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Zero point energy of hydrogen atom

  1. Jan 19, 2009 #1
    Zero point energy and hydrogen atom

    In quantum mechanics, the lowest energy level of Simple Harmonic Oscilator is not zero and equal to hw/2. It is called the zero point energy.
    Is the energy level at ground state of hydrogen atom the zero point energy also?
     
    Last edited: Jan 19, 2009
  2. jcsd
  3. Jan 20, 2009 #2
    good question
     
  4. Jan 20, 2009 #3
    No. The ground state of the H atom does however have a non-zero energy. The energy eigenvalues of the H atom are yielded by this formula. At first glance, it looks like it should have a zero energy in the ground state, as n=0 would give E=0. But it's a quirk of the maths that n is an integer greater than zero, or you don't have a solution of the Schroedinger equation.
     
  5. Jan 20, 2009 #4

    Avodyne

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    If we define the zero-point energy as the difference between the smallest possible energy in the quantum theory and the smallest possible energy in the classical theory, then the zero-point energy of hydrogen is infinite. This is because the classical energy can be made as negative as we like by putting the electron in a classical orbit with an arbitrarily small radius. By the virial theorem, the potential energy is always -2 times the kinetic energy, and the potential energy goes like one over the radius, so the smaller the radius, the more negative the total energy.
     
  6. Jan 20, 2009 #5

    Avodyne

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    Actually, n=0 would give E = minus infinity.
     
  7. Jan 20, 2009 #6
    Yes, yes it would. Silly error.
    I've never seen that definition of the zero point energy before, but it makes sense. You learn something new every day... (or at least you'd hope to, being a student :tongue:)
     
  8. Jan 20, 2009 #7
    I don't think zero point energy means n must be zero. e.g. in the case of particle in a box, zero-point energy is at n=1, since n cannot be zero. So I think zero-point energy has nothing to do with n=0

    http://en.wikipedia.org/wiki/Particle_in_a_box
     
  9. Jan 21, 2009 #8

    malawi_glenn

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    "In physics, the zero-point energy is the lowest possible energy that a quantum mechanical physical system may have and is the energy of the ground state. "

    One can define it as this
     
  10. Jan 21, 2009 #9
    From Wikipedia, the free encyclopedia http://en.wikipedia.org/wiki/Zero-point_energy

    In physics, the zero-point energy is the lowest possible energy that a quantum mechanical physical system may have and is the energy of the ground state. The quantum mechanical system that encapsulates this energy is the zero-point field. The concept was first proposed by Albert Einstein and Otto Stern in 1913. The term "zero-point energy" is a calque of the German Nullpunktenergie. All quantum mechanical systems have a zero point energy. The term arises commonly in reference to the ground state of the quantum harmonic oscillator and its null oscillations.
     
  11. Jan 21, 2009 #10
    So this post is incorrect. And the ground state energy of hydrogen is supplied by zero-point energy.
     
  12. Jan 21, 2009 #11
    I think he means that the ground state energy of hydrogen is itself a kind of zero point energy.
     
  13. Jan 22, 2009 #12
    Re: Zero point energy and hydrogen atom

    Yes. It's the lowest allowed energy you can have. At zero temperature (K=0), the hydrogen atom will still have this energy remaining. Zero point energy as far as I'm aware, just means any energy that will still be contained in the system at the point of zero temperature. Classically, as has already been said, the system could reach lower and lower energy simply by allowing the electron to get as close as it wants to the proton, releasing a constant stream of Bremsstrahlung radiation as it accelerates inwards. However, the laws of quantum mechanics forbid this event (fortunately) and so there's a point at which the electron can get no lower in energy. This is known as the zero point energy.

    Why anyone thinks that this energy can be harnessed in any way is beyond me. That electron is not going to get lower in energy. It's almost like someone harnessing the gravitational potential of the earth. Once you get to the centre, you're not gonna get anymore out. Correct me if I'm wrong, but since it's a conservative field, you can set your potential energy at any arbitrary value. So we could say that this centre of the earth is at a potential energy of 1000000000000000000 Joules, but trying to get this energy out at the end would be stupid. Sorry, I'm just a little frustrated by these perpetual motion fanatics trying to get blood out of a stone.
     
  14. Jan 22, 2009 #13
    n=1 is generally (according to the usual notation) taken to be the lowest value. It all depends on how you define the terms in the maths. I could replace all the (n)'s in the above formula for (n+2) and then use n=-1 for the zero point. The maths would be the same, but the index, n, would take on different values.
     
  15. Jan 22, 2009 #14
    I gather they are thinking of zero point energy as a sort of heat energy of space itself.
     
  16. Jan 22, 2009 #15
    It's possible to use ocean wave to generate electricity. So why not zero point energy?
     
  17. Jan 22, 2009 #16

    malawi_glenn

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    if you take the hbar omega/2 from the harmonic oscilator for instance, where would that state go into? It is already in the ground state, there is not state below the ground state which it can go to, hence you can "steal" that energy.
     
  18. Jan 22, 2009 #17
    In the case of particle in a box, n is the number of wavelength that can fit in the box.
    Try set n=0 and see what happens to the energy level of <particle in a box>?
    If you set n=-1, does it mean that "negative wavelength" can fit in the box?
    http://en.wikipedia.org/wiki/Particle_in_a_box
     
    Last edited: Jan 22, 2009
  19. Jan 22, 2009 #18
    That's only because they've taken [tex]sin(n \pi x/ \lambda)[/tex] (or whatever it is) as the solution to the wave equation. If they took [tex]sin( (n+2) \pi x/ \lambda)[/tex] that would work just as well, but only for n>-1. That's what I meant by replacing all the terms by (n+2). The maths is the same, it's just translated to a different index. It changes nothing physically. Some sort of symmetry I guess. Invariance under a change of index?
     
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