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How do you know that there's an electron in the hydrogen atom?

  1. May 16, 2006 #1

    reilly

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    Among the reasons for asking this nominally "no-brainer" question is that it touches directly upon the notion of virtual particles.

    So, how do we know there's that tiny little electron in a hydrogen atom -- and, since we know there is one there, is the electron real or virtual?

    Regards,
    Reilly
     
  2. jcsd
  3. May 16, 2006 #2
    Well you may ascertain the presence in two ways - experimentally & mathematically. In my opinion they complement each other. One good example is using spectral anlysis of excited H atom and then analysing the result mathematically. Quantam mathematics can also be used to show this -as in Heisenberg's model of a simple H atom.
    I do not agree with the word virtual. I prefer to swear by deBrogli's theory of wave particle duality. If electron is virtual then every enrgy in this universe is virtual. Let us not make things "unscientifically abstract".
    The theories which I mentioned above should satisfy your doubts, please go through them. :smile:
    Thank you
     
  4. May 16, 2006 #3
    Too bad, it's standard terminology. It doesn't mean what you think it means either; see http://en.wikipedia.org/wiki/Virtual_particle

    In response to reilly - the hydrogenic electron is real, not virtual. It most definitely has energy, 511 keV's worth in fact (minus the 13.6 eV binding energy). This is testable; you can release all of this energy by throwing an anti-electron (positron) at it, and letting them annihilate each other; this is the principle of PET scans in nuclear medicine. In contrast, virtual particles are only short-lived fluctuations due to uncertainty in energy at short time scales, [itex]\Delta E \Delta t \geq \hbar[/itex] (Heisenberg). If they have "energy" Delta E, it's only for a timescale of Delta T. Real electrons, like those in atoms, are stable.
     
    Last edited: May 16, 2006
  5. May 17, 2006 #4
    How? Write out the Hamiltonian for the system, and explain to me what every term corresponds to. And there's your answer.
     
  6. May 17, 2006 #5
    Well, is it on or off mass shell ?

    Problem solved

    regards
    marlon
     
  7. May 20, 2006 #6
    And how do we know the electron is a tiny little thing? Does it have a shape? Does it have a size? :confused:
     
  8. May 20, 2006 #7

    vanesch

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    And this brings us, as usual, to the measurement problem and all what's involved with it. Virtual particles become real when you don't have the need anymore for them to be in superpositions which are off shell. According to your personal religion, this may be when you switch to "classical physics", when you do an "observation", or... never. The fact that these different interpretations (religions ?) may exist one next to the other is that, from a certain point on, the different viewpoints don't make a difference anymore, because no significant quantum interference occurs.
    "virtual particles" in Feynman graphs are virtual, because their off-shell contributions contribute a lot to the end result. If you'd keep them on shell, you'd have a totally different result (usually a trivial and wrong result). You could, if you'd like, consider the electron in a hydrogen atom also as a superposition of different off-shell contributions, but it wouldn't make much of a difference by replacing the integral over these off-shell states by just the contribution of the on-shell state. The better this apparoximation works, the more the particle is "real".
    So, all particles can be considered as virtual, but some have a high degree of reality, and others less. The first can then be called "real" without much of a problem, and doing so (in other words, stopping the superpositions there) will not lead to a significant error in the calculation.

    Well, we don't, really. We only see that if we model the hydrogen atom as a thing that contains an electron, that calculations work out well, and compare well to experiment.

    Now, this answer can be considered ridiculous, and for hydrogen, it is, because of the excellent quality of the hypothesis. But if you'd consider atoms where the binding energy exceeds by far 511 KeV, then the idea is not so silly anymore. And this is, btw, what happens inside protons and neutrons: the binding energy is by far greater than the rest energy of the components (quarks), hence the fuzzy "virtual soup".

    So, "there's a little electron in the hydrogen atom and it's more or less real" because of the huge energy difference between the states of the hydrogen atom and the rest energy of an electron, so that the "on shell contribution" is dominant.
     
