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What does a particle actually look like - if a person wanted a realistic image of it in their head?


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I have seen some pictures when I was in school where they used coloured balls to represent sub atomic particles. I have also seen on Wikipedia the probability clouds of where a particle may most likely be but that doesn't tell you really what a particle looks like. So what does it actually look like? How do I start thinking of a subatomic particle in my head?

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I'll start but I'm sure you'll get much better answers from others.

It depends on which particle you are talking about. When an electron is viewed as a particle it has no shape at all as it is a 'point' particle and has no dimensions.

A proton on the other hand has a definite diameter and its shape is that of a sphere, although it is of course not a solid ball.

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Assuming of course you are 'using' visible light to picture the particle in your head ...

It will look like a 'fuzzy' volume.                     
Denser 'fuzz' near the center, more rarefied 'fuzz' toward the edges, but never actually ending ( no abrupt cut-off )

You can shrink this fuzzy volume down, by imagining it with higher energy x-rays, or gamma rays, but you can never bring it into 'focus'.
It will always be 'fuzzy'.

Edited by MigL
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15 hours ago, Alex Mercer said:

So what does it actually look like?

The problem here is that our sense of ‘seeing’ is a purely classical process - it’s light of certain wavelengths being reflected off macroscopic objects that simultaneously have well defined positions and momenta. But subatomic particles are not classical objects in that same way - so your question is, in some sense, a category mistake; quantum objects don’t ‘look like’ anything, because they don’t obey the classical principles which underlie our visual sense. If anything, you’d have to turn the question around and ask: what would the rest of the universe look like if you were somehow able to piggy-back along on an elementary particle? And I’m afraid I don’t have a good answer for that one.

15 hours ago, Alex Mercer said:

How do I start thinking of a subatomic particle in my head?

Don’t think of it visually at all - think of it as an abstraction, similar to how an emoji can be an abstraction of someone’s mental state. The essence of an elementary particle is that it is a representation of a set of fundamental symmetries, nothing more. In tech speak: it is an irreducible representation of a symmetry group. So the best and most accurate way to think of elementary particles isn’t as ‘things’ at all, no matter how tempting that may be, but as abstract expressions of symmetry. 

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17 hours ago, Alex Mercer said:

I have seen some pictures when I was in school where they used coloured balls to represent sub atomic particles. I have also seen on Wikipedia the probability clouds of where a particle may most likely be but that doesn't tell you really what a particle looks like. So what does it actually look like? How do I start thinking of a subatomic particle in my head?

Good morning, Alex.

I will keep this short in case you can't be bothered to come back here to look at answers.

Why do you expect to be able to see sub atomic or even atomic particles ?
Can you see the wind?

Do you understand how black and wet photographs work ?
The light produces a monochrome image of the scene, we can see but is not the same as if we viewed the scene directly with our eyes.

It is possible to see black and white (and yes add 'false colour' as in your opening post) images but suitable techniques.

Here are a couple.

parpics1.jpg.e56e1f2f2530f1ef2585739114c124b4.jpg

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Excellent answers. I particularly like Markus' one, as I feel it goes to the crux of the matter. Particles have no colour, nor are they matte or shiny, so they don't "look like" anything. The attribute of "looking like" something is ultimately explained in terms of particles, so trying to tell what they look like is as hopeless as trying to figure out whether they're smiling or not. Electrons don't smile.

Surprising? May be. But electrons don't look like anything that looks like something.

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My thinking also John.
The OP wasn't "What is an elementary particle ?", but "What would it look like ?".

And we can use higher and higher energy probing to get a reasonable 'image' of it.
( of course HUP would then deny us other information )

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13 hours ago, John Cuthber said:

This is neat, but it isn’t really what the OP was referring to. 
For one thing, the object in question here is an entire strontium atom - not an elementary particle. The other thing of course is that this isn’t a visual image of the actual atom, but merely diffuse re-emitted light, after exciting the two valence electrons of the outer shell with a laser. That’s not the same thing at all. The atom itself has 38 electrons in five shells, none of which is spherical - I wasn’t able to find a good 3D diagram of the orbital configuration, but suffice to say it is pretty non-trivial.

So the picture in the link is quite an astonishing feat (kudos to the guy who took it) - but it’s not a “photo of the atom” in the sense I understand the OP to mean.

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2 hours ago, Markus Hanke said:

For one thing, the object in question here is an entire strontium atom - not an elementary particle.

I'm sorry Markus to be pernickety but the OP specified a sub atomic particle not an elementary particle.

 

On 2/19/2021 at 5:19 PM, Alex Mercer said:

How do I start thinking of a subatomic particle in my head?

