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Normalization conditions in quantum mechanics

  1. Dec 3, 2004 #1
    I am familiar with the normalization


    Because we want to normalize the probability to 1. However if a state vector isn't in the x basis and is just a general vector in Hilbert space, we can take the normalization condition to be:


    Correct? I guess this is making more and more sense to me. I guess I just never thought about it before. But since we can put a completeness relation in the middle of the latter equation then these two are basically equivalent. I guess I'm just having difficultly seeing how probability over all states is 1 in the second case.

    Please tell me if this is incorrect.

  2. jcsd
  3. Dec 3, 2004 #2


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    It is perfectly correct.Though the logic should contain the same ideas,but in a different order.From the general condition of normalization of vectors in a separable Hilbert space u have to make several assumptions (that is sentences which gradually make u lose generality) in order to obtain the integral over [tex] R^{3n} [/tex] in case of an "n" point-like particles quantum system.
    Hopefully u're reading a "decent" ook which respects the logic mentioned above.

    Both Schroedinger's wave mechanics and Heisenberg-Born-Jordan's matrix mechanics can be marvelously derived from the abstract theory by Dirac.A common example of showing that is the QM linear harmonic oscillator.
  4. Dec 3, 2004 #3


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    That's correct. Since every observable [itex]Q[/itex] has a complete set of orthonormal eigenvectors [itex]|q_n\rangle[/itex] we have:
    [tex]\sum_n |q_n\rangle \langle q_n|= 1[/tex]
    and [itex]|\Psi \rangle = \sum_n c_n|q_n \rangle[/itex] with [itex]c_n=\langle q_n|\Psi\rangle[/itex].
    [tex]<\Psi|\Psi>=<\Psi|1|\Psi>=\sum_n \langle\Psi|q_n\rangle \langle q_n|\Psi \rangle=\sum_n |c_n|^2=1[/tex].
    Last edited: Dec 3, 2004
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