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Very nice, but: what am I looking at? A probability density? And what do the control represent? Some sort of energy level? An isosurface of the modulus of the w

Very nice, but: what am I looking at? A probability density? And what do the control represent? Some sort of energy level? An isosurface of the modulus of the wave function? If so what does the number mean from an intuitive perspective?

ypcx 6y ago
Seems like the relevant blog entry is here: https://blog.c0nrad.io/posts/hydrogen-pt2/ https://blog.c0nrad.io/posts/hydrogen-pt2/ Edit: the Reddit discussion has more info: https://www.reddit.com/r/Physics/comments/gt1set/interactive_electron_orbitals_of_hydrogen/ https://www.reddit.com/r/Physics/comments/gt1set/interactive...
nsxwolf 6y ago
Is this really what hydrogen "looks" like in space, or is this just another impossible for the layperson to understand abstraction? I never even know how to ask the question I want to ask, just like it's impossible to ask if the colors in the photograph of a planet are "real" or not. I've just given up and decided that all of space is in black and white except for Earth.
- throwaway_pdp09 6y ago
  An excellent point (the first sentence I mean). I'd appreciate a physicist's opinion.
  - antepodius 6y ago
    It's a data visualisation of a property of a model of hydrogen. The dots are denser where the absolute value of the 'wavefunction' (which is just a function that takes in space and time coordinates and returns a complex value) is higher. It's not 'real' in the same way a simulation of a tennis ball flying through the air, rendered with dots isn't, but worse: that's a visualisation of a model, too, but if you imagine it being a simulation of possible sensory data it makes more sense- the world is set up so it's possible for a tennis ball to be seen by a conscious being, but that isn't true for a hydrogen atom. In a model of a tennis ball, you might have a black screen, and then draw a dot where the tennis ball is, according to the model. In this model of the hydrogen atom, you have a black screen, and then you draw a cloud with denser and less dense regions to represent where the electron 'is', according to the model. The problem is that electrons aren't point particles; in this model, an electron is a cloud- it's described not by some vector describing its position, but by the aforementioned wavefunction. It's a cloud in space, (except every point is complex-valued- they're taking the magnitude for this rendering) that changes (or doesn't) over time. There's layers here. To what degree is a simplified model 'real'? To what degree is a visualisation of a model a picture of a 'real' thing, even if that model were true and complete?
    - throwaway_pdp09 6y ago
      I wasn't clear - sorry. It obviously can't be seen, and I (just about) get it's a probability cloud. My question is, is the underlying probability cloud real, in the sense that we can think of it existing and in it's actual peculiar shape, so if we could probe it we would indeed find something shaped like that, or is it just an abstraction/model 'that just works'? It's not even an easy question to ask, come to think of it.
      - cinntaile 6y ago
        Yes, the probability cloud tells you how the hydrogen atom (H-O-H) will be shaped. The atoms will be repelled by the electrons that create this probability cloud and this will force the atoms into a geometry with minimal energy (which is the shape you are probably familiar with). Just remember that it's a probability so the shape isn't set in stone, it's more like the most common shape. A common way to learn this is by implementing the Hartree-Fock algorithm. https://en.wikipedia.org/wiki/Hartree%E2%80%93Fock_method https://en.wikipedia.org/wiki/Hartree%E2%80%93Fock_method
      - mncharity 6y ago
        As commented elsewhere, the OP renders look perhaps not quite right. But in general... Atoms are little balls. With electron density that's mostly spherically symmetric. It's very high at the nucleus, and falls off exponentially outward. Down by several orders of magnitude by the time you reach distances at which atoms hang out together. A 2D analogue might be a stereotypic volcano, if height were density. I wish I knew of a better one. Few familiar objects have this degree of fuzzy. Diffusing smells, but you can't see those. The common representations with a solidish surface at some large distance from the nucleus is... useful when doing chemistry, but badly misrepresents the physical object. Electron density manifests clearly and concretely. For example, you can poke at it with the vibrating tip of an Atomic Force Microscope. Electron states, orbitals, seem less often encountered that directly. Rather than at one step remove - seeing density, or some other phenomena, and explaining it with states. Though there's a fun STM image I'm just now failing to quickly find. An STM scans a tip across a sample, measuring and mapping the tunneling current between them. Usually with a boringly symmetric ball of a tip, so the interestingness is all sample. In this case however, the sample had a grid of boring s states, and tip conduction was through a tilted f state. So the sample repeatedly mapped the tip. The image is thus a grid of little images of the tip's f state, laid out like rows of little buoys. EDIT: Maybe this is it: https://arxiv.org/pdf/cond-mat/0305103.pdf https://arxiv.org/pdf/cond-mat/0305103.pdf fig 6 page 12 (though it doesn't entirely match what I'm remembering). Just to be clear, representing density by dots, rather than say by a color gradient on voxels, is purely a data-visualization rendering choice. Like old newspapers using halftone images. Which I mention only because there are misconceptions around the individual dots themselves representing something about the electron.
