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rasko Jul14-06 09:19 AM

outer product
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}
\end{array} \right)[/tex]

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[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.

rasko Jul16-06 08:02 AM

Thank you!

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