Meaning of Wave Function Collapse

[stevendaryl]

I do not see a pair of particles, let alone entagled.

Yes, I know. Electrons and positrons are too tiny to see. But the point of saying that their spins are entangled is that you distant measurements that are correlated. You have a source of particle/antiparticle pairs. Out of each pair, Alice measures the spin of one particle, and Bob measures the spin of the other particle. Empirically, if you want to eliminate mentioning things that are not visible, the way things look is like this (simplified)

A spin measurement device has a dial that can be set to any number between 0 and 360. It has two lights, one on the left and one on right.

One "round" of the EPR experiment has the following steps:

1. Alice picks a number $\alpha$ and sets her device.
2. Bob picks a number $\beta$ and sets his device.
3. Charlie, halfway between them, presses a button (what it does can't be seen by you, so I won't mention it)
4. Either Alice's left light comes on, or her right light comes on.
5. Either Bob's left light comes on, or his right light comes on.
6. (Realistically, there are other possibilities, such as neither light coming on, but I'm oversimplifying)

The facts for the EPR experiment are these:

• If Alice and Bob choose the same number, then they always get opposite results.
• If they choose different numbers, then a fraction of the time $cos^2(\frac{\theta}{2})$, they get opposite results, and a fraction of the time $sin^2(\frac{\theta}{2})$ they get the same result (where $\theta = \beta - \alpha$).

So Alice's and Bob's results are strongly correlated. According to Bell's theorem, the correlation cannot be explained in terms of local hidden variables, but it can be explained in terms of entangled wave functions.

 

[stevendaryl]

So in a quantum state measure, you are 100% sure to get a certain result, just like, knowing the initial conditions of a cannonball, can you predict exactly where it ends?

In the famous EPR experiment, Alice and Bob are guaranteed 100% correlated results, but their individual results are completely unpredictable.

If Alice and Bob choose the same detector setting, then it is 100% certain that they will get opposite results: If Alice gets spin-up, Bob gets spin-down, and vice-versa. But it is completely unpredictable who gets which result.

 

[stevendaryl]

Suppose an apparatus similar to that of the Stern-Gerlach experiment. Suppose that you count from a beam of electrons those that have spin up (where the axis is chosen arbitrarily) and those with spin down. How many electrons will you count with spin up and how many with spin down? Or rather, the question is: how likely is (in your opinion) to find a spin up and to find a spin down. If you made a bet with a large sum where would you put it: on spin up or down?

To see the effects of entanglement, you have to have a source of entangled electron/positron pairs, and two different Stern-Gerlach devices. Then the statistics will be that:

• Each device will measure half of the particles to have spin-up and half to have spin-down.
• For any pair of particles, if one device measures spin-up for one of the particles, then the other device will measure spin-down for the other particle.

These facts by themselves don't imply that the particles are entangled. But the effects of entanglement are seen when the two Stern-Gerlach devices are not given the same orientation. Then the statistics are such that it is impossible to explain them using local hidden variables.

 

[stevendaryl]

Here's a game that summarizes the strangeness of EPR:

• Charlie, the dealer, deals out three cards to Alice, a left card, a middle card and a right card.
• He similarly deals out three cards to Bob.
• After the cards are dealt, Alice picks one card and Bob picks another. The remaining cards are left face-down.
• If Alice and Bob both pick the same position (left, middle or right), then their cards have opposite colors: If Alice's is red, Bob's is black, and vice-versa.
• If Alice and Bob pick different positions, then their cards have opposite colors 25% of the time and the same colors 75% of the time.

There is no way for Charlie to do this without either:

• Reading Alice's and Bob's minds to know which card they will pick, or
• Having trick cards that change color
• Charlie does some other trick (like switching Bob's cards around after Alice picks her card)

If Charlie tried to do it with regular cards and no tricks, then he would have to give one person two blacks and one red, and give the other person two reds and one black. But if he did that, then the probability that they would have opposite colors when they pick different positions is 1/3, not 1/4.

 

[Boing3000]

So in a quantum state measure, you are 100% sure to get a certain result,

Well, kind of. QM is verifyied 100% but only to the extend that you make many (many many) measures. It is a stochastic theory, meaning one outcome is very unpredictable. And that is not the case for classical mechanic.

just like, knowing the initial conditions of a cannonball, can you predict exactly where it ends?

