Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Quantum Harmonic Oscillator

  1. Sep 27, 2008 #1
    I am working with the following harmonic oscillator formula.

    \psi_n \left( y \right) = \left( \frac {\alpha}{{\pi}} \right) ^ \frac{1}{{4}} \frac{1}{{\sqrt{2^nn!}}}H_n\left(y\right)e^{\frac{-y^2}{{2}}}

    y = \sqrt{\alpha} x

    \alpha = \frac{m\omega}{{\hbar}}
    I can not find a non-circular definition for [tex]\omega[/tex] or k.
    For a quantum harmonic oscillator the only definitions for [tex]\omega[/tex] and k that I have found are

    \omega = \sqrt\frac{k}{{mass}}

    k = mass*\omega^2

    which are circular.

    Does anybody have a different definition for [tex]\omega[/tex] or k that works in the above quantum harmonic oscillator?
  2. jcsd
  3. Sep 27, 2008 #2


    User Avatar
    Science Advisor
    Gold Member

    For linear systems [tex]\omega[/tex] and [tex]k[/tex] are direcly related, and basically contain the same information. They are never independent variables.

    What exacly are you trying to show?
  4. Sep 27, 2008 #3


    User Avatar
    Science Advisor

    w has the interpretation "temporal frequency".

    k has the interpretation "spatial frequency".

    Every material or system possesses a property called the dispersion relation, which is a relationship between w and k, which may be linear or nonlinear. The aim of quantum modelling is to figure out the dispersion relations of different systems.

    The dispersion relation is related to the energy-momentum formula because of the de Broglie hypothesis that E=hf, and p=hk for massless and massive particles. Because E2=p2c2+mo2c4, it can be said that massless and massive particles have different dispersion relations.
  5. Sep 27, 2008 #4
    Hummmm, looks like I can not get around the linear or non-linear relationship between [tex]\omega[/tex] and k.
    So then let’s approach the problem from a different perspective.

    I am trying to determine the values of the harmonic oscillator so that I can generate a wavefunction plot for the first couple of energy levels.

    However, I am stuck on how to determine the value for [tex] \alpha [/tex] which is defined in terms of [tex] \omega [/tex]
  6. Sep 27, 2008 #5


    User Avatar
    Science Advisor


    Looking at the Hamiltonian in wikipedia, it seems that m and w are input parameters, so we cannot plot anything in the variable x unless those are specified. If we plot in the normalized variable y, it seems that a convenient choice may be [tex]\alpha = \frac{m\omega}{{\hbar}}=1[/tex]?
  7. Sep 27, 2008 #6
    Yelp, I had considered setting [tex]\alpha[/tex] = 1 but felt uneasy in doing so because each of the wavefunctions becomes wider as one increments up the energy level. Meaning that [tex]\alpha[/tex] changes for each energy level thus changing the wavefunction. This is supported by ATYY’s comment about de Broglie’s function which is in terms of either frequency or inverse wavelength.

    Earlier, I did make a couple of plots using alpha as a constant and a couple of the curves had become flat at 0 or at 1. So that is why I am pursuing the actual value of [tex]\alpha[/tex].
  8. Sep 27, 2008 #7
    You don't need to find the exact values of all these parameters to plot the wavefunctions unless you want exact plots. The key is to look at the local and asymptotic behavior of the wavefunction, their maxima/minima in the well, etc.
  9. Sep 28, 2008 #8


    User Avatar
    Science Advisor
    Gold Member

    There is no definite "value" of [tex]\alpha[/tex]. The harmonic oscillator is one of the most "general" system in physics, simply because quadratic or nearly-quadratic potentials are so common.
    Hence, in the system I work on (solid-state devices) [tex]\alpha[/tex] has one meaning, for someone working on atomic systems another etc. The neat thing is of course that the solutions LOOK the same when they have the same value of [tex]\alpha[/tex], which is why I frequently can use methods and solutions directly from other fields of physics.

