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Absolute Stillness?

  1. Dec 11, 2009 #1
    Hi everyone,

    I am new to this forum - happy to meet you! :-)

    Background to my questions: According to quantum physics, there is no "absolute stillness" (but always some "quantum noise"). This seems to be one of the implications of Heisenberg's uncertainty principle (because if there was stillness, you could know both the position and momentum of a particle).

    My questions:

    1.) Are there also other theories/derivations which conclude that there cannot be absolute stillness (besides the Heisenberg one)?

    2.) There are areas in physics where stillness still seems to be a part (e.g. "rest mass of an atom"). Is this one of the conflict points of quantum theory and relativity theory?

    A short response would be fantastic! :-)

    Thank you,

    Nick
     
  2. jcsd
  3. Dec 11, 2009 #2
    The uncertainty principle is caused by the idea of the probability density in the Schrodinger equation.
    First, the Schrodinger equation is a non-relativistic theory, so it doen't contain the relativistic mass (like "rest mass"). On the other hand, the Dirac equation which is a relativistic quantum theory doesn't contain the probability density. So it's difficult and complicated to explain these phenomina.

    For example, the relativistic effects of the electron's movements are actually observed.
    See this Wikipedia.

    --------------------------
    A nucleus with a large charge will cause an electron to have a high velocity. A higher electron velocity means an increased electron relativistic mass, as a result the electrons will be near the nucleus more of the time and thereby contract the radius for small principal quantum numbers.....
    ---------------------------

    This fact means the electron's actual movement. But if the electron is actually moving obeying the probability density of the Schrodinger equation, some problems will occur.

    The probability density of the Scrodinger equation is complicated, so it is difficult to explain why the moving electron doesn't radiate energy. And to be precise, the probability density near the point at infinity (for example in the hydrogen) is not zero.

    (On the other hand, the Bohr model, which used the same idea of de Broglie's wavelength and the Coulomb force in calculating the energy level as the Schrodinger equation, could contain the relativistic effect (like the Sommerfeld fine structure) and explain the energy radiation problem by getting the orbital length a integer times the de Broglie's wavelength.
    Actually for the calculation of the contracted radius (of the upper wikipedia), "the Bohr radius" is now used.)
     
  4. Dec 12, 2009 #3

    alxm

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    1) 'Stillness' is vague. Or rather the concept of 'motion' as we know it from classical physics does not apply at the quantum level. Things don't have trajectories, what you have is a probability distribution telling you where the particle is more or less likely to be detected. Measuring the location of a particle can give a different result each time - in that sense, the particle is 'moving'.

    On the other hand, the probability distribution itself may not change at all with time.

    2) No, that bit is special relativity, which has been integrated into quantum mechanics (the Schrödinger equation becomes the Klein-Gordon equation). The problem between QM and relativity is with general relativity, which concerns gravity.
     
  5. Dec 12, 2009 #4

    alxm

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    Yes it does. [tex]\psi^\dagger\psi[/tex] is the probability density with the Dirac equation.

    Problems? It just means the S.E. is an approximation, valid for slow electrons. In the case of atoms, this is a decent approximation for elements up to the first row of transition metals or thereabouts.

    Not this again. It's very simple to explain this. The ground-state energy of the electron is its lowest possible energy state. This is mathematically proven by the variational theorem. Insofar quantum mechanics is valid, which most of us believe, then the reason is quite simple.

    Talk about putting lipstick on a pig. Nobody cares about the Bohr model, it's a broken theory.

    Also, relativistic mass of electrons is not the only relativistic effect significant in atoms.

    Which is just a matter of being the choice of length unit for atomic units. Nothing to do with the Bohr model.
     
  6. Dec 12, 2009 #5
    Then can you show the concrete image of [tex]\psi^\dagger\psi[/tex] of Dirac equation like the hydogen probability distribution of Schrodinger equation?
    As far as I know, the Dirac equation do not mean the probability amplitude.

    In page 110 (the Story of Spin)
    -------------------------------
    The Dirac equation is also the relativistic field equation for the electron and it cannot be considered to be an equation of probability amplitude in x,y,z space. They insisted that a concept like "the probability of a particle to be at x in space" is meaningless for relativistic particles- be they electrons, photons .....
    ------------------------------------

    The Bohr model used the same idea of de Broglie's wavelength and the Coulomb force in calculating the energy level as the Schrodinger equation. So, for example, the energy levels of the hydrogen is the same value as that of the Schrodinger equation.
    (See this Bohr model thread, which is proper for further discussion about this.)

    The differense is the electron's property. In Schrodinger equation + spinor,
    1 There is a electron spin. But how can you discribe the "two-valued" spinor rotation and the spin speed?
    2 The orbital angular momentum of S-state is zero.
    3 The probability density near the point at infinity of the hydorogen S-state is not zero. But this doesn't mean the free particle, because also near the point at infinity, the electron has the ground state energy.

    If you say these are the "mathematical things", what is real ? "What is real" is another thing?
     
  7. Dec 13, 2009 #6
  8. Dec 14, 2009 #7

    alxm

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    I'd say the whole point with how the Dirac equation is derived is to recover the probability amplitude. It's true that the Dirac equation is not an exact description either in the field-theoretical sense, since it doesn't allow for creation and annhilation of particles. View it as an approximation that neglects this, and you have your probability density. This is done all the time in practice, and it works.

    The de Broglie wavelength inspired the development of the Schrödinger equation, but is not part of its derivation. The Bohr model relies fundamentally on a lot more assumptions, which are a lot more ad-hoc than quantum mechanics.

    Well no, it's not. It's the same as the S.E. assuming the Born-Oppenheimer approximation and neglecting spin-orbit coupling, nuclear coupling, and relativistic effects. Since you talked about relativistic corrections, it's worth mentioning that the clamped-nucleus alone gives you an error several orders of magnitude larger than relativistic effects. Spin-orbit coupling is typically on the same order as relativistic corrections.

    More importantly, the Bohr model doesn't explain why spin-orbit coupling would exist, nor does it even begin to explain many-electron systems. And it is simply at odds with experiment in too many ways to enumerate. Such as the simple fact that the electronic densities corresponding to S.E. solutions are the ones seen experimentally.

    Spinor rotation is explained in the textbooks. If it doesn't make sense to you, I'd consider the possibility you may not have understood it, rather than leap to the conclusion quantum mechanics is wrong. Trust me, smarter people than you or I have spent a lot more time studying this.

    Spin is not rotation. Also, why would you assume the validity of classical mechanics, anyway? Classical mechanics could never explain an atom. Quantum mechanics has 80 years of successfully explaining and predicting atomic and subatomic phenomena.

    Yes and that's an experimentally observable property. That's another one of the many failures of the Bohr model, not of quantum mechanics, which gets this correct.

    If you measured where the particle is you don't know it's momentum, hence its energy.
     
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