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Electron size

  1. Nov 30, 2007 #1

    Dale

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    On another recent thread a claim was made about the size of an electron being particularly small (maybe a point). I was unclear what that means. I thought that an electron was rather "smeared" in an orbital or wavefunction.

    So does the "size" of the electron refer to its size after the wavefunction collapses? Does the concept of an electron's size even make sense before the collapse?

    How about with particles like a proton that have a definite size? What does their size refer to?

    -Thanks
     
  2. jcsd
  3. Nov 30, 2007 #2

    blechman

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    The proton has a definite "size" in the sense that it is a bound state of quarks, and we can measure this size explicitly in experiments; that is: the average distance between the quarks. It is roughly 1 fm == [itex]10^{-15}[/itex] m).

    One might ask if the electron also has a physical "size". That is: if we probe with a short enough wavelength probe (high enough energy) can we start to see sub-structure (the analog of quarks in the proton)? So far, it seems that the electron does NOT have a size - that is to say that it has no substructure at the scales that we can probe. There is a naive theoretical argument from classical E&M that you might imagine that the electron is a sphere of radius

    [tex]r_e=K\frac{e^2}{mc^2}[/tex] (K being some O(1) number I don't remember)

    called the "classical electron radius". However, we have probed at these distances and the electron still looks "pointlike" (no substructure - just a lump of probability at a given point).
     
    Last edited: Nov 30, 2007
  4. Nov 30, 2007 #3

    blechman

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    One way you might choose to define "size" of a quantum system is the following:

    Typically, the wavefunction of a bound state system ultimately falls off exponentially like [itex]e^{-r/a}[/itex] for some a that depends on the details of the system as r gets large. For example, in Hydrogen, a is the Bohr radius. In a finite square well, a is related to the size of the well. Anyway, this is a calculable number (at least in principle), and it can be used to define a "size" of a quantum bound state.
     
  5. Dec 1, 2007 #4

    ZapperZ

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    The "smearing" has more to do with the electron's position, rather than its size. You can make a crude classical estimate of its size, but in the Standard Model of the electron, its size is not defined. Thus, until that happens, QED considers it as a point particle.

    Zz.
     
  6. Dec 1, 2007 #5
    It seems to me I remember someone trying to prove that point particles are like little black holes. Then perhaps the size could be related to the horizon radius of these small black holes.
     
  7. Dec 1, 2007 #6

    cristo

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    I would urge you not to take that person's word as scientific knowledge-- he's wrong. For example, a photon is taken to be a point particle, but is massless-- try and explain that with the little black hole theory!

    I should probably point you to the Forum Rules in an attempt to guide this away from a disallowed avenue if discussion.
     
  8. Dec 1, 2007 #7

    Dale

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    Thanks for all of the replies so far!

    What does it mean experimentally to "see sub-structure"? Does it mean that the energies are high enough to knock quarks off (or other pieces), or does it mean that it somehow scatters short wavelengths differently than a single point would?
     
  9. Dec 1, 2007 #8

    Dale

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    I thought that the wavefunction was more than just our inability to know the position, but that the electron in some sense is actually physically in all of the positions in the wavefunction. For example, the double-slit experiment only makes sense to me if each electron actually goes through both slits. In that sense it is at least as big as the separation between slits (and simultaneously at least as small as the thickness of the slit). I guess, as usual, mundane concepts such as "size" are difficult to translate to the QM world.
     
  10. Dec 1, 2007 #9
    In which sense?
     
    Last edited: Dec 1, 2007
  11. Dec 1, 2007 #10

    malawi_glenn

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    Well, sort of. When Rutherford did his famous scattering of alpha particles on Gold, the distribution of the alpha's was such that you could not explained the scattering-pattern by assuming that the charges of the gold atoms was equally distributed. The model before Rutherford did this, was that electrons and protons were mixed homogenous in the atom (as raisins in a bun). But the pattern you saw was that the alphas was scattered as if they haved "hit" a very very small object of positive charge. And that was the birth of our modern picture of the atom and its nucleus.

    Later you probed the nucleus with electrons, and you saw that the nucleus is not a big charged ball, but has a certain distribution of protons (and neutrons).

    And finally, they have probed the NUCLEONS, to see charged distribution in them; the quarks :)

    What you see in your scattered beam, is a sample of the charge distribution, easy speeking. So depending on what distribution you have of charges, you get certain scattering results. The scattering pattern is connected to the Form Factor, which is the fourier transform of the charge distribution. The form factor for a delta distribution of charges (i.e. a point) becomes unity. And this is what you see when you do scattering experiments on electrons.
     
  12. Dec 1, 2007 #11

    ZapperZ

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    Er... no. If you look at the operator being used on the wave function, there's no operator for "size". There is a position operator.

    Zz.
     
  13. Dec 1, 2007 #12

    blechman

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    Malawi_glenn has it right. When you scatter a "probe particle" (alpha particle for Rutherford; electron for "deep inelastic scattering" on nucleons), the scattering distribution ("scattering cross section" is the technical term if you're not familiar) is computable and gives you direct information as to the structure of your scatterer. The same thing happens when you "see" things - then the way the photons are scattered off the object and into your eye tells you what the object "looks" like. If your probe has enough energy (short enough wavelength) it can probe more and more details about the structure. And when people started doing DIS experiments in the 1960s onward, they found that the protons (and other hadrons) had substructure. They literally "saw" the quarks!
     
  14. Dec 3, 2007 #13

    Dale

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    Thanks again for all of the answers. Sometime I will have to actually get my hands dirty and learn QM! But for now, this answers my immediate question.
     
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