  9. May 31, 2006 #8

    reilly

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    As usual, my approach is simple minded. That is, break a hydrogen atom, and you will find that H-> proton+electron+ other stuff -- destructive testing if you will. And, of course, at low energy. the other stuff will be photons/radiation.Thomson -> Rutherford -> Bohr provided a generally accepted basis for the hydrogen atom = proton+electron,( Rutherford's nuclear atom),which basis is admittedly circumstantial. All the stuff with Hamiltonians came later, and provided a more detailed explanation on the structure of the H atom, including the Zeeman and Stark effects, fine strucure of levels from spin-orbit interactions, and the Lanb Effect, all of which are nicely described by the Dirac Eq. for Hydrogen, along with basic photon-current interactions. Theory, here, follows experiment, not the other way around. (See, for example, Condon and Shortley's Theory of Atomic Spectra, Dirac's Quantum Mechanics for starters) Also, at low energies, the reaction e+p->H + photon, photoproduction of hydrogen, occurs.

    What about electrons? Well, in countless experiments no evidence for electron structure, as in form factors, has been found. All this means is, that there is an upper bound for the electromagnetic radius of the electron which, so far is beyond our capability to measure -- unlike the case for nucleons. That is, experimental evidence can be explained by considering the electron to be a point particle, clearly an approximation, but a good one in practice.

    The virtual-real distinction is dicey. If, for example, we convert the Schrodinger Eq. into, say, the Lippman-Schwinger Eq. and expand in a perturbation series of powers of the charge, e, then the intermediate states are virtual; they are free particle states, not eigenstates of the full Hamiltonian -- see Collision Theory by Goldberger and Watson. The same is true for the Bethe-Salpeter Eq. But, convergence is an issue, as bound states show up as poles in the T or scattering matrix. This indicates to me, as I've said before, that virtual states, and virtual particles are artifacts of perturbation theory, and they do not show up in exact solutions -- as in the central force problem described by the Schrodinger Eq.

    Can any one indicate where the virtual particles are in the Laguerre polynomial solutions for Hydrogen?

    Virtual or real? It all depends on your description. Given that the e and P in Hydrogen are not strongly coupled, I vote for the electron and proton as real. But, I can certainly see how one might prefer to go with virtual. Does it really make any difference which you choose?
    Regards,
    Reilly Atkinson
     
  10. May 31, 2006 #9
    Misleading question. Its not that we "know" that tiny electrons are in atoms, its that experimentation is consistent with the model of such electrons and atoms.

    Pete
     
  11. May 31, 2006 #10
    Why should I care which way round history did it? If I did, I'd take about 6000 years learning everything people learnt before me in the order that they did.

    I'd rather stand on the oft-quoted shoulders, not spend my life climbing up to them.
     
  12. May 31, 2006 #11

    reilly

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    Ah, the impatience of youth.(I'm just a year shy of 70) However, why do you suppose as eminent a physicist as Steven Weinberg provides a history, albeit brief, of QFT as the first chapter of his monumental work on QFT? Or that, equally eminent Abraham Pais has written scientific biographies of Bohr, Einstein and Oppenheimer, or the distinguished professor Silvan Schweber has written QED and the Men Who made it. Then there's Sir. E.T.Whittaker's History of the Aether and Electricity, and on and on. These books have been well received are highly regarded in the trade, and are highly valuable.

    I challenge you to read Schweber's book or Pais, say, on Einstein, and conclude your time was at best marginally spent -- we are talking less than 50-60 hours.

    And, of course, virtually any physicist past the age of 30 would say your response to the history of physics is, shall we say, short sighted. After all, those who neglect history are doomed to repeat it. Further, I suggest that concentration on the last 150 years is appropriate.
    A study of the history of physics, particularly since 1900 is crucial to the development of a solid and deep understanding of physics. And, no way is 6000 hours realistic; it is a vastly overestimate of the time needed.

    The ladder needed to stand on the oft-cited shoulders is built from a strong knowledge of physics, past and present, including history that indicates those whose shoulders you should stand on.

    Regards,
    Reilly Atkinson
    (Old Fashioned)

    PS Save your post, and return to it in ten years. Chances are good your tune will have changed -- such was the case with some of my better students. The less able ones, tended to stay stubborn and a bit myopic.
     
