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A point particle is a point in space of a fixed co-ordinate. However, the wavefunction must always be spread out in space to be normalizable to unity.

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8 Answers 8

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A point particle is a point in space of a fixed co-ordinate

That is not what a point particle is in modern physics. A point particle is a particle with no structure as measured by scattering experiments. An electron fits that requirement to the smallest scales currently measurable.

The scattering mentioned above refers to "deep inelastic scattering" experiments, which in turn were based on Rutherford's work on elastic scattering. Basically, Rutherford derived a formula for elastic scattering off a point particle. If the observed scattering matches that then it is either a point particle or a particle with a smaller structure than the wavelength used in the experiment. If the particle has sub-structure then the actual scattering will deviate from Rutherford's formula. This is how quarks were proven en.wikipedia.org/wiki/Deep_inelastic_scattering

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    $\begingroup$ I think this is the clearest answer so far. But what do you mean exactly by "a particle with no structure as measured by scattering experiments?" $\endgroup$ Commented Apr 11, 2022 at 11:14
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    $\begingroup$ @AccidentalTaylorExpansion that comment was referencing the "deep inelastic scattering" experiments, which in turn were based on Rutherford's work on elastic scattering. Basically, Rutherford derived a formula for elastic scattering off a point particle. If the observed scattering matches that then it is either a point particle or a particle with a smaller structure than the wavelength used in the experiment. If the particle has sub-structure then the actual scattering will deviate from Rutherford's formula. This is how quarks were proved en.wikipedia.org/wiki/Deep_inelastic_scattering $\endgroup$
    – Dale
    Commented Apr 11, 2022 at 12:49
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    $\begingroup$ @Dale that's a great explanation, I'd suggest adding it to the answer $\endgroup$
    – llama
    Commented Apr 11, 2022 at 18:11
  • $\begingroup$ @llama thanks, that is a good suggestion. Done! $\endgroup$
    – Dale
    Commented Apr 11, 2022 at 22:19
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You are hitting the essence of particle duality, Egg Man.

According to Feynman (in his recorded lectures on Young's experiment, although I cannot find the link now) particles are particles in the sense they make clicks, or spots, on the detector, and never extended signal. In fact, this is what is typically the definition of modern understanding of "particle" based on many-body quantum mechanics but may go beyond simple matter particles such as electrons or atoms: this idea includes photons in photocounting experiments and quasiparticles in condensed matter systems.

Wave behaviour is the simplest model to explain a number of phenomena that quantum particles exhibit, such as interference and tunneling. Perhaps a model that does not postulate exitence of any waves could account for quantum phenomena; although it would have to explain the phenomena already succesfully explained by quantum mechanics that are historically associated with wave behaviour. Historically, wave mechanics was very successful because it derives Heisenberg's uncertainty principle as a mathematical fact of Fourier analysis. Modern approaches embed Heisenberg's uncertainty on purely algebraic (rather than analytic) grounds, by assuming certain commutativity relations between observables.

To clarify, quantum waves are waves in coordinate space, not necessarily real space. This means that as variables they take degrees of freedom of the system rather than necessarily physical position or momentum; for example, a relative position of two particles, or a sum of positions of a group of particles. Both of those are coordinates which can be variables of a quantum wave but not neccesarily as position or momentum of a quantum object. In that sense, it is not particles that are waves (as they remain particles in Feynman sense explained above at all times), but rather their kinematic properties (coordinates as measured in a reference frame) or properties of their ensembles and many-body systems they comprise.

Ps. By the way, neither of the statements you made are actually correct. The first sentence is false by inverting the relation of identity (point particles are points in space (...), so the reverse must be true too). If this is the case then points in space are point particles. This is clearly not true as there exists empty space. You need to add extra information about which points in space are particles to the definition to make it correct.

Second, wavefunctions are in general distributions that simple function; so physicsist use Dirac delta distribution which is in fact localised to a point and normalizable at the same time, so not spread out.

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    $\begingroup$ Unless your detector has infinite precision (and it doesn't), all detections are extended. In fact, you can say that it was detected everywhere with some minuscule intensity. No? $\endgroup$
    – Juan Perez
    Commented Apr 10, 2022 at 2:18
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    $\begingroup$ @JuanPerez No, detections are measurements and measurements cause wave function collapse, meaning the particles behave like point particles when detected. Here the particle is forced to choose one position out of the infinite possibilities that the wave function allows. It is detected at a single physical point, and the smaller and more tightly resolved our detectors become they continue to behave like point particles under measurement. This is how the particle part of the wave/particle duality manifests. $\endgroup$
    – J...
    Commented Apr 10, 2022 at 18:53
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The troubling answer to your question is that the electron is neither a point particle nor a wave function.

