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Eigenstates of the momentum operator

  1. Dec 27, 2012 #1
    For the free particle the solution to the SE are eigenstates of the momentum.
    You get something like:
    ψ = Aexp(ik(x-vt)) + Bexp(-ik(x+vt)) , where k is a constant
    And my book then says that first term represents a wave travelling to the right and the second a wave travelling to the left. But I have a problem with the second statement, because doesn't the minus sign on the second term, that is the minus sign in front of (x+vt), change anything. Certainly the waves are not identical in shap because of that minus sign?
     
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  3. Dec 27, 2012 #2

    mfb

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    Eigenstates of the momentum are solutions, but not the only solutions - every superposition of them is a solution, too.

    Your equation can be written as ψ = Aexp(ikx-ikvt) + Bexp(ik(-x)-ikvt)), which is just a substitution x -> -x, so the wave is mirrored.
     
  4. Dec 27, 2012 #3
    but I dont understand then. Isn't x+vt the mirror of x-vt, not as you say -x+vt mirror of x-vt.
     
  5. Dec 27, 2012 #4

    mfb

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    Just put them into the SE to see which signs you need :).
     
  6. Dec 27, 2012 #5
    what? I already put them in the SE and the general solution is:

    ψ(x,t) = Aexp(ik(x-vt)) + Bexp(-ik(x+vt))

    What I dont understand is just that the form a wave travelling to the left is for me exp(ik(x+vt)) not exp(-ik(-x+vt)) - what is the essential difference between these two waveforms?
     
  7. Dec 27, 2012 #6

    mfb

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    They have a different relation between phase and x-coordinate, while their time-evolution of the phase is the same (just determined by k and v).
    All this is multiplied with i and in an exponential - it is just a phase.
     
  8. Dec 27, 2012 #7
    so you say its just a phasefactor? That would make sense I guess but if you multiply
    exp(-ik(x+vt)) by a phasefactor exp(i) you get exp(k(x+vt)) - clearly that is not what you wanted is it?
     
  9. Dec 27, 2012 #8

    mfb

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    If you multiply exp(-ik(x+vt)) by exp(i) the result is exp(-ik(x+vt) + i) = exp(-ik(x+vt-1/k))
    Which is equivalent to a shift by 1/k in x-direction or 1/(kv) in time.

    The whole exp()-part is just a phase.
     
  10. Dec 27, 2012 #9
    man I still dont get it. My multiplication of the exponential was wrong but I still dont see the answer to why:

    exp(-ik(x+vt)) is the same as exp(ik(x-vt)) except for a phase.
     
  11. Dec 27, 2012 #10
    Well, you agree that, in general, complex exponentials with the same coefficients differ only by a phase factor, yes? If so, then surely you can see how this is simply a particular case of that general rule.
     
  12. Dec 27, 2012 #11
    but the the function in the exponential is not the same! x-vt≠x+vt
     
  13. Dec 27, 2012 #12
    Sure, and that's the whole point. There is a *phase* by which they differ, though the *amplitudes* of these particular plane waves are equal. In particular, their phases differ such that they propagate in opposite directions (+x and -x).
     
  14. Dec 27, 2012 #13
    ugh I see but normally you have exp(ik(x-vt)) and exp(ik(x+vt)) where you in this case instead of the last term have exp(ik(-x+vt)) - my question is really just what difference this -x instead of x does.
     
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