No, very unlike the cannonball. Even if the is prediction is actually more difficult in reality that one may think (because of chaos). But with idealized cannonball (in vacuum etc... you can obtain very small margin of error).

 

[member 648030]

Well, kind of. QM is verifyied 100% but only to the extend that you make many (many many) measures.

Exactly like like throwing a coin or a nut ...

 

[member 648030]

Yes, I know. Electrons and positrons are too tiny to see. But the point of saying that their spins are entangled is that you distant measurements that are correlated. You have a source of particle/antiparticle pairs. Out of each pair, Alice measures the spin of one particle, and Bob measures the spin of the other particle. Empirically, if you want to eliminate mentioning things that are not visible, the way things look is like this (simplified)
...
So Alice's and Bob's results are strongly correlated. According to Bell's theorem, the correlation cannot be explained in terms of local hidden variables, but it can be explained in terms of entangled wave functions.

Ok everything's right what you say, but I do not understand why you talk about pairs of particles. I am considering the case of a single particle, if you want, the classic Schrodinger equation of the electron in the hydrogen atom.

 

[Boing3000]

Exactly like like throwing a coin or a nut ...

Exactly unlike the throwing of a coin which is at every single moment in a precise state. There is no stochastics involved.
The fact that you arrange the coin to be thrown in a random way (like using a hand) is where you get the illusion that the coin is in an unknown state and need a stochastic approach.

I hope you won't count on a "collapse" to win again future gambling robot. Because if they get to throw the coin, or even to bet after having watched you trough it ... you'll lose 99% of the time.
Meanwhile, you'll be as powerful as any robot to bet on spin of particles (because quanta behave very differently)

 

[member 648030]

Exactly unlike the throwing of a coin which is at every single moment in a precise state. There is no stochastics involved.
The fact that you arrange the coin to be thrown in a random way (like using a hand) is where you get the illusion that the coin is in an unknown state and need a stochastic approach.

I hope you won't count on a "collapse" to win again future gambling robot. Because if they get to throw the coin, or even to bet after having watched you trough it ... you'll lose 99% of the time.
Meanwhile, you'll be as powerful as any robot to bet on spin of particles (because quanta behave very differently)

So when Einstein (I repeat) saying the famous phrase "God does not play dice!" (evidently referring to the probabilistic character of the QM), Bohr should have answered: "Albert, but a dice is not a random object! But did you study classical physics? Did you give it the general physics exam?

 

[PeterDonis]

The fact that a state has a value and another value

This is not the case. The state has only one value.

 

[PeterDonis]

I am considering the case of a single particle

If your quantum system consists of only a single particle, there are no entangled states, and none of the issues being discussed in this thread arise. That doesn't mean those issues don't exist. It just means you have picked the wrong system to illustrate them.

 

[Boing3000]

So when Einstein (I repeat) saying the famous phrase "God does not play dice!" (evidently referring to the probabilistic character of the QM), Bohr should have answered: "Albert, but a dice is not a random object! But did you study classical physics? Did you give it the general physics exam?

Why should he have done that ? Don't you know that they were both top physicists ? Do you think he was doing applied theology or either illustrating its frustration with the most common example of pseudo randomness?
Should they have bothered that hundred years later some random people (pun intended) on the internet were going to make wide claims that coin are "truly" in superposition of state or that God is a gambler ?

 

[member 648030]

If your quantum system consists of only a single particle, there are no entangled states, and none of the issues being discussed in this thread arise. That doesn't mean those issues don't exist. It just means you have picked the wrong system to illustrate them.

The title of the topic is "The meaning of the collapse of the wave function" (I do not see what entagled states have to do with.) Carefully review the topic title, Mr. administrator of the forum!

 

[PeterDonis]

The title of the topic is "The meaning of the collapse of the wave function" (I do not see what entagled states have to do with.)

Then you apparently have not read the actual thread. Dealing with wave function collapse is easy if you only talk about a single particle. All of the issues that give rise to questions like the one being asked here only arise when you consider multiple particles, and if those particles aren't entangled, then even that case ends up being about as easy as the case of a single particle. Only when you deal with systems containing multiple particles that are entangled does wave function collapse become a real issue. Which is exactly what this thread discussion has been about.

Carefully review the topic title, Mr. administrator of the forum!