    Last edited: May 31, 2006
  13. May 31, 2006 #12

    reilly

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    Pete -- As I said above, the evidence is circumstantial. As lawyers might put it, the preponderance of evidence says that theory correctly describes many experiments, not the other way around. And, as I'm sure you know, I tend to use colloquial language, unless precision is necessary -- which is not the case in these circumstances. And, note that the phrase, "hydrogen is composed of an electron and a proton" is commonplace in physics and chemistry.
    Regards,
    Reilly

     
  14. May 31, 2006 #13
    Allow me to quote our eminent Weinberg: "Our immersion in the present state of physics makes it hard for us to understand the difficulties of physicists... or to profit from their experience. At the same time, a knowledge of our history is a mixed blessing -- it can stand in the way of the logical reconstruction of physical theory that seems to be continually necessary."

    My point is the second part of his paragraph. He goes on to say that he wishes to emphasize the deductive trail of reasoning that gives us S.R., Q.M. etc. He then goes on to say that since field theories grew out of wave equations, his point-of-view was that starting from Wigner's later work on defining particles as representations of the inhom. Lorentz group is better.

    I'm not saying, chuck all the history in the (recycle) bin. I'm saying there's no need to worry about how it happened in history. Relevant historical works should not be excluded, but saying "Theory here followed experiment, not the other way round," is irrelevant to me. If one can explain the problem in simpler terms, and still emphasise all the subtlety involved, then one should use a clearer approach to the problem. That is a true lesson of physics. I shall demonstrate this at the bottom of the post.

    I didn't say 6000 hours, I said 6000 years. I'm talking about calculus, arithmetic etc, since maths and physics drive each other; historically this is all relevant. But I'm not interested in the history of physics. I'm interested in the physics! In any case, one's suggestion of 150 years seems arbitrary. By doing this, one is excluding the Lagrangian/Hamiltonian formalisms, which elucidate many modern approaches to physics (e.g. the canonical Q.M., Lagrangian field theory &c.)

    I do admit, that my initial post may have been too brief; but I never meant to forget what past physicists have learnt; instead to realise that their approaches to problems are old; there are other, often better ways, to get to the same result.

    Here is my "demonstration at the bottom of the post": If one explains to me all the terms in the Hamiltonian of the system, one will say (if one includes the kinetic energy of the proton, or explicitly tells me that one has transformed to a frame where it is at rest) something like, "this term is the K.E. of the electron, this is the K.E. of the proton, this is the Coulomb interaction, and this is the spin-orbit coupling." And that's how I know there is an electron in this model.

    Remember, we construct models of reality, and then compare predictions of our models (ss someone has pointed out). We don't know there is truly an electron in a hydrogen atom, but all we can say is that if we construct a model where there is an electron in a hydrogen atom, it's predictions agree remarkably with the reality.

    As one is older than me, one can always play the "we older know better" card; but it is patronising. I will always respect my elders, but note that it's the ambitious young physicists that mostly make the big discoveries; rarely the wizened old.

    Masud.

    P.S. My use of "one" instead of "you" is not meant to patronise; I am sure it could be taken that way. Instead, it is meant to make the post less personal. One of the things I learnt was that formally addressing someone is less personal, and there is almost always less scope for any offence to be taken to my remarks.
     
    Last edited: May 31, 2006
  15. May 31, 2006 #14
    So what happens to the electron then? Does that mean there is one less electron in the universe?

    Don't mean to hijack your post Reilly, but I was just wondering. :smile:
     
  16. May 31, 2006 #15
    Yes. And two new photons.
     
  17. May 31, 2006 #16
    Theoretically then, it should be possible to create new electrons in some way?
     
  18. May 31, 2006 #17

    SpaceTiger

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    Sure, just pass a high-energy photon by a nucleus -- you can generate an electron-positron pair. Usually the electron and positron annihilate soon afterwards, but they can be separated. There is no conservation of "electron number" in nature, but the standard model does include conservation of "lepton number". In the above example, the photon has zero lepton number, while the electron and positron have 1 and -1, respectively. Both the initial and final states have 0 net lepton number.
     
