people out of one, the evolution of the wave function would be quadratic i.e. nonlinear which would conflict with quantum mechanics (and its no-xerox theorem), too.

These many-worlds apologists, including Brian Greene, often like to say (see e.g. The Hidden Reality) that the proper Copenhagen interpretation doesn't allow us to treat macroscopic objects by the very same rules of quantum mechanics with which the microscopic objects are treated and that's why they promote the many worlds. This proposition is what I call chutzpah.

In reality, the claim that right after the measurement by one person, there suddenly exist several people is in a striking contradiction with facts that may be easily extracted from quantum mechanics applied to a system of people. The quantum mechanical laws – laws meticulously followed by the Copenhagen school, regardless of the size and context – still imply that the total mass is conserved, at least at a 1-kilogram precision, so it is simply impossible for one person to evolve into two. It's impossible because of the very same laws of quantum mechanics that, among many other things, protect Nature against the violation of charge conservation in nuclear processes. It's them, the many-worlds apologists, who are totally denying the validity of the laws of quantum mechanics for the macroscopic objects. In reality, quantum mechanics holds for all systems and for macroscopic objects, one may prove that classical physics is often a valid approximation, as the founding fathers of quantum mechanics knew and explicitly said. The validity of this approximation, as they also knew, is also a necessary condition for us to be able to make any "strict valid statements" of the classical type. The condition is hugely violated by interfering quantum microscopic (but, in principle, also large) objects before they are measured so one can't talk about the state of the system before the measurement in any classical language.

In Nature, all observables (as well as the S-matrix and other evolution operators) are expressed by linear operators acting on the Hilbert space and Schrödinger's equation describing the evolution of any physical system has to be linear, too. Even if you use the density matrix, it evolves according to the "mixed Schrödinger equation" which is also linear:\[

i\hbar \ddfrac{}{t}\hat\rho = [\hat H(t),\hat \rho(t)].

\] It's extremely important that the density matrix \(\hat \rho\) enters linearly because \(\hat \rho\) is the quantum mechanical representation of the probability distribution, even the initial one. And the probabilities of final states are always linear combinations of the probabilities of the initial states. This claim follows from pure logic and will hold in any physical system, regardless of its laws.

Why? Classically, the probabilities of final states \(P({\rm final}_j)\) are always given by\[

P({\rm final}_j) = \sum_{i=1}^N P({\rm initial}_i) P({\rm evolution}_{i\to j})

\] whose right hand side is linear in the probabilities of the initial states and the left hand side is linear in the probabilities of the final states. Regardless of the system, these dependences are simply linear. Quantum mechanics generalizes the probability distributions to the density matrices which admit states arising from superpositions (by having off-diagonal elements) and which are compatible with the non-zero commutators between generic observables. However, whenever your knowledge about a system may be described classically, the equation above strictly holds. It is pure maths; it is as questionable or unquestionable (make your guess) as \(2+2=4\). There isn't any "alternative probability calculus" in which the final probabilities would depend on the initial probabilities nonlinearly.

If you carefully study the possible consistent algorithms to calculate the probabilities of various final outcomes or observations, you will find out that it is indeed the case that the quantum mechanical evolution still has to be linear in the density matrix. The Gertruda-Poisson equation fails to obey this condition so it violates totally basic rules of the probability calculus.

Just to connect the density matrix discussion with a more widespread formalism, let us mention that quantum mechanics allows you to decompose any density matrix into a sum of terms arising from pure states,\[

\hat\rho = \sum_{k=1}^M p_k \ket{\psi_k}\bra{\psi_k}

\] and it may study the individual terms, pure states, independently of others. When we do so, and we often do, we find out that the evolution of \(\ket\psi\), the pure states, has to be linear as well. The linear maps \(\ket\psi\to U\ket\psi\) produce \(\hat\rho\to U\hat\rho \hat U^\dagger\) for \(\hat\rho=\ket\psi\bra\psi\) which is still linear in the density matrix, as required. If you had a more general, nonlinear evolution – or if you represented observables by non-linear operators etc. – then these nonlinear rules for the wave function would get translated to nonlinear rules for the density matrix as well. And nonlinear rules for the density matrix would contradict some completely basic "linear" rules for probabilities that are completely independent of any properties of the laws of physics, such as\[

P(A\text{ or }B) = P(A)+P(B) - P(A\text{ and }B).

