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Do Electron's move around an atom?

  1. Nov 4, 2005 #1
    The majority of the science teachers in my school agree that electrons do move in orbits around an atom in a physical way but I disagree I see them as wave functions/ clouds that are stationary. Who's right?
     
  2. jcsd
  3. Nov 4, 2005 #2
    Well your certainly right that they do not move in orbits around the atom. That's was Bohr's model of the atom and it is incorrect, but it is still taught because it is easy to work with and useful for some problems. Electrons do exist as probability wave functions, although I'm not sure if stationary is the correct word to use. It gets complicated, and I'm certainly not the best person to try to go into details on this, but at least in a hydrogen atom, the sum of the probabilty functions for the states with the same principal quantum number produces a spherically symmetric distribution, so I suppose you could think of the probability function as being stationary relative to the atom.
     
  4. Nov 4, 2005 #3

    Danger

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    And the energy levels are referred to as orbitals, not orbits, for the very reason that they're indistinct as to position.
     
  5. Nov 5, 2005 #4
    Electrons do not move around the atoms. In Bohr model, Bohr had to postulate (3rd postulate) that electrons do not radiate electromagnetic energy according to the classical electrodynamic laws because they would loose energy and fall to the core in (I dont know exactly now) about 10^-15 s.

    But it is interesting that in for example hydrogen atom you get the same result of electron energy levels using Schroedinger equation and 3rd Bohr postulate (using hypothesis that electron really moves in circle and that it can have angular momentum just n x h trans, n=1,2,.....)
     
  6. Nov 5, 2005 #5
    When you measure the elctron and the wave function collapses the electron randomly picks a place to go, and then a different place next time; So does that mean it is moving in between checks?
     
  7. Nov 5, 2005 #6
    No. It means that it goes back into the superposition of eigenstates it was in before you made the measurement it.
     
  8. Nov 5, 2005 #7
    when you stop measuring the atom does it instantly go back to it's wavefunction form?
     
  9. Nov 5, 2005 #8

    pervect

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    Because the electronic wavefunctions have orbital angular momentum, I consider them to be "moving".

    The question as stated doesn't have a really definite ansewr, because the observables in QM are position and momentum, not velocity.

    Another argument for moving electrons - relativity matters to quantum mechanics, for instance relativistic calculations of gold give its correct color, while non-relativistic calculations give a silver color. The fact that relativistic corrections exist is somewhat easier to explain if one thinks of the electrons as moving.
     
  10. Nov 5, 2005 #9

    Danger

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    It doesn't exactly ever leave that form; it just 'chooses' which of the many possible forms to adopt. I'm probably not stating this properly, because it's not one of my specialties. Everything exists in all possible forms (eigenstates) at once until acted upon by an external influence such as observation. The act of observing forces a decision as to which is 'real', and that 'reality' is what we observe. It's still a wave function, but a single one rather than many. Since the act of observation alters the condition of that state, it is undetermined again immediately following the observation.
    Hope I didn't make matters worse with that input.
     
  11. Nov 5, 2005 #10
    This partly answers a question I was going to ask but perhaps I have got this wrong .I read that in the p,d,and f orbitals the node seems to pass through the nucleus , which is the 1 place where the electron has a zero probability of being found. So It cannot be moving round the nucleus in a path which passes through this point to get to the other lobes of its orbital.Therfore it cannot be orbiting the nucleus.
     
  12. Nov 5, 2005 #11
    When you say "moving" do you mean the electron is moving like planets around the sun or the wave function itself is moving?
     
    Last edited: Nov 5, 2005
  13. Nov 5, 2005 #12
    The wave function is not a physical object, it can't move.
     
  14. Nov 5, 2005 #13
    If electrons do move, won't then the collapses of wave functions not be random any more?
     
  15. Nov 5, 2005 #14

    Danger

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    One point about that is that in order to know that they're moving, you have to observe them. Back to interference. It's like if you fire a BB gun into a swarm of gnats. When you recover the BB and find a squished gnat stuck to it, you know that the gnat was once somewhere along the BB's trajectory. It no longer is, though.
    Okay, that was a pretty lame analogy. I'm tired and hung over, so gimme a break. :tongue:
     
  16. Nov 5, 2005 #15
    or like sending a rocket through an asteroid field
     
  17. Nov 5, 2005 #16
    Well... hopefully the pilot could avoid the asteroids.
     
  18. Nov 5, 2005 #17

    pervect

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    Neither one. Treating electrons like planets invites problems, and the wave function itself is a mathematical abstraction.

    [Some of] the electrons have orbital angular momentum, which is a bit hard to explain if you think of all of them as "standing still".