- ars 6y ago
  The concept of "look" doesn't apply to hydrogen. It doesn't have a fixed, defined, shape. To define "look" you must say "look how", i.e. with what sense are you looking. For atoms you can use gravity to look at them, or the electromagnetic force, or the strong force. And the atoms will look different each way. I suppose you could define look as "how strongly will this test particle interact with the atom at this distance, using this force". But notice "how strongly" - there is no fixed boundary, the interaction just gets weaker (or less likely) as things get farther.
  - 8bitsrule 6y ago
    Heisenberg (1958): "What we observe is not nature itself, but nature exposed to our method of questioning."
- hrjfigkt 6y ago
  If you want to go deeper, the hydrogen atom doesn't even exists in the way we think of that word, because particles are just an illusion, only quantum fields are real, just like a yellow dot you see on a computer screen is not really there, instead it's just a simultaneous excitation in the G and B fields of your screen. For the planets in our solar system the colors are real.
plus 6y ago
What's being plotted is the amplitude of a "hydrogenic" orbitals Ψ_{n,l,m}(x, y, z)^2. Hydrogenic orbitals are the eigenfunctions (wavefunctions) of the 1-electron 1-nucleus Hamiltonian (many-electron wavefunctions are visually similar to this but due to electron-electron correlation become waaaaay more mathematically complex). This is a nice system because the result is analytic (ignoring relativistic effects and assuming a point-like nucleus). Specifically, Ψ_{n,l,m}(r, θ, φ) = L_n(r) * Y_l^m(θ, φ), where L_n(r) is the n-th order Laguerre polynomial and Y_l^m(θ, φ) is the m-th spherical harmonic with angular momentum l. From a chemical perspective, n indicates which electron "shell" you are in, l indicates which type of orbital (l=0 is an s-orbital, l=1 is a p-orbital, l=2 is a d-orbital, etc.), and m indicates which of the different orbitals within that shell having that angular momentum (e.g. p_x vs p_y vs p_z).
- ur-whale 6y ago
  I think you misunderstood my question: 1. The fact that the function is easy to compute because there is an analytical solution to the ODE when the atom is simple enough tells precious little about what the picture actually represents. 2. The fact that the function you talk about has 6 parameters and this is a 3D visualization (3 degrees of freedom) is confusing. 3. The chemistry lesson about orbitals is also an interesting fact but still not properly correlated to the interactive depiction. Notoriously missing: where are m,n,l actually depicted in the story? Am I looking at one specific choice for those? What are the menu entries? I think there is something that would truly help: if one would take a volume integral over a infinitesimal cube of the 3D interactive representation, what physical units would the result be in?
  - antepodius 6y ago
    The 3-d space shown is just physical space. n,l,m are the triplet of numbers you can select in the top right. They're called quantum numbers, and they describe the state of this particular system, in particular the state of the electron. Basically: n is how much energy the electron has (the higher, the further from the nucleus). l and m further describe which orbital the electron is in; l is related to the electron's angular momentum, m is related to its angle to the x-y plane in this visualisation. Only certain combinations of these numbers are allowed by physics (angular momentum, l, has to be smaller than energy, n, for example). n,l,m together describe the state of the electron inside the hydrogen atom. So what are the dots? Basically, they're meant to represent a cloud. Where the cloud is denser, the electron has more measure; the electron is more there than in other places. Practically speaking, if you made a measurement to see where the electron was, your results would probabilistically correlate with the density of the cloud. The process whereby the electron goes from being a probabilistic cloud to a point particle interacting with your test particle back to a probabilistic cloud is called 'wave function collapse' in the Copenhagen interpretation, or, more generally, 'magic'. (Or it's just how the universal wavefunction's branches look from the inside.) A volume integral would be unitless, by definition: the value of the square-absolute-value of the wavefunction at any point (what's represented by this graph) is the probability of finding the electron at that point per cubic metre. A volume integral from negative to positive infinity in x, y, and z gives 1 (no units).

