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Momentum eigenfunctions proof and Fourier Transform question

  1. Aug 27, 2015 #1
    I have the following problem:
     

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  2. jcsd
  3. Aug 27, 2015 #2

    DrDu

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    How do you define the operator ##\Psi##?
     
  4. Aug 27, 2015 #3
    the equations are from this link

    https://en.wikipedia.org/wiki/Position_and_momentum_space


    But, is it a matter of how you define things? Isn't there a right way to do it that does not depend on how you define the operator? Excuse me if these questions are moronic, but I haven't been far into quantum mechanics yet(I am a self learner at this point).
     
  5. Aug 27, 2015 #4

    blue_leaf77

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    I think you misunderstood the terms here. The first expression is the eigenfunction of momentum operator in position representation, and the second expression is the eigenfunction of position in momentum representation. They are not related by Fourier transformation.
     
  6. Aug 27, 2015 #5

    DrDu

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    All I know from your post is that ##\Psi## apparently is an operator which has some eigenfunctions. So it seems appropriate to as for some more details before answering your questions.
     
  7. Aug 27, 2015 #6
    check this link by Wikipedia please

    https://en.wikipedia.org/wiki/Position_and_momentum_space
     
  8. Aug 27, 2015 #7
    its all in the Wikipedia link that I have provided you with.
     
  9. Aug 27, 2015 #8

    DrDu

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    I can't find any operator ##\Psi## in this link.
     
  10. Aug 27, 2015 #9
    I think its r(hat)=i*d/dk
    its in the link under "Functions and operators in momentum space"
     
  11. Aug 27, 2015 #10

    DrDu

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    You think so?!? So basically, we shall not only answer your question, but also guess what your question might be?
     
  12. Aug 27, 2015 #11
    I found it in the link(by Wikipedia) and I also have given you the link and told you where to look for it. I "think so" because I have not seen operators in my self-stude of quantum mechanics yet, I am a beginner as I warned you.
     
  13. Aug 27, 2015 #12
    in the link, it writes: "position operator r(hat)".
     
  14. Aug 27, 2015 #13

    blue_leaf77

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    Note the presence of subscripts in ##\psi_{\mathbf{k}}(\mathbf{r})## and ##\phi_{\mathbf{r}}(\mathbf{k})##, their presence give them different meaning from ##\psi## and ##\phi## without subscripts. It's true that ##\psi(\mathbf{r})## and ##\phi(\mathbf{k})## are related through Fourier transform, because both functions actually represent the same state, they are just represented in different space. On the other hand, ##\psi_{\mathbf{k}}(\mathbf{r})## and ##\phi_{\mathbf{r}}(\mathbf{k})## are not related through Fourier transform as they correspond to different state (more precisely because they are eigenfunctions of different operators).
    For beginners, Wikipedia and other online sources are often not very good resources to learn because they are generally not designed to initiate learning. It's advisable to start your adventure from standard textbooks.
     
  15. Aug 27, 2015 #14
    yes, sorry for the misunderstanding.
    But, why does Ψk have a plus sign on the exponent while Φr has a minus sign?
     
  16. Aug 27, 2015 #15

    blue_leaf77

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    ##\psi_{\mathbf{k}}(\mathbf{r})## is an eigenfunction of momentum operator ##\hat{\mathbf{p}} = -i\hbar\frac{\partial}{\partial \mathbf{r}}##, so, to calculate ##\psi_{\mathbf{k}}(\mathbf{r})## you need to solve
    $$
    \mathbf{p} \psi_{\mathbf{k}}(\mathbf{r}) = -i\hbar\frac{\partial}{\partial \mathbf{r}} \psi_{\mathbf{k}}(\mathbf{r})
    $$
     
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