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

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23 hours ago, studiot said:

We cannot do this without an external agent.

Indeed, and that’s the crux - there simply is no meaningful notion of the “shape” of the atom until such time when it is interacted with in some way.

19 hours ago, John Cuthber said:

But the sum of the trio of P orbitals (one on each axis) is symmetrical.

Yes, this makes sense now, and it is essentially what I was thinking about in my last post.
Also, it’s important to remember that the wave function is a probability density distribution, so it needs to be volume-integrated first in order to become a probability distribution; and for an isolated atom in free space, one is free to choose the orientation of the volume form in whatever way one wants.

18 hours ago, swansont said:

You can't say which p orbital the electron is in, so it's in all of them until you measure.

Indeed.

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11 minutes ago, Markus Hanke said:

Indeed, and that’s the crux - there simply is no meaningful notion of the “shape” of the atom until such time when it is interacted with in some way.

Yes, this makes sense now, and it is essentially what I was thinking about in my last post.
Also, it’s important to remember that the wave function is a probability density distribution, so it needs to be volume-integrated first in order to become a probability distribution; and for an isolated atom in free space, one is free to choose the orientation of the volume form in whatever way one wants.

Indeed.

It's also worth remembering that the Schroedinger equation is about the motion of the electron in the field of the nucleus.

And the field of the nucleus is spherically symmetrical.

This can be derived from the solution to the equation being varaibles separable ie the product of independent functions of R theta and phi.

You can have the math if you like.

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5 hours ago, studiot said:

It's also worth remembering that the Schroedinger equation is about the motion of the electron in the field of the nucleus.

Not motion, as such. There’s no trajectory information there. Location and momentum (probabilities) and energy and angular momentum (eigenvalues) can be found.

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11 minutes ago, swansont said:

Not motion, as such. There’s no trajectory information there. Location and momentum (probabilities) and energy and angular momentum (eigenvalues) can be found.

$\frac{{{d^2}\psi }}{{d{x^2}}} + \frac{{8{\pi ^2}m}}{{{h^2}}}\left( {E - U} \right)\psi = 0$

??

And yes, I know this is only the one dimensional version, but it's easier to write.

Edited by studiot
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9 minutes ago, studiot said:

d2ψdx2+8π2mh2(EU)ψ=0

??

And yes, I know this is only the one dimensional version, but it's easier to write.

Yes. Energy terms. Position. Not trajectories - not motion.

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3 hours ago, swansont said:

Yes. Energy terms. Position. Not trajectories - not motion.

So you wouldn't describe the equation for the motion of a falling weight

Kinetic energy gained = loss of potential energy

as an equation of motion?

Fair enough.

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4 minutes ago, studiot said:

So you wouldn't describe the equation for the motion of a falling weight

Kinetic energy gained = loss of potential energy

as an equation of motion?

Fair enough.

Nice bait and switch there.
The thing about "falling" is that it implies that the direction is "down" so there is information about a trajectory.

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1 minute ago, studiot said:

So you wouldn't describe the equation for the motion of a falling weight

Kinetic energy gained = loss of potential energy

as an equation of motion?

Fair enough.

In classical physics, one can often ascribe a trajectory to the solution using energy. Not so much with QM.

The kinematics equation s = v0t + 1/2 at^2 explicitly has a velocity in it, but there is no corresponding QM equation.

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15 minutes ago, John Cuthber said:

Nice bait and switch there.

Nasty little snide remark there.

I did not say that equations of motion do not contain information about trajectory.

Swansont and I were discussing affably the possibility of such an equation containing information about energy.

Not every conceivable equation of motion contains trajectory information or for that matter information about every conceivable variable.

So the equation

distance = speed times time contains no trajectory information and has other 'missing' variables as does the equation

acceleration = time rate of change of speed.

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Wow it got really technical, now I don't understand!! Oh well...

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16 minutes ago, Alex Mercer said:

Wow it got really technical, now I don't understand!! Oh well...

I havn't followed this thread, but I reckon the first three answers are pretty good.

Physicist: Actual pictures of atoms aren't actually pictures at all. There are a few good rules of thumb in physics. Among the best is: light acts like you'd expect on scales well above its wavelength and acts weird on scales below.

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23 hours ago, studiot said:

It's also worth remembering that the Schroedinger equation is about the motion of the electron in the field of the nucleus.

And the field of the nucleus is spherically symmetrical.

This can be derived from the solution to the equation being varaibles separable ie the product of independent functions of R theta and phi.

You can have the math if you like.

Thanks! I do have a copy of Griffiths “Introduction to Quantum Mechanics” here, he goes through the maths in quite some detail.
I’d look at the SE as an eigenvalue equation for the system’s Hamiltonian though, not so much as an equation of motion.

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17 hours ago, Alex Mercer said:

Wow it got really technical, now I don't understand!! Oh well...

Lol, I feel ya bro.

I agree with beecee, first three answers are good for an amateur.

If you do try to understand the technical stuff, Swansont is the person to stick with. No offence to the others, but Swansont is top dog around here.

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What does a particle actually look like?

Imagine a stone that's much smaller than a mountain?

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18 hours ago, dimreepr said:

Imagine a stone that's much smaller than a mountain?

The problem is that both the stone and the mountain are classical objects and as such share the same fundamental properties. The same is not true, however, for the chair I am sitting on and the elementary particles of which it is ultimately composed - you can’t describe the properties and interactions of these particles with Newtonian mechanics, and conversely the chair as a whole will not exhibit any quantum effects. So these are distinct categories of objects, even though there is a definite relationship between them.

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On 2/25/2021 at 8:11 AM, Markus Hanke said:

I’d look at the SE as an eigenvalue equation for the system’s Hamiltonian though, not so much as an equation of motion.

Hamilton's equations of Motion

Lagrange's equations of Motion

They are all equations of motion, realising that even a stationary state is a state of motion.

I will leave the last word to Hamilton himself.

Quote

Hamilton

The theoretical of the laws of motion of bodies is of such interest and importance, that it has engaged the of all the most eminent mathematicians, since the invention of Dynamics as a mathematical science by Galileo.

Amongst the successors, Lagrange has done more than perhaps than any other analystto give extent and harmony to such researches, by showing that the most varied consequences respecting the motions of systems of bodies, may be derived from one radical formula; the beauty of the method so suiting the dignity of the results, as to make his great work a kind of scientific poem.

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19 hours ago, studiot said:

Lagrange's equations of Motion

Well ok, point taken
I guess it’s ultimately just a convention of terminology anyway.

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On 2/23/2021 at 3:00 PM, swansont said:

What it "looks" like really only makes sense for objects where diffraction is not important.

Exactly. And even for macroscopic objects, what they look like strongly depends on the kind of radiation we use to look at them:

When you look at this person covered by a plastic bag, couldn't we say that "normal" reflected light is deceiving you, by offering a picture of the object that is actually less faithful than emitted infrared light?

In order to get an infrared picture of the object, we also need to map these colours somewhere in our visual cortex. So there is a neurological aspect about the whole question too.

Most of the light we see is reflected light, and the way our brains process this information gives rise to the usual "palette" of visual concepts, such as shiny, matte, red or blue, fuzzy, etc.

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