ypcx 6y ago

Seems like the relevant blog entry is here: https://blog.c0nrad.io/posts/hydrogen-pt2/ https://blog.c0nrad.io/posts/hydrogen-pt2/ Edit: the Reddit discussion has more info: https://www.reddit.com/r/Physics/comments/gt1set/interactive_electron_orbitals_of_hydrogen/ https://www.reddit.com/r/Physics/comments/gt1set/interactive...

nsxwolf 6y ago

Is this really what hydrogen "looks" like in space, or is this just another impossible for the layperson to understand abstraction? I never even know how to ask the question I want to ask, just like it's impossible to ask if the colors in the photograph of a planet are "real" or not. I've just given up and decided that all of space is in black and white except for Earth.

throwaway_pdp09 6y ago
An excellent point (the first sentence I mean). I'd appreciate a physicist's opinion.
- antepodius 6y ago
  It's a data visualisation of a property of a model of hydrogen. The dots are denser where the absolute value of the 'wavefunction' (which is just a function that takes in space and time coordinates and returns a complex value) is higher. It's not 'real' in the same way a simulation of a tennis ball flying through the air, rendered with dots isn't, but worse: that's a visualisation of a model, too, but if you imagine it being a simulation of possible sensory data it makes more sense- the world is set up so it's possible for a tennis ball to be seen by a conscious being, but that isn't true for a hydrogen atom. In a model of a tennis ball, you might have a black screen, and then draw a dot where the tennis ball is, according to the model. In this model of the hydrogen atom, you have a black screen, and then you draw a cloud with denser and less dense regions to represent where the electron 'is', according to the model. The problem is that electrons aren't point particles; in this model, an electron is a cloud- it's described not by some vector describing its position, but by the aforementioned wavefunction. It's a cloud in space, (except every point is complex-valued- they're taking the magnitude for this rendering) that changes (or doesn't) over time. There's layers here. To what degree is a simplified model 'real'? To what degree is a visualisation of a model a picture of a 'real' thing, even if that model were true and complete?
  - throwaway_pdp09 6y ago
    I wasn't clear - sorry. It obviously can't be seen, and I (just about) get it's a probability cloud. My question is, is the underlying probability cloud real, in the sense that we can think of it existing and in it's actual peculiar shape, so if we could probe it we would indeed find something shaped like that, or is it just an abstraction/model 'that just works'? It's not even an easy question to ask, come to think of it.
    - cinntaile 6y ago
      Yes, the probability cloud tells you how the hydrogen atom (H-O-H) will be shaped. The atoms will be repelled by the electrons that create this probability cloud and this will force the atoms into a geometry with minimal energy (which is the shape you are probably familiar with). Just remember that it's a probability so the shape isn't set in stone, it's more like the most common shape. A common way to learn this is by implementing the Hartree-Fock algorithm. https://en.wikipedia.org/wiki/Hartree%E2%80%93Fock_method https://en.wikipedia.org/wiki/Hartree%E2%80%93Fock_method
    - mncharity 6y ago
      As commented elsewhere, the OP renders look perhaps not quite right. But in general... Atoms are little balls. With electron density that's mostly spherically symmetric. It's very high at the nucleus, and falls off exponentially outward. Down by several orders of magnitude by the time you reach distances at which atoms hang out together. A 2D analogue might be a stereotypic volcano, if height were density. I wish I knew of a better one. Few familiar objects have this degree of fuzzy. Diffusing smells, but you can't see those. The common representations with a solidish surface at some large distance from the nucleus is... useful when doing chemistry, but badly misrepresents the physical object. Electron density manifests clearly and concretely. For example, you can poke at it with the vibrating tip of an Atomic Force Microscope. Electron states, orbitals, seem less often encountered that directly. Rather than at one step remove - seeing density, or some other phenomena, and explaining it with states. Though there's a fun STM image I'm just now failing to quickly find. An STM scans a tip across a sample, measuring and mapping the tunneling current between them. Usually with a boringly symmetric ball of a tip, so the interestingness is all sample. In this case however, the sample had a grid of boring s states, and tip conduction was through a tilted f state. So the sample repeatedly mapped the tip. The image is thus a grid of little images of the tip's f state, laid out like rows of little buoys. EDIT: Maybe this is it: https://arxiv.org/pdf/cond-mat/0305103.pdf https://arxiv.org/pdf/cond-mat/0305103.pdf fig 6 page 12 (though it doesn't entirely match what I'm remembering). Just to be clear, representing density by dots, rather than say by a color gradient on voxels, is purely a data-visualization rendering choice. Like old newspapers using halftone images. Which I mention only because there are misconceptions around the individual dots themselves representing something about the electron.
ars 6y ago
The concept of "look" doesn't apply to hydrogen. It doesn't have a fixed, defined, shape. To define "look" you must say "look how", i.e. with what sense are you looking. For atoms you can use gravity to look at them, or the electromagnetic force, or the strong force. And the atoms will look different each way. I suppose you could define look as "how strongly will this test particle interact with the atom at this distance, using this force". But notice "how strongly" - there is no fixed boundary, the interaction just gets weaker (or less likely) as things get farther.
- 8bitsrule 6y ago
  Heisenberg (1958): "What we observe is not nature itself, but nature exposed to our method of questioning."
hrjfigkt 6y ago
If you want to go deeper, the hydrogen atom doesn't even exists in the way we think of that word, because particles are just an illusion, only quantum fields are real, just like a yellow dot you see on a computer screen is not really there, instead it's just a simultaneous excitation in the G and B fields of your screen. For the planets in our solar system the colors are real.