\] So the linearity of the evolution equations in the density matrix (and, consequently, also the linearity in the state vector which is a polotovar for the density matrix) is totally necessary for the internal consistency of a theory that predicts probabilities, whatever the internal rules that yield these probabilistic predictions are! That's why two pieces of the wave function (or the density matrix) can never attract each other or otherwise interact with each other. As long as they're orthogonal, they're mutually exclusive possibilities of what may happen.

They can never be interpreted as objects that simultaneously exist at the same moment. The product of their probabilities (and anything that depends on its being nontrivial) is zero because at least one of them equals zero. And the wave functions and density matrix cannot be interpreted as classical objects because it's been proven, by the most rudimentary experiments, that these objects are probabilistic distributions or their polotovars rather than observables. These statements depend on no open questions at the cutting edge of the modern physics research; they're parts of the elementary undergraduate material that has been understood by active physicists since the mid 1920s. It now trivially follows that all the people who study Schrödinger-Newton equations are profoundly deluded, moronic crackpots.

And that's the memo.

Single mom: totally off-topic

Totally off-topic. I had to click somewhere, not sure where (correction: e-mail tip from Tudor C.), and I was led to this "news article"; click to zoom in.

Single mom Amy Livingston of Plzeň, 87, is making $14,000 a month. That's not bad. First of all, not every girl manages to become a mom at the age of 87. Second of all, it is impressive for a mom with such a name – who probably doesn't speak Czech at all – to survive in my hometown at all. Her having 12 times the average salary makes her achievements even more impressive. ;-)


  1. Lubos,

    Your points are well taken: the Schrodinger-Newton equation is fundamentally flawed.

    Expanding on these issues, I'd like to know your views on the validity of:

    1)the WKB approximation,
    2)semiclassical gravity,
    3)quantum chaos and quantization of classically chaotic dynamical systems?



  2. Dear Ervin, thanks for your listening. All the entries in your systems are obviously legitimate and interesting approximations (1,2) or topics that may be studied (3). That doesn't mean that all people say correct things about them and use them properly, of course. ;-)

    The WKB approximation is just the "leading correction coming from quantum mechanics" to classical physics. Various simplified Ansaetze may be written down in various contexts.

    Semiclassical gravity either refers to general relativity with the first (one-loop) quantum corrections; or it represents the co-existence of quantized matter fields with non-quantized gravitational fields. This is only legitimate if the gravitational fields aren't affected by the matter fields - if the spacetime geometry solve the classical Einstein equations with sources that don't depend on the microscopic details of the matter fields and particles which are studied in the quantum framework. The matter fields propagate on a fixed classical background in this approximation but they don't affect the background by their detailed microstates.

    Indeed, if the dependence of the gravitational fields on the properties of the matter fields is substantial or important, there's no way to use the semiclassical approximation. Some people would evolve the gravitational fields according to the expectation values of the stress-energy tensor but that's the same mistake as discussed in this article in the context of the Poisson-Hossenfelder equation.

    Classical systems may be chaotic - unpredictable behavior very sensitive on initial conditions. Quantum chaos is about the research of the complicated wave functions etc. in systems that are analogous to (hatted) classically chaotic systems.

  3. Thanks Lubos.

    I also take classical approximations with a grain of salt. For instance, mixing classical gravity with quantum behavior is almost always questionable a way or another.

    Here is a follow up question. What would you say if experiments on carefully prepared quantum systems could be carried out in highly accelerated frames of references? Could this be a reliable way of falsifying predictions of semiclassical gravity, for example?