Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Ket Notation - Effects of the Projection Operator

  1. Jan 31, 2012 #1
    Ket Notation -- Effects of the Projection Operator

    1. The problem statement, all variables and given/known data
    From Sakurai's Modern Quantum Mechanics (Revised Edition), it is just deriving equation 1.3.12.


    2. Relevant equations
    [tex] \begin{eqnarray*}\langle \alpha |\cdot (\sum_{a'}^N |a'\rangle \langle a'|) \cdot|\alpha \rangle \end{eqnarray*} [/tex]
    3. The attempt at a solution
    The summation can be moved to the left, so everything is being summed from a' to N, but does an alpha bra inner product with a' (or <α|a'>) does the sum of this from all a' to N equal Ʃ<a'|α>? maybe this is simple and I just can't see it?
     
  2. jcsd
  3. Jan 31, 2012 #2

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Re: Ket Notation -- Effects of the Projection Operator

    Check (1.2.12), which is a fundamental property of the inner product. That property holds even when the bras and kets correspond to different bases.
     
  4. Jan 31, 2012 #3
    Re: Ket Notation -- Effects of the Projection Operator

    So, is it in this particular case that they are equal because we are considering the eigenkets of A, a hermitian operator? Because these are the eigenkets of A, does that mean that all operators on it are real? Even though it is the operator <α| acting on it and not A?
     
  5. Jan 31, 2012 #4

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Re: Ket Notation -- Effects of the Projection Operator

    There's no assumption here that [itex]\langle \alpha | a'\rangle[/itex] is real, just that [itex]\langle \alpha | a'\rangle = \langle a'| \alpha\rangle^*.[/itex] That is why the absolute value appears in the formula.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook