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Proving the Schwarz Inequality

  1. Sep 1, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove the schwarz inequality:

    [tex]|<\alpha|\beta>|^2\leq<\alpha|\alpha><\beta|\beta>[/tex]

    2. Relevant equations

    [tex]<\alpha|\alpha>\geq 0[/tex]

    [tex]|\alpha>=|\beta>-\left(\frac{<\alpha|\beta>}{<\alpha|\alpha>}\right )|\alpha>[/tex]

    3. The attempt at a solution

    The first step would obviously be to evaluate the first equation using the second:

    [tex] <\alpha|\alpha>=\left <|\beta>-\left(\frac{<\alpha|\beta>}{<\alpha|\alpha>}\right )|\alpha>\middle | |\beta>-\left(\frac{<\alpha|\beta>}{<\alpha|\alpha>}\right )|\alpha>\right>[/tex]

    ..And from here I am kind of stumped. I am familiar with the identity [tex]<a+b|c>=<a|c>+<b|c>[/tex], however what would the identity be for [tex]<a+b|a+b>[/tex]? Am I even going in the right direction here?


    In Shanker's Principles of Quantum Mechanics 2nd ed. Pg 17, it says that the next step is:

    [tex]=<\beta|\beta>-\frac{<\alpha|\beta><\beta|\alpha>}{<\alpha|\alpha>}-\frac{<\alpha|\beta>^*<\alpha|\beta>}{<\alpha|\alpha>}[/tex]

    I am not understanding this logic. I know that [tex]<\alpha|\beta>[/tex] represents the inner product of [tex]\alpha[/tex] and [tex]\beta[/tex], respectfully, but I do not understand how he gets to that step.
     
  2. jcsd
  3. Sep 1, 2009 #2
    For a start I don't understand your second relevant equation; you've got alpha on both sides. Are you choosing a beta (resp. alpha) such that that equation holds for each given alpha (resp. beta)?

    I think you want to define a new quantity [tex]|\gamma \rangle [/tex]:

    [tex]|\gamma \rangle=|\beta \rangle-\left(\frac{\langle \alpha|\beta \rangle}{\langle \alpha|\alpha \rangle}\right )|\alpha \rangle [/tex].

    Now we want to take the adjoint; that is essentially replace all bras with kets and all complex numbers with their complex conjugates: some examples
    If
    [tex] |\gamma \rangle = i |\alpha \rangle [/tex]
    then
    [tex] \langle \gamma | = -i \langle \alpha | [/tex]

    Or If
    [tex] |\gamma \rangle = \langle \alpha | \beta \rangle |\alpha \rangle [/tex]
    then
    [tex] \langle \gamma | = \langle \beta | \alpha \rangle \langle \alpha | [/tex]

    (Why? Remember changing the order of the inner product gives the complex conjugate).

    Your linearity equations are right. You should be able to follow Shankar if you rewrite the equation for [tex]\langle \gamma | \gamma \rangle [/tex] correctly.
     
  4. Sep 1, 2009 #3
    Thanks for the reply fantispug!

    Yes; I apologize I meant for that to be gamma instead of alpha. So are you staying I should take the adjoint of the new quantity, or the adjoint of the inner product of the new quantity with itself?

    Also: did you mean:

    [tex] |\gamma \rangle =i|\gamma \rangle[/tex]

    instead of

    [tex] |\gamma \rangle = i |\alpha \rangle [/tex]

    ?

    So if I rewrite what I had in my original post with gamma instead of alpha, we have:

    [tex]
    \langle \gamma|\gamma \rangle=\left < |\beta \rangle-\left(\frac{\langle\alpha|\beta\rangle}{\langle\alpha|\alpha\rangle}\right )|\alpha\rangle\middle | |\beta\rangle-\left(\frac{\langle\alpha|\beta\rangle}{\langle\alpha|\alpha\rangle}\right )|\alpha\rangle\right>
    [/tex]

    And taking the adjoint (I think Im doing this right):


    [tex]
    \langle \gamma|\gamma \rangle=\left < \langle\beta|^*-\left(\frac{\langle\beta|\alpha\rangle}{\langle\alpha|\alpha\rangle}\right )\langle\alpha|\middle | \langle\beta|^*-\left(\frac{\langle\beta|\alpha\rangle}{\langle\alpha|\alpha\rangle}\right )\langle\alpha|\right>
    [/tex]

    ?
     
    Last edited: Sep 1, 2009
  5. Sep 1, 2009 #4

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    I'm pretty sure he meant take the adjoint of the Ket [itex]\vert\gamma\rangle[/itex], in order to obtain the Bra [itex]\langle\gamma\vert[/itex], since (as you should have learned)

    [tex]\vert\gamma\rangle^\dagger=\langle\gamma\vert[/tex]

    for any Ket [itex]\vert\gamma\rangle[/itex]

    Again, I'm pretty sure he meant exactly what he posted....he is simply providing an example for the ket defined by the equation [itex] |\gamma \rangle \equiv i |\alpha \rangle [/itex] (this is a different gamma than the one you will use in your problem, it is just an example to demonstrate how to turn kets into bras)

    For this example,

    [tex]\langle\gamma\vert=\vert\gamma\rangle^\dagger=\left(i\vert\alpha\rangle\right)^\dagger=-i\langle\alpha\vert[/tex]

    No, if [itex]|\gamma \rangle=|\beta \rangle-\left(\frac{\langle \alpha|\beta \rangle}{\langle \alpha|\alpha \rangle}\right )|\alpha \rangle [/itex]. then

    [tex]\langle\gamma\vert=|\gamma \rangle^\dagger=\left(|\beta \rangle-\left(\frac{\langle \alpha|\beta \rangle}{\langle \alpha|\alpha \rangle}\right )|\alpha \rangle \right)^\dagger[/tex]
     
  6. Sep 2, 2009 #5
    Got it, thanks yall!
     
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