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Questions in wave function

  1. Nov 7, 2014 #1
    why is psi = cos (k r - w t) + i sin ( k r - w t) = e^ [ i ( k r - w t)]?
    my question precisely is why not:
    1. psi = sin (k r - w t) + i cos ( k r - w t) ?
    2. psi = sin (k r - w t) + i sin ( k r - w t) ?
    3. psi = cos (k r - w t) + i cos ( k r - w t) ?

    why not any of these three? is there a reason it is only the first form?

    this means that there always has to be reflection between the real and imaginary parts of the wave function and i don't know why??


    when solving the Schrodinger's equation we assume that
    PSI (r,t) = psi (r) X phi (t)

    Why this form in particular not another form including addition or subtraction wouldn't this be possible and why???

  2. jcsd
  3. Nov 7, 2014 #2
    You should clarify what you are talking about. Are you talking about harmonic oscillators(seeminlgy not but just for example), steppotencials(aswell not I know) or any other wavefunction?
  4. Nov 7, 2014 #3
    I am talking about the wave function form used in Schrodinger's equation
  5. Nov 7, 2014 #4
    Also I am sorry but my background in Chemical Engineering so I didn't study quantum at all before so please give me any simple detail that can help me

  6. Nov 7, 2014 #5
    K is the wavenumber and one has simply reformed the basic exponetntial with the wavenumber k. I don't know if that helps but it is a start.
  7. Nov 7, 2014 #6


    Staff: Mentor

    The precise form has to do with the solution of the free particle Schroedinger equation. Substitute into it to see which are possible solutions:

    There are also key existence theorems in Partial differential equations that if you find solutions in one form, other forms don't exist. Best to consult a book on PDE's if that interests you.

    You next question probably is why the Schroedinger equation and complex numbers. That's a deep issue. The use of imaginary numbers is an almost magical part of QM.

    Check out the following:

    Also get a hold of a copy of Ballentine - Quantum Mechanics - A Modern Development:

    Give the first three chapters a read.

    But the deepest reason of all is the requirement of continuous transformations between so called pure states:

    Last edited by a moderator: May 7, 2017
  8. Nov 7, 2014 #7

    Simon Bridge

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    Homework Helper

    ... looks like you are asking, if the first one is a solution (to the time dependent Schrodinger equation) then why are the others not solutions? Is this correct?

    I'd answer with a question: why would you expect the others to be solutions?
    Maybe they are. Did you substitute the others into the equation and see?

    Note: the Scrodinger equation is usually set up for a particular situation - the specifics of the situation can rule out some wave-functions that may otherwise be valid solutions. The first line you wrote is not the only possible solution. Context is everything.

    ... it is not assumed, it is a property of the solution you first wrote down: $$\Psi(\vec r, t) = e^{i(\vec k\cdot \vec r - \omega t)} = e^{i\vec k\cdot \vec r} e^{- i\omega t} = \psi(\vec r)\chi(t)$$ It's a bit like asking why the equation for a line is assumed to be: ax+by=c ... why not some other form including multiplications and divisions?
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