plus 6y ago

What's being plotted is the amplitude of a "hydrogenic" orbitals Ψ_{n,l,m}(x, y, z)^2. Hydrogenic orbitals are the eigenfunctions (wavefunctions) of the 1-electron 1-nucleus Hamiltonian (many-electron wavefunctions are visually similar to this but due to electron-electron correlation become waaaaay more mathematically complex). This is a nice system because the result is analytic (ignoring relativistic effects and assuming a point-like nucleus). Specifically, Ψ_{n,l,m}(r, θ, φ) = L_n(r) * Y_l^m(θ, φ), where L_n(r) is the n-th order Laguerre polynomial and Y_l^m(θ, φ) is the m-th spherical harmonic with angular momentum l. From a chemical perspective, n indicates which electron "shell" you are in, l indicates which type of orbital (l=0 is an s-orbital, l=1 is a p-orbital, l=2 is a d-orbital, etc.), and m indicates which of the different orbitals within that shell having that angular momentum (e.g. p_x vs p_y vs p_z).

ur-whale 6y ago
I think you misunderstood my question: 1. The fact that the function is easy to compute because there is an analytical solution to the ODE when the atom is simple enough tells precious little about what the picture actually represents. 2. The fact that the function you talk about has 6 parameters and this is a 3D visualization (3 degrees of freedom) is confusing. 3. The chemistry lesson about orbitals is also an interesting fact but still not properly correlated to the interactive depiction. Notoriously missing: where are m,n,l actually depicted in the story? Am I looking at one specific choice for those? What are the menu entries? I think there is something that would truly help: if one would take a volume integral over a infinitesimal cube of the 3D interactive representation, what physical units would the result be in?
- antepodius 6y ago
  The 3-d space shown is just physical space. n,l,m are the triplet of numbers you can select in the top right. They're called quantum numbers, and they describe the state of this particular system, in particular the state of the electron. Basically: n is how much energy the electron has (the higher, the further from the nucleus). l and m further describe which orbital the electron is in; l is related to the electron's angular momentum, m is related to its angle to the x-y plane in this visualisation. Only certain combinations of these numbers are allowed by physics (angular momentum, l, has to be smaller than energy, n, for example). n,l,m together describe the state of the electron inside the hydrogen atom. So what are the dots? Basically, they're meant to represent a cloud. Where the cloud is denser, the electron has more measure; the electron is more there than in other places. Practically speaking, if you made a measurement to see where the electron was, your results would probabilistically correlate with the density of the cloud. The process whereby the electron goes from being a probabilistic cloud to a point particle interacting with your test particle back to a probabilistic cloud is called 'wave function collapse' in the Copenhagen interpretation, or, more generally, 'magic'. (Or it's just how the universal wavefunction's branches look from the inside.) A volume integral would be unitless, by definition: the value of the square-absolute-value of the wavefunction at any point (what's represented by this graph) is the probability of finding the electron at that point per cubic metre. A volume integral from negative to positive infinity in x, y, and z gives 1 (no units).