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Does a phonon have a location?

  1. Mar 15, 2010 #1

    ghc

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    Question 1: If we consider that each atom in a cristal solid is a harmonic oscillator with the energy eigenvalues [tex](n+1/2) \hbar \omega[/tex] and if one oscillator has for example an energy [tex](1+1/2) \hbar \omega[/tex], does this mean that there is a phonon located on this atom?

    Question 2: If this is true, can we define a wave function, [tex]\psi (x)[/tex] for example, as the sum (or the integral) of plane waves that describes the amplutide of finding the phonon somewhere on the [tex]x[/tex] axis?

    Question 3: Does a single phonon in a solid cristal behave in the same manner as a free electron in a box for exemple?

    Thank you.
     
  2. jcsd
  3. Mar 15, 2010 #2
    A phonon is localized to the lattice in which it causes vibrations. When each atom is considered a harmonic oscillator this means they vibrate independently. In essence, the phonon is localized on the atom, however in real crystals atoms don't vibrate independently, and hence a phonon will exist throughout the crystal.
     
  4. Mar 16, 2010 #3

    ghc

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    Thank you.

    If oscillators are not coupled, each phonon will keep the same location which means that every oscillator will keep the same energy?

    And if oscillators are coupled the phonons will propagate through the crystal.
    In this case can we write a wave function which its modulus squared gives the probability density of finding the phonon in a position (x,y,z) at instant t?
     
  5. Mar 16, 2010 #4
    To your first question. Not every oscillator will have the same energy. For a given temperature you will have a distribution of harmonic oscillator energies.

    To your second question. A phonon is a quasi-particle just like a hole, but it isn't a "real" particle like an electron. So it would be meaningless to try to describe it with a wave function.

    modey3
     
  6. Mar 16, 2010 #5

    Cthugha

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    The idea of phonons was more or less to describe the collective oscillations of all atoms in the lattice. The phonon picture is therefore not extremely sensible when considering independent oscillators. So in the basic case every atom contributes to a photon and it is accordingly usually not localized. However it can be localized in special cases (nanostructures, phonon lasers, etc.).

    So hole wave functions and exciton wave functions are nonsense? And this paper is nonsense, too:
    "Visualizing the phonon wave function", S.C. Johnson and T.D. Gutierrez, Am. J. Phys. v.70, p. 227 (2002)?

    I do not think so.
     
  7. Mar 17, 2010 #6
    The original question was if a wave function for a phonon could be deduced, which could describe it in a way analogous to the electronic wave function. Since a phonon isn't a particle in the same manner as an electron its location cannot be described using a wave function. The paper by S.C. Johnson in no way describes the phonon as a particle. In fact, the wave function described the coordinates of the atomic degree-of-freedom (DOF), where the probability-density described the probable location of the atoms. This wave function is the same wave-function you would use in the many-body Schrodinger Equation in order to describe the atomic DOF when the Born-Oppenheimer Approximation is not implemented. The existence of a phonon is manifested by the vibrations of the atoms so how is it a particle like an atom or electron? After all, you cannot shoot a phonon through a vacuum!

    modey3
     
  8. Mar 17, 2010 #7

    Cthugha

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    In QM the notion of a particle means nothing else, but that it is always detected transfering discrete chunks of energy at some well defined location. Electrons fulfill that definition, delocalized electrons fulfill that definition, photons fulfill that definition and also phonons fulfill that definition. After all this concept is exactly the same for phonons and electrons: Knowing it, you can deduce the probability to have the interaction occur at some location. A phonon wavefunction is in some respects similar to that of e.g. a delocalized electron in a periodic potential.

    I would agree that this shows what a particle is, if we were in the classical mechanics section, where the meaning of a particle is a different one. However, we are not and a particle in QM is not necessarily localized.
     
    Last edited: Mar 18, 2010
  9. Mar 19, 2010 #8

    ghc

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    If a phonon carry an energy, does this energy have a location? i.e. is there a probability density to find this energy on an atom?

    The energy of all atoms is everywhere on the crystal because they interact. Still oscillators have discrete energy levels? And do these energy levels have a contribution to the energy that a phonon carry?

    I'm very confused.
     
  10. Mar 19, 2010 #9

    Cthugha

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    No, usually this energy is not localized.

    The crucial point lies in seeing the difference between one and many oscillators. "More is different" as Anderson said.

    Take the easiest example: An atom is some kind of oscillator and has some oscillatory resonances. However, a molecule consisting of two of such atoms will show different oscillatory resonances. In particular, there will be two in the direction of the bond. If the molecule looks like that:
    O---O

    you will get this oscillation:

    _O---O
    ___O---O
    ____O---O
    ___O---O
    __O---O
    _O---O
    O---O
    _O---O
    __O---O

    and this one:
    ___O---O
    ____O-O
    ___O---O
    __O-----O
    ___O---O

    The normal oscillation modes of a coupled ensemble of oscillators are significantly different from the normal modes of the single oscillators as long as the coupling strength is strong enough.
     
    Last edited: Mar 19, 2010
  11. Mar 19, 2010 #10
    A phonon has exactly the same "shape" as the corresponding classical vibrational mode of the system. Unfortunately, field theory demotes position from an operator to a parameter, so it doesn't make sense to talk about a [itex]\psi (x)[/itex]. However, you can write down a wavefunction that depends on the the mode's quadratures (i.e., the real and imaginary part of its amplitude). The real quadrature is analogous to position and the imaginary one is analogous to momentum, so you would write down the wavefunction as [itex]\psi (x_i)[/itex], where [itex]x_i[/itex] is defined as the real part of the amplitude of the ith classical mode.
     
  12. Mar 20, 2010 #11
    I just wanted to thank ghc for asking the question, and Cthugha et al here for their explanations. I had a poor understanding of phonons, and this is helping to fill out their behaviour in my mind. Just reading threads here is fantastic.
     
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