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What’s Dispersion got to do with Schroedinger?

  1. Mar 31, 2009 #1

    mysearch

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    Gold Member

    Hi,
    In my ignorance of quantum mechanics, which I am trying to address, I had initially assumed that any resolution of the wave-particle duality of photon and electrons led to some sort of common probability equation. Therefore, I am trying to better understand how different dispersion relationships seem to lead to lead to different wave equations and how to interpret the implications. Up to now I had only understood dispersion as a wave attribute within optics, e.g. spectrum spread through a prism due to different propagation velocities, within a given media, associated with each frequency.

    [1] [tex] A = A_0sin(kx-wt) [/tex]

    Citing equation [1] as a classical starting point for a wave function, which leads to the following equation based on the 2nd derivative and the relationship [tex][c=w/k = f \lambda ] [/tex], where [c] is the speed of light.

    [2] [tex] \frac { \delta^2 \Psi}{ \delta t^2} = c^2 \frac { \delta^2 \Psi}{ \delta x^2} [/tex]

    So my first question is whether equation [2] is still a valid baseline for light, when being described as a wave and a photon particle within quantum mechanics?

    The reason for asking this question is based on the apparent fact that a photon [c=w/k] and a particle have different dispersion relationships. In a pseudo-derivation of Schroedinger’s wave equation, which I won’t detail here, you are led to the dispersive relationship for a free particle of mass [m] that originates with deBroglie’s wavelength equation:

    [3] [tex] w = \frac {\hbar k^2}{2m}[/tex]

    This is also described as a dispersion relationship, but one which then seems to prevent the previous method, as used for light, of equating the 2nd derivatives with respect to [x] and [t]. However, the equivalent solution to equation [2] has the following form and was described as Schroedinger’s wave equation for a free particle, ignoring the issue of potential energy [V] for the moment.

    [4] [tex] -\frac { \hbar ^2}{2m} \frac { \delta^2 \Psi}{ \delta x^2} = i \hbar \frac { \delta \Psi}{ \delta t} = E \Psi [/tex]

    As such, I am assuming that equation [4] only applies to particles with rest mass [m] and therefore cannot be applied to a photon. So:

    Is there another Schroedinger’s wave equation for photons?
    How do I interpret the meaning of equation [3]?


    Here is my current assumption, not a statement of fact:
    In the case of a photon, the dispersion relationship [c=w/k] corresponds to the propagation velocity equalling the phase velocity [vp], while by re-arranging equation [3], the propagation velocity seems to correspond to the group velocity [vg]:

    [5] [tex] v_g = \frac {dw}{dk} = \frac {\hbar k}{2m} [/tex]

    On the basis of deBroglie’s wavelength [tex] \lambda = h/mv [/tex] being substituted back into [5]

    [6] [tex] v_g = \frac {\hbar k}{2m} = \frac {h}{2m \lambda} = \frac {mv}{2m} [/tex]

    I am slightly confused by the 2 in the denominator of [6], which comes from [3] and suggests that [vg =v/2] after the mass [m] is cancelled:

    Does the 2 have anything to do with kinetic energy E=mv/2?
    Have I just made a mistake?
    Is mass [m] subject to relativistic effects, e.g
    [tex]E^2=p^2c^2+m_0^2c^4[/tex] ?


    Would appreciate correction of any wrong assumptions on my part. Thanks
     
  2. jcsd
  3. May 8, 2009 #2
    That 2 comes from a mistake doing the derivative in [5] :P
    What you should get is Vg = V, the velocity of the particle. Mind you, I'm also trying to understand this whole thing as well...
     
  4. May 8, 2009 #3

    alxm

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    Science Advisor

    The Schrödinger equation is non-relativistic and photons are as relativistic as a particle can get. So it simply can't be used at all to describe them.

    However, if you forget about photons and work with a classical electromagnetic field, you can calculate optical properties of matter by studying a system with a varying electromagnetic field (=light) as a time-dependent perturbation. So we're talking about the S.E. + Maxwell's equations, and it's a pretty difficult subject. See a textbook on quantum optics.

    A proper (and modern) derivation of the Schrödinger equation (e.g. see http://en.wikipedia.org/wiki/Mathem...um_mechanics#Postulates_of_quantum_mechanics") doesn't make use of de Broglie's semi-empirical result, etc. The latter becomes a result of the former, not vice-versa.
     
    Last edited by a moderator: Apr 24, 2017
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