1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Interpretation Schrödinger's formulae

  1. May 8, 2005 #1
    Dear Friends,

    Does anybodi knows the meaning, or anything related to the term:

    [tex] \Psi \nabla \Psi^* [/tex]

    or

    [tex] \Psi \nabla \Psi^* - \Psi^* \nabla \Psi [/tex]

    Is the representation of something in the reality?

    Best reggards.
     
  2. jcsd
  3. May 8, 2005 #2

    Galileo

    User Avatar
    Science Advisor
    Homework Helper

    If you have a particle having a wavefunction [itex]\psi(\vec r, t)[/itex], then:

    [tex]\vec J(\vec r,t)=\frac{\hbar}{2mi}(\psi^* \nabla \psi-\psi\nabla \psi^*)[/tex]

    is the so-called probability current. It represents the flow of probability density, like electrical current is the flow of charge density.
     
  4. May 8, 2005 #3
    Spinor?


    Thanks, Galileo, but I'm trying to "imagine" what is, for example, one of the 2 terms:

    [tex] \psi^* \nabla \psi[/tex]

    Has it any meaning? Is a rotor of the nabla operator?

    :smile:
     
  5. May 8, 2005 #4

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Nope,it's just the ~ to the integral nucleus of the momentum operator.

    Daniel.
     
  6. May 8, 2005 #5
    Sorry if this is really basic, i'm no quantum guru yet :tongue:, but could you say that the (classical) velocity is in the direction where [tex]\vec J[/tex] has global max at a given [tex]t[/tex]? Is it possible somehow to calculate [tex]\vec v[/tex] from [tex]\vec J[/tex]?
     
    Last edited: May 8, 2005
  7. May 9, 2005 #6

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    "(Classical) velocity" has nothing to do with the probability current density...

    Daniel.
     
  8. May 9, 2005 #7
    empirical

    empirical experiments verify this formulae is ok?
     
  9. May 9, 2005 #8

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Which formulae...?We can't measure [itex] \vec{j}\left(\vec{r},t\right) [/itex],but only probabilities.

    Daniel.
     
  10. May 9, 2005 #9
    Chr

    Daniel,

    I mean that if exists any experiment or example in real word that confirms that formula or some of its components. For example, if the probability to find a particle in some place or time has this formula...
     
  11. May 9, 2005 #10

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    I'm not an experimentalist and never will be,but i can assure that this simple part of QM has been fully checked and confirmed.We can't measure certain abstract things.Since QM is a probabilistic theory,all we can do is statistics.

    Daniel.
     
  12. May 9, 2005 #11
    tks Daniel.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Interpretation Schrödinger's formulae
  1. The Formula (Replies: 3)

  2. Interpreting Results? (Replies: 1)

Loading...