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Zero point energy

  1. Feb 10, 2005 #1
    Imagine that a system with two coupled dipole is diagonalized, so that the symmetric and anti-symmetric states are chosen to be the bases. Why does the Zero point energy equal
    [tex] \frac{1}{2}\hbar (\omega_s + \omega_a) [/tex]
     
  2. jcsd
  3. Feb 10, 2005 #2

    dextercioby

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    What's the Hamiltonian...?

    Daniel.
     
  4. Feb 10, 2005 #3
    [tex] H = H_0 - \frac {e^2 x_1 x_2}{2 \pi \epsilon_0 R^3} [/tex]

    where Ho is the sum of of two unperturbed SHO, x1 and x2 are respectively the distance between positive and negative ends of the dipoles, and R is the inter-dipole distance
     
  5. Feb 10, 2005 #4

    dextercioby

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    Compute the matrix element of the perturbation wrt fundamental states and show it is zero.Therefore,the vacuum (zero point energy) doesn't have 1-st order corrections.So the total energy is the one stated by you in the first post.

    Daniel.
     
  6. Feb 10, 2005 #5

    I can see why the vacuum (zero point energy) doesn't have 1-st order corrections, but shouldn't half h-bar (omega)_s alone has lower energy than the one in the first post?
     
  7. Feb 10, 2005 #6

    dextercioby

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    Yes,of course,but why did u reject the contribution due to antisymmetric wave functions...?

    Daniel.
     
  8. Feb 10, 2005 #7
    But isn't zero point energy defined as the lowest possible energy?
    I am a little confused. Or maybe there is some good reference regarding this specific example?
     
  9. Feb 10, 2005 #8

    dextercioby

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    I wouldn't know,it's the first time i saw it.Where did u get it...?

    Daniel.
     
  10. Feb 10, 2005 #9
    It's from one of the problem sheet for the condensed matter course. I just wonder if the zero point energy is defined to be as mentioned.
     
  11. Feb 11, 2005 #10
    Zero point energy of a single dipole oscillator is defined (derived by Planck) to be
    [tex] \frac{1}{2}\hbar (\omega) [/tex]

    Creator :biggrin:
     
  12. Feb 11, 2005 #11
    Of course, of course.
    And my question is why NOT exclude the anti-symmetric mode, which has higher energy than the symmetric mode alone?
     
  13. Feb 11, 2005 #12
    I think you had it right the first time.....
    For a coupled harmonic oscillator the zero point energy is defined to be the sum of the lowest energy of each oscillator.

    However, note:

    [tex] \frac{1}{2}\hbar (\omega_s) \neq \frac{1}{2}\hbar(\omega_o) [/tex]

    [tex] \frac{1}{2}\hbar (\omega_a) \neq \frac{1}{2}\hbar(\omega_o) [/tex]

    ...where [tex]\omega_o[/tex] is the frequency of the uncoupled oscillator. IOW, the lowest energy of each coupled oscillator is different from the lowest energy of each when uncoupled.

    More specifically, [tex]\frac{1}{2}\hbar(\omega_s + \omega_a) \ll \frac{1}{2}\hbar(\omega_o) + \frac{1}{2}\hbar(\omega_o) [/tex]

    IOW, for the coupled oscillators the total zero point energy is lower than the sum of two uncoupled oscillators. :cool:

    Creator :biggrin:
     
    Last edited: Feb 11, 2005
  14. Feb 11, 2005 #13
    Please note I've made an addition to the previous post to Secret2 for clarity.
     
  15. Feb 12, 2005 #14
    Thank you!
     
  16. Feb 12, 2005 #15
    Now I think I understand. When diagonalizing the system, we turned the two INTERACTING SHO's into two non-physical NON-INTERACTING symmetric and anti-symmetric modes, and they just act like any other SHO's. The zero point energy of the system is the sum of the zero point energy of the two non-physical SHO's, and quantum mechanics dictates that none of the individual (ie [itex] \frac{1}{2} \hbar \omega_s [/itex] or [itex] \frac{1}{2} \hbar \omega_a [/itex]) could be zero, hence the answer :)
     
    Last edited: Feb 12, 2005
  17. Feb 12, 2005 #16
    A question: What was the proposed cause of the dipole in this question? Thanks.
     
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