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

Physical meaning for wavefunction

  1. Nov 29, 2007 #1
    I faced a weird question:what is the requirement of having a wave equation for the QM systems?

    I think unless we have it,we cannot predict the probabilistic time and space evolution of the wave function subjected to potential constraints.(dynamicity of the wave function will be lost).
    However,this does not help us to see the effect of performing measurements on the wave function...

    Any point missing???

    Another thing:

    The essential features of this equations should be:

    (i) Should be consistent with statistical interpretation of wave function
    (ii) [tex]\psi[/tex] should be normalized all the way
    (iii)The equation should be consistent with correspodence principle

    Any suggestion???
  2. jcsd
  3. Nov 29, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    As about half of the posts on the quantum physics forum here (I didn't do statistics, just a guess), you're talking about the interpretational issues of quantum theory (what does the wavefunction mean ? What does it physically represent ? ...)
    There are several views on the issue, none of which can be said to be the undisputable consensus of the entire physics community. Each view has its advantages and disadvantages. We're close to 80 years of dispute...
  4. Nov 29, 2007 #3


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes. Once one accepts that quantum mechanics involves a wave, the obvious question is: how does this wave evolves with time. In other words, one needs a wave equation.
    That's right. The time evolution (which is unitary) is one thing. The measurement process is a totally separate question and then one has to deal with interpretations of quantum mechanics. In the Copenhagen interpretation, the measurement leads to a collapse which is non-unitary. In other interpretation, the measurement process does not involve collapse but entanglement of the object's wavefunction with the wavefunction of the observing apparatus and this process is still unitary.
    Right. But I am not sure why these are listed separately, It seems to me that they are the same thing (i.e. the only requirement coming from the statistical interpretation is that the wavefunction must remain normalized)
    I am not sure if you mean the correspondence principle in the sense of "for large quantum numbers the classical results are recovered" inw hich case I don't see this as directly connected to the wave equation. But I would say that the de Broglie relations must be recovered from the wave equation in the special (unphysical) case of a pure plane wave. as far as I know, this is the argument that leads to obtaining Schrodinger's equation.
  5. Nov 29, 2007 #4
    Can you please clarify somewhat more?
  6. Nov 29, 2007 #5


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    EDIT: I corrected a typo

    I mean that p = h/ lambda and E = h f must be recovered when psi describes a plane wave. taking a plane wave to be of the form [itex] e^{i kx - \omega t} [/itex], one sees that [tex]\frac{p^2}{2m} \psi [/tex] must be given by [tex] -\frac{ \hbar^2}{2m} \frac{\partial^2}{\partial x^2} \psi(x)[/tex] and so on when psi is a plane wave.

    Now, imposing that [tex] E \psi = \frac{p^2}{2m} \psi + V \psi [/tex]must be valid for any wave (not only plane waves), one gets Schrodinger's equation.
    Last edited: Nov 29, 2007
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook