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-   -   Outer product (http://www.physicsforums.com/showthread.php?t=126060)

 rasko Jul14-06 09:19 AM

outer product

In Sakurai's book, page 22:

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

How can people get it? Following is my idea:

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

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

Is the STEP* right? I'm not sure if i have understood the ruls of ket and bra.

 Gokul43201 Jul14-06 10:16 AM

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$\langle \alpha | \beta \rangle = \sum_{a'} \langle \alpha | a' \rangle \langle a' | \beta \rangle$.

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.

 rasko Jul16-06 08:02 AM

Thank you!

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