The basic tenet is that we strongly believe that the electron exists and we have seen that it behaves in certain ways in the experiments we have done so far. We have a number of ways of trying to condense down our knowledge into mathematical (or other form) that we call a model.

The models we have today closely corresponds with our experimental data. The wavefunction model is one of these models that closely describes some of the outcomes of our experiments. In a larger picture, the so called standard model closely describes several of the experiments extremely well.

But the important point is to not confuse the model with the reality. Reality will continue to surprise us and force us to change our models in the future.

So, currently, until we will know better, the electron is best described by models that allows both an interpretation as a point charge/particle and as a wave function. In the future we may very well find that none of our current models describes every aspect of the outcome of the experiments we do and we will then have to create new models. Until then we will have to use both models, and select the correct one in different circumstances.

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    $\begingroup$ Everything else aside, this is the most important aspect of the answer, IMO. We have no idea what anything really "is"... this has been true in the time of Plato as well as today. $\endgroup$
    – AnoE
    Commented Apr 11, 2022 at 11:29
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    $\begingroup$ It's implied by your answer but I think it's worth noting that both 'points' and 'wavefunctions' are mathematical constructs exist only in our imaginations. There's a hard limit to measurement precision and we cannot measure/observe a waveform directly (because it 'collapses'.) The best we can do is infer a given waveform statistically by repeating the same experiments many times. $\endgroup$
    – JimmyJames
    Commented Apr 11, 2022 at 13:54
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Both "point particle" (think bullets) and "wave" (think oceans and ripple tanks) are metaphors.

An electron is not one of these.

An electron is an electron.

Under certain conditions, an electron may be described in a simplified way by the "particle" or "wave" metaphors, but it is always an electron.

Saying, "Jack is a firecracker," doesn't mean that he has a fuse, explosives, and a paper wrapper. But, under the right circumstance, the metaphor may be useful.

It is important in physics, as in life, to use metaphors both for simplified explanation, and to remember that the thing is not the metaphor.

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How can an electron be a point particle but also a wavefunction?

See here to learn the definition of a wave function for a quantum mechanical entity

Each particle is represented by a wavefunction $Ψ(position,time)$ such that $Ψ^*Ψ$= probability of finding the particle at that position at that time

bold mine

If you search the link you will see that this is an axiomatic statement in the quantum mechanics postulates.

So the electron is not a wavefunction, it is modeled by functions of wave equations representing waves in probability space.

The electron and the other elementary particles in the table seen in the standard model of particle physics are modeled as point particles in the very successful field theory standard model , successful meaning that the theory fits the existing data and is predictive of new data.

To get a feeling/intuition of how an electron can be a particle and have a probability wave this answer of mine may help.

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Perhaps consider the double slit experiment performed with electrons. From it we can observe how electrons exhibit behavior both like particles, and like waves.

In our macroscopic world, we expect 'particles' to behave in a certain way. Throw some tennis balls at two tennis ball sized slits in a wall, and you'd expect to see on a detection screen behind, two lines where the tennis balls went through the slits.

Similarly, if you send a plane wave of water at a wall with two slits, we expect the two wave sources on the other side to interfere, and create an interference pattern on a detection screen through the constructive and destructive interference of the two wave sources.

These are macroscopic behaviors that we expect in our everyday lives. But things get weird when dealing with fundamental particles. Even though we call them particles, what happens when the electrons are sent to the double slit?

We see an interference pattern - just like the plane wave of water would make. The hypothesis? The electron has gone through both slits at once, and interfered with itself. How is this possible when we consider an electron as a particle?

The answer is a foundational element of quantum mechanics: superposition. The electron exists in a state of all possible outcomes simultaneously, where it can exhibit wave like properties.

But the oddest thing happens when we decide to detect which slit an electron has passed through. The wave like behavior then reverts to that of a classical particle and the interference pattern is no more. Only when the electron position is unknown, can it be in a superposition of states.

This is something that lead to the foundation of the theory of quantum mechanics in that: how do we explain this behavior of electrons as a facet of nature?

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    $\begingroup$ The electron does not "exist in all possible states simultaneously." A quantum system exists in one state, which can be expressed as a linear combination of some basis states. $\endgroup$
    – Sandejo
    Commented Apr 12, 2022 at 4:07
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It's quite simple.

A point particle is what an electron is, the wave function is where the electron is.