The thread title is not the same as the entire thread discussion, taken in context.

 

[member 648030]

xactly unlike the throwing of a coin which is at every single moment in a precise state. There is no stochastics involved.
The fact that you arrange the coin to be thrown in a random way (like using a hand) is where you get the illusion that the coin is in an unknown state and need a stochastic approach.

I hope you won't count on a "collapse" to win again future gambling robot. Because if they get to throw the coin, or even to bet after having watched you trough it ... you'll lose 99% of the time.
Meanwhile, you'll be as powerful as any robot to bet on spin of particles (because quanta behave very differently)

Perfectly agree: you can know the state of a coin in every moment of time (as Lagrange teaches): more, you can predict exactly the result: just calculate the force impressed on the coin, (taking into account the exact moment that the finger of the hand imprints to it), the shape, the weight, the dimensions and the possible deformations of the coin to the minimum detail, as variations of the density of the material etc, to set the equations of the motion (rotary and translatory) in the gravitational field, taking into account the frictions with the air, calculate the forces that the surface in the impact with the coin will impress it, the moments etc, the dissipation by friction, and with some equations of Euler-Lagrange etc you can determine without error if it will come out head or cross. Without involving the "statistic" you will be able to win any amount at the casino: what are you waiting for !

 

[member 648030]

Then you apparently have not read the actual thread. Dealing with wave function collapse is easy if you only talk about a single particle. All of the issues that give rise to questions like the one being asked here only arise when you consider multiple particles, and if those particles aren't entangled, then even that case ends up being about as easy as the case of a single particle. Only when you deal with systems containing multiple particles that are entangled does wave function collapse become a real issue. Which is exactly what this thread discussion has been about.
The thread title is not the same as the entire thread discussion, taken in context.

Is it stochastic ... or "chaotic"?
bye

 

[PeterDonis]

Is it stochastic ... or "chaotic"?

I don't know what you're asking about. Neither of those words appear in what you quoted from me.

 

[stevendaryl]

Ok everything's right what you say, but I do not understand why you talk about pairs of particles. I am considering the case of a single particle, if you want, the classic Schrodinger equation of the electron in the hydrogen atom.

Because entanglement is about a relationship between two different particles (or two different systems). And entanglement is a way to show how quantum probabilities are different from classical probabilities. You can't show this for a single particle.

 

[stevendaryl]

The title of the topic is "The meaning of the collapse of the wave function" (I do not see what entagled states have to do with.)

Because if you have a way of interpreting wave function collapse for a single particle that doesn't work for entangled particles, then that means it's wrong.

 

[richrf]

I find it very uncomfortable discussing "particles" in the context of QM. There is no such thing much less a single particle. There is no Interpretation of QM that I know of that suggests there are free particles floating around in space.

 

[edmund cavendish]

As i opined in an earlier posting this is exactly the area that sir roger penrose is developing. He has lectured recently about the importance of gravity in observer independent/ indifferent wave function collapse. Having approached this from a philosophical angle I am following mentor reading suggestions to uplift my maths. He has lectured to consciousness seminars in the us and there are plans for a uk based centre. I am aware that his views are not mai stream but am aware that one function of the proposed centre is to seek scientific/experimental underpinning. Hope this is useful.
Edmund

 

[vanhees71]

Because entanglement is about a relationship between two different particles (or two different systems). And entanglement is a way to show how quantum probabilities are different from classical probabilities. You can't show this for a single particle.

That's also not true. Entanglement is between observables. So first of all you have to say which observables are entangled. A very well-known example for entanglement concerning one-particle observables is the Stern-Gerlach experiment. In the original experiment they used silver atoms, running through an inhomogeneous magnetic field. You can as well do this experiment with, e.g., neutrons. In terms of modern quantum mechanics the SG apparatus leads to the preparation of states describing the entanglement between the position of the particle and the spin component in direction of the magnetic field.

 

[stevendaryl]

That's also not true. Entanglement is between observables. So first of all you have to say which observables are entangled. A very well-known example for entanglement concerning one-particle observables is the Stern-Gerlach experiment. In the original experiment they used silver atoms, running through an inhomogeneous magnetic field. You can as well do this experiment with, e.g., neutrons. In terms of modern quantum mechanics the SG apparatus leads to the preparation of states describing the entanglement between the position of the particle and the spin component in direction of the magnetic field.

