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Technical question about Nikolic' Quantum Myths

  1. Jun 14, 2008 #1
    I am referring to the http://arxiv.org/abs/quant-ph/0609163" [Broken] discussed at length in other threads with a purely technical question to help me understand more of the paper.

    On page 12, formula 26 defines the norm of a vector as

    [tex]\langle\psi|\psi\rangle = \psi_1^*\psi_1 + \psi_2^*\psi_2 .[/tex]

    My question is: isn't there a square root missing to get a norm. When I look up my lecture notes on functional analysis, a norm on a Hilbert space is defined by

    [tex]||f|| = \sqrt{(f,f)}[/tex]

    where [itex](\cdot,\cdot)[/itex] is the scalar product. If the above formula 26 would nevertheless be correct, I would end up with a scalar product with a square in, which is not linear and therefore not a scalar product.

    What am I confusing here?

    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jun 14, 2008 #2
    Square root is indeed missing. Or he meant to write norm squared.
  4. Jun 16, 2008 #3


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    Yes he did. :wink:

    Even the the mass squared is sometimes called mass by physicists, especially relativists. :smile:
  5. Jun 18, 2008 #4
    technical question about Nikolic' Quantum Myths (2nd round)

    Thanks for the answers so far. Reading on in http://arxiv.org/abs/quant-ph/0609163" [Broken], I next stumble over formula 28. There I get

    [tex]p_1=|\langle \phi_1|\psi\rangle|^2 = |\sqrt{1^*\psi_1 + 0\psi_2}|^2 = |\sqrt{\psi_1}|^2 =
    |\psi_1| = \sqrt{\psi_1^*\psi_1} .[/tex]

    This, however, contradicts the equation in the paragraph before formula 24, where it reads [itex]p_1 = \psi_1^*\psi_1[/itex].

    My question is, whether formula 28 should rather start with [itex]\sqrt{p_1}[/itex] or whether actually it should read [itex]p_1 = \sqrt{\psi_1^*\psi1}[/itex] in the paragraph before 24?

    Last edited by a moderator: May 3, 2017
  6. Jun 18, 2008 #5


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    This is wrong. The correct statement is
    |\langle \phi_1|\psi\rangle|^2 = |1^*\psi_1 + 0\psi_2|^2
    To repeat, <a|a> denotes the norm SQUARED.
  7. Jun 18, 2008 #6
    Arrrgh, this missing square of the norm got me all messed up. I should have read my own initial post:confused:.

    Thanks for getting me back on the right track.
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