Understanding |\beta><\alpha| in Sakurai's Book

[rasko]

In Sakurai's book, page 22:

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

How can people get it? Following is my idea:

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

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

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

 

[Gokul43201]

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]

Thank you!