Yes, entanglement is a general fact about observables. But the significance for demonstrating that quantum probabilities are inherently different from classical probabilities requires entanglements between observations taken at a distance. Two entangled properties of the same particle can be as easily explained by hidden variables as a single property.

 

[bhobba]

The fact that a state has a value and another value means, from the point of view of the experimenter, one of two things:
1. the state is defined, but we do not know it
2. they are co-present multiple states simultaneously.

You are tying yourself in logical knots because of one simple thing - you are trying to think in terms of an unobserved system. We have no idea at all about a system when its not observed. We have conjectures - but nothing is known for sure.

Peter has posted the problem with QM - what is this thing called an observation in terms of QM itself. QM is theory about observations in a common-sense classical world. But QM is a theory that is supposed to explain that classical world. How can it be when it assumes it in the first place? As it stands now it, in Peters words, is incomplete - or you can be kinder and be like me and say some issues still remain - but that is just semantics. The observation problem has not been solved - but a lot of progress has been made. For example a coarse graining argument shows how the classical world emerges, and using that we can understand observations better. This is the Decoherent Histories approach. But the program is not complete yet and key theorems are still not done eg as mentioned before it relies on partitioning a system into what's doing the observing, what's being observed, and the environment. Does the result depend on that decomposition or is it independent of it. The answer will crucially affect what the interpretation is telling us and if the observation issue is resolved or not.

Thanks
Bill

 

[Sophrosyne]

Here is a related question:

Molecules attach to each other via their outer orbital shell. The electrons in these outer shells are not, I presume, really collapsed wave functions of position and momentum. They are still in a superposition state. And yet molecules maintain very particular shapes based on the probability cloud of these electrons around the nuclei. Are the electrons in molecules collapsed wave functions where they can hold these other nuclei in the particular shapes they do, or just uncollapsed probability waves, and the nuclei are just sort of held in position by that probability wave?

 

[bhobba]

Molecules attach to each other via their outer orbital shell. The electrons in these outer shells are not, I presume, really collapsed wave functions of position and momentum. They are still in a superposition state. And yet molecules maintain very particular shapes based on the probability cloud of these electrons around the nuclei. Are the electrons in molecules collapsed wave functions where they can hold these other nuclei in the particular shapes they do, or just uncollapsed probability waves, and the nuclei are just sort of held in position by that probability wave?

Again you are getting confused about what QM says. When not observed you can't say anything other than predict probabilities if you observed it. When we say a molecule has a certain shape its simply a heuristic about that probability distribution. We can say, for example, there are negligible probabilities, if you did observe it to lie outside a certain region, hence, in that sense it has a certain shape. But it simply is a heuristic.

Thanks
Bill

 

[Sophrosyne]

Again you are getting confused about what QM says. When not observed you can't say anything other than predict probabilities if you observed it. When we say a molecule has a certain shape its simply a heuristic about that probability distribution. We can say, for example, there are negligible probabilities, if you did observe it to lie outside a certain region, hence, in that sense it has a certain shape. But it simply is a heuristic.

Thanks
Bill

But these molecules have very distinct shapes, with very real macro-world consequences: a snowflake, for example. When you look at a snowflake crystal, that whole thing is a very distinct shape, that obviously requires all the water molecules to have a very particular shape, which in turn means all the orbitals connecting the hydrogens to the oxygen have to have a very distinct shape at any given time. Does that shape come into being just because you are looking at it and observing it, or is it just that the hydrogen atoms are held in a "general" position around the oxygen and sort of acting as "observers" of its electrons, collapsing their wave functions, within the uncertainty principle levels of uncertainty, in a general sense?

 

[Grinkle]

When we say a molecule has a certain shape its simply a heuristic about that probability distribution.

What is the dividing line between a macro object and a quantum object?

If asked, I'd have said (with no foundation for saying) a molecule is as macro an object as a chair or a table with behavior that can be accurately described by classical physics.

 

[PeterDonis]

But these molecules have very distinct shapes, with very real macro-world consequences: a snowflake, for example.

A snowflake is not a molecule, or even close to one. It is made of something like $10^{23}$ water molecules.

When you look at a snowflake crystal, that whole thing is a very distinct shape, that obviously requires all the water molecules to have a very particular shape

No, it doesn't; it just requires the water molecules to be arranged in a particular shape. But that shape, the snowflake shape, is only one of many, many possible arrangements of water molecules, and doesn't tell you anything useful about the shapes of water molecules themselves.

