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Why doesn't the electron stick on the proton in the L=0 states?

  1. May 4, 2006 #1
    In The L=0 states in Hydrogen atom, the electron has a maximum probablity to exist at r=0 ( at the location of the proton ) Is this logical? What prevents the electron then from falling on the proton? There is no angular momentum to prevent that !!
     
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  3. May 4, 2006 #2

    ZapperZ

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    Please read our FAQ in the General Physics forum.

    Zz.
     
  4. May 4, 2006 #3

    nrqed

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    The probability density is [itex] \vert \psi(\vec r,t) \vert^2 dV [/itex], not [itex] \vert \psi(\vec r,t) \vert^2 [/itex]. Because of the r^2 in the volume element dV you actually find that the probaility density is suppressed near the origin! The probability density actually peaks at soem distance from the orgin (to find the actual distance, just find the value of r for which [itex] \vert \psi(\vec r,t) \vert^2 r^2 [/itex] is maximum).

    Patrick
     
  5. May 4, 2006 #4
    Patrick is right: the probability density peaks at the Bohr radius.

    What prevents the fuzzy position of the electron relative to the proton to get less fuzzy is the fuzziness of the electron's momentum relative to the proton. If a fuzzy position propagates (in time if not in space) with a sharp momentum, the measure of its fuzziness (e.g., the standard deviation of the corresponding probability distribution) remains unchanged. If it propagates with a fuzzy momentum, it becomes fuzzier. As a result you have a stable equilibrium between two tendencies of the internal relative position of atomic hydrogen:
    • the tendency to become less fuzzy due to the electric attraction between proton and electron, and
    • the tendency to become more fuzzy due to the fuzziness of the momentum mandated by the uncertainty principle (which is a mistranslation of the German original Unschärfe Prinzip, the correct translation of which is "fuzziness principle").
     
  6. May 5, 2006 #5
    nrqed, calculate the probabiliyt density with respect to another frame (whose origin not at the proton) and you will find that [itex] \vert \psi(\vec r,t) \vert^2 [/itex] has a maximum at the proton while the volume element is not zero there.


    Koantum, I know the force that attracts the electron to the proton : it's
    Coulomb's force. But what is this force that tries to impose the
    Fuzziness principle ? which one of the four force is it?
     
  7. May 5, 2006 #6

    ZapperZ

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    Have you done this?

    Zz.
     
    Last edited: May 5, 2006
  8. May 5, 2006 #7

    nrqed

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    I'd like to see your calculation for this.

    Notice that the full Schrodinger equation involves both the proton and the electron kinetic energies. What one does is the standard treatment is to decouple the equation in the center of mass frame from the overall constant velocity translation of the center of mass frame (this is pretty much the same as in classical mechanics). The meaning of the Schrodinger equation for hydrogen that is usually given in textbooks is only the part hat corresponds to the equation in the center of mass frame (and the mass that is there should really be the reduced mass, not the electron mass). So, by definition, if you change frame, this equation will not change, you will just be adding momentum to the center of mass.
    See any textbook, for example Griffiths (Intro to quantum mechanics, section 5.1 and especially Problem 5.1 which shows all the steps)
     
    Last edited by a moderator: May 5, 2006
  9. May 5, 2006 #8
    None of the four forces. You know the first force from classical physics. You don’t know the second because it's purely quantum. Besides, it's not a force but a tendency that every relative position has: to grow fuzzier unless counterbalanced by an attractive force that gives rise to the opposite tendency. Nothing needs to act on a fuzzy relative position to make it grow fuzzier. This tendency, encapsulated by the uncertainty principle (to stick to the conventional name) is part and parcel of the quantum behavior of matter.
     
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