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A little Bra-Ket notation theorem that I don't get

  1. Feb 27, 2013 #1
    I'm continuing through Dirac's book, The Principles of Quantum Mechanics. You can view this as a google book in the link below.

    http://books.google.com.au/books?id=...page&q&f=false [Broken]

    On page 28-29 he proves this Theorem:

    If ξ is a real linear operator and

    ξm|P> = 0 (1)

    for a particular ket |P>, m being a positive integer, then

    ξ|P> = 0

    ...

    And the proof is fine by me except for the first part where he says:

    put m = 2 then (1) gives

    <P|ξ2|P> = 0 (2)

    and goes on to prove it by induction. But I don't see how (2) is 'given' from (1).

    Can anyone explain?

    If he is relying on assuming (1) is true for some ket |P> and positive integer m, then I just pick m = 1 and it's not a theorem. And besides, in this case he never demonstrates a definite example of the statement being true, which you need to do in induction.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Feb 27, 2013 #2
    By the way I do understand that because ξ is a real linear operator that the conjugate to ξ|P> is <P|ξ so that you can write <P|ξξ|P>, but I don't see that it follows this is equal to zero.
     
  4. Feb 27, 2013 #3

    micromass

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    Link doesn't work for me.

    Are there any conditions on [itex]\xi[/itex]? If you only assume it to be a real linear operator, then I think I can find counterexamples to the statement.
     
  5. Feb 27, 2013 #4
  6. Feb 27, 2013 #5

    CompuChip

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    The assumption is that for some positive integer m, equation (8) holds.
    Clearly, if m = 1 the statement is trivial, so he considers m = 2 as the base case.

    So first he proves that if ξ²|P> = 0, then ξ|P> = 0. The reasoning is as follows: assume ξ²|P> = 0. <P| is a linear operator, so <P|0 = 0. Hence, <P|ξ²|P> = 0. However, instead of looking at that expression as <P| applied to ξ²|P>, you can also view it as <P|ξ applied to ξ|P>. Since ξ is real, <P|ξ = <P|ξ*, so what you have is really || ξ|P> ||² (the length of ξ|P>) which can only be zero if ξ|P> is zero.

    Then he shows that if it is not given that ξ²|P> = 0 but ξm|P> = 0 for some higher m, it can be reduced to the m = 2 case from which we have just shown that the m = 1 case follows.
     
  7. Feb 27, 2013 #6

    micromass

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    That is not really true, is it??

    Anyway, a counterexample is given by

    [tex]A=\left(\begin{array}{cc} 0 & 1\\ 0 & 0\end{array}\right)[/tex]

    Then [itex]A^2 = 0[/itex], but [itex]A\neq 0[/itex]. So take, for example, |P> = (0,1) and the theorem is false.
     
  8. Feb 27, 2013 #7

    CompuChip

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    Good point.
    It's been a while, wasn't it ξ|P> = <P|ξ?
    So you need A = A := (A*)tr, in other words, for ξ to be Hermitian.
     
  9. Feb 27, 2013 #8
    Compuchip,

    So you mean to say that the theorem should specify m > 1 since otherwise you allow the trivial solution? It might be a silly detail but I find things more confusing when they are not stated rigorously.

    In any case I think I get it now thanks to you.
     
  10. Feb 27, 2013 #9

    CompuChip

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    It doesn't have to: it is also true for m = 1, but trivially so, as you say. However, if you want to prove it for all m then this case doesn't help you, so you need to consider m = 2 as the base case.
     
  11. Feb 27, 2013 #10

    micromass

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    Yes, you need [itex]\xi[/itex] to be Hermitian. Or since [itex]\xi[/itex] is already real, you need it to be symmetric.
    Or a little more general, if [itex]\xi[/itex] is normal (= commutes with its adjoint), then it holds. But it fails for more general operators.
     
  12. Feb 27, 2013 #11

    micromass

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    Oh, apparently Dirac uses the term "real linear operator" to mean "self-adjoint operator". In that case, the theorem holds true, of course. But it's a weird terminology...
     
  13. Feb 27, 2013 #12
    If ξ is real then it equals it's own adjoint, and on page 28 Dirac remarks that if ξ is real than so is ξξ, so if you can take that at face value then ξ must commute with it's adjoint.
     
  14. Feb 27, 2013 #13
    Ah, you got in before me.

    Ok thanks both of you for helping me out.

    Cheers
     
  15. Feb 27, 2013 #14

    dextercioby

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    Hi micromass,

    Dirac wrote his QM book in 1930, before the terminology imposed in functional analysis through M.H.Stone's 1932 book appeared. He did review his QM book and 3 more editions appeared, but the mathematical terminology was not adjusted.
     
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