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This upper limit is wrong I need a help

  1. Dec 25, 2008 #1
    this upper limit is wrong...!! I need a help

    I have some concerns about this text:


    it can easily be shown that F satisfies all the criteria of a vector space. As an example, we demostrate that if


    [tex]\psi_{1}(r)[/tex] and [tex]\psi_{2}(r)[/tex] [tex]\in[/tex] F

    both can be cosidered to be complex functions
    then:

    [tex]\psi(r) [/tex] [tex] = \lambda_{1}\psi_{1}(r) + \lambda_{2}\psi_{2}(r)[/tex] [tex]\in[/tex] F (A-1)


    where [tex]\lambda_{1}[/tex] and [tex]\lambda_{2}[/tex] are two arbitrary complex numbers

    In order to show that [tex]\psi(r)^{2}[/tex] is square integrable

    expand [tex]\psi(r)^{2}[/tex] :

    [tex]\psi(r)^{2}[/tex] [tex] = |\lambda_{1}|^{2}|\psi_{1}(r)|^{2} + |\lambda_{2}|^{2} |\psi_{2}(r)|^{2} + \lambda_{1}^\ast\lambda_2\psi_1^\ast(r)\psi_2(r) + \lambda_1\lambda_2^\ast\psi_1(r)\psi_2^\ast(r)[/tex] (A-2)

    (*) simbol means complex-conjugate

    The last two terms of (A-2) have the same modulus, wich has as an upper limit:

    ----- [tex]|\lambda_{1}||\lambda_{2}| . [ |\psi_{1}(r)|^{2} + |\psi_{2}(r)|^{2}][/tex] -----


    [tex]|\psi(r)|^{2}[/tex] is therefore smaller than a function whose integral converges, since [tex]\psi_{1}(r)[/tex] and [tex]\psi_{2}(r)[/tex] are square-integrable

    The questions are:

    Why has the author used the above expression as "upper limit" term?

    How did he obtain this expression?

    What is the relation of this question with "triangular inequality" referred to complex-variable?

    see --- Churchill complex variables and applications ----
    can anybody explain me it a little better?

    all suggestions will be welcome



    thank you all
     
  2. jcsd
  3. Dec 25, 2008 #2
    Re: this upper limit is wrong...!! I need a help

    As for why: He used that to show that each term is integrable, by showing that |ψ(r)|2 is smaller than a combination of integrable functions, so that |ψ(r)|2 is integrable.

    As for how: That follows from the fact that 2ab ≤ a2 + b2. (This can be proven by noting that 0 ≤ (a - b)2 = a2 + b2 -2ab.) In this case, a = |ψ1(r)| and b = |ψ2(r)|. I don't see any immediate connection with the triangle inequality.
     
  4. Dec 27, 2008 #3
    Re: this upper limit is wrong...!! I need a help

    Hi adriank
    I am grateful for your help
    really there is no connection with the triangle inequality
    I can set the last two terms of (A-3) equation to constant a and b respectively
    hence using the 2ab < a^2 + b^2 relation I can obtain the upper limit:

    [tex]|\lambda_{1}||\lambda_{2}| . [ |\psi_{1}(r)|^{2} + |\psi_{2}(r)|^{2}][/tex]

    since these terms have the same modulus

    thank you
     
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