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Why this combination and not another

  1. May 6, 2014 #1
    Why did the author of the book choose the combinations in (9.66) and (9.67) as follows? Why didn't he choose some other combination; say 1/sqrt(3) and the other 2/sqrt(3) instead of 1/sqrt(2) and 1/sqrt(2)?

    Please see attachment. Note that this is problem on stark effect.
     

    Attached Files:

  2. jcsd
  3. May 6, 2014 #2

    Matterwave

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    Different combinations lead to different vectors...the author found the "eigenvectors" corresponding to the specific eigenvalues, these eigenvectors for non-degenerate eigenvalues are unique up to normalization.
     
  4. May 6, 2014 #3

    ChrisVer

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    I guess a physical explanation is that because they both build up [itex]\psi[/itex] state with equal contributions...
     
  5. May 7, 2014 #4
    Then it is ok to choose whatever normalization I want? I can go for the 1/sqrt(3) and the other 2/sqrt(3) here?
     
  6. May 7, 2014 #5

    DrClaude

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    No, all eigenstates need to have the same nomalization, and in most cases you want the norm to be 1.
     
  7. May 7, 2014 #6

    Bill_K

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    You need to look back in your textbook or your notes and review what an eigenvector is.
     
  8. May 7, 2014 #7
    I know what an eigenvector is and I know how to normalize it. Let's say we have |+> I know it is already normalized. When I have (1 2)^T I know that 1/sqrt*5 is the normalization factor. But how would I know how to normalize |2 0 0>? |2 1 0>?
    It is probably the notation of the ket including n, l, m that I am not being capable to translate into something familiar like a column vector....
     
  9. May 7, 2014 #8

    DrClaude

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    For the kets ##|a,b,c \rangle## to form a basis, they must have the same norm. Therefore, to form a new state ##\psi## as ##|a,b,c \rangle + |a',b',c' \rangle## with the same norm requires
    $$
    \psi = \frac{1}{\sqrt{2}} \left( |a,b,c \rangle + |a',b',c' \rangle \right)
    $$
     
  10. May 7, 2014 #9

    Fredrik

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    I haven't looked at this specific case, but it's conventional to use notations like |210> only for vectors with norm 1. So these guys are almost certainly already normalized. If you're given a linear combination like ##|200\rangle +|210\rangle##, and you want to normalize it, you need to keep in mind what it means to normalize a vector. It means to find a vector in the same 1-dimensional subspace that has norm 1. The vector ##a|200\rangle +b|210\rangle## isn't in the same 1-dimensional subspace as ##|200\rangle +|210\rangle## unless a=b. So you need to ask yourself this: For what values of ##a## does ##a\big(|200\rangle +|210\rangle\big)## have norm 1? The first step is of course to calculate the norm of this vector. Do you know how to do that? (Assume that |200> and |210> are already normalized).

    There are infinitely many complex numbers ##a## that get the job done, but only one of them has imaginary part zero and a positive real part. It's convenient to choose that one.
     
  11. May 7, 2014 #10
    Oh great!!!! Thank you for making this so clear Dr Claude!
     
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