Outer product

  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. Gokul43201

    Gokul43201 11,141
    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. Thank you!
     
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