    (It's been a while so I had to double-check this - it turns out some quantum states of the electron do have zero angular orbital momentum. I do think it's fair to think of the electrons with no linear momentum and no angular momentum as "not moving").

    Because you can't assign a definite position to an electron, you can't find it's velocity, which is the rate of change of position with respect to time. But you can still assign it a momentum and an angular momentum.

    Thus the closest quantum mechanical equivalent of "motion"" is momentum. It's a mistake to think of quantum phenomenon as being classical - Bell's theorem shows the pitfalls in assuming that quantum particles are classical ones with "hidden variables". If it weren't for the pitfalls involved with hidden variables, one could assume that the electron always had a definite position, but that it wasn't known. Unfortunately this application of hidden variables leads to contradictions as Bell's theorem illustrates. Therfore one cannot assume an electron has a definite position, and neither can one assume that it has a velocity.
     
    Last edited: Nov 6, 2005
  19. Nov 7, 2005 #18

    SpaceTiger

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    Some excellent responses so far. Here's my understanding of the situation:

    The answer here is pretty clearcut -- you're absolutely right. The only minor ambiguity is in our definition of stationary, as was pointed out later in the thread. In the rest frame of the nucleus, the ground state wave function that describes the electron cloud does not change with time unless acted upon by outside forces (such as a photon). Whether or not the electron itself is stationary...well, let's step back a bit and look at the big picture.

    One of the things that made quantum mechanics such a revolutionary theory was the idea that, as far as our experiments were concerned, particles could not always be said to have definite values of position and momentum. This led to the idea that, in general, a particle's position or momentum was not a single number, but instead a superposition of states, each with a different value of position or momentum (depending on which set of states you're looking at). Again, in general, this means that a particle is not said to have this position or this velocity, but rather some combination of positions and/or velocities. This idea is expressed by the wave function which, when its norm is squared, will give you the probability that you measure a particular position or velocity for the particle.

    So what does this mean for our electron cloud? Well, the "cloud" is basically the wave function I mentioned before, so it tells us how likely a particle is to be in a particular place. Another way of looking at this is that it gives us a superposition of answers to the question, "where is the electron?". Since most of the answers to this question are relegated to the space very near the nucleus of the atom, we won't notice this superposition unless our instruments are very sensitive. That is, in the classical limit, we'll just give the electron a single position.

    The same is true for the momentum. In general, a stationary state of an atom (that is, one where the electron's wave function doesn't change with time) will have a "cloud" of momenta and, since velocity is just the momentum over the mass, a "cloud" of velocities. Be cautious in interpreting this, however, because it's not a cloud in space (that is, a given position is not associated with a particular velocity), it's a cloud in momentum space. Anyway, this gives us a means of answering -- or generating a superposition of answers to -- the question, "Is the electron moving?" In general, the state of the electron will include both "yes" and "no" answers.

    This has the additional caveat that, from the theoretical point of view, the space of possible velocities is continuous. This means that the "no" answer actually takes up an infinitesimally small portion of the space, so for all intents and purposes, the answer would be yes. On the other hand, no instrument of measurement is perfect, so there is not an infinitesimally small portion of the space that would be consistent with stationarity within the errors. I fear, however, that we're now delving too much into technicalities.


    Quantum mechanics includes not only a means of generating the wave function for a particular situation, but also evolving it with time. Whenever we "measure" the position of an electron, we "collapse" its wave function. If we neglect the instrumental errors, this means that we force it into a particular position state and generate a single answer to the question, "Where is the electron?". Remember, however, that the single answer that we get will not be exactly predictable; that is, it could have been a range of other answers.

    If, after generating this answer, we stop measuring (and interfering with) the electron's position, the wave function will be able to evolve -- spread out again to create a superposition answers. There may or may not be a final stationary state to this evolution, but the point is that it will not spread out instantaneously. Roughly speaking, the sooner we make our second measurement, the less likely we are to get an answer very discrepant from our original one.

    I hope this is helpful.
     
  20. Nov 7, 2005 #19
    Wait a minute. I was under the impression that if you measured an operator, after the wavefunction collapses to the eigenvalue measured, it then evolves according to the Schrodinger equation. So, if you were measuring the position, you'd obtain a certain value, and the wavefunction would collapse to a delta function centered at that point. Then, it would evolve as if the delta function was its intial condition. Is this not correct?
     
  21. Nov 7, 2005 #20

    SpaceTiger

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    Well, you don't really measure the operator. It's more like the act of measurement can be represented by the application of the operator to the eigenstate. The eigenvalue would be the value of the observable that you measure in a particular eigenstate.


    It wouldn't be an exact delta function because no measurement is perfect, but otherwise that seems correct. Did some part of my post seem to contradict this?
     
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