Does that shape come into being just because you are looking at it and observing it, or is it just that the hydrogen atoms are held in a "general" position around the oxygen and sort of acting as "observers" of its electrons, collapsing their wave functions, within the uncertainty principle levels of uncertainty, in a general sense?

A water molecule in a normal environment, i.e., surrounded by lots and lots of other water molecules, not to mention molecules and atoms of many other substances, is constantly being decohered, so the uncertainty principle does not play any significant role in its behavior.

If you isolated a single water molecule from all external influences, you could run quantum experiments on it, such as the double slit (which has not, to my knowledge, been done with water molecules, but it has been done with buckyballs, which have 60 atoms), to show that the water molecule does have similar quantum properties to other quantum objects. But I don't know what experiments you would run to test whether the water molecule has a single definite "shape" or whether there is uncertainty about its shape at the quantum level.

What is the dividing line between a macro object and a quantum object?

Quantum mechanics does not tell us that. In practice, physicists put the line wherever they need to to make accurate predictions.

f asked, I'd have said (with no foundation for saying) a molecule is as macro an object as a chair or a table with behavior that can be accurately described by classical physics.

This is obviously false, since, as I mentioned above, we have done quantum experiments like the double slit with buckyballs, which have 60 atoms each.

 

[Nugatory]

What is the dividing line between a macro object and a quantum object?

Suppose I were to ask you what was the dividing line between a cloud of gas (described by the ideal gas law $PV=nRT$) and a collection of gas molecules bouncing around (described by Newton's laws for particle motion)? For most systems the answer would be obvious: A mole of nitrogen in a ten-liter flask is obviously a classical gas with a classical temperature and pressure and three stray nitrogen molecules are just as obviously not. But there's no hard and fast dividing line; it's possible to construct systems for which the answer will be "it depends" or "how accurate of an answer do you need?".

Your question is similar: There are things like bound electrons that are clearly quantum. There are also things like cats and cannonballs for which any quantum effects are totally negligible and we accept classical mechanics as an an accurate description of their behavior. But there's also a wide gray area in between - whether a 60-atom buckyball is a classical object or a quantum object very much depends on what you're trying to do with it.

In general, the harder it is to maintain something in a coherent superposition, the more likely it is that a classical description will be accurate.

 

[bhobba]

But these molecules have very distinct shapes, with very real macro-world consequences:

I am not denying that - I am clarifying what shape means. There is nothing collapsed. The electrons are entangled with the nucleus - most notably with the EM field of the nucleus. This creates a certain probability distribution for the electrons if you observe it. That probability distribution has a shape. To be clear there is no denying its shape - I am simply making clear what that shape means.

With a scanning tunneling microscope you can see that shape - but how it works is not a simple matter:
https://en.wikipedia.org/wiki/Scanning_tunneling_microscope

Its an interaction of the electrons in the atom and the electrons in the microscope:
'Knowing the wave function allows one to calculate the probability density for that electron to be found at some location. In the case of tunneling, the tip and sample wave functions overlap such that when under a bias, there is some finite probability to find the electron in the barrier region and even on the other side of the barrier'

And yes if you observe it with the microscope that is an observation and you can 'see' things like the shape of a probability distribution. They are also interacting with other molecules near them and that too has a effect such as forming a lattice structure etc.

What may be complicating your understanding is here in the macro world things have an actual shape, are solid etc. How this happens is not trivial. For example you may think, and even some elementary textbooks or not very knowledgeable teachers have told you, the reason objects are solid is the outer electrons of the molecule or atom repel the other outer electrons. That is wrong as shown by Dyson in a fundamental paper written in 1967 (The Stability Of Matter) - its the Pauli Exclusion principle:
http://www.ams.org/journals/bull/2013-50-01/S0273-0979-2011-01366-0/S0273-0979-2011-01366-0.pdf

In the quantum world various heuristics etc we have, or were taught in elementary classes at school, are often wrong.

That's why its important to understand what QM really says. Its a theory about the probability of the results of observations. Thinking in terms of shapes etc in the usual classical sense, except as a heuristic, is not compatible with the basis of QM. When in doubt its QM that is invariably correct - the heuristic wrong. If you can not figure out why in a particular situation then post here and we will get to the bottom of it.

