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Schwarz inequality with bra-ket notation

  1. Mar 13, 2015 #1

    Maylis

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    1. The problem statement, all variables and given/known data
    upload_2015-3-13_22-6-26.png

    2. Relevant equations


    3. The attempt at a solution
    Hello,

    I just want to make sure I am doing this right
    $$<a|b> = a_{x}^{*}b_{x} + a_{y}^{*}b_{y} + a_{z}^{*}b_{z}$$
    $$= [(1-i)|x>][-i|x>] + (2 |y>)(-3 |y>) + (0|z>)(|z>)$$
    $$=(-i + i^{2})|x> - 6 |y> + 0|z>$$
    $$=(-1-i)|x> - 6 |y> $$

    Then to find ##\mid<a|b> \mid^{2}##, I need to take the complex conjugate since this is a complex vector
    $$\mid<a|b> \mid^{2} = [(-1-i)|x> - 6|y> + 0|z>][(-1+i)|x> - 6|y> + 0|z>]$$
    $$= 2|x> + 36 |y> + 0|z>$$

    Now I want to make sure that this is even right before I spend time evaluating the other inner products to determine if Schwarz inequality is true.
     
    Last edited: Mar 13, 2015
  2. jcsd
  3. Mar 13, 2015 #2

    Dick

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    The inner product is just a number not a vector. Try that again. In your formula ##a_x## is just the coefficient of ##(1+i)|x>##, i.e. ##(1+i)##.
     
  4. Mar 13, 2015 #3

    Maylis

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    ok, so then
    $$<a|b> = 1 - i^{2} - 6 = -4$$
    $$<a|a> = 1 + 1 + 4 = 6$$
    $$<b|b> = 1 + 9 + 1 = 11$$

    So clearly ##16 \le 6(11)##

    Thanks
     
  5. Mar 13, 2015 #4

    Dick

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    Better slow down a bit. I don't think <a|b> comes out to -4. Want to show the steps that lead to that?
     
  6. Mar 13, 2015 #5

    Maylis

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    Sure,

    $$(1-i)(-i) + 2(-3) + 0(1)$$
    $$=-i+i^{2} - 6$$
    $$=-i-7$$
    So then the absolute value squared is ##(-i - 7)(i - 7) = -i^{2} + 49 = 50##

    So ##50 \le 66##
     
  7. Mar 13, 2015 #6

    Dick

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    That's better.
     
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