In most expositions of physical theory found in textbooks, there isn't often a clear and neat separation made between the language used to talk about the "what" and the "where" of things - we often tend to see them thrown together like at one point we'll be talking about the motion of a center of mass, then we'll talk about "points on the body", and the like, but if we want to really do this question justice, we'll need to clean that up and make the separation explicit.

You can regard the theoretical description of a material body as basically consisting of two parts: one of these is its kinematic state, which describes where it is located and how it is moving at a given point in time. In classical mechanics in one dimension, this is typically a pair of real numbers, or better a point and a vector both of dimension one, but those things are "similar" enough you can "elide" the distinction (but watch out! That's what gets you into this trouble!) - $(x, p)$, called the position and the momentum. In three dimensions, these get upgraded to 3D points and vectors, $(X, \mathbf{p})$ and when we take account of rotation, we must introduce two more parameters, $\Theta$ and $\mathbf{L}$, or orientation and angular momentum.

The second part, however, of the theoretical description, is the pure shape of the object, which I would suggest be called its avatar. The avatar is not the actual region of physical space occupied by the object, but rather the "stock" shape that we would put in a conceptual "rigging space" or "modelling space", and that can be applied via the kinematic state to determine the set of points that is actually occupied aka. the "points on the body" as you usually see in textbooks. For lack of real notation existing, we can denote the avatar by $\mathrm{Av}(O)$, where $O$ is the object.

As to what "rigging space" is, it could be considered either a copy of physical space but I'd actually say (at least in cases with a Euclidean background space), it's better to consider it the real vector space, because that is mathematically much neater: to get the "set of points on the body" we can then just write $\mathrm{Occ}(O) = X + \mathrm{Av}(O)$ (meaning "occupied set"), which is elementwise application, since you can add vectors affinely to a point, but you cannot add points, and you'd need some additional gymnastics to make that work if we gave it in the "data type" of points.

So what does it mean to be a "point body"? Simple: $\mathrm{Av}(O)$ contains only one point (vector, actually). Usually that is origin or, better, the zero vector $\mathbf{0}$, because we like to center the avatar on the center of mass (we don't have to, but then if you want your rotational motions to be physically plausible, you have to go through some extra gymnastics). Hence, for an electron, that's it: $\mathrm{Av}(e) = \{ \mathbf{0} \}$.

Now whether that avatar is a good representation of a real electron is, of course, open for experimental testing - but if I remember right it has been verified down to $10^{-22}\ \mathrm{m}$.

And so now we're talking about electrons - on to quantum mechanics. What quantum mechanics does is it changes the description of the physical properties of an object. How this applies to the avatars of elementary objects is still uncertain - I said above that it is a point, which means actually there is no change(*) - what we're after is the wave function in the textbooks: $\psi$ or better, $\psi_X$, the wave function of position, because the point we want to make is that it $\psi_X$ does not replace $\mathrm{Av}(e)$, it replaces the $(X, \mathbf{p})$ part - at least, as a crude way of putting things, because really, to be accurate we want to talk of a density operator, which gives the most general phase-space description, i.e. taking account of $X$ and $\mathbf{p}$ simultaneously in all possible ways one can have meaningful probability distributions on them; wave functions only apply in certain restricted cases.

So that's how it can be a point-sized particle while also having a wave function. The two do different jobs in the theory.


(*) Well, there might be, e.g. in string theory, but that's not what we want to get into here because the point of this post is about understanding standing theory, not inquiring about speculative theory.

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  • $\begingroup$ So you mean the Avatar and the position can be completely independent things? Being a point doesn't imply that you occupy a point? $\endgroup$
    – Egg Man
    Commented Apr 12, 2022 at 2:16
  • $\begingroup$ @Egg Man: They are independent as mathematical objects, but a complete description of a physical object must include both, just as it's nonsense to say an object has a position but no momentum (not zero momentum, but $\mathbf{p}$ simply erased from the description), so a point particle in classical mechanics is both a point and occupies a point. In quantum mechanics, it's just that the designator for which point is occupied has changed form - it's no longer a single real number, it's instead an amplitude distribution, or even better, a so-called "density operator", $\hat{\rho}$. $\endgroup$ Commented Apr 12, 2022 at 7:44
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People mix up definitions and models.

Point particle, as in freshman physics lectures is a model. Mobile phone is not a point particle but it falling from a window can be modelled with point particle model. You can calculate the trajectories of a cannonball with point particle model, but neither mobile phone, nor the cannonball are point particles.

Particle is another model to describe the behaviour of something. Wave is yet another model. Two different models can be used to describe the same thing with different settings. Using particle model for electron helps with understanding Compton scattering or the photoelectric effect, but you need to use the wave model to understand double slit experiment or tunnelling.

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