Thanks
Bill

 

[vanhees71]

What is the dividing line between a macro object and a quantum object?

If asked, I'd have said (with no foundation for saying) a molecule is as macro an object as a chair or a table with behavior that can be accurately described by classical physics.

There is no dividing line beetween a macro object and an quantum object. As far as we know, all phenomena regarding matter and radiation is described by quantum theory, i.e., relativistic quantum field theory. At least there's not one exception found for any system ever observed which violates the fundamental principles of QT.

The disinction between macroscopic objects from quantum objects, i.e., the apparent behavior of macroscopic objects according to classical mechanics and classical field theory is just the amazing number of microscopic degrees of freedom, which are rather irrelevant to effectively describe the behavior of such systems. E.g., you understand a whole lot about the motion of the planets around the Sun by just making the very much simplifying an abstract assumption that the planets and the sun can be described as classical "point particles". It's just not so relevant that these are extended objects as far as the motion of the planets around the sun is concerned.

It's of course different if it comes to an understanding of the Sun or the planet itself. Then it might be interesting to now, how it is composed what is its intrinsic mechanics etc. Of course, here we'd also not use quantum theory to describe every little detail, in the extreme the constitution of the system in terms of quarks and electrons together with the four interactions holding these systems together. E.g., to understand the Sun, it's enough to use (magneto-)hydrodynamics and Newtonian gravity etc.

In other words, we use (quantum) statistical physics to reduce the zillions of zillions of microscopic degrees of freedom to the effective description of a few macroscopic observables, depending on the problem you want to understand choosing appropriate relevant macroscopic degrees of freedom. The state of the system is then described sufficiently by these relevant degrees of freedom, which can be defined as averages over many microscopic degrees of freedom.

On the other hand, if you wish to explore more and more details about large objects down to their quantum behavior, it becomes more and more complicated to prepare these systems in states, where quantum effects become relevant. That's due to decoherence. One funny example is the double-slit experiment with the rather large bucky-ball molecules, i.e., a bound state of 60 carbon atoms. These are not that many degrees of freedom yet (I'd call it a mesoscopic system), but it's already difficult to isolate the balls enough from the environment that one gets coherent enough matter-wave like states to do the double-slit experiment. This was done by Zeilinger et al some years ago, and first of all they had to cool down the bucky-balls to sufficiently low temperatures to bring them to low energies of their intrinsic states. Otherwise the bucky-balls would produce a lot of thermal rather soft photons. First of all Zeilinger an his team indeed could successfully demonstrate the paradigmatic example for quantum behavior in the double-slit experiment with C60 molecules. But they could also "heat the molecules up" in a controlled way, so that they were still pretty cold, but warm enough to emit a small amount of photons on their way to the double slit. As expected emitting just a few photons randomized the state of the bucky-balls enough to let the contrast of the interference pattern get worse, and only if they got warm enough to emit some more photons the interference pattern was gone, and they got the distribution as expected from classical objects on the screen. It's not that the bucky-balls weren't behaving according to classical physics all of a sudden, invalidating the quantum description, but the emission of a few thermal photons in random directions was enough "decoherence" to let the outcome of the experiment look classical.

 

[member 648030]

I find it very uncomfortable discussing "particles" in the context of QM. There is no such thing much less a single particle. There is no Interpretation of QM that I know of that suggests there are free particles floating around in space.

No, but there is a Schrodinger equation for a single particle in the presence of a potential V
And the solutions are not only correct, but perfectly coherent with the experiments, when V takes particular forms, for example: potential of the Coulambian type, or potential barrier: see hydrogen atom, tunnel effect, etc. Well known quantum effects.
These solutions do not resort to entaglement. The energy levels of the hydrogen atom spectrum do not use entagled states

 

[vanhees71]

The non-relativistic hydrogen-atom energy eigenstates are highly position-entangled states of a proton and an electron :-))):

https://arxiv.org/abs/quant-ph/9709052
https://doi.org/10.1119/1.18977

What's unentangled are the states of a highly abstract object, i.e., a quasi particle with the effective mass $\mu=m_e m_p/(m_e+m_p)$ describing the "relative motion" after separating off the center-of-mass motion (and this works in a strict sense also only in the non-relativistic limit!).