I admit I introduced bigger particles before John did.
But that was to make my point that none of these, neither sub nor atomic sized particles can be 'seen' directly.

I introduced the wind as an example of something familiar that also cannot be seen directly.

The wind is a good example because it can be felt, and no one is suprised about this.
In a similar way we can 'feel' the impact of particles, so we can (and do) sense them.

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I have to agree with Markus again. I take "subatomic particles" to loosely mean "elementary particles", ie. more elementary than atoms. It's true that protons are in no way elementary, because they have multipolar distributions of charge, but still.

The question of what is elementary can be pretty slippery, especially when QCD is involved.

But let me go back to the OP's words. They are:

Quote

What does a particle actually look like - if a person wanted a realistic image of it in their head?

Quite a different matter it would be if they had asked, eg, "Can we see an elementary particle in any way?" Then I would no doubt adhere to @MigL and @John Cuthber's posture on how to address this question.

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4 hours ago, Markus Hanke said:

For one thing, the object in question here is an entire strontium atom

If you are going to try to be petty: get it right. It's a strontium ion.

 

" The photo shows a pinprick of a positively charged strontium atom "

If it was neutral, it wouldn't be held in an ion trap.

4 hours ago, Markus Hanke said:

The atom itself has 38 electrons in five shells, none of which is spherical

Yes they are.

Those funny looking dumbbell shapes you see are poor representations of the probability density. The actual functions sum to a spherically symmetrical distribution (unless there's one outside influence on them)

But the biggest problem is this

 

4 hours ago, Markus Hanke said:

The other thing of course is that this isn’t a visual image of the actual atom, but merely diffuse re-emitted light, after exciting the two valence electrons of the outer shell with a laser.

If I look at the end of my finger, what I see are photons that were emitted by virtual stated populated from the ground sates by incoming photons.

The process is the same- albeit that the energy levels are different.

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19 hours ago, John Cuthber said:

If you are going to try to be petty: get it right. It's a strontium ion.

Fair enough. The point though was to contrast it against “subatomic particles” as mentioned in the OP, which clearly this isn’t.

19 hours ago, John Cuthber said:

The actual functions sum to a spherically symmetrical distribution (unless there's one outside influence on them)

This really isn’t my area of expertise (I’m much more of a relativity guy), but I question if this is actually true.
Assuming for a minute that this is a non-relativistic situation, the solutions to Schroedinger’s equation for a 3D potential well with electrons that themselves interact electromagnetically would need to involve products of associated Laguerre polynomials and spherical harmonics, which in the general case don’t yield anything like a spherical distribution. The issue I have is that the analytic expression for this can be derived only for hydrogen, and even then only \(\Psi_{100}\) appears to be spherically symmetric - so how do we know that the distribution is spherically symmetric for something as complicated as strontium? I’m not saying you’re wrong, I’m just trying to understand how you know this.

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The simple answer is that, according to our experiments, atoms are round.


The case for non hydrogen-like atoms is complex, but the simple answer is that the orbitals we get have to be linear combinations of the corresponding orbitals for hydrogen (in order to meet the same momentum + energy criteria).
So they look like these
https://commons.wikimedia.org/wiki/File:D_orbitals.svg

But what is never explained in those diagrams is that the actual sum of the functions is (So I was told as a student) spherically symmetrical.

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The electric dipole moment of a neutral atom in zero field is zero

https://journals.aps.org/pr/abstract/10.1103/PhysRev.174.125

 

"The observation of an electric dipole moment (EDM) in an atomic system of well-defined angular momentum would be direct evidence for violations of both parity and time-reversal invariances."

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Doesn't the spherical symmetry arise like this:

 

When you choose the actual direction of the X axis you automatically set the diections of the Y and Z axes.

Then you can solve (Markus, other atoms have been solved numerically and so called 'hydrogen like' atoms can be solved analytically) you obtain the various wavefunction plots.

But your choice of the initial direction of X is arbitrary and has spherical symmetry, because whatever direction you choose you will get the same set of solutions along your particular X,Y and Z axes. It is the whole set of solutions that has 'spherical symmetry'.

So in a normal sized aggregate of atoms you will 'see' an average sphere.

Note this is not the same situation as the Principal Axes in the stress and inertia tensors where the X, Y and Z directions are set by an external influence such as a load, there are no such principal axes in the solution to the quantum wave equation.

Edited by studiot
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2 hours ago, studiot said:

So in a normal sized aggregate of atoms you will 'see' an average sphere.

It's more than that.

The atom can not know what axes you chose for your coordinate system and must therefore have an shape that is independent of your choice.
The only way it can do that is to be spherical.

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20 minutes ago, John Cuthber said:

It's more than that.

