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Properly normalized

  1. Aug 15, 2012 #1

    I don't see how they know this is properly normalized. None of the values are specified, not psi, a sub i, E sub i, a sub k, or p sub i.
  2. jcsd
  3. Aug 15, 2012 #2
    They're not saying that this equation follows from anything that has been given so far. They're introducing a label, "properly normalized", which can be applied to a state if and only if the condition in 1.31 is true.

    Equation 1.29 tells you that any quantum state can be thought of as a weighted combination of energy states, whose weights are given by [itex]a_i[/itex]. What they're saying is that if someone hands you some arbitrary state, you can find those values by using 1.30. If the values that you find happen to satisfy 1.31, then you can call the state a normalized state. If not, then you can't call it that. In the case of a non-normalized state, you can make it normalized by dividing every [itex]a_i[/itex] by the total magnitude, which is what they're doing in 1.32. You can check to confirm that doing this will ensure that [itex]\langle \psi | \psi \rangle = 1[/itex].
  4. Aug 15, 2012 #3
    thanks for clearing that up for me
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