  19. Jun 1, 2006 #18
    Hi Reilly!

    I agree that the evidence indicates that the theory is correct and it is in that sense that I say that there is an electron "orbiting" a proton in an H atom. I was addressing the concept you pointed to when you asked "how do we know". The only way that we can come close to knowing about nature is by making observations, forwarding an hypothesis to explain the observations and then construct an experiment to determine if nature conforms to other observations predicted by the hypothesis. So when you ask "how do we know" then I point to theory and experimentation and the correlation of predictions observed by the theory.

    I'm thinking of Willis E. Lamb's article "Anti-photon" when I think of such questions. In that paper Lamb claims that photons do not exit even though almost all physicists would say that photons do indeed exist. When we do experiments we only see what the experiment was designed to see and there has only been a finite amount of experiments done reggardinng the experimentation regarding electrons. Who knows? Perhaps one day we will have to modify the concept of the electron when physics learns more about particles. Perhaps the things we called electrons and envisioned as "particles" is really a tiny string. Isn't that what string theory is about?

    I don't know much about virtual particles beyond that of the layman so I have nothing to contribute to that point.

    Pete
     
  20. Jun 2, 2006 #19

    vanesch

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    These are (as often) very wise words.

    What you essentially mean is that what we want to calculate is:
    ( A | B ), and when we do that, using a "basis expansion of unity", then:

    [tex]
    <A | B > = <A|1|B> = <A| \sum_k |k><k| B> = \sum_k <A|k> <k|B>
    [/tex]

    and we call then the |k> states, virtual states through which our initial state A goes to reach B, and that the choice of this basis |k> is totally arbitrary and just a way to help us organize our calculation. Often, one even doesn't use the FULL set of |k> but limits oneself to a subset (as in the cutoff of a series devellopment).

    If you're smart enough to calculate directly (A|B) without needing such a (truncated) basis to help you, then of course you wouldn't have a clue what that other guy is talking about, as "virtual" particles or anything.

    However, what I wanted to point out was the following. In all generality, you can never say what you'd like to calculate. For one experimenter, (A|B) is important, and he'll calculate (A|B), and then the probability of obtaining B starting from A, using the good old |(A|B)|^2 ;
    but for his peer, it might be more important to measure (A|C). Now, it might very well be, that C is "further down the line" than B, and then we could think of having |B1>, |B2> ... |Bk> as a basis which contains the B cases (in the case the first experimenter is going to calculate several possible outcomes of his experiment, and not just a yes/no answer)

    And now, what was the "end" result for the first experimenter, (namely the B states), are now the "virtual states" of the calculation of the second one:

    [tex] <A|C> = \sum_k <A|B_k> <B_k|C> [/tex]

    Now, when working out
    [tex] |<A|C>|^2 = \sum_k \sum_l \left((<A|B_k><B_l|A>)(<C|B_k><B_l|C>)\right) =
    \sum_k \sum_l \left((<B_l|A><A|B_k>)(<B_l|C><C|B_k>)\right) =
    \sum_k \sum_l A_{lk}C_{lk}[/tex]

    When the off-diagonal contributions in the matrix representation of A and C in the basis B can somehow be neglected, one sees that this sum reduces to:
    [tex] \sum_k A_{kk} C_{kk}[/tex]

    In other words, to go from A to C one goes, in a classical way from A to B_k (with probability A_{kk}), and then from B_k to C with probability C_{kk}, and one sums over all these intermediate results. This is when I say that one could consider the B states as real.

    However, if the off-diagonal terms in A_{kl} and C_{kl} are not neglegible in the sum above, then one needs to consider the B states as "virtual" because they cause quantum interference (in other words, neglecting them will give you an error on (A|C) calculated as if B_k were real states, using classical probability).

    When the situation is such, that for all thinkable C, we can go through the last sum, then B is, for all practical purposes, a real state.
     
    Last edited: Jun 2, 2006
  21. Jun 8, 2006 #20

    reilly

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    vanesch -- Wonderful post; wise and on target. Let's hope that everybody will now understand virtual states -- but I'm not going to hold my breath.

    Your discussion is along the lines I had in mind in the "Challenge" post, to which I will return soon.

    As I was wont to say in my youth, "Right on, Baby"

    Best Regards,
    Reilly
     
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