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Angular momentum and the hydrogenic atom

  1. Dec 20, 2011 #1
    This is a problem that's been bothering me for a while. For electrons "in" hydrogenic atoms, do the electrons in states for which l=0 actually orbit the nucleus, or would they only ever be observed to be moving radially (assuming they could be directly observed in that way).

    Here's my logic. I understand that the solutions of the Schrodinger equation for the hdyrogenic atom are also eigenfunctions of the total angular momentum squared operator, [itex]L^2 \psi = \hbar^2 l(l+1) \psi[/itex]. The eigenvalue is [itex]\hbar^2 l(l+1)[/itex]. As a result the expectation value for the magnitude of the angular momentum of the electron about the nucleus is [itex]\hbar \sqrt{l(l+1)}[/itex].

    So, if I understand eigenvalues and eigenfunctions, which I probably don't, that means the angular momentum for states where [itex]l=0[/itex] the angular momentum will be observed to be zero, the electron will never be "orbiting". Even though there is probability density all around the nucleus. So it could be observed to change angular position, but it would never observed to be moving angularly. This seems peculiar to me. Please help.
     
  2. jcsd
  3. Dec 20, 2011 #2
    Hello. You seem to be thinking that the psi function is a time-average? After all, you seem to claim that since the electron has non-zero probabilities for being in positions with differing angular positions, that the electron must somewhere, in its time evolution, change its angle. So if I'm correct, you're assuming the electron is somewhere hiding below the wave function and the wave function simply gives a probability distribution based on time-averaging the unknown trajectory of the electron, something like that?

    But this is not what quantum mechanics is saying (if you're staying with the orthodox interpretation, and I advise you to do that (in relation to this issue, at least), because although other interpretations might say something else, none of them say what you are saying, so I don't think it would help your confusion (and only add to it, for the moment!) by dragging in other interpretations). Nowhere in your QM book will it say that the psi function is a result of some sort of time-averaging. As a result, if different positions both have non-zero probabilities of having the electron there when measuring it, this does not mean that the electron most ever go from the one position to the other. All it means is the first part of the previous sentence: it gives the probabilities of measuring it there. All the rest, you're reading into the equations yourself.

    At this point you might be wondering "but what does the electron actually do then, if we're not measuring it" and sadly this is not settled yet (or maybe better: some people regard it as a nonsensical question, so for them it is settled, but for others it is not) and if you really want to think more about this question, I suggest you take a look at the other interpretations of quantum mechanics. That being said, I advise you to first stick to the mathematics, to first understand the mathematics, for example what the mathematical theory itself really says about the electron (as discussed earlier in this post), and only look at the interpretation of it once you grasp that.

    Fyi, the "orhodox interpretation" says that there is actually nothing but the psi function, so no "electron hiding beneath it". Consequently, the wave function is not just "the probability of it being measured somewhere" but has become the electron itself. Maybe some "orthodox interpreters" might disagree with the previous statement, because the orthodox interpretation is so popular there are a lot of different versions that differ from each other slightly. But the main point I want to get across with this is to show you that your rather naive idea about the electron actually moving from point to point and the psi function only being a time average is far from evident and actually wrong, if taken at face-value. The interpretation closest to this view (and not necessarily wrong) is DeBroglie-Bohm, but let me state clearly that it is not exactly the view expressed in the previous sentence, it simply comes closest.

    I hope this helps somewhat?
     
  4. Dec 20, 2011 #3

    Bill_K

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    In other words, electrons do not "orbit" the nucleus. Each electron state is time-independent, and there is no moving going on, even when L ≠ 0. (Early mechanistic models of the atom pictured the L = 0 states as bouncing radially off the nucleus, boing, boing boing... :eek:)
     
  5. Dec 20, 2011 #4
    That being said, when you measure the velocity, it doesn't have to be zero :)
     
  6. Dec 21, 2011 #5
    the electron in the ground state still has angular momentum it just dosnt have any orbital angular momentum
     
  7. Dec 21, 2011 #6
    Thanks everyone! This has been really helpful. (:
     
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