The atom can not know what axes you chose for your coordinate system and must therefore have an shape that is independent of your choice.
The only way it can do that is to be spherical.

I think we are saying basically the same thing, with a different emphasis.

Perhaps a more formal version of what you are saying might help.

https://pubs.acs.org/doi/pdf/10.1021/ed042p397

There is also an interesting discussion about this point on StackExchange.

https://physics.stackexchange.com/questions/610064/are-all-atoms-spherically-symmetric-if-so-why-are-atoms-with-half-filled-fille

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8 hours ago, John Cuthber said:

It's more than that.

The atom can not know what axes you chose for your coordinate system and must therefore have an shape that is independent of your choice.
The only way it can do that is to be spherical.

This is as interesting as it is confusing to me - maybe I should just stick to my good old simple GR, atomic physics is too complicated :) 
Let’s take the normalised wave function of the H atom for example:

\[\Psi _{nlm}( r,\theta ,\phi ) =\sqrt{\left(\frac{2}{na}\right)^{3}\frac{( n-l-1) !}{2n[ n+l) !]^{3}}} e^{-\frac{r}{na}}\left(\frac{2r}{na}\right)^{l}\left[ L^{2l+1}_{n-l-1}\frac{2r}{na}\right] Y^{m}_{l}( \theta ,\phi )\]

wherein L are the associated Laguerre polynomials, and Y are the spherical harmonics, as usual. When you plot this function for some possible choices of n,l,m (see e.g. Griffiths) then it is pretty obvious that only \(\Psi_{100}\) and \(\Psi_{200}\) are actually spherically symmetric. So when you say that the “shape of the atom” is spherically symmetric, then you can’t mean this analytic wave function.  

But as you quite rightly say, obtaining (and plotting) this wave function implicitly involves a specific choice of coordinate system; since there are no preferred coordinate choices in the real world, and since the components of the angular momentum and spin vectors don’t commute, the overall atom cannot have any specific shape until we effectively impose a coordinate system by measuring any which one of the angular momentum components as well as the total angular momentum (since each of the vector components commutes with the magnitude of the vector). So to make a long story short, the atom exists in a linear superposition of all possible “shapes” (which would add up to something that is approximately spherical) until we perform a suitable measurement on it that establishes a definite orientation in space - at which point the wave function resolves into a definite shape as in the plots above, which won’t in general be spherical.

Is this the right way to look at it? I can’t make the conceptual connection at the moment, so help needed here please.

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15 minutes ago, Markus Hanke said:

This is as interesting as it is confusing to me - maybe I should just stick to my good old simple GR, atomic physics is too complicated :) 
Let’s take the normalised wave function of the H atom for example:

\Psi _{nlm}( r,\theta ,\phi ) =\sqrt{\left(\frac{2}{na}\right)^{3}\frac{( n-l-1) !}{2n[ n+l) !]^{3}}} e^{-\frac{r}{na}}\left(\frac{2r}{na}\right)^{l}\left[ L^{2l+1}_{n-l-1}\frac{2r}{na}\right] Y^{m}_{l}( \theta ,\phi )

 

 

wherein L are the associated Laguerre polynomials, and Y are the spherical harmonics, as usual. When you plot this function for some possible choices of n,l,m (see e.g. Griffiths) then it is pretty obvious that only \Psi_{100} and \Psi_{200} are actually spherically symmetric. So when you say that the “shape of the atom” is spherically symmetric, then you can’t mean this analytic wave function.  

But as you quite rightly say, obtaining (and plotting) this wave function implicitly involves a specific choice of coordinate system; since there are no preferred coordinate choices in the real world, and since the components of the angular momentum and spin vectors don’t commute, the overall atom cannot have any specific shape until we effectively impose a coordinate system by measuring any which one of the angular momentum components as well as the total angular momentum (since each of the vector components commutes with the magnitude of the vector). So to make a long story short, the atom exists in a linear superposition of all possible “shapes” (which would add up to something that is approximately spherical) until we perform a suitable measurement on it that establishes a definite orientation in space - at which point the wave function resolves into a definite shape as in the plots above, which won’t in general be spherical.

Is this the right way to look at it? I can’t make the conceptual connection at the moment, so help needed here please.

We are back to the old question

What direction does the zero vector point in ?

Or. as I said earlier,

What are the principle axes of the stress tensor when the imposed load is zero ?

until we effectively impose a coordinate system by measuring any which one of the angular momentum components as well as the total angular momentum

We cannot do this without an external agent.

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3 hours ago, Markus Hanke said:

This is as interesting as it is confusing to me - maybe I should just stick to my good old simple GR, atomic physics is too complicated :) 
Let’s take the normalised wave function of the H atom for example:

 

Ψnlm(r,θ,ϕ)=(2na)3(nl1)!2n[n+l)!]3erna(2rna)l[L2l+1nl12rna]Yml(θ,ϕ)

 

wherein L are the associated Laguerre polynomials, and Y are the spherical harmonics, as usual. When you plot this function for some possible choices of n,l,m (see e.g. Griffiths) then it is pretty obvious that only Ψ100 and Ψ200 are actually spherically symmetric. So when you say that the “shape of the atom” is spherically symmetric, then you can’t mean this analytic wave function.  

But as you quite rightly say, obtaining (and plotting) this wave function implicitly involves a specific choice of coordinate system; since there are no preferred coordinate choices in the real world, and since the components of the angular momentum and spin vectors don’t commute, the overall atom cannot have any specific shape until we effectively impose a coordinate system by measuring any which one of the angular momentum components as well as the total angular momentum (since each of the vector components commutes with the magnitude of the vector). So to make a long story short, the atom exists in a linear superposition of all possible “shapes” (which would add up to something that is approximately spherical) until we perform a suitable measurement on it that establishes a definite orientation in space - at which point the wave function resolves into a definite shape as in the plots above, which won’t in general be spherical.

Is this the right way to look at it? I can’t make the conceptual connection at the moment, so help needed here please.

The spherical harmonics are not, themselves, spherically symmetrical.
But the sum of the trio of P orbitals (one on each axis) is symmetrical.

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Not only are the trio of orbitals symmetric, but you don't have a preferred axis in the absence of some external field. So the orbitals can be oriented in any direction in space. You can't say which p orbital the electron is in, so it's in all of them until you measure.

IOW, you don't have an intrinsic electric dipole moment, as I posted earlier.

 

Going back to the OP, these particles don't really "look" like anything since we can't form an image of them. Even the Sr ion doesn't "look" like anything based on the light it's giving off. A neutral Sr atom is a few hundred pm in radius, but the light being emitted is around half a micron. You wouldn't be able to determine the size based on that light. What it "looks" like really only makes sense for objects where diffraction is not important.  

We can use other particles that can have much shorter wavelengths, e.g. electron microscopes, in order to form an image. We can get other information by reconstructing what happens with e.g. scattering experiments. But because of quantum mechanical effects, everything will have a wave behavior, and the notion of what it "looks" like loses meaning. We talk about how it behaves — how it interacts — and what its properties are.

Elementary particles are depicted as balls in diagrams because that's a way to visualize interactions. You could look at Feynman diagrams to see various interactions between particles. But it's like a schematic of a circuit — that's not necessarily what the actual circuit looks like,

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49 minutes ago, swansont said:

We can use other particles that can have much shorter wavelengths, e.g. electron microscopes, in order to form an image. We can get other information by reconstructing what happens with e.g. scattering experiments. But because of quantum mechanical effects, everything will have a wave behavior, and the notion of what it "looks" like loses meaning. We talk about how it behaves — how it interacts — and what its properties are.

Which is exactly HOW we see things.
By reconstructing ( in our brain ) what happens when light scatters off those things.

Using the same method, one could 'visualize' how any particle looks by scattering increasingly shorter wavelength/higher energy particles off it.

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12 minutes ago, MigL said:

Which is exactly HOW we see things.
By reconstructing ( in our brain ) what happens when light scatters off those things.

Using the same method, one could 'visualize' how any particle looks by scattering increasingly shorter wavelength/higher energy particles off it.

And there's a limit to what we can "see" and that may or may not tell us what something "looks like" (which is what the OP asked)

e.g. the size of electron from scattering experiments is smaller than (IIRC) 10^-18m. We can't measure a smaller number

We can get an image from sonar or radar, but again, the size and shape we reconstruct may not reflect reality. It wouldn't tell us what color the object is (Something that doesn't preferentially absorb a particular wavelength range of photons isn't going to have a color). It doesn't tell us what it looks like, which is a visual limitation.

"What does it look like" is a quasi-classical inquiry, and like other aspects of classical physics, it loses meaning at small scales.

 

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3 minutes ago, swansont said:

We can get an image from sonar or radar, but again, the size and shape we reconstruct may not reflect reality. It wouldn't tell us what color the object is (Something that doesn't preferentially absorb a particular wavelength range of photons isn't going to have a color). It doesn't tell us what it looks like, which is a visual limitation.

And things like 'color' of an object don't really exist.
They are an interpretation of a particular property of the object.

Images we get from sonar or long wave radar are 'fuzzy' compared to shorter wavelength imaging, but they still depict a 'representation' of the object at that wavelength.

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