1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Outer product

  1. Jul 14, 2006 #1
    In Sakurai's book, page 22:

    [tex]|\beta><\alpha| \doteq
    \left( \begin{array}{ccc}
    <a^{(1)}|\beta><a^{(1)}|\alpha>^{*} & <a^{(1)}|\beta><a^{(2)}|\alpha>^{*} & \ldots \\
    <a^{(2)}|\beta><a^{(1)}|\alpha>^{*} & <a^{(2)}|\beta><a^{(2)}|\alpha>^{*} & \ldots \\
    \vdots & \vdots & \ddots
    \end{array} \right)[/tex]

    How can people get it? Following is my idea:

    [tex]|\beta><\alpha|\\= |\beta> (\sum_{a'}|a'><a'|)<\alpha|\\
    =\sum_{a'}(<a'|\beta>)(<\alpha|a'>) [STEP *][/tex]

    then we get
    [tex]\doteq(<a^{(1)}|\alpha>^{*}, <a^{(2)}|\alpha>^{*} ,\ldots)\cdot
    \left( \begin{array}{c}
    <a^{(1)}|\beta>\\
    <a^{(2)}|\beta>\\
    \vdots
    \end{array} \right)[/tex]

    Is the STEP* right? I'm not sure if i have understood the ruls of ket and bra.
     
  2. jcsd
  3. Jul 14, 2006 #2

    Gokul43201

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    STEP* is incorrect. The LHS is a matrix (operator) while the RHS is a number (a scalar). What you have actually calculated (your error is in not being careful with the order) is the inner product[itex]\langle \alpha | \beta \rangle = \sum_{a'} \langle \alpha | a' \rangle \langle a' | \beta \rangle [/itex].

    For the outer product, you are (post)multiplying a row vector with a column vector (in that order). Reversing the order gives the inner product, a scalar.
     
  4. Jul 16, 2006 #3
    Thank you!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Similar Discussions: Outer product
  1. Pair Production (Replies: 0)

  2. Pair production (Replies: 2)

  3. Products of Interia (Replies: 1)

Loading...