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Possible Logical Flaw in Sakurai

  1. Jan 6, 2015 #1
    So I'm trying to work my way through Sakurai's quantum in prep for grad school, and I'm so tied up by one of his steps in a theorem that I can't help but think the logic is flawed. It's theorem 1.1 on page 17 of the second edition. The theorem is as follows: The eigenvalues of a Hermitian operator A are real; the eigenkets of A corrosponding to different eigenvalues are orthogonal. Starting from the following point (as I follow his logic up until here), he progresses exactly as so:

    [tex](a' - a''^*)\langle a''|a'\rangle = 0 \hspace{35pt} (1.3.3)[/tex] Now a' and a'' can be taken to be either the same or different. Let us first choose them to be the same; we then deduce the reality condition (the first half of the theorem)[tex]a' = a'^* \hspace{92pt} (1.3.4)[/tex] where we have used the fact that |a'> is not a null ket. Let us now assume a' and a'' to be different. Because of the just-proved reality condition, the difference a' - a''* that appears in (1.3.3) is equal to a' - a'', which cannot vanish, by assumption. The inner product must then vanish: [tex]\langle a''|a'\rangle = 0, (a' \neq a'') \hspace{35pt} (1.3.5)[/tex] which proves the orthogonality property (the second half of the theorem.


    Now, my problem is this: he chooses the situation where the two eigenvalues are equal, and the fact that the inner product of their respective eigenkets must then be equal to the unity, I assume, to show the reality condition (1.3.4). However, he uses this reality condition (which has ONLY been shown to be true when the eigenvalues are the same) to deduce the orthogonality principle when the two eigenvalues are different. In order to do this, shouldn't one have to prove the reality condition for the case when the two eigenvalues are different (which I'm not sure how to do)? It would seem that he's using case 1 to prove the second part of the theorem which concerns case 2.

    I'm sure there's a nuance here that's very subtle, preventing me from getting this.
     
  2. jcsd
  3. Jan 6, 2015 #2

    atyy

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    ##|a' \rangle## is an arbitrary eigenvector of the Hermitian operator ##A##, so what has been shown in part 1 is that the eigenvalue of any eigenvector is real.
     
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