    Btw, the equations you wrote down that relate [tex]\omega[/tex] and [tex]k[/tex] are still valid even when there is no "real" mass involved, the reason is that one can always define a "generalized mass" which in e.g. the system I work on is inversely proportional to the capacitance of the circuit (try solving the Schroedinger equation for a LC-oscillator and you will see why).
  10. Sep 28, 2008 #9
    I am trying to describe the harmonic wavefunction of nucleons. It is possible to create several incremental energy level plots showing the relationships between the curves without knowing the exact value of [tex]\alpha[/tex] . However, I would like to be able to explain to my readers the quantum harmonic oscillator formula and properly define the terms used therein. I could present the circular definition of [tex]\alpha[/tex] and k but I am sure they would point out the problem to me as I have flagged it here. Since I am working within a specific domain (nucleons) I would expect [tex]\alpha[/tex] has been defined somewhere. I can not use the [tex]\alpha[/tex] defined in the finite and infinite potential well examples because there is no potential defined in a harmonic oscillator. I have got to believe that somewhere [tex]\alpha[/tex] has been quantitative described or determined for nucleons.

    Secondly, if you look at the quantum harmonic oscillator waveform at the top right of the Wikipedia link:
    http://en.wikipedia.org/wiki/Image:HarmOsziFunktionen.jpg [Broken]

    you should observe a parabola overlaying the wavefunctions. The definition of this parabola is [tex] \frac{1}{{2}}kx^2 [/tex] . The parabola defines the point where the wavefunction changes and attenuates to its energy level. Without k I will not be able to duplicate this parabola. Since someone created the parabola they must have had some knowledge of the value of k.

    In summary, it appears that I really need to quantitatively determine the value of either [tex]\alpha[/tex] , k.

    --- I think I have figured out the answer to my question. While trying to preview this message the Message Preview or LaTex database kept going down. So, I went back and took a look at de Broglie’s work. So, what I am going to do is present this possible solution and I really need you folks to look over the concept to make sure I didn’t make any logic or math error.

    First though, let me introduce the equation that defines the harmonic oscillator energy level defined in terms of n where n = 0, 1, 2 etc.
    [tex]E = \hbar\omega \open( n+\frac{1}{{2}}\close) [/tex]

    As you can see the energy is defined in terms of [tex]\omega[/tex].

    However, [tex]\omega = 2\pi\upsilon [/tex]

    Next I’ll introduce de Broglie’s work. He started with the energy equation of a wave:
    [tex]E = \hbar\omega [/tex]

    Then he substituted for angular frequency [tex]\omega = 2 \pi \upsilon [/tex]

    This would yield [tex]E=\hbar 2\pi\upsilon [/tex]

    Then he substituted for frequency [tex]\upsilon = \frac{c}{{\lambda}}[/tex]

    This then produced the equation I am interested in [tex]E=\hbar 2 \pi \frac{c}{{\lambda}} [/tex]

    If you compare the harmonic oscillator energy equation and the ending de Broglie equation you should see the two are equal except one is in terms of [tex]\omega[/tex] and the other [tex]\frac{c}{{\lambda}}[/tex]

    The bottom wavefunction in the harmonic oscillator graph represents the ground state. Thus everything above is considered excited states. Scientists have determined the ground state wavelength of the proton and neutron and it is called the Compton Wavelength. The following is the ground state wavelength of the two nucleons:

    The Compton Wavelength for the proton is 1.3214E-15m
    The Compton Wavelength for the proton is 1.3195E-15m

    With these values I can determine the ground state energy level and all incremental energy levels.

    Finally, I also found that k was defined as the wavenumber given by the following formula
    [tex]k \equiv\frac{2\pi}{{\lambda}}[/tex]

    Making the wavelength subsitutions I can solve for [tex]\alpha [/tex] and create the oscillator wavefunctions plots.

    The only problem I have now is that [tex]\lambda [/tex] changes with each excited level. This change would then in turn change k which would create a family of parabolas which does not appear to be consistent with the Wikipedia graph.

    If anybody has any comments on the logic or mathematics or if anybody has a solution to the multiple parabola problem I would be interested in hearing your thoughts.
    Last edited by a